Precalculus

Precalculus




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ISBN-10 1938168348


ISBN-13 978-1-938168-34-5


Revision PR-1-000-RS




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Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 1: Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9


1.1 Functions and Function Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
1.3 Rates of Change and Behavior of Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
1.4 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
1.5 Transformation of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
1.6 Absolute Value Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
1.7 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151


Chapter 2: Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
2.1 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
2.2 Graphs of Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
2.3 Modeling with Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
2.4 Fitting Linear Models to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247


Chapter 3: Polynomial and Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
3.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
3.2 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
3.3 Power Functions and Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 311
3.4 Graphs of Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
3.5 Dividing Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362
3.6 Zeros of Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
3.7 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
3.8 Inverses and Radical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421
3.9 Modeling Using Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438


Chapter 4: Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 459
4.1 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
4.2 Graphs of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482
4.3 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
4.4 Graphs of Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513
4.5 Logarithmic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538
4.6 Exponential and Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
4.7 Exponential and Logarithmic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569
4.8 Fitting Exponential Models to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 590


Chapter 5: Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625
5.1 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626
5.2 Unit Circle: Sine and Cosine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651
5.3 The Other Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674
5.4 Right Triangle Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693


Chapter 6: Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715
6.1 Graphs of the Sine and Cosine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 716
6.2 Graphs of the Other Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 739
6.3 Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764


Chapter 7: Trigonometric Identities and Equations . . . . . . . . . . . . . . . . . . . . . . . . . 789
7.1 Solving Trigonometric Equations with Identities . . . . . . . . . . . . . . . . . . . . . . . . 790
7.2 Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803
7.3 Double-Angle, Half-Angle, and Reduction Formulas . . . . . . . . . . . . . . . . . . . . . . 821
7.4 Sum-to-Product and Product-to-Sum Formulas . . . . . . . . . . . . . . . . . . . . . . . . 835
7.5 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844
7.6 Modeling with Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862


Chapter 8: Further Applications of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . 903
8.1 Non-right Triangles: Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904
8.2 Non-right Triangles: Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923
8.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939
8.4 Polar Coordinates: Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955
8.5 Polar Form of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 978
8.6 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992
8.7 Parametric Equations: Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008




8.8 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022
Chapter 9: Systems of Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 1055


9.1 Systems of Linear Equations: Two Variables . . . . . . . . . . . . . . . . . . . . . . . . 1056
9.2 Systems of Linear Equations: Three Variables . . . . . . . . . . . . . . . . . . . . . . . . 1077
9.3 Systems of Nonlinear Equations and Inequalities: Two Variables . . . . . . . . . . . . . . 1090
9.4 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104
9.5 Matrices and Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116
9.6 Solving Systems with Gaussian Elimination . . . . . . . . . . . . . . . . . . . . . . . . . 1130
9.7 Solving Systems with Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1145
9.8 Solving Systems with Cramer's Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1161


Chapter 10: Analytic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1185
10.1 The Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186
10.2 The Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1208
10.3 The Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1230
10.4 Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1248
10.5 Conic Sections in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1266


Chapter 11: Sequences, Probability and Counting Theory . . . . . . . . . . . . . . . . . . . . 1289
11.1 Sequences and Their Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1290
11.2 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1307
11.3 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1320
11.4 Series and Their Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1331
11.5 Counting Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1346
11.6 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358
11.7 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366


Chapter 12: Introduction to Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387
12.1 Finding Limits: Numerical and Graphical Approaches . . . . . . . . . . . . . . . . . . . 1388
12.2 Finding Limits: Properties of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1404
12.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416
12.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432


A Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1467
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1473


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PREFACE
Welcome to Precalculus, an OpenStax College resource. This textbook has been created with several goals in mind:
accessibility, customization, and student engagement—all while encouraging students toward high levels of academic
scholarship. Instructors and students alike will find that this textbook offers a strong foundation in precalculus in an
accessible format.
About OpenStax College
OpenStax College is a non-profit organization committed to improving student access to quality learning materials. Our
free textbooks go through a rigorous editorial publishing process. Our texts are developed and peer-reviewed by educators
to ensure they are readable, accurate, and meet the scope and sequence requirements of today’s college courses. Unlike
traditional textbooks, OpenStax College resources live online and are owned by the community of educators using them.
Through our partnerships with companies and foundations committed to reducing costs for students, OpenStax College is
working to improve access to higher education for all. OpenStax College is an initiative of Rice University and is made
possible through the generous support of several philanthropic foundations. OpenStax College textbooks are used at many
colleges and universities around the world. Please go to https://openstaxcollege.org/pages/adoptions to see our rapidly
expanding number of adoptions.
About OpenStax College’s Resources
OpenStax College resources provide quality academic instruction. Three key features set our materials apart from others:
they can be customized by instructors for each class, they are a "living" resource that grows online through contributions
from educators, and they are available free or for minimal cost.
Customization
OpenStax College learning resources are designed to be customized for each course. Our textbooks provide a solid
foundation on which instructors can build, and our resources are conceived and written with flexibility in mind. Instructors
can select the sections most relevant to their curricula and create a textbook that speaks directly to the needs of their classes
and student body. Teachers are encouraged to expand on existing examples by adding unique context via geographically
localized applications and topical connections.
Precalculus can be easily customized using our online platform (http://cnx.org/content/col11667/latest/). Simply select
the content most relevant to your current semester and create a textbook that speaks directly to the needs of your class.
Precalculus is organized as a collection of sections that can be rearranged, modified, and enhanced through localized
examples or to incorporate a specific theme to your course. This customization feature will ensure that your textbook truly
reflects the goals of your course.
Curation
To broaden access and encourage community curation, Precalculus is “open source” licensed under a Creative Commons
Attribution (CC-BY) license. The mathematics community is invited to submit feedback to enhance and strengthen the
material and keep it current and relevant for today’s students. Submit your suggestions to info@openstaxcollege.org, and
check in on edition status, alternate versions, errata, and news on the StaxDash at http://openstaxcollege.org.
Cost
Our textbooks are available for free online, and in low-cost print and e-book editions.
About Precalculus
Precalculus is intended for college-level precalculus students. Since precalculus courses vary from one institution to the
next, we have attempted to meet the needs of as broad an audience as possible, including all of the content that might be
covered in any particular course. The result is a comprehensive book that covers more ground than an instructor could likely
cover in a typical one- or two-semester course; but instructors should find, almost without fail, that the topics they wish to
include in their syllabus are covered in the text.
Many chapters of Openstax College Precalculus are suitable for other freshman and sophomore math courses such as
College Algebra and Trigonometry; however, instructors of those courses might need to supplement or adjust the material.


Preface 1




Openstax will also be releasing College Algebra and Algebra and Trigonometry titles tailored to the particular scope,
sequence, and pedagogy of those courses.
Coverage and Scope
Precalculus contains twelve chapters, roughly divided into three groups.
Chapters 1-4 discuss various types of functions, providing a foundation for the remainder of the course.


Chapter 1: Functions
Chapter 2: Linear Functions
Chapter 3: Polynomial and Rational Functions
Chapter 4: Exponential and Logarithmic Functions


Chapters 5-8 focus on Trigonometry. In Precalculus, we approach trigonometry by first introducing angles and the unit
circle, as opposed to the right triangle approach more commonly used in College Algebra and Trigonometry courses.


Chapter 5: Trigonometric Functions
Chapter 6: Periodic Functions
Chapter 7: Trigonometric Identities and Equations
Chapter 8: Further Applications of Trigonometry


Chapters 9-12 present some advanced Precalculus topics that build on topics introduced in chapters 1-8. Most Precalculus
syllabi include some of the topics in these chapters, but few include all. Instructors can select material as needed from this
group of chapters, since they are not cumulative.


Chapter 9: Systems of Equations and Inequalities
Chapter 10: Analytic Geometry
Chapter 11: Sequences, Probability and Counting Theory
Chapter 12: Introduction to Calculus


All chapters are broken down into multiple sections, the titles of which can be viewed in the Table of Contents.
Development Overview
Openstax Precalculus is the product of a collaborative effort by a group of dedicated authors, editors, and instructors whose
collective passion for this project has resulted in a text that is remarkably unified in purpose and voice. Special thanks is due
to our Lead Author, Jay Abramson of Arizona State University, who provided the overall vision for the book and oversaw
the development of each and every chapter, drawing up the initial blueprint, reading numerous drafts, and assimilating field
reviews into actionable revision plans for our authors and editors.
The first eight chapters are a derivative work, built on the foundation of Precalculus: An Investigation of Functions,
by David Lippman and Melonie Rasmussen. Chapters 9-12 were written and developed from by our expert and highly
experienced author team. All twelve chapters follow a new and innovative instructional design, and great care has been
taken to maintain a consistent voice from cover to cover. New features have been introduced to flesh out the instruction,
all of the graphics have been re-done in a more contemporary style, and much of the content has been revised, replaced, or
supplemented to bring the text more in line with mainstream approaches to teaching Precalculus.
Accuracy of the Content
We have taken great pains to ensure the validity and accuracy of this text. Each chapter’s manuscript underwent at least
two rounds of review and revision by a panel of active Precalculus instructors. Then, prior to publication, a separate team
of experts checked all text, examples, and graphics for mathematical accuracy; multiple reviewers were assigned to each
chapter to minimize the chances of any error escaping notice. A third team of experts was responsible for the accuracy of the
Answer Key, dutifully re-working every solution to eradicate any lingering errors. Finally, the editorial team conducted a
multi-round post-production review to ensure the integrity of the content in its final form. The Solutions Manual, which was
written and developed after the Student Edition, has also been rigorously checked for accuracy following a process similar
to that described above. Incidentally, the act of writing out solutions step-by-step served as yet another round of validation
for the Answer Key in the back of the Student Edition.
In spite of the efforts described above, we acknowledge the possibility that—as with any textbook—some errata have
slipped past the guards. We encourage users to report errors via our Errata (https://cnx.org/content/https://
openstaxcollege.org/errata/latest/) page.


2 Preface


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Pedagogical Foundations and Features
Learning Objectives
Each chapter is divided into multiple sections (or modules), each of which is organized around a set of learning objectives.
The learning objectives are listed explicitly at the beginning of each section, and are the focal point of every instructional
element
Narrative text
Narrative text is used to introduce key concepts, terms, and definitions, to provide real-world context, and to provide
transitions between topics and examples. Throughout this book, we rely on a few basic conventions to highlight the most
important ideas:


Key terms are boldfaced, typically when first introduced and/or when formally defined
Key concepts and definitions are called out in a blue box for easy reference.
Key equations, formulas, theorems, identities, etc. are assigned a number, which appears near the right margin.
Occasionally the text may refer back to an equation or formula by its number.


Example
Each learning objective is supported by one or more worked examples, which demonstrate the problem-solving approaches
that students must master. Typically, we include multiple Examples for each learning objective in order to model different
approaches to the same type of problem, or to introduce similar problems of increasing complexity. All told, there are more
than 650 Examples, or an average of about 55 per chapter.
All Examples follow a simple two- or three-part format. First, we pose a problem or question. Next, we demonstrate the
Solution, spelling out the steps along the way. Finally (for select Examples), we conclude with an Analysis reflecting on the
broader implications of the Solution just shown.
Figures
Openstax Precalculus contains more than 2000 figures and illustrations, the vast majority of which are graphs and diagrams.
Art throughout the text adheres to a clear, understated style, drawing the eye to the most important information in each
figure while minimizing visual distractions. Color contrast is employed with discretion to distinguish between the different
functions or features of a graph.


Supporting Features
Four small but important features, each marked by a distinctive icon, serve to support Examples.


Preface 3




A “How To” is a list of steps necessary to solve a certain type of problem. A How To typically precedes an
Example that proceeds to demonstrate the steps in action.


A “Try It” exercise immediately follows an Example or a set of related Examples, providing the student with
an immediate opportunity to solve a similar problem. In the Online version of the text, students can click an Answer link
directly below the question to check their understanding. In other versions, answers to the Try-It exercises are located in the
Answer Key.


A Q&A may appear at any point in the narrative, but most often follows an Example. This feature pre-empts
misconceptions by posing a commonly asked yes/no question, followed by a detailed answer and explanation.


The “Media” icon appears at the conclusion of each section, just prior to the Section Exercises. This icon
marks a list of links to online video tutorials that reinforce the concepts and skills introduced in the section.
Disclaimer: While we have selected tutorials that closely align to our learning objectives, we did not produce these tutorials,
nor were they specifically produced or tailored to accompany Openstax Precalculus. We are deeply grateful to James Sousa
for compiling his incredibly robust and excellent library of video tutorials, which he has made available to the public under a
CC-BY-SA license at http://mathispower4u.yolasite.com/. Most or all of the videos to which we link in our “Media” feature
(plus many more) are found in the Algebra 2 and Trigonometry video libraries at the above site.
Section Exercises
Each section of every chapter concludes with a well-rounded set of exercises that can be assigned as homework or used
selectively for guided practice. With over 5900 exercises across the 12 chapters, instructors should have plenty to choose
from[1].
Section Exercises are organized by question type, and generally appear in the following order:


Verbal questions assess conceptual understanding of key terms and concepts.
Algebraic problems require students to apply algebraic manipulations demonstrated in the section.
Graphical problems assess students’ ability to interpret or produce a graph.
Numeric problems require the student perform calculations or computations.
Technology problems encourage exploration through use of a graphing utility, either to visualize or verify algebraic
results or to solve problems via an alternative to the methods demonstrated in the section.
Extensions pose problems more challenging than the Examples demonstrated in the section. They require students
to synthesize multiple learning objectives or apply critical thinking to solve complex problems.
Real-World Applications present realistic problem scenarios from fields such as physics, geology, biology,
finance, and the social sciences.


Chapter Review Features
Each chapter concludes with a review of the most important takeaways, as well as additional practice problems that students
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1 | FUNCTIONS


Figure 1.1 Standard and Poor’s Index with dividends reinvested (credit "bull": modification of work by Prayitno Hadinata;
credit "graph": modification of work by MeasuringWorth)


Chapter Outline
1.1 Functions and Function Notation
1.2 Domain and Range
1.3 Rates of Change and Behavior of Graphs
1.4 Composition of Functions
1.5 Transformation of Functions
1.6 Absolute Value Functions
1.7 Inverse Functions


Introduction
Toward the end of the twentieth century, the values of stocks of internet and technology companies rose dramatically. As a
result, the Standard and Poor’s stock market average rose as well. Figure 1.1 tracks the value of that initial investment of
just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed
up to about $1100 by the beginning of 2000. That five-year period became known as the “dot-com bubble” because so many
internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew
too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at
the end of 2000.
Notice, as we consider this example, that there is a definite relationship between the year and stock market average. For
any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore
these kinds of relationships and their properties.


Chapter 1 Functions 9




1.1 | Functions and Function Notation
Learning Objectives


In this section, you will:
1.1.1 Determine whether a relation represents a function.
1.1.2 Find the value of a function.
1.1.3 Determine whether a function is one-to-one.
1.1.4 Use the vertical line test to identify functions.
1.1.5 Graph the functions listed in the library of functions.


A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child
increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that
we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set of the first components of each ordered pair is called the domain and the set
of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first
numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.


{(1,  2),  (2,  4),  (3,  6),  (4,  8),  (5,  10)}
The domain is {1,  2,  3,  4,  5}. The range is {2,  4,  6,  8,  10}.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the
lowercase letter  x. Each value in the range is also known as an output value, or dependent variable, and is often labeled
lowercase letter  y.
A function   f   is a relation that assigns a single value in the range to each value in the domain. In other words, no x-
values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation
is a function because each element in the domain, {1,  2,  3,  4,  5}, is paired with exactly one element in the range,
{2,  4,  6,  8,  10}.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would
appear as


{(odd,  1),  (even,  2),  (odd,  3),  (even,  4),  (odd,  5)}
Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1,  2,  3,  4,  5}.
For example, the term “odd” corresponds to three values from the domain, {1,  3,  5} and the term “even” corresponds to
two values from the range, {2,  4}. This violates the definition of a function, so this relation is not a function.
Figure 1.2 compares relations that are functions and not functions.


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Figure 1.2 (a) This relationship is a function because each input is associated with a single output. Note that input  q  and  r 
both give output  n.  (b) This relationship is also a function. In this case, each input is associated with a single output. (c) This
relationship is not a function because input   q  is associated with two different outputs.


Function
A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a
function of the input.”
The input values make up the domain, and the output values make up the range.


Given a relationship between two quantities, determine whether the relationship is a function.
1. Identify the input values.
2. Identify the output values.
3. If each input value leads to only one output value, classify the relationship as a function. If any input value
leads to two or more outputs, do not classify the relationship as a function.


Example 1.1
Determining If Menu Price Lists Are Functions


The coffee shop menu, shown in Figure 1.3 consists of items and their prices.
a. Is price a function of the item?
b. Is the item a function of the price?


Figure 1.3


Solution
a. Let’s begin by considering the input as the items on the menu. The output values are then the prices. See


Figure 1.4.


Chapter 1 Functions 11




Figure 1.4


Each item on the menu has only one price, so the price is a function of the item.
b. Two items on the menu have the same price. If we consider the prices to be the input values and the
items to be the output, then the same input value could have more than one output associated with it. See
Figure 1.5.


Figure 1.5


Therefore, the item is a not a function of price.


Example 1.2
Determining If Class Grade Rules Are Functions


In a particular math class, the overall percent grade corresponds to a grade point average. Is grade point average
a function of the percent grade? Is the percent grade a function of the grade point average? Table 1.1 shows a
possible rule for assigning grade points.


Percent
grade 0–56 57–61 62–66 67–71 72–77 78–86 87–91 92–100


Grade point
average 0.0 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Table 1.1


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1.1


Solution
For any percent grade earned, there is an associated grade point average, so the grade point average is a function
of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average.
In the grading system given, there is a range of percent grades that correspond to the same grade point average.
For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging
from 78 all the way to 86. Thus, percent grade is not a function of grade point average.


Table 1.2[1] lists the five greatest baseball players of all time in order of rank.
Player Rank


Babe Ruth 1


Willie Mays 2


Ty Cobb 3


Walter Johnson 4


Hank Aaron 5


Table 1.2


a. Is the rank a function of the player name?
b. Is the player name a function of the rank?


Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we
can understand and use them, and sometimes also so that we can program them into computers. There are various ways of
representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables h for height and a for age. The
letters   f ,  g, and  h  are often used to represent functions just as we use x,  y, and z to represent numbers and A,  B, and
C to represent sets.


h is f of a We name the function f ; height is a function of age.


h = f (a) We use parentheses to indicate the function input.


f (a) We name the function f ; the expression is read as “ f of a.”


Remember, we can use any letter to name the function; the notation  h(a)  shows us that  h  depends on  a. The value  a must
be put into the function  h  to get a result. The parentheses indicate that age is input into the function; they do not indicate
multiplication.
We can also give an algebraic expression as the input to a function. For example   f (a + b) means “first add a and b, and the
result is the input for the function f.” The operations must be performed in this order to obtain the correct result.


1. http://www.baseball-almanac.com/legendary/lisn100.shtml. Accessed 3/24/2014.


Chapter 1 Functions 13




1.2


Function Notation
The notation  y = f (x)  defines a function named   f . This is read as  “y  is a function of  x.” The letter  x  represents the
input value, or independent variable. The letter  y,  or   f (x),   represents the output value, or dependent variable.


Example 1.3
Using Function Notation for Days in a Month


Use function notation to represent a function whose input is the name of a month and output is the number of
days in that month.


Solution
The number of days in a month is a function of the name of the month, so if we name the function f , we write
days = f (month) or d = f (m). The name of the month is the input to a “rule” that associates a specific number
(the output) with each input.


Figure 1.6


For example,   f (March) = 31,   because March has 31 days. The notation  d = f (m)  reminds us that the number
of days,  d  (the output), is dependent on the name of the month,  m  (the input).


Analysis
Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of
geometric objects, or any other element that determines some kind of output. However, most of the functions we
will work with in this book will have numbers as inputs and outputs.


Example 1.4
Interpreting Function Notation


A function  N = f (y)  gives the number of police officers,  N,   in a town in year  y. What does   f (2005) = 300 
represent?


Solution
When we read   f (2005) = 300,  we see that the input year is 2005. The value for the output, the number of police
officers  (N),   is 300. Remember,  N = f (y). The statement   f (2005) = 300  tells us that in the year 2005 there
were 300 police officers in the town.


Use function notation to express the weight of a pig in pounds as a function of its age in days  d.


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Instead of a notation such as  y = f(x),  could we use the same symbol for the output as for the function,
such as  y = y(x), meaning “y is a function of x?”
Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering.
However, in exploring math itself we like to maintain a distinction between a function such as   f ,  which is a rule
or procedure, and the output  y we get by applying   f   to a particular input  x.  This is why we usually use notation
such as  y = f (x), P = W(d),   and so on.


Representing Functions Using Tables
A common method of representing functions is in the form of a table. The table rows or columns display the corresponding
input and output values. In some cases, these values represent all we know about the relationship; other times, the table
provides a few select examples from a more complete relationship.
Table 1.3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number
of days in that month. This information represents all we know about the months and days for a given year (that is not a leap
year). Note that, in this table, we define a days-in-a-month function   f  where  D = f (m)  identifies months by an integer
rather than by name.


Month number,
 m  (input) 1 2 3 4 5 6 7 8 9 10 11 12


Days in month,
 D  (output) 31 28 31 30 31 30 31 31 30 31 30 31


Table 1.3


Table 1.4 defines a function  Q = g(n). Remember, this notation tells us that  g  is the name of the function that takes the
input  n  and gives the output  Q .


n 1 2 3 4 5


Q 8 6 7 6 8


Table 1.4


Table 1.5 displays the age of children in years and their corresponding heights. This table displays just some of the data
available for the heights and ages of children. We can see right away that this table does not represent a function because
the same input value, 5 years, has two different output values, 40 in. and 42 in.


Age in years, a (input) 5 5 6 7 8 9 10


Height in inches, h (output) 40 42 44 47 50 52 54


Table 1.5


Given a table of input and output values, determine whether the table represents a function.
1. Identify the input and output values.
2. Check to see if each input value is paired with only one output value. If so, the table represents a function.


Chapter 1 Functions 15




Example 1.5
Identifying Tables that Represent Functions


Which table, Table 1.6, Table 1.7, or Table 1.8, represents a function (if any)?
Input Output


2 1


5 3


8 6


Table 1.6


Input Output


–3 5


0 1


4 5


Table 1.7


Input Output


1 0


5 2


5 4


Table 1.8


Solution
Table 1.6 and Table 1.7 define functions. In both, each input value corresponds to exactly one output value.
Table 1.8 does not define a function because the input value of 5 corresponds to two different output values.
When a table represents a function, corresponding input and output values can also be specified using function
notation.
The function represented by Table 1.6 can be represented by writing


f (2) = 1, f (5) = 3, and f (8) = 6


Similarly, the statements
g(−3) = 5,  g(0) = 1, and g(4) = 5


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1.3


represent the function in Table 1.7.
Table 1.8 cannot be expressed in a similar way because it does not represent a function.


Does Table 1.9 represent a function?
Input Output


1 10


2 100


3 1000


Table 1.9


Finding Input and Output Values of a Function
When we know an input value and want to determine the corresponding output value for a function, we evaluate the
function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output
value.
When we know an output value and want to determine the input values that would produce that output value, we set the
output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different
input values can produce the same output value.
Evaluation of Functions in Algebraic Forms
When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function
  f (x) = 5 − 3x2   can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.


Given the formula for a function, evaluate.
1. Replace the input variable in the formula with the value provided.
2. Calculate the result.


Example 1.6
Evaluating Functions at Specific Values


Evaluate   f (x) = x2 + 3x − 4  at
a. 2
b. a
c. a + h


d. f (a + h) − f (a)
h


Chapter 1 Functions 17




Solution
Replace the  x  in the function with each specified value.
a. Because the input value is a number, 2, we can use simple algebra to simplify.


  f (2) = 22 + 3(2) − 4
= 4 + 6 − 4
= 6


b. In this case, the input value is a letter so we cannot simplify the answer any further.
f (a) = a2 + 3a − 4


c. With an input value of  a + h,  we must use the distributive property.
f (a + h) = (a + h)2 + 3(a + h) − 4


= a2 + 2ah + h2 + 3a + 3h − 4
d. In this case, we apply the input values to the function more than once, and then perform algebraic
operations on the result. We already found that


f (a + h) = a2 + 2ah + h2 + 3a + 3h − 4


and we know that
f (a) = a2 + 3a − 4


Now we combine the results and simplify.


f (a + h) − f (a)
h


= (a
2 + 2ah + h2 + 3a + 3h − 4) − (a2 + 3a − 4)


h


= 2ah + h
2 + 3h


h


= h(2a + h + 3)
h


Factor out h.


= 2a + h + 3 Simplify.


Example 1.7
Evaluating Functions


Given the function  h(p) = p2 + 2p,   evaluate  h(4). 


Solution
To evaluate  h(4),  we substitute the value 4 for the input variable  p  in the given function.


h(p) = p2 + 2p


h(4) = (4)2 + 2(4)
= 16 + 8
= 24


Therefore, for an input of 4, we have an output of 24.


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1.4


1.5


Given the function  g(m) = m − 4,   evaluate  g(5).


Example 1.8
Solving Functions


Given the function  h(p) = p2 + 2p,   solve for  h(p) = 3.


Solution
h(p) = 3


p2 + 2p = 3 Substitute the original function h(p) = p2 + 2p.


p2 + 2p − 3 = 0 Subtract 3 from each side.
(p + 3)(p − 1) = 0 Factor.


If  ⎛⎝p + 3⎞⎠⎛⎝p − 1⎞⎠ = 0,   either  ⎛⎝p + 3⎞⎠ = 0  or  ⎛⎝p − 1⎞⎠ = 0  (or both of them equal 0). We will set each factor equal
to 0 and solve for  p  in each case.


(p + 3) = 0, p = − 3
(p − 1) = 0, p = 1


This gives us two solutions. The output  h(p) = 3 when the input is either  p = 1  or  p = − 3. We can also verify
by graphing as in Figure 1.7. The graph verifies that  h(1) = h(−3) = 3  and  h(4) = 24.


Figure 1.7


Given the function  g(m) = m − 4,   solve  g(m) = 2.


Chapter 1 Functions 19




Evaluating Functions Expressed in Formulas
Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the
function output with a formula involving the input quantity, then we can define a function in algebraic form. For example,
the equation  2n + 6p = 12  expresses a functional relationship between  n  and  p. We can rewrite it to decide if  p  is a
function of  n.


Given a function in equation form, write its algebraic formula.
1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an
expression that involves only the input variable.


2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity
to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.


Example 1.9
Finding an Equation of a Function


Express the relationship  2n + 6p = 12  as a function  p = f (n),   if possible.


Solution
To express the relationship in this form, we need to be able to write the relationship where  p  is a function of  n,
which means writing it as  p = [expression involving  n].


2n + 6p = 12
  6p = 12 − 2n Subtract 2n from both sides.
 p = 12 − 2n


6
Divide both sides by 6 and simplify.


 p = 12
6


− 2n
6


 p = 2 − 1
3
n


Therefore,  p  as a function of  n  is written as


p = f (n) = 2 − 1
3
n


Analysis
It is important to note that not every relationship expressed by an equation can also be expressed as a function
with a formula.


Example 1.10
Expressing the Equation of a Circle as a Function


Does the equation  x2 + y2 = 1  represent a function with  x  as input and  y  as output? If so, express the
relationship as a function  y = f (x).


Solution


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1.6


First we subtract  x2   from both sides.
y2 = 1 − x2


We now try to solve for  y  in this equation.


y = ± 1 − x2


= + 1 − x2 and − 1 − x2


We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function
 y = f (x).


If  x − 8y3 = 0,   express  y  as a function of  x.


Are there relationships expressed by an equation that do represent a function but which still cannot be
represented by an algebraic formula?
Yes, this can happen. For example, given the equation  x = y + 2y,   if we want to express  y  as a function of  x,  
there is no simple algebraic formula involving only  x  that equals  y. However, each  x  does determine a unique
value for  y,   and there are mathematical procedures by which  y  can be found to any desired accuracy. In this
case, we say that the equation gives an implicit (implied) rule for  y  as a function of  x,   even though the formula
cannot be written explicitly.


Evaluating a Function Given in Tabular Form
As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions,
and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with
them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember
up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than
30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16
hours.
The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table.
See Table 1.10.[2]


2. http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014.


Chapter 1 Functions 21




Pet Memory span in hours


Puppy 0.008


Adult dog 0.083


Cat 16


Goldfish 2160


Beta fish 3600


Table 1.10


At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P.
The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s
memory span lasts. We can evaluate the function  P  at the input value of “goldfish.” We would write P(goldfis ) = 2160.
Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the
pertinent row of the table. The tabular form for function  P  seems ideally suited to this function, more so than writing it in
paragraph or function form.


Given a function represented by a table, identify specific output and input values.
1. Find the given input in the row (or column) of input values.
2. Identify the corresponding output value paired with that input value.
3. Find the given output values in the row (or column) of output values, noting every time that output value
appears.


4. Identify the input value(s) corresponding to the given output value.


Example 1.11
Evaluating and Solving a Tabular Function


Using Table 1.11,
a. Evaluate  g(3).
b. Solve  g(n) = 6.


n 1 2 3 4 5


g(n) 8 6 7 6 8


Table 1.11


Solution


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1.7


a. Evaluating g(3) means determining the output value of the function g for the input value of n = 3. The
table output value corresponding to n = 3 is 7, so g(3) = 7.


b. Solving g(n) = 6 means identifying the input values, n, that produce an output value of 6. Table 1.12
shows two solutions: n = 2 and n = 4.


n 1 2 3 4 5


g(n) 8 6 7 6 8


Table 1.12
When we input 2 into the function  g,   our output is 6. When we input 4 into the function  g,   our output is also 6.


Using Table 1.12, evaluate  g(1).


Finding Function Values from a Graph
Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in
this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all
instances of the given output value on the graph and observing the corresponding input value(s).


Example 1.12
Reading Function Values from a Graph


Given the graph in Figure 1.8,
a. Evaluate   f (2).
b. Solve   f (x) = 4.


Chapter 1 Functions 23




Figure 1.8


Solution
a. To evaluate   f (2),   locate the point on the curve where  x = 2,   then read the y-coordinate of that point.
The point has coordinates  (2, 1),   so   f (2) = 1.  See Figure 1.9.


Figure 1.9


b. To solve   f (x) = 4, we find the output value  4  on the vertical axis. Moving horizontally along the
line  y = 4,  we locate two points of the curve with output value  4: (−1, 4)  and  (3, 4). These points
represent the two solutions to   f (x) = 4: x = −1  or  x = 3. This means   f (−1) = 4  and   f (3) = 4,   or
when the input is  −1  or 3,  the output is  4.  See Figure 1.10.


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1.8


Figure 1.10


Using Figure 1.8, solve   f (x) = 1.


Determining Whether a Function is One-to-One
Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart
shown in Figure 1.1 at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there
were five different input values that all resulted in the same output value of $1000.
However, some functions have only one input value for each output value, as well as having only one output for each input.
We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal
equivalents, as listed in Table 1.13.


Letter grade Grade point average


A 4.0


B 3.0


C 2.0


D 1.0


Table 1.13


This grading system represents a one-to-one function, because each letter input yields one particular grade point average
output and each grade point average corresponds to one input letter.


Chapter 1 Functions 25




1.9


1.10


To visualize this concept, let’s look again at the two simple functions sketched in Figure 1.2(a) and Figure 1.2(b). The
function in part (a) shows a relationship that is not a one-to-one function because inputs  q  and  r  both give output  n. The
function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.


One-to-One Function
A one-to-one function is a function in which each output value corresponds to exactly one input value.


Example 1.13
Determining Whether a Relationship Is a One-to-One Function


Is the area of a circle a function of its radius? If yes, is the function one-to-one?


Solution
A circle of radius  r  has a unique area measure given by  A = πr2, so for any input,  r,   there is only one output,
A. The area is a function of radius  r.
If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any
area measure  A  is given by the formula  A = πr2. Because areas and radii are positive numbers, there is exactly
one solution: r = Aπ . So the area of a circle is a one-to-one function of the circle’s radius.


a. Is a balance a function of the bank account number?
b. Is a bank account number a function of the balance?
c. Is a balance a one-to-one function of the bank account number?


Evaluate the following:


a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of
the percent grade?


b. If so, is the function one-to-one?


Using the Vertical Line Test
As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-
output pairs in a small space. The visual information they provide often makes relationships easier to understand. By
convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the
vertical axis.
The most common graphs name the input value  x  and the output value  y,   and we say  y  is a function of  x,   or  y = f (x) 
when the function is named   f . The graph of the function is the set of all points  (x, y)  in the plane that satisfies the equation
y = f (x).  If the function is defined for only a few input values, then the graph of the function is only a few points, where
the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For
example, the black dots on the graph in Figure 1.11 tell us that   f (0) = 2  and   f (6) = 1. However, the set of all points
 (x, y)  satisfying  y = f (x)  is a curve. The curve shown includes  (0, 2)  and  (6, 1)  because the curve passes through those
points.


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Figure 1.11


The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that
intersects a graph more than once, then the graph does not define a function because a function has only one output value
for each input value. See Figure 1.12.


Figure 1.12


Given a graph, use the vertical line test to determine if the graph represents a function.
1. Inspect the graph to see if any vertical line drawn would intersect the curve more than once.
2. If there is any such line, determine that the graph does not represent a function.


Example 1.14
Applying the Vertical Line Test


Which of the graphs in Figure 1.13 represent(s) a function  y = f (x)?


Chapter 1 Functions 27




Figure 1.13


Solution
If any vertical line intersects a graph more than once, the relation represented by the graph is not a function.
Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of
Figure 1.13. From this we can conclude that these two graphs represent functions. The third graph does not
represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as
shown in Figure 1.14.


Figure 1.14


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1.11 Does the graph in Figure 1.15 represent a function?


Figure 1.15


Using the Horizontal Line Test
Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use
the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once,
then the graph does not represent a one-to-one function.


Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one
function.


1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.
2. If there is any such line, determine that the function is not one-to-one.


Example 1.15
Applying the Horizontal Line Test


Consider the functions shown in Figure 1.13(a) and Figure 1.13(b). Are either of the functions one-to-one?


Solution
The function in Figure 1.13(a) is not one-to-one. The horizontal line shown in Figure 1.16 intersects the graph
of the function at two points (and we can even find horizontal lines that intersect it at three points.)


Chapter 1 Functions 29




1.12


Figure 1.16


The function in Figure 1.13(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.


Is the graph shown in Figure 1.14 one-to-one?


Identifying Basic Toolkit Functions
In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas,
and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic,
we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements.
We call these our “toolkit functions,” which form a set of basic named functions for which we know the graph, formula, and
special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions
we will use  x  as the input variable and  y = f (x)  as the output variable.
We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently
throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name,
formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in
Table 1.14.


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Toolkit Functions
Name Function Graph


Constant f (x) = c,  where c is aconstant


Identity f (x) = x


Absolute
value f (x) = |x|


Table 1.14


Chapter 1 Functions 31




Toolkit Functions
Name Function Graph


Quadratic f (x) = x2


Cubic f (x) = x3


Reciprocal f (x) = 1x


Table 1.14


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Toolkit Functions
Name Function Graph


Reciprocal
squared


f (x) = 1
x2


Square root f (x) = x


Cube root f (x) = x3


Table 1.14


Chapter 1 Functions 33




Access the following online resources for additional instruction and practice with functions.
• Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction)
• Vertical Line Test (http://openstaxcollege.org/l/vertlinetest)
• Introduction to Functions (http://openstaxcollege.org/l/introtofunction)
• Vertical Line Test on Graph (http://openstaxcollege.org/l/vertlinegraph)
• One-to-one Functions (http://openstaxcollege.org/l/onetoone)
• Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone)


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1.


2.


3.


4.


5.


6.


7.


8.


9.


10.


11.


12.


13.


14.


15.


16.


17.


18.


19.


20.


21.


22.


23.


24.


25.


26.


27.


28.


29.


30.


31.


32.


33.


34.


35.


36.


1.1 EXERCISES
Verbal
What is the difference between a relation and a


function?
What is the difference between the input and the output


of a function?
Why does the vertical line test tell us whether the graph


of a relation represents a function?
How can you determine if a relation is a one-to-one


function?
Why does the horizontal line test tell us whether the


graph of a function is one-to-one?


Algebraic
For the following exercises, determine whether the relation
represents a function.


{(a, b), (c, d), (a, c)}


{(a, b), (b, c), (c, c)}


For the following exercises, determine whether the relation
represents  y  as a function of  x. 


5x + 2y = 10


y = x2


x = y2


3x2 + y = 14


2x + y2 = 6


y = − 2x2 + 40x


y = 1x


x =
3y + 5
7y − 1


x = 1 − y2


y = 3x + 5
7x − 1


x2 + y2 = 9


2xy = 1


x = y3


y = x3


y = 1 − x2


x = ± 1 − y


y = ± 1 − x


y2 = x2


y3 = x2


For the following exercises, evaluate the function   f   at the
indicated values f (−3), f (2), f (−a), − f (a), f (a + h).


f (x) = 2x − 5


f (x) = − 5x2 + 2x − 1


f (x) = 2 − x + 5


f (x) = 6x − 1
5x + 2


f (x) = |x − 1| − |x + 1|


Given the function  g(x) = 5 − x2,   evaluate
 g(x + h) − g(x)


h
,  h ≠ 0.


Given the function  g(x) = x2 + 2x,   evaluate
 g(x) − g(a)x − a ,  x ≠ a.


Given the function  k(t) = 2t − 1 :
a. Evaluate  k(2).
b. Solve  k(t) = 7.


Given the function   f (x) = 8 − 3x :
a. Evaluate   f ( − 2).
b. Solve   f (x) = − 1.


Given the function  p(c) = c2 + c :


Chapter 1 Functions 35




37.


38.


39.


40.


41.


42.


43.


44.


45.


a. Evaluate  p( − 3).
b. Solve  p(c) = 2.


Given the function   f (x) = x2 − 3x :
a. Evaluate   f (5).
b. Solve   f (x) = 4.


Given the function   f (x) = x + 2 :
a. Evaluate   f (7).
b. Solve   f (x) = 4.


Consider the relationship  3r + 2t = 18.
a. Write the relationship as a function  r = f (t).
b. Evaluate   f ( − 3).
c. Solve   f (t) = 2.


Graphical
For the following exercises, use the vertical line test to
determine which graphs show relations that are functions.


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46.


47.


48.


49.


50.


51.


52.


53.


Given the following graph,
• Evaluate   f (−1).
• Solve for   f (x) = 3.


Given the following graph,
• Evaluate   f (0).
• Solve for   f (x) = −3.


Chapter 1 Functions 37




54.


55.


56.


57.


58.


59.


60.


61.


62.


63.


64.


Given the following graph,
• Evaluate   f (4).
• Solve for   f (x) = 1.


For the following exercises, determine if the given graph is
a one-to-one function.


Numeric
For the following exercises, determine whether the relation
represents a function.


{(−1, −1), (−2, −2), (−3, −3)}


{(3, 4), (4, 5), (5, 6)}





⎨(2, 5), (7, 11), (15, 8), (7, 9)⎫⎭⎬


For the following exercises, determine if the relation
represented in table form represents  y  as a function of  x.


x 5 10 15


y 3 8 14


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65.


66.


67.


68.


69.


70.


71.


72.


73.


74.


75.


76.


77.


78.


79.


80.


81.


x 5 10 15


y 3 8 8


x 5 10 10


y 3 8 14


For the following exercises, use the function   f   represented
in Table 1.15.


x f(x)


0 74


1 28


2 1


3 53


4 56


5 3


6 36


7 45


8 14


9 47


Table 1.15


Evaluate   f (3).


Solve   f (x) = 1.


For the following exercises, evaluate the function   f   at the
values f (−2),   f ( − 1),   f (0),   f (1), and   f (2).


f (x) = 4 − 2x


f (x) = 8 − 3x


f (x) = 8x2 − 7x + 3


f (x) = 3 + x + 3


f (x) = x − 2
x + 3


f (x) = 3x


For the following exercises, evaluate the expressions, given
functions f ,   g, and  h :


• f (x) = 3x − 2


• g(x) = 5 − x2


• h(x) = − 2x2 + 3x − 1


3 f (1) − 4g(−2)


f ⎛⎝
7
3

⎠− h(−2)


Technology
For the following exercises, graph  y = x2   on the given
viewing window. Determine the corresponding range for
each viewing window. Show each graph.


[ − 0.1, 0.1]


[ − 10, 10]


[ − 100, 100]


For the following exercises, graph  y = x3   on the given
viewing window. Determine the corresponding range for
each viewing window. Show each graph.


[ − 0.1, 0.1]


[ − 10, 10]


[ − 100, 100]


For the following exercises, graph  y = x  on the given
viewing window. Determine the corresponding range for
each viewing window. Show each graph.


Chapter 1 Functions 39




82.


83.


84.


85.


86.


87.


88.


89.


90.


91.


92.


[0, 0.01]


[0, 100]


[0, 10,000]


For the following exercises, graph y = x3 on the given
viewing window. Determine the corresponding range for
each viewing window. Show each graph.


[−0.001, 0.001]


[−1000, 1000]


[−1,000,000, 1,000,000]


Real-World Applications
The amount of garbage,  G,   produced by a city with


population  p  is given by  G = f (p).  G  is measured in
tons per week, and  p  is measured in thousands of people.
a. The town of Tola has a population of 40,000 and
produces 13 tons of garbage each week. Express
this information in terms of the function   f .


b. Explain the meaning of the statement   f (5) = 2.


The number of cubic yards of dirt,  D,   needed to cover
a garden with area  a  square feet is given by  D = g(a).
a. A garden with area 5000 ft2 requires 50 yd3 of dirt.
Express this information in terms of the function
 g.


b. Explain the meaning of the statement  g(100) = 1.


Let   f (t)  be the number of ducks in a lake  t  years after
1990. Explain the meaning of each statement:
a. f (5) = 30
b. f (10) = 40


Let  h(t)  be the height above ground, in feet, of a
rocket  t  seconds after launching. Explain the meaning of
each statement:
a. h(1) = 200
b. h(2) = 350


Show that the function   f (x) = 3(x − 5)2 + 7  is not
one-to-one.


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1.2 | Domain and Range
Learning Objectives


In this section, you will:
1.2.1 Find the domain of a function defined by an equation.
1.2.2 Graph piecewise-defined functions.


If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all
time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1.17 shows the amount, in dollars, each
of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice
that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies
by year. In creating various functions using the data, we can identify different independent and dependent variables, and we
can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for
determining the domain and range of functions such as these.


Figure 1.17 Based on data compiled by www.the-numbers.com.[3]


Finding the Domain of a Function Defined by an Equation
In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we
will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges,
we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the
horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include
any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or
in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.
We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as
another “holding area” for the machine’s products. See Figure 1.18.


3. The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-
numbers.com/market/genre/Horror. Accessed 3/24/2014


Chapter 1 Functions 41




Figure 1.18


We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers.
In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the
endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would
need to express the interval that is more than 0 and less than or equal to 100 and write  (0, 100]. We will discuss interval
notation in greater detail later.
Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain
of such functions involves remembering three different forms. First, if the function has no denominator or an even root,
consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude
values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that
would make the radicand negative.
Before we begin, let us review the conventions of interval notation:


• The smallest term from the interval is written first.
• The largest term in the interval is written second, following a comma.
• Parentheses, ( or ), are used to signify that an endpoint is not included, called exclusive.
• Brackets, [ or ], are used to indicate that an endpoint is included, called inclusive.


See Figure 1.19 for a summary of interval notation.


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Figure 1.19


Example 1.16
Finding the Domain of a Function as a Set of Ordered Pairs


Find the domain of the following function:  {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)} .


Solution
First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions,
as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.


{2, 3, 4, 5, 6}


Chapter 1 Functions 43




1.13


1.14


Find the domain of the function:



⎨(−5, 4), (0, 0), (5, −4), (10, −8), (15, −12)⎫⎭⎬


Given a function written in equation form, find the domain.
1. Identify the input values.
2. Identify any restrictions on the input and exclude those values from the domain.
3. Write the domain in interval form, if possible.


Example 1.17
Finding the Domain of a Function


Find the domain of the function   f (x) = x2 − 1.


Solution
The input value, shown by the variable  x  in the equation, is squared and then the result is lowered by one. Any
real number may be squared and then be lowered by one, so there are no restrictions on the domain of this
function. The domain is the set of real numbers.
In interval form, the domain of   f   is  (−∞, ∞).


Find the domain of the function:   f (x) = 5 − x + x3.


Given a function written in an equation form that includes a fraction, find the domain.
1. Identify the input values.
2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the
denominator equal to zero and solve for  x  . If the function’s formula contains an even root, set the
radicand greater than or equal to 0, and then solve.


3. Write the domain in interval form, making sure to exclude any restricted values from the domain.


Example 1.18
Finding the Domain of a Function Involving a Denominator


Find the domain of the function   f (x) = x + 1
2 − x


.


Solution
When there is a denominator, we want to include only values of the input that do not force the denominator to be
zero. So, we will set the denominator equal to 0 and solve for  x.


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1.15


2 − x = 0
   − x = − 2
          x = 2


Now, we will exclude 2 from the domain. The answers are all real numbers where  x < 2  or  x > 2. We can
use a symbol known as the union,   ∪ , to combine the two sets. In interval notation, we write the solution:
(−∞, 2) ∪ (2, ∞).


Figure 1.20


In interval form, the domain of   f   is  (−∞, 2) ∪ (2, ∞).


Find the domain of the function:   f (x) = 1 + 4x
2x − 1


.


Given a function written in equation form including an even root, find the domain.
1. Identify the input values.
2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set
the radicand greater than or equal to zero and solve for  x.


3. The solution(s) are the domain of the function. If possible, write the answer in interval form.


Example 1.19
Finding the Domain of a Function with an Even Root


Find the domain of the function   f (x) = 7 − x.


Solution
When there is an even root in the formula, we exclude any real numbers that result in a negative number in the
radicand.
Set the radicand greater than or equal to zero and solve for  x.


7 − x ≥ 0
    − x ≥ − 7
          x ≤ 7


Chapter 1 Functions 45




1.16


Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or
equal to  7,   or  ( − ∞, 7].


Find the domain of the function   f (x) = 5 + 2x.


Can there be functions in which the domain and range do not intersect at all?
Yes. For example, the function   f (x) = − 1x   has the set of all positive real numbers as its domain but the set of all
negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely
different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance
chart), in such cases the domain and range have no elements in common.


Using Notations to Specify Domain and Range
In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities,
or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation.
For example,  {x|10 ≤ x < 30}  describes the behavior of  x  in set-builder notation. The braces  {}  are read as “the set of,”
and the vertical bar | is read as “such that,” so we would read  {x|10 ≤ x < 30}  as “the set of x-values such that 10 is less
than or equal to  x,   and  x  is less than 30.”
Figure 1.21 compares inequality notation, set-builder notation, and interval notation.


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Figure 1.21


To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier
examples, we use the union symbol,   ∪ , to combine two unconnected intervals. For example, the union of the sets
{2, 3, 5}  and  {4, 6}  is the set  {2, 3, 4, 5, 6}.  It is the set of all elements that belong to one or the other (or both) of
the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending
order of numerical value. If the original two sets have some elements in common, those elements should be listed only once
in the union set. For sets of real numbers on intervals, another example of a union is


{x| |x| ≥ 3} = (−∞, − 3] ∪ [3, ∞)


Set-Builder Notation and Interval Notation
Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form
{x| statement about x} which is read as, “the set of all  x  such that the statement about  x  is true.” For example,


{x|4 < x ≤ 12}


Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not
be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or
parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For
example,


(4, 12]


Chapter 1 Functions 47




1.17


Given a line graph, describe the set of values using interval notation.
1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line.
2. At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each
excluded end value (open dot).


3. At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for
each excluded end value (open dot).


4. Use the union symbol   ∪   to combine all intervals into one set.


Example 1.20
Describing Sets on the Real-Number Line


Describe the intervals of values shown in Figure 1.22 using inequality notation, set-builder notation, and interval
notation.


Figure 1.22


Solution
To describe the values,  x,   included in the intervals shown, we would say, “ x  is a real number greater than or
equal to 1 and less than or equal to 3, or a real number greater than 5.”


Inequality 1 ≤ x ≤ 3 or x > 5


Set-builder notation ⎧⎩⎨x|1 ≤ x ≤ 3 or x > 5⎫⎭⎬


Interval notation [1, 3] ∪ (5, ∞)


Remember that, when writing or reading interval notation, using a square bracket means the boundary is included
in the set. Using a parenthesis means the boundary is not included in the set.


Given Figure 1.23, specify the graphed set in
a. words
b. set-builder notation
c. interval notation


Figure 1.23


Finding Domain and Range from Graphs
Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of
possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of


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possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the
graph we can see, the domain and range may be greater than the visible values. See Figure 1.24.


Figure 1.24


We can observe that the graph extends horizontally from  −5  to the right without bound, so the domain is  ⎡⎣−5, ∞).  The
vertical extent of the graph is all range values  5  and below, so the range is  (−∞, 5⎤⎦. Note that the domain and range are
always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of
the graph for range.


Example 1.21
Finding Domain and Range from a Graph


Find the domain and range of the function   f   whose graph is shown in Figure 1.25.


Chapter 1 Functions 49




Figure 1.25


Solution
We can observe that the horizontal extent of the graph is –3 to 1, so the domain of   f   is  (−3, 1].
The vertical extent of the graph is 0 to –4, so the range is  [−4, 0).  See Figure 1.26.


Figure 1.26


Example 1.22
Finding Domain and Range from a Graph of Oil Production


Find the domain and range of the function   f  whose graph is shown in Figure 1.27.


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1.18


Figure 1.27 (credit: modification of work by the U.S. Energy
Information Administration)[4]


Solution
The input quantity along the horizontal axis is “years,” which we represent with the variable  t  for time. The
output quantity is “thousands of barrels of oil per day,” which we represent with the variable  b  for barrels. The
graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is
visible, we can determine the domain as  1973 ≤ t ≤ 2008  and the range as approximately  180 ≤ b ≤ 2010.
In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range,
we approximate the smallest and largest values since they do not fall exactly on the grid lines.


Given Figure 1.28, identify the domain and range using interval notation.


Figure 1.28


Can a function’s domain and range be the same?
Yes. For example, the domain and range of the cube root function are both the set of all real numbers.


Finding Domains and Ranges of the Toolkit Functions
We will now return to our set of toolkit functions to determine the domain and range of each.


4. http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A.


Chapter 1 Functions 51




Figure 1.29 For the constant function   f (x) = c,   the
domain consists of all real numbers; there are no restrictions on
the input. The only output value is the constant  c,   so the range
is the set  {c}  that contains this single element. In interval
notation, this is written as  [c, c],   the interval that both begins
and ends with  c.


Figure 1.30 For the identity function   f (x) = x,   there is no
restriction on  x. Both the domain and range are the set of all
real numbers.


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Figure 1.31 For the absolute value function   f (x) = |x|,  
there is no restriction on  x. However, because absolute value is
defined as a distance from 0, the output can only be greater than
or equal to 0.


Figure 1.32 For the quadratic function   f (x) = x2,   the
domain is all real numbers since the horizontal extent of the
graph is the whole real number line. Because the graph does not
include any negative values for the range, the range is only
nonnegative real numbers.


Chapter 1 Functions 53




Figure 1.33 For the cubic function   f (x) = x3,   the domain
is all real numbers because the horizontal extent of the graph is
the whole real number line. The same applies to the vertical
extent of the graph, so the domain and range include all real
numbers.


Figure 1.34 For the reciprocal function   f (x) = 1x ,  we
cannot divide by 0, so we must exclude 0 from the domain.
Further, 1 divided by any value can never be 0, so the range also
will not include 0. In set-builder notation, we could also write
{x| x ≠ 0}, the set of all real numbers that are not zero.


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Figure 1.35 For the reciprocal squared function
  f (x) = 1


x2
, we cannot divide by 0, so we must exclude 0


from the domain. There is also no x that can give an output of
0, so 0 is excluded from the range as well. Note that the output
of this function is always positive due to the square in the
denominator, so the range includes only positive numbers.


Figure 1.36 For the square root function   f (x) = x,  we
cannot take the square root of a negative real number, so the
domain must be 0 or greater. The range also excludes negative
numbers because the square root of a positive number  x  is
defined to be positive, even though the square of the negative
number  − x  also gives us  x.


Chapter 1 Functions 55




Figure 1.37 For the cube root function   f (x) = x3 ,   the
domain and range include all real numbers. Note that there is no
problem taking a cube root, or any odd-integer root, of a
negative number, and the resulting output is negative (it is an
odd function).


Given the formula for a function, determine the domain and range.
1. Exclude from the domain any input values that result in division by zero.
2. Exclude from the domain any input values that have nonreal (or undefined) number outputs.
3. Use the valid input values to determine the range of the output values.
4. Look at the function graph and table values to confirm the actual function behavior.


Example 1.23
Finding the Domain and Range Using Toolkit Functions


Find the domain and range of   f (x) = 2x3 − x.


Solution
There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result.
The domain is  (−∞, ∞)  and the range is also  (−∞, ∞).


Example 1.24
Finding the Domain and Range


Find the domain and range of   f (x) = 2
x + 1


.


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1.19


Solution
We cannot evaluate the function at  −1  because division by zero is undefined. The domain is
 (−∞, −1) ∪ (−1, ∞). Because the function is never zero, we exclude 0 from the range. The range is
 (−∞, 0) ∪ (0, ∞).


Example 1.25
Finding the Domain and Range


Find the domain and range of   f (x) = 2 x + 4.


Solution
We cannot take the square root of a negative number, so the value inside the radical must be nonnegative.


x + 4 ≥ 0 when x ≥ − 4


The domain of   f (x)  is  [ − 4, ∞).
We then find the range. We know that   f (−4) = 0,   and the function value increases as  x  increases without any
upper limit. We conclude that the range of   f   is  ⎡⎣0, ∞).


Analysis
Figure 1.38 represents the function   f .


Figure 1.38


Find the domain and range of   f (x) = − 2 − x.


Graphing Piecewise-Defined Functions
Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example,
in the toolkit functions, we introduced the absolute value function   f (x) = |x|. With a domain of all real numbers and a
range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number


Chapter 1 Functions 57




value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater
than or equal to 0.
If we input 0, or a positive value, the output is the same as the input.


f (x) = x if x ≥ 0
If we input a negative value, the output is the opposite of the input.


f (x) = − x if x < 0
Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function.
A piecewise function is a function in which more than one formula is used to define the output over different pieces of the
domain.
We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain
“boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is
discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise
functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional
income is taxed at 20%. The tax on a total income  S would be  0.1S  if  S ≤ $10,000  and  $1000 + 0.2(S − $10,000)  if
 S > $10,000.


Piecewise Function
A piecewise function is a function in which more than one formula is used to define the output. Each formula has its
own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:


f (x) =




formula 1 if x is in domain 1
formula 2 if x is in domain 2
formula 3 if x is in domain 3


In piecewise notation, the absolute value function is


|x| =




x if x ≥ 0
−x if x < 0


Given a piecewise function, write the formula and identify the domain for each interval.
1. Identify the intervals for which different rules apply.
2. Determine formulas that describe how to calculate an output from an input in each interval.
3. Use braces and if-statements to write the function.


Example 1.26
Writing a Piecewise Function


A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of
10 or more people. Write a function relating the number of people,  n,   to the cost,  C.


Solution
Two different formulas will be needed. For n-values under 10,  C = 5n.  For values of  n  that are 10 or greater,
 C = 50.


C(n) =




5n if 0 < n < 10
50 if n ≥ 10


Analysis


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The function is represented in Figure 1.39. The graph is a diagonal line from  n = 0  to  n = 10  and a constant
after that. In this example, the two formulas agree at the meeting point where  n = 10,   but not all piecewise
functions have this property.


Figure 1.39


Example 1.27
Working with a Piecewise Function


A cell phone company uses the function below to determine the cost,  C,   in dollars for  g  gigabytes of data
transfer.


C(g) =





25 if 0 < g < 2


25 + 10(g − 2) if g ≥ 2


Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.


Solution
To find the cost of using 1.5 gigabytes of data,  C(1.5),  we first look to see which part of the domain our input
falls in. Because 1.5 is less than 2, we use the first formula.


C(1.5) = $25


To find the cost of using 4 gigabytes of data,  C(4),  we see that our input of 4 is greater than 2, so we use the
second formula.


C(4) = 25 + 10(4 − 2) = $45


Analysis
The function is represented in Figure 1.40. We can see where the function changes from a constant to a shifted
and stretched identity at  g = 2. We plot the graphs for the different formulas on a common set of axes, making
sure each formula is applied on its proper domain.


Chapter 1 Functions 59




Figure 1.40


Given a piecewise function, sketch a graph.
1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain.
2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that
piece. Do not graph two functions over one interval because it would violate the criteria of a function.


Example 1.28
Graphing a Piecewise Function


Sketch a graph of the function.


f (x) =




x2 if x ≤ 1
3 if 1 < x ≤ 2
x if x > 2


Solution
Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine
graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we
draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality;
we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-
to inequality.
Figure 1.41 shows the three components of the piecewise function graphed on separate coordinate systems.


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1.20


Figure 1.41 (a)   f (x) = x2 if x ≤ 1;   (b)   f (x) = 3 if 1< x ≤ 2;   (c)   f (x) = x if x > 2


Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure
1.42.


Figure 1.42


Analysis
Note that the graph does pass the vertical line test even at  x = 1  and  x = 2  because the points (1, 3) and (2, 2)
are not part of the graph of the function, though (1, 1) and (2,  3) are.


Graph the following piecewise function.


f (x) =




x3 if x < − 1
−2 if −1 < x < 4
x if x > 4


Can more than one formula from a piecewise function be applied to a value in the domain?
No. Each value corresponds to one equation in a piecewise formula.


Chapter 1 Functions 61




Access these online resources for additional instruction and practice with domain and range.
• Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot)
• Determining Domain and Range (http://openstaxcollege.org/l/determinedomain)
• Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph)
• Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable)
• Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/
l/drcoordinate)


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93.
94.


95.


96.


97.


98.


99.


100.


101.


102.


103.


104.


105.


106.


107.


108.


109.


110.


111.


112.


113.


114.


115.


116.


117.


118.


119.


120.


121.


1.2 EXERCISES
Verbal
Why does the domain differ for different functions?
How do we determine the domain of a function defined


by an equation?


Explain why the domain of   f (x) = x3   is different
from the domain of   f (x) = x.


When describing sets of numbers using interval
notation, when do you use a parenthesis and when do you
use a bracket?
How do you graph a piecewise function?


Algebraic
For the following exercises, find the domain of each
function using interval notation.


f (x) = − 2x(x − 1)(x − 2)


f (x) = 5 − 2x2


f (x) = 3 x − 2


f (x) = 3 − 6 − 2x


f (x) = 4 − 3x


f (x) = x2 + 4


f (x) = 1 − 2x
3


f (x) = x − 1
3


f (x) = 9
x − 6


f (x) = 3x + 1
4x + 2


f (x) = x + 4
x − 4


f (x) = x − 3
x2 + 9x − 22


f (x) = 1
x2 − x − 6


f (x) = 2x
3 − 250


x2 − 2x − 15


5
x − 3


2x + 1
5 − x


f (x) = x − 4
x − 6


f (x) = x − 6
x − 4


f (x) = xx


f (x) = x
2 − 9x


x2 − 81


Find the domain of the function   f (x) = 2x3 − 50x 
by:
a. using algebra.
b. graphing the function in the radicand and
determining intervals on the x-axis for which the
radicand is nonnegative.


Graphical
For the following exercises, write the domain and range of
each function using interval notation.


Chapter 1 Functions 63




122.


123.


124.


125.


126.


127.


128.


129.


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130.


131.


132.


133.


134.


135.


136.


137.


138.


139.


140.


141.


142.


143.


144.


145.


146.


147.


148.


149.


150.


151.


152.


For the following exercises, sketch a graph of the piecewise
function. Write the domain in interval notation.


f (x) =




x + 1 if x < − 2
−2x − 3 if x ≥ − 2


f (x) =




2x − 1 if x < 1
1 + x if x ≥ 1


f (x) =




x + 1  if  x < 0
x − 1  if   x > 0


f (x) =




3 if x < 0
x if x ≥ 0


f (x) =



⎨x


2 if x < 0
1 − x if x > 0


f (x) =



⎨ x


2


x + 2
  if     x < 0
if     x ≥ 0


f (x) =




x + 1 if x < 1


x3 if x ≥ 1


f (x) =



⎨|x|
1
   if   x < 2
   if   x ≥ 2


Numeric
For the following exercises, given each function f ,
evaluate f (−3),   f (−2),   f (−1), and f (0).


f (x) =




x + 1 if x < − 2
−2x − 3 if x ≥ − 2


f (x) =




1 if x ≤ − 3
0 if x > − 3


f (x) =



⎨−2x


2 + 3 if x ≤ − 1
5x − 7 if x > − 1


For the following exercises, given each function   f ,  
evaluate f (−1),   f (0),   f (2),   and   f (4).


f (x) =




7x + 3 if x < 0
7x + 6 if x ≥ 0


f (x) =



⎨ x


2 − 2 if x < 2
4 + |x − 5| if x ≥ 2


f (x) =




5x if x < 0
3 if 0 ≤ x ≤ 3


x2 if x > 3


For the following exercises, write the domain for the
piecewise function in interval notation.


f (x) =




x + 1      if  x < − 2
−2x − 3  if  x ≥ − 2


f (x) =



⎨x


2 − 2      if  x < 1
−x2 + 2  if  x > 1


f (x) =




2x − 3


−3x2
  if   x < 0
if   x ≥ 2


Technology


Graph  y = 1
x2


  on the viewing window
 [−0.5, −0.1]  and  [0.1, 0.5]. Determine the
corresponding range for the viewing window. Show the
graphs.


Graph  y = 1x   on the viewing window  [−0.5, −0.1] 
and  [0.1, 0.5]. Determine the corresponding range for the
viewing window. Show the graphs.


Extension
Suppose the range of a function   f   is  [−5, 8]. What


is the range of  | f (x)| ?


Create a function in which the range is all
nonnegative real numbers.


Create a function in which the domain is  x > 2.


Real-World Applications
The height  h  of a projectile is a function of the time


 t  it is in the air. The height in feet for  t  seconds is given
by the function h(t) = −16t2 + 96t. What is the domain of
the function? What does the domain mean in the context of
the problem?


Chapter 1 Functions 65




153. The cost in dollars of making  x  items is given by the
function  C(x) = 10x + 500.
a. The fixed cost is determined when zero items are
produced. Find the fixed cost for this item.


b. What is the cost of making 25 items?
c. Suppose the maximum cost allowed is $1500. What
are the domain and range of the cost function,
 C(x)?


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1.3 | Rates of Change and Behavior of Graphs
Learning Objectives


In this section, you will:
1.3.1 Find the average rate of change of a function.
1.3.2 Use a graph to determine where a function is increasing, decreasing, or constant.
1.3.3 Use a graph to locate local maxima and local minima.
1.3.4 Use a graph to locate the absolute maximum and absolute minimum.


Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1.17[5] lists the average cost,
in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year.


y 2005 2006 2007 2008 2009 2010 2011 2012


C(y) 2.31 2.62 2.84 3.30 2.41 2.84 3.58 3.68


Table 1.17


If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per
gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at
how much the price changed per year. In this section, we will investigate changes such as these.
Finding the Average Rate of Change of a Function
The price change per year is a rate of change because it describes how an output quantity changes relative to the change
in the input quantity. We can see that the price of gasoline in Table 1.17 did not change by the same amount each year, so
the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of
change over the specified period of time. To find the average rate of change, we divide the change in the output value by
the change in the input value.


Average rate of change =
Change in output
Change in input


=
Δy
Δx


=
y2 − y1
x2 − x1


=
f (x2) − f (x1)


x2 − x1


The Greek letterΔ  (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in  y 
divided by the change in  x. ” Occasionally we write  Δ f   instead of  Δy,  which still represents the change in the function’s
output value resulting from a change to its input value. It does not mean we are changing the function into some other
function.
In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was


Δy
Δx


= $1.37
7 years


≈ 0.196 dollars per year


On average, the price of gas increased by about 19.6¢ each year.
Other examples of rates of change include:


5. http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014.


Chapter 1 Functions 67




1.21


• A population of rats increasing by 40 rats per week
• A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes)
• A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon)
• The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage
• The amount of money in a college account decreasing by $4,000 per quarter


Rate of Change
A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a
rate of change are “output units per input units.”
The average rate of change between two input values is the total change of the function values (output values) divided
by the change in the input values.


(1.1)Δy
Δx


=
f (x2) − f (x1)


x2 − x1


Given the value of a function at different points, calculate the average rate of change of a function for the
interval between two values  x1   and  x2.


1. Calculate the difference y2 − y1 = Δy.
2. Calculate the difference x2 − x1 = Δx.


3. Find the ratio  Δy
Δx


.


Example 1.29
Computing an Average Rate of Change


Using the data in Table 1.17, find the average rate of change of the price of gasoline between 2007 and 2009.


Solution
In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is


Δy
Δx


=
y2 − y1
x2 − x1


= $2.41 − $2.84
2009 − 2007


= −$0.43
2 years


= − $0.22 per year


Analysis
Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when
the output decreases as the input increases or when the output increases as the input decreases.


Using the data in Table 1.17, find the average rate of change between 2005 and 2010.


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Example 1.30
Computing Average Rate of Change from a Graph


Given the function  g(t)  shown in Figure 1.43, find the average rate of change on the interval  [−1, 2].


Figure 1.43


Solution
At t = − 1, Figure 1.44 shows g(−1) = 4. At  t = 2, the graph shows g(2) = 1.


Figure 1.44


The horizontal change  Δt = 3  is shown by the red arrow, and the vertical change Δg(t) = − 3 is shown by the
turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of


1 − 4
2 − (−1)


= −3
3


= −1


Analysis
Note that the order we choose is very important. If, for example, we use  y2 − y1x1 − x2,  we will not get the correct
answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as  (x1, y1)  and
 (x2, y2).


Example 1.31


Chapter 1 Functions 69




Computing Average Rate of Change from a Table


After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values
are shown in Table 1.18. Find her average speed over the first 6 hours.


t (hours) 0 1 2 3 4 5 6 7


D(t) (miles) 10 55 90 153 214 240 282 300


Table 1.18


Solution
Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of


292 − 10
6 − 0


= 282
6


= 47


The average speed is 47 miles per hour.


Analysis
Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2,3], the
average speed is 63 miles per hour.


Example 1.32
Computing Average Rate of Change for a Function Expressed as a Formula


Compute the average rate of change of f (x) = x2 − 1x on the interval [2, 4].


Solution
We can start by computing the function values at each endpoint of the interval.


f (2) = 22 − 1
2


f (4) = 42 − 1
4


= 4 − 1
2


= 16 − 1
4


= 7
2


= 63
4


Now we compute the average rate of change.


Average rate of change =
f (4) − f (2)


4 − 2


=
63
4


− 7
2


4 − 2


=
49
4
2


= 49
8


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1.22 Find the average rate of change of f (x) = x − 2 x on the interval [1,  9].


Example 1.33
Finding the Average Rate of Change of a Force


The electrostatic force  F, measured in newtons, between two charged particles can be related to the distance
between the particles  d, in centimeters, by the formula  F(d) = 2


d2
. Find the average rate of change of force if


the distance between the particles is increased from 2 cm to 6 cm.


Solution
We are computing the average rate of change of  F(d) = 2


d2
  on the interval  [2, 6].


Average rate of change = F(6) − F(2)
6 − 2


                                                           =
2


62
− 2


22


6 − 2
Simplify.


                                                           =
2
36


− 2
4


4


                                                           = −
16
36
4


Combine numerator terms.


                                                           = − 1
9


Simplify


The average rate of change is  − 1
9
  newton per centimeter.


Example 1.34
Finding an Average Rate of Change as an Expression


Find the average rate of change of g(t) = t2 + 3t + 1 on the interval [0,  a]. The answer will be an expression
involving a.


Solution
We use the average rate of change formula.


Chapter 1 Functions 71




1.23


Average rate of change =
g(a) − g(0)


a − 0
Evaluate.


= (a
2 + 3a + 1) − (02 + 3(0) + 1)


a − 0
Simplify.


= a
2 + 3a + 1 − 1


a Simplify and factor.


= a(a + 3)a
Divide by the common factor a.


= a + 3


This result tells us the average rate of change in terms of  a  between  t = 0  and any other point  t = a.  For
example, on the interval  [0, 5],   the average rate of change would be  5 + 3 = 8.


Find the average rate of change of f (x) = x2 + 2x − 8 on the interval [5,  a].


Using a Graph to Determine Where a Function is Increasing,
Decreasing, or Constant
As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways.
We say that a function is increasing on an interval if the function values increase as the input values increase within that
interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over
that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing
function is negative. Figure 1.45 shows examples of increasing and decreasing intervals on a function.


Figure 1.45 The function   f (x) = x3 − 12x  is increasing on
 (−∞, − 2) ∪ (2,  ∞)  and is decreasing on  ( − 2, 2).


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While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input
where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is
called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where
a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form
is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function.
(The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term
local.
Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither
increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as
defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.
For the function whose graph is shown in Figure 1.46, the local maximum is 16, and it occurs at  x = −2. The local
minimum is  −16  and it occurs at  x = 2.


Figure 1.46


To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its
highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function
is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at
neighboring points. Figure 1.47 illustrates these ideas for a local maximum.


Figure 1.47 Definition of a local maximum


These observations lead us to a formal definition of local extrema.


Chapter 1 Functions 73




Local Minima and Local Maxima
A function   f   is an increasing function on an open interval if   f (b) > f (a)  for any two input values  a  and  b  in the
given interval where  b > a.
A function   f   is a decreasing function on an open interval if   f (b) < f (a)  for any two input values  a  and  b  in the
given interval where  b > a.
A function f has a local maximum at  x = b if there exists an interval  (a, c) with a < b < c such that, for any x
in the interval (a, c), f (x) ≤ f (b). Likewise, f has a local minimum at x = b if there exists an interval (a, c) with
a < b < c such that, for any x in the interval (a, c), f (x) ≥ f (b).


Example 1.35
Finding Increasing and Decreasing Intervals on a Graph


Given the function  p(t)  in Figure 1.48, identify the intervals on which the function appears to be increasing.


Figure 1.48


Solution
We see that the function is not constant on any interval. The function is increasing where it slants upward as we
move to the right and decreasing where it slants downward as we move to the right. The function appears to be
increasing from  t = 1  to  t = 3  and from  t = 4  on.
In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval
(4, ∞).


Analysis
Notice in this example that we used open intervals (intervals that do not include the endpoints), because the
function is neither increasing nor decreasing at  t = 1 ,  t = 3 , and  t = 4  . These points are the local extrema (two
minima and a maximum).


Example 1.36
Finding Local Extrema from a Graph


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Graph the function   f (x) = 2x + x3. Then use the graph to estimate the local extrema of the function and to
determine the intervals on which the function is increasing.


Solution
Using technology, we find that the graph of the function looks like that in Figure 1.49. It appears there is a
low point, or local minimum, between  x = 2  and  x = 3,   and a mirror-image high point, or local maximum,
somewhere between  x = −3  and  x = −2.


Figure 1.49


Analysis
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure
1.50 provides screen images from two different technologies, showing the estimate for the local maximum and
minimum.


Figure 1.50


Based on these estimates, the function is increasing on the interval  ( − ∞, − 2.449)  and  (2.449,∞). Notice
that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals
due to the differing approximation algorithms used by each. (The exact location of the extrema is at  ± 6,   but
determining this requires calculus.)


Chapter 1 Functions 75




1.24 Graph the function   f (x) = x3 − 6x2 − 15x + 20  to estimate the local extrema of the function. Use these
to determine the intervals on which the function is increasing and decreasing.


Example 1.37
Finding Local Maxima and Minima from a Graph


For the function   f  whose graph is shown in Figure 1.51, find all local maxima and minima.


Figure 1.51


Solution
Observe the graph of   f . The graph attains a local maximum at  x = 1  because it is the highest point in an open
interval around  x = 1. The local maximum is the  y -coordinate at  x = 1,  which is  2.
The graph attains a local minimum at x = −1 because it is the lowest point in an open interval around  x = −1. 
The local minimum is the y-coordinate at   x = −1,   which is   −2.


Analyzing the Toolkit Functions for Increasing or Decreasing
Intervals
We will now return to our toolkit functions and discuss their graphical behavior in Figure 1.52, Figure 1.53, and Figure
1.54.


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Figure 1.52


Figure 1.53


Chapter 1 Functions 77




Figure 1.54


Use A Graph to Locate the Absolute Maximum and Absolute
Minimum
There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally)
and locating the highest and lowest points on the graph for the entire domain. The  y- coordinates (output) at the highest and
lowest points are called the absolute maximum and absolute minimum, respectively.
To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it
highest and lowest points on the domain of the function. See Figure 1.55.


Figure 1.55


Not every function has an absolute maximum or minimum value. The toolkit function   f (x) = x3   is one such function.


Absolute Maxima and Minima
The absolute maximum of   f   at  x = c  is   f (c) where   f (c) ≥ f (x)  for all  x  in the domain of   f .
The absolute minimum of   f   at  x = d  is   f (d) where   f (d) ≤ f (x)  for all  x  in the domain of   f .


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Example 1.38
Finding Absolute Maxima and Minima from a Graph


For the function   f   shown in Figure 1.56, find all absolute maxima and minima.


Figure 1.56


Solution
Observe the graph of   f . The graph attains an absolute maximum in two locations,  x = −2  and  x = 2,   because
at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the
y-coordinate at  x = −2  and  x = 2,  which is  16.
The graph attains an absolute minimum at  x = 3,   because it is the lowest point on the domain of the function’s
graph. The absolute minimum is the y-coordinate at  x = 3, which is−10.


Access this online resource for additional instruction and practice with rates of change.
• Average Rate of Change (http://openstaxcollege.org/l/aroc)


Chapter 1 Functions 79




154.


155.


156.


157.


158.


159.


160.


161.


162.


163.


164.


165.


166.


167.


168.


169.


170.


171.


172.


173.


1.3 EXERCISES
Verbal


Can the average rate of change of a function be
constant?


If a function   f   is increasing on  (a, b)  and
decreasing on  (b, c),   then what can be said about the local
extremum of   f   on  (a, c)?  


How are the absolute maximum and minimum similar
to and different from the local extrema?


How does the graph of the absolute value function
compare to the graph of the quadratic function,  y = x2,   in
terms of increasing and decreasing intervals?


Algebraic
For the following exercises, find the average rate of change
of each function on the interval specified for real numbers
 b  or  h.


f (x) = 4x2 − 7  on  [1, b]


g(x) = 2x2 − 9  on  ⎡⎣4, b⎤⎦


p(x) = 3x + 4  on  [2, 2 + h]


k(x) = 4x − 2  on  [3, 3 + h]


f (x) = 2x2 + 1  on  [x, x + h]


g(x) = 3x2 − 2  on  [x, x + h]


a(t) = 1
t + 4


  on  [9, 9 + h]


b(x) = 1
x + 3


  on  [1, 1 + h]


j(x) = 3x3   on  [1, 1 + h]


r(t) = 4t3   on  [2, 2 + h]


f (x + h) − f (x)
h


  given   f (x) = 2x2 − 3x  on
 [x, x + h]


Graphical
For the following exercises, consider the graph of   f   shown
in Figure 1.57.


Figure 1.57


Estimate the average rate of change from  x = 1  to
 x = 4.


Estimate the average rate of change from  x = 2  to
 x = 5.
For the following exercises, use the graph of each function
to estimate the intervals on which the function is increasing
or decreasing.


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174.


175.


176.


177.


178.


179.For the following exercises, consider the graph shown in
Figure 1.58.


Figure 1.58


Estimate the intervals where the function is increasing
or decreasing.


Estimate the point(s) at which the graph of   f   has a
local maximum or a local minimum.
For the following exercises, consider the graph in Figure
1.59.


Figure 1.59


If the complete graph of the function is shown,
estimate the intervals where the function is increasing or
decreasing.


If the complete graph of the function is shown,
estimate the absolute maximum and absolute minimum.


Numeric
Table 1.19 gives the annual sales (in millions of


dollars) of a product from 1998 to 2006. What was the
average rate of change of annual sales (a) between 2001 and
2002, and (b) between 2001 and 2004?


Chapter 1 Functions 81




180.


181.


182.


183.


184.


185.


186.


187.


Year Sales (millions of dollars)


1998 201


1999 219


2000 233


2001 243


2002 249


2003 251


2004 249


2005 243


2006 233


Table 1.19


Table 1.20 gives the population of a town (in
thousands) from 2000 to 2008. What was the average rate
of change of population (a) between 2002 and 2004, and (b)
between 2002 and 2006?


Year Population (thousands)


2000 87


2001 84


2002 83


2003 80


2004 77


2005 76


2006 78


2007 81


2008 85


Table 1.20


For the following exercises, find the average rate of change
of each function on the interval specified.


f (x) = x2   on  [1, 5]


h(x) = 5 − 2x2   on  [−2, 4]


q(x) = x3   on  [−4, 2]


g(x) = 3x3 − 1  on  [−3, 3]


y = 1x   on  [1, 3]


p(t) =

⎝t
2 − 4⎞⎠(t + 1)


t2 + 3
  on  [−3, 1]


k(t) = 6t2 + 4
t3


  on  [−1, 3]


Technology
For the following exercises, use a graphing utility to
estimate the local extrema of each function and to estimate
the intervals on which the function is increasing and
decreasing.


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188.


189.


190.


191.


192.


193.


194.


195.


196.


197.


198.


199.


200.


f (x) = x4 − 4x3 + 5


h(x) = x5 + 5x4 + 10x3 + 10x2 − 1


g(t) = t t + 3


k(t) = 3t
2
3 − t


m(x) = x4 + 2x3 − 12x2 − 10x + 4


n(x) = x4 − 8x3 + 18x2 − 6x + 2


Extension
The graph of the function   f   is shown in Figure


1.60.


Figure 1.60


Based on the calculator screen shot, the point
 (1.333, 5.185)  is which of the following?
A. a relative (local) maximum of the function
B. the vertex of the function
C. the absolute maximum of the function
D. a zero of the function


Let f (x) = 1x . Find a number  c  such that the
average rate of change of the function   f   on the interval
 (1, c)  is  − 1


4
.


Let   f (x) = 1x . Find the number  b  such that the
average rate of change of   f   on the interval  (2, b)  is
 − 1


10
.


Real-World Applications
At the start of a trip, the odometer on a car read


21,395. At the end of the trip, 13.5 hours later, the odometer
read 22,125. Assume the scale on the odometer is in miles.
What is the average speed the car traveled during this trip?


A driver of a car stopped at a gas station to fill up his
gas tank. He looked at his watch, and the time read exactly
3:40 p.m. At this time, he started pumping gas into the tank.
At exactly 3:44, the tank was full and he noticed that he had
pumped 10.7 gallons. What is the average rate of flow of
the gasoline into the gas tank?


Near the surface of the moon, the distance that an
object falls is a function of time. It is given by
 d(t) = 2.6667t2,  where  t  is in seconds and  d(t)  is in
feet. If an object is dropped from a certain height, find the
average velocity of the object from  t = 1  to  t = 2.


The graph in Figure 1.61 illustrates the decay of a
radioactive substance over  t  days.


Figure 1.61


Use the graph to estimate the average decay rate from
 t = 5  to  t = 15.


Chapter 1 Functions 83




1.4 | Composition of Functions
Learning Objectives


In this section, you will:
1.4.1 Combine functions using algebraic operations.
1.4.2 Create a new function by composition of functions.
1.4.3 Evaluate composite functions.
1.4.4 Find the domain of a composite function.
1.4.5 Decompose a composite function into its component functions.


Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will
depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the
year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends
on the day.
Using descriptive variables, we can notate these two functions. The function  C(T)  gives the cost  C  of heating a house for
a given average daily temperature in  T   degrees Celsius. The function  T(d)  gives the average daily temperature on day  d
of the year. For any given day,  Cost = C⎛⎝T(d)⎞⎠ means that the cost depends on the temperature, which in turns depends on
the day of the year. Thus, we can evaluate the cost function at the temperature  T(d).  For example, we could evaluate  T(5) 
to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that
temperature. We would write  C⎛⎝T(5)⎞⎠.


By combining these two relationships into one function, we have performed function composition, which is the focus of
this section.
Combining Functions Using Algebraic Operations
Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic
operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations
with the function outputs, defining the result as the output of our new function.
Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period
of years, with the result being their total household income. We want to do this for every year, adding only that year’s
incomes and then collecting all the data in a new column. If  w(y)  is the wife’s income and  h(y)  is the husband’s income in
year  y,   and we want  T   to represent the total income, then we can define a new function.


T(y) = h(y) + w(y)


If this holds true for every year, then we can focus on the relation between the functions without reference to a year and
write


T = h + w


Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that
have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers
so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add
and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.


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1.25


For two functions   f (x)  and  g(x) with real number outputs, we define new functions f + g,   f − g,   f g,   and   fg   by the
relations


( f + g)(x) = f (x) + g(x)
( f − g)(x) = f (x) − g(x)
( f g)(x) = f (x)g(x)





f
g

⎠(x) =


f (x)
g(x)


Example 1.39
Performing Algebraic Operations on Functions


Find and simplify the functions  ⎛⎝g − f ⎞⎠(x)  and  ⎛⎝
g
f

⎠(x),   given   f (x) = x − 1  and  g(x) = x2 − 1. Are they the


same function?


Solution
Begin by writing the general form, and then substitute the given functions.


(g − f )(x) = g(x) − f (x)


(g − f )(x) = x2 − 1 − (x − 1)


= x2 − x
= x(x − 1)





g
f

⎠(x) =


g(x)
f (x)





g
f

⎠(x) =


x2 − 1
x − 1


= (x + 1)(x − 1)
x − 1


where x ≠ 1


= x + 1


No, the functions are not the same.


Note: For  ⎛⎝
g
f

⎠(x),   the condition  x ≠ 1  is necessary because when  x = 1,   the denominator is equal to 0, which


makes the function undefined.


Find and simplify the functions  ⎛⎝ f g⎞⎠(x)  and  ⎛⎝ f − g⎞⎠(x).
f (x) = x − 1 and g(x) = x2 − 1


Are they the same function?


Create a Function by Composition of Functions
Performing algebraic operations on functions combines them into a new function, but we can also create functions by
composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that
takes a day as input and yields a cost as output. The process of combining functions so that the output of one function
becomes the input of another is known as a composition of functions. The resulting function is known as a composite
function. We represent this combination by the following notation:


Chapter 1 Functions 85





⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠


We read the left-hand side as “ f   composed with  g  at  x,” and the right-hand side as “ f   of  g  of  x.” The two sides of the
equation have the same mathematical meaning and are equal. The open circle symbol  ∘   is called the composition operator.
We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring
to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much
as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function
composition with multiplication because, as we learned above, in most cases f (g(x)) ≠ f (x)g(x).
It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention
with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above,
the function  g  takes the input  x  first and yields an output  g(x). Then the function   f   takes  g(x)  as an input and yields an
output   f ⎛⎝g(x)⎞⎠.


In general,   f ∘g  and  g ∘ f   are different functions. In other words, in many cases   f ⎛⎝g(x)⎞⎠ ≠ g⎛⎝ f (x)⎞⎠  for all  x. We will also
see that sometimes two functions can be composed only in one specific order.
For example, if   f (x) = x2   and  g(x) = x + 2, then


  f (g(x)) = f (x + 2)
= (x + 2)2


= x2 + 4x + 4


but
g( f (x)) = g(x2)


= x2 + 2


These expressions are not equal for all values of  x,   so the two functions are not equal. It is irrelevant that the expressions
happen to be equal for the single input value  x = − 1


2
.


Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside
function. Less formally, the composition has to make sense in terms of inputs and outputs.


Composition of Functions
When the output of one function is used as the input of another, we call the entire operation a composition of functions.
For any input  x  and functions   f   and  g,   this action defines a composite function, which we write as   f ∘g  such that


(1.2)⎛⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠
The domain of the composite function   f ∘g  is all  x  such that  x  is in the domain of  g  and  g(x)  is in the domain of   f .
It is important to realize that the product of functions   f g  is not the same as the function composition   f ⎛⎝g(x)⎞⎠,   because,
in general,   f (x)g(x) ≠ f ⎛⎝g(x)⎞⎠.


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Example 1.40
Determining whether Composition of Functions is Commutative


Using the functions provided, find   f ⎛⎝g(x)⎞⎠  and  g⎛⎝ f (x)⎞⎠. Determine whether the composition of the functions is
commutative.


f (x) = 2x + 1             g(x) = 3 − x


Solution
Let’s begin by substituting  g(x)  into   f (x).


f (g(x)) = 2(3 − x) + 1


= 6 − 2x + 1
= 7 − 2x


Now we can substitute   f (x)  into  g(x).
g( f (x)) = 3 − (2x + 1)
= 3 − 2x − 1
= − 2x + 2


We find that  g( f (x)) ≠ f (g(x)),   so the operation of function composition is not commutative.


Example 1.41
Interpreting Composite Functions


The function  c(s)  gives the number of calories burned completing  s  sit-ups, and  s(t)  gives the number of sit-ups
a person can complete in  t minutes. Interpret  c(s(3)).


Solution
The inside expression in the composition is  s(3). Because the input to the s-function is time,  t = 3  represents 3
minutes, and  s(3)  is the number of sit-ups completed in 3 minutes.
Using  s(3)  as the input to the function  c(s)  gives us the number of calories burned during the number of sit-ups
that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).


Example 1.42
Investigating the Order of Function Composition


Suppose   f (x)  gives miles that can be driven in  x  hours and  g(y)  gives the gallons of gas used in driving  y 
miles. Which of these expressions is meaningful:   f ⎛⎝g(y)⎞⎠  or  g⎛⎝ f (x)⎞⎠?


Chapter 1 Functions 87




1.26


Solution
The function  y = f (x)  is a function whose output is the number of miles driven corresponding to the number of
hours driven.


number of miles = f  (number of hours)
The function  g(y)  is a function whose output is the number of gallons used corresponding to the number of miles
driven. This means:


number of gallons = g (number of miles)
The expression  g(y)  takes miles as the input and a number of gallons as the output. The function   f (x)  requires a
number of hours as the input. Trying to input a number of gallons does not make sense. The expression   f ⎛⎝g(y)⎞⎠ 
is meaningless.
The expression   f (x)  takes hours as input and a number of miles driven as the output. The function  g(y)  requires
a number of miles as the input. Using   f (x)  (miles driven) as an input value for  g(y),  where gallons of gas
depends on miles driven, does make sense. The expression  g⎛⎝ f (x)⎞⎠ makes sense, and will yield the number of
gallons of gas used,  g,   driving a certain number of miles,   f (x),   in  x  hours.


Are there any situations where   f(g(y))  and  g( f(x)) would both be meaningful or useful expressions?
Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce
different new functions. In real-world problems, functions whose inputs and outputs have the same units also may
give compositions that are meaningful in either order.


The gravitational force on a planet a distance r from the sun is given by the function G(r). The
acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition
of these two functions, and explain what it means.


Evaluating Composite Functions
Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain.
We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as
inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use
the inner function’s output as the input for the outer function.
Evaluating Composite Functions Using Tables
When working with functions given as tables, we read input and output values from the table entries and always work from
the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to
the outside function.


Example 1.43
Using a Table to Evaluate a Composite Function


Using Table 1.21, evaluate   f (g(3))  and  g( f (3)).


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1.27


x f(x) g(x)


1 6 3


2 8 5


3 3 2


4 1 7


Table 1.21


Solution
To evaluate   f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression  g(3) 
using the table that defines the function  g :   g(3) = 2. We can then use that result as the input to the function
  f ,   so  g(3)  is replaced by 2 and we get   f (2). Then, using the table that defines the function   f ,  we find that
  f (2) = 8.


           g(3) = 2
    f (g(3)) = f (2) = 8


To evaluate  g( f (3)), we first evaluate the inside expression   f (3)  using the first table:   f (3) = 3. Then, using the
table for  g, we can evaluate


g( f (3)) = g(3) = 2


Table 1.22 shows the composite functions   f ∘g  and  g ∘ f   as tables.


x g(x) f ⎛⎝g(x)⎞⎠ f (x) g⎛⎝ f (x)⎞⎠


3 2 8 3 2


Table 1.22


Using Table 1.21, evaluate   f (g(1))  and  g( f (4)).


Evaluating Composite Functions Using Graphs
When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process
we use for evaluating tables. We read the input and output values, but this time, from the  x- and y- axes of the graphs.


Chapter 1 Functions 89




Given a composite function and graphs of its individual functions, evaluate it using the information
provided by the graphs.


1. Locate the given input to the inner function on the  x- axis of its graph.
2. Read off the output of the inner function from the  y- axis of its graph.
3. Locate the inner function output on the  x- axis of the graph of the outer function.
4. Read the output of the outer function from the  y- axis of its graph. This is the output of the composite
function.


Example 1.44
Using a Graph to Evaluate a Composite Function


Using Figure 1.62, evaluate   f (g(1)).


Figure 1.62


Solution
To evaluate   f (g(1)),  we start with the inside evaluation. See Figure 1.63.


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Figure 1.63


We evaluate  g(1)  using the graph of  g(x),   finding the input of 1 on the  x- axis and finding the output value of
the graph at that input. Here,  g(1) = 3. We use this value as the input to the function   f .


f (g(1)) = f (3)


We can then evaluate the composite function by looking to the graph of   f (x),   finding the input of 3 on the x-
axis and reading the output value of the graph at this input. Here,   f (3) = 6,   so   f (g(1)) = 6.


Analysis
Figure 1.64 shows how we can mark the graphs with arrows to trace the path from the input value to the output
value.


Chapter 1 Functions 91




1.28


Figure 1.64


Using Figure 1.62, evaluate  g( f (2)).


Evaluating Composite Functions Using Formulas
When evaluating a composite function where we have either created or been given formulas, the rule of working from the
inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a
numerical value, a variable name, or a more complicated expression.
While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that
will calculate the result of a composition   f ⎛⎝g(x)⎞⎠. To do this, we will extend our idea of function evaluation. Recall that,
when we evaluate a function like   f (t) = t2 − t,  we substitute the value inside the parentheses into the formula wherever
we see the input variable.


Given a formula for a composite function, evaluate the function.
1. Evaluate the inside function using the input value or variable provided.
2. Use the resulting output as the input to the outside function.


Example 1.45
Evaluating a Composition of Functions Expressed as Formulas with a Numerical
Input


Given   f (t) = t2 − t  and  h(x) = 3x + 2,   evaluate   f (h(1)).


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1.29


Solution
Because the inside expression is  h(1),  we start by evaluating  h(x)  at 1.


h(1) = 3(1) + 2
h(1) = 5


Then   f (h(1)) = f (5),   so we evaluate   f (t)  at an input of 5.
f (h(1)) = f (5)


f (h(1)) = 52 − 5


f (h(1)) = 20


Analysis
It makes no difference what the input variables  t  and  x were called in this problem because we evaluated for
specific numerical values.


Given   f (t) = t2 − t  and  h(x) = 3x + 2,   evaluate


a. h( f (2))
b. h( f ( − 2))


Finding the Domain of a Composite Function
As we discussed previously, the domain of a composite function such as   f ∘g  is dependent on the domain of  g  and the
domain of   f .  It is important to know when we can apply a composite function and when we cannot, that is, to know the
domain of a function such as   f ∘g. Let us assume we know the domains of the functions   f   and  g  separately. If we write
the composite function for an input  x  as   f ⎛⎝g(x)⎞⎠,  we can see right away that  x must be a member of the domain of  g  in
order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However,
we also see that  g(x) must be a member of the domain of   f ,   otherwise the second function evaluation in   f ⎛⎝g(x)⎞⎠  cannot
be completed, and the expression is still undefined. Thus the domain of   f ∘g  consists of only those inputs in the domain of
 g  that produce outputs from  g  belonging to the domain of   f . Note that the domain of   f   composed with  g  is the set of all
 x  such that  x  is in the domain of  g  and  g(x)  is in the domain of   f .


Domain of a Composite Function
The domain of a composite function   f ⎛⎝g(x)⎞⎠  is the set of those inputs  x  in the domain of  g  for which  g(x)  is in the
domain of   f .


Given a function composition   f(g(x)), determine its domain.
1. Find the domain of  g.
2. Find the domain of   f .
3. Find those inputs  x  in the domain of  g  for which  g(x)  is in the domain of   f . That is, exclude those
inputs  x  from the domain of  g  for which  g(x)  is not in the domain of   f . The resulting set is the domain
of   f ∘g.


Chapter 1 Functions 93




Example 1.46
Finding the Domain of a Composite Function


Find the domain of

⎝ f ∘g⎞⎠(x) where       f (x) = 5x − 1 and   g(x) =


4
3x − 2


Solution
The domain of  g(x)  consists of all real numbers except  x = 2


3
,   since that input value would cause us to divide


by 0. Likewise, the domain of   f   consists of all real numbers except 1. So we need to exclude from the domain of
 g(x)  that value of  x  for which  g(x) = 1.


4
3x − 2


= 1


  4 = 3x − 2
  6 = 3x
  x = 2


So the domain of   f ∘g  is the set of all real numbers except  2
3
  and  2. This means that


x ≠ 2
3
  or  x ≠ 2


We can write this in interval notation as

⎝−∞,


2
3

⎠ ∪


2
3
, 2⎞⎠ ∪ (2, ∞)


Example 1.47
Finding the Domain of a Composite Function Involving Radicals


Find the domain of

⎝ f ∘g⎞⎠(x) where       f (x) = x + 2 and   g(x) = 3 − x


Solution
Because we cannot take the square root of a negative number, the domain of  g  is  (−∞, 3]. Now we check the
domain of the composite function



⎝ f ∘g⎞⎠(x) = 3 − x + 2 or ⎛⎝ f ∘g⎞⎠(x) = 5 − x


The domain of this function is  (−∞, 5⎤⎦. To find the domain of   f ∘g,  we ask ourselves if there are any further
restrictions offered by the domain of the composite function. The answer is no, since  (−∞, 3]  is a proper subset
of the domain of   f ∘g. This means the domain of   f ∘g  is the same as the domain of  g,   namely,  (−∞, 3].


Analysis


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1.30


1.31


This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful
in finding the domain of a composite function. It also shows that the domain of   f ∘g  can contain values that are
not in the domain of   f ,   though they must be in the domain of  g.


Find the domain of

⎝ f ∘g⎞⎠(x) where       f (x) = 1x − 2 and   g(x) = x + 4


Decomposing a Composite Function into its Component Functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition
of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the
decomposition that appears to be most expedient.


Example 1.48
Decomposing a Function


Write   f (x) = 5 − x2  as the composition of two functions.


Solution
We are looking for two functions,  g  and  h,   so   f (x) = g(h(x)). To do this, we look for a function inside a
function in the formula for   f (x). As one possibility, we might notice that the expression  5 − x2   is the inside of
the square root. We could then decompose the function as


h(x) = 5 − x2 and g(x) = x


We can check our answer by recomposing the functions.
g(h(x)) = g⎛⎝5 − x


2⎞
⎠ = 5 − x


2


Write   f (x) = 4
3 − 4 + x2


  as the composition of two functions.


Access these online resources for additional instruction and practice with composite functions.
• Composite Functions (http://openstaxcollege.org/l/compfunction)
• Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot)
• Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph)
• Decompose Functions (http://openstaxcollege.org/l/decompfunction)
• Composite Function Values (http://openstaxcollege.org/l/compfuncvalue)


Chapter 1 Functions 95




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1.4 EXERCISES
Verbal


How does one find the domain of the quotient of two
functions,   fg ?  


What is the composition of two functions,   f ∘g?


If the order is reversed when composing two
functions, can the result ever be the same as the answer in
the original order of the composition? If yes, give an
example. If no, explain why not.


How do you find the domain for the composition of
two functions,   f ∘g?


Algebraic


Given   f (x) = x2 + 2x and g(x) = 6 − x2, find
  f + g,   f − g,   f g, and   fg .Determine the domain for
each function in interval notation.


Given   f (x) = − 3x2 + x and g(x) = 5,   find
  f + g,   f − g,   f g,   and   fg . Determine the domain for
each function in interval notation.


Given   f (x) = 2x2 + 4x and g(x) = 1
2x


, find


  f + g,   f − g,   f g, and fg . Determine the domain for
each function in interval notation.


Given   f (x) = 1
x − 4


  and  g(x) = 1
6 − x


,   find


  f + g,   f − g,   f g, and fg . Determine the domain for
each function in interval notation.


Given   f (x) = 3x2   and  g(x) = x − 5,   find
  f + g,   f − g,   f g, and fg . Determine the domain for
each function in interval notation.


Given   f (x) = x  and  g(x) = |x − 3|,   find  gf .
Determine the domain of the function in interval notation.


Given   f (x) = 2x2 + 1  and  g(x) = 3x − 5,   find the
following:
a. f (g(2))
b. f (g(x))


c. g( f (x))
d. (g ∘g)(x)
e. ⎛⎝ f ∘ f ⎞⎠(−2)


For the following exercises, use each pair of functions to
find   f ⎛⎝g(x)⎞⎠  and  g⎛⎝ f (x)⎞⎠.  Simplify your answers.


f (x) = x2 + 1,  g(x) = x + 2


f (x) = x + 2,  g(x) = x2 + 3


f (x) = |x|,  g(x) = 5x + 1


f (x) = x3 ,  g(x) = x + 1
x3


f (x) = 1
x − 6


,  g(x) = 7x + 6


f (x) = 1
x − 4


,  g(x) = 2x + 4


For the following exercises, use each set of functions to
find   f ⎛⎝g⎛⎝h(x)⎞⎠⎞⎠.  Simplify your answers.


f (x) = x4 + 6,   g(x) = x − 6,   and  h(x) = x


f (x) = x2 + 1,   g(x) = 1x ,   and  h(x) = x + 3


Given   f (x) = 1x   and  g(x) = x − 3,   find the
following:
a. ( f ∘g)(x)
b. the domain of  ( f ∘g)(x)  in interval notation
c. (g ∘ f )(x)
d. the domain of  (g ∘ f )(x) 


e. ⎛⎝
f
g

⎠x


Given   f (x) = 2 − 4x  and  g(x) = − 3x ,   find the
following:
a. (g ∘ f )(x)
b. the domain of  (g ∘ f )(x)  in interval notation


Given the functions
f (x) = 1 − xx  and g(x) = 11 + x2,   find the following:


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244.


a. (g ∘ f )(x)
b. (g ∘ f )(2)


Given functions  p(x) = 1x   and  m(x) = x2 − 4,  
state the domain of each of the following functions using
interval notation:


a. p(x)
m(x)


b. p(m(x))
c. m(p(x))


Given functions  q(x) = 1x   and  h(x) = x2 − 9,  
state the domain of each of the following functions using
interval notation.


a. q(x)
h(x)


b. q⎛⎝h(x)⎞⎠
c. h⎛⎝q(x)⎞⎠


For   f (x) = 1x   and  g(x) = x − 1,  write the domain
of  ( f ∘g)(x)  in interval notation.


For the following exercises, find functions   f (x)  and  g(x) 
so the given function can be expressed as  h(x) = f ⎛⎝g(x)⎞⎠.


h(x) = (x + 2)2


h(x) = (x − 5)3


h(x) = 3
x − 5


h(x) = 4
(x + 2)2


h(x) = 4 + x3


h(x) = 1
2x − 3


3


h(x) = 1
(3x2 − 4)−3


h(x) = 3x − 2
x + 5


4


h(x) =


8 + x3


8 − x3



4


h(x) = 2x + 6


h(x) = (5x − 1)3


h(x) = x − 1
3


h(x) = |x2 + 7|
h(x) = 1


(x − 2)3


h(x) = ⎛⎝
1


2x − 3



2


h(x) = 2x − 1
3x + 4


Graphical
For the following exercises, use the graphs of   f , shown in
Figure 1.65, and  g, shown in Figure 1.66, to evaluate
the expressions.


Figure 1.65


Figure 1.66


f ⎛⎝g(3)⎞⎠


f ⎛⎝g(1)⎞⎠


g⎛⎝ f (1)⎞⎠


Chapter 1 Functions 97




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253.


254.


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257.


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259.


260.


g⎛⎝ f (0)⎞⎠


f ⎛⎝ f (5)⎞⎠


f ⎛⎝ f (4)⎞⎠


g⎛⎝g(2)⎞⎠


g⎛⎝g(0)⎞⎠


For the following exercises, use graphs of   f (x), shown
in Figure 1.67,  g(x), shown in Figure 1.68, and  h(x),
shown in Figure 1.69, to evaluate the expressions.


Figure 1.67


Figure 1.68


Figure 1.69


g⎛⎝ f (1)⎞⎠


g⎛⎝ f (2)⎞⎠


f ⎛⎝g(4)⎞⎠


f ⎛⎝g(1)⎞⎠


f ⎛⎝h(2)⎞⎠


h⎛⎝ f (2)⎞⎠


f ⎛⎝g⎛⎝h(4)⎞⎠⎞⎠


f ⎛⎝g⎛⎝ f (−2)⎞⎠⎞⎠


Numeric
For the following exercises, use the function values for
  f and g  shown in Table 1.23 to evaluate each
expression.


x f(x) g(x)


0 7 9


1 6 5


2 5 6


3 8 2


4 4 1


5 0 8


6 2 7


7 1 3


8 9 4


9 3 0


Table 1.23


f ⎛⎝g(8)⎞⎠


f ⎛⎝g(5)⎞⎠


g⎛⎝ f (5)⎞⎠


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g⎛⎝ f (3)⎞⎠


f ⎛⎝ f (4)⎞⎠


f ⎛⎝ f (1)⎞⎠


g⎛⎝g(2)⎞⎠


g⎛⎝g(6)⎞⎠


For the following exercises, use the function values for
  f and g  shown in Table 1.24 to evaluate the expressions.


x f(x) g(x)


-3 11 -8


-2 9 -3


-1 7 0


0 5 1


1 3 0


2 1 -3


3 -1 -8


Table 1.24


( f ∘g)(1)


( f ∘g)(2)


(g ∘ f )(2)


(g ∘ f )(3)


(g ∘g)(1)


( f ∘ f )(3)


For the following exercises, use each pair of functions to
find   f ⎛⎝g(0)⎞⎠  and  g⎛⎝ f (0)⎞⎠.


f (x) = 4x + 8,  g(x) = 7 − x2


f (x) = 5x + 7,  g(x) = 4 − 2x2


f (x) = x + 4,  g(x) = 12 − x3


f (x) = 1
x + 2


,  g(x) = 4x + 3


For the following exercises, use the functions
  f (x) = 2x2 + 1  and  g(x) = 3x + 5  to evaluate or find
the composite function as indicated.


f ⎛⎝g(2)⎞⎠


f ⎛⎝g(x)⎞⎠


g⎛⎝ f ( − 3)⎞⎠


(g ∘g)(x)


Extensions
For the following exercises, use   f (x) = x3 + 1  and
 g(x) = x − 13 .


Find  ( f ∘g)(x)  and  (g ∘ f )(x). Compare the two
answers.


Find  ( f ∘g)(2)  and  (g ∘ f )(2).


What is the domain of  (g ∘ f )(x)?


What is the domain of  ( f ∘g)(x)?


Let   f (x) = 1x .
a. Find  ( f ∘ f )(x).
b. Is  ( f ∘ f )(x)  for any function   f   the same result as
the answer to part (a) for any function? Explain.


For the following exercises, let  F(x) = (x + 1)5,  
f (x) = x5,   and  g(x) = x + 1.


True or False:  (g ∘ f )(x) = F(x).


True or False:  ( f ∘g)(x) = F(x).


For the following exercises, find the composition when
  f (x) = x2 + 2  for all  x ≥ 0  and  g(x) = x − 2.


( f ∘g)(6);  (g ∘ f )(6)


(g ∘ f )(a);  ( f ∘g)(a)


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289.


290.


291.


292.


293.


294.


295.


296.


297.


( f ∘g)(11);  (g ∘ f )(11)


Real-World Applications
The function  D(p)  gives the number of items that


will be demanded when the price is  p. The production cost
 C(x)  is the cost of producing  x  items. To determine the
cost of production when the price is $6, you would do
which of the following?
a. Evaluate  D⎛⎝C(6)⎞⎠.
b. Evaluate  C⎛⎝D(6)⎞⎠.
c. Solve  D(C(x)) = 6.
d. Solve  C⎛⎝D(p)⎞⎠ = 6.


The function  A(d)  gives the pain level on a scale of 0
to 10 experienced by a patient with  d milligrams of a pain-
reducing drug in her system. The milligrams of the drug in
the patient’s system after  t minutes is modeled by  m(t). 
Which of the following would you do in order to determine
when the patient will be at a pain level of 4?
a. Evaluate  A(m(4)).
b. Evaluate  m(A(4)).
c. Solve  A(m(t)) = 4.
d. Solve  m⎛⎝A(d)⎞⎠ = 4.


A store offers customers a 30% discount on the price
 x  of selected items. Then, the store takes off an additional
15% at the cash register. Write a price function  P(x)  that
computes the final price of the item in terms of the original
price  x.  (Hint: Use function composition to find your
answer.)


A rain drop hitting a lake makes a circular ripple. If
the radius, in inches, grows as a function of time in minutes
according to  r(t) = 25 t + 2,   find the area of the ripple as
a function of time. Find the area of the ripple at  t = 2.


A forest fire leaves behind an area of grass burned in
an expanding circular pattern. If the radius of the circle of
burning grass is increasing with time according to the
formula  r(t) = 2t + 1,   express the area burned as a
function of time,  t  (minutes).


Use the function you found in the previous exercise to
find the total area burned after 5 minutes.


The radius  r,   in inches, of a spherical balloon is
related to the volume,  V ,   by  r(V) = 3V



3


. Air is pumped


into the balloon, so the volume after  t  seconds is given by
 V(t) = 10 + 20t.
a. Find the composite function  r(V(t)).
b. Find the exact time when the radius reaches 10
inches.
The number of bacteria in a refrigerated food product


is given by N(T) = 23T 2 − 56T + 1,   3 < T < 33,
where  T is the temperature of the food. When the food is
removed from the refrigerator, the temperature is given by
T(t) = 5t + 1.5, where t is the time in hours.
a. Find the composite function  N(T(t)).
b. Find the time (round to two decimal places) when
the bacteria count reaches 6752.


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1.5 | Transformation of Functions
Learning Objectives


In this section, you will:
1.5.1 Graph functions using vertical and horizontal shifts.
1.5.2 Graph functions using reflections about the  x -axis and the  y -axis.
1.5.3 Determine whether a function is even, odd, or neither from its graph.
1.5.4 Graph functions using compressions and stretches.
1.5.5 Combine transformations.


Figure 1.70 (credit: "Misko"/Flickr)


We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt
the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror?
Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or
vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or
processes in the real world. In this section, we will take a look at several kinds of transformations.
Graphing Functions Using Vertical and Horizontal Shifts
Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and
equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given
scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to
solve.
Identifying Vertical Shifts
One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest
shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative
constant to the function. In other words, we add the same constant to the output value of the function regardless of the input.
For a function  g(x) = f (x) + k,   the function   f (x)  is shifted vertically  k  units. See Figure 1.71 for an example.


Chapter 1 Functions 101




Figure 1.71 Vertical shift by  k = 1  of the cube root function
  f (x) = x3 .


To help you visualize the concept of a vertical shift, consider that  y = f (x). Therefore,   f (x) + k  is equivalent to  y + k. 
Every unit of  y  is replaced by  y + k,   so the  y- value increases or decreases depending on the value of  k. The result is a
shift upward or downward.


Vertical Shift
Given a function f (x), a new function g(x) = f (x) + k, where  k is a constant, is a vertical shift of the function
f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will
shift down.


Example 1.49
Adding a Constant to a Function


To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure
1.72 shows the area of open vents  V   (in square feet) throughout the day in hours after midnight,  t. During the
summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open
vents by 20 square feet throughout the day and night. Sketch a graph of this new function.


Figure 1.72


Solution


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We can sketch a graph of this new function by adding 20 to each of the output values of the original function.
This will have the effect of shifting the graph vertically up, as shown in Figure 1.73.


Figure 1.73


Notice that in Figure 1.73, for each input value, the output value has increased by 20, so if we call the new
function  S(t), we could write


S(t) = V(t) + 20


This notation tells us that, for any value of  t, S(t)  can be found by evaluating the function  V   at the same input
and then adding 20 to the result. This defines  S  as a transformation of the function  V ,   in this case a vertical shift
up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change.
See Table 1.25.


t 0 8 10 17 19 24


V(t) 0 0 220 220 0 0


S(t) 20 20 240 240 20 20


Table 1.25


Given a tabular function, create a new row to represent a vertical shift.
1. Identify the output row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.


Example 1.50
Shifting a Tabular Function Vertically


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1.32


A function   f (x)  is given in Table 1.26. Create a table for the function  g(x) = f (x) − 3.


x 2 4 6 8


f(x) 1 3 7 11


Table 1.26


Solution
The formula  g(x) = f (x) − 3  tells us that we can find the output values of  g  by subtracting 3 from the output
values of   f .  For example:


f (2) = 1 Given


g(x) = f (x) − 3 Given transformation
g(2) = f (2) − 3


= 1 − 3
= − 2


Subtracting 3 from each   f (x)  value, we can complete a table of values for  g(x)  as shown in Table 1.27.


x 2 4 6 8


f(x) 1 3 7 11


g(x) −2 0 4 8


Table 1.27


Analysis
As with the earlier vertical shift, notice the input values stay the same and only the output values change.


The function  h(t) = − 4.9t2 + 30t  gives the height  h  of a ball (in meters) thrown upward from the
ground after  t  seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new
height function  b(t)  to  h(t),   and then find a formula for  b(t).


Identifying Horizontal Shifts
We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to
input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of
the function left or right in what is known as a horizontal shift, shown in Figure 1.74.


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Figure 1.74 Horizontal shift of the function   f (x) = x3 . Note
that  h = + 1  shifts the graph to the left, that is, towards
negative values of  x.


For example, if   f (x) = x2,   then  g(x) = (x − 2)2   is a new function. Each input is reduced by 2 prior to squaring the
function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2
units to yield the same output value as given in   f .


Horizontal Shift
Given a function   f ,   a new function  g(x) = f (x − h),  where  h  is a constant, is a horizontal shift of the function   f . 
If  h  is positive, the graph will shift right. If  h  is negative, the graph will shift left.


Example 1.51
Adding a Constant to an Input


Returning to our building airflow example from Figure 1.72, suppose that in autumn the facilities manager
decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier.
Sketch a graph of the new function.


Solution
We can set  V(t)  to be the original program and  F(t)  to be the revised program.


             V(t) = the original venting plan
F(t) = starting 2 hrs sooner


In the new graph, at each time, the airflow is the same as the original function  V  was 2 hours later. For example,
in the original function  V ,   the airflow starts to change at 8 a.m., whereas for the function  F,   the airflow starts to
change at 6 a.m. The comparable function values are  V(8) = F(6).  See Figure 1.75. Notice also that the vents
first opened to  220 ft2   at 10 a.m. under the original plan, while under the new plan the vents reach  220 ft2   at 8
a.m., so  V(10) = F(8).
In both cases, we see that, because  F(t)  starts 2 hours sooner,  h = − 2. That means that the same output values
are reached when  F(t) = V(t − (−2)) = V(t + 2).


Chapter 1 Functions 105




Figure 1.75


Analysis
Note that  V(t + 2)  has the effect of shifting the graph to the left.
Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and
often seem counterintuitive. The new function  F(t)  uses the same outputs as  V(t),   but matches those outputs to
inputs 2 hours earlier than those of  V(t).  Said another way, we must add 2 hours to the input of  V   to find the
corresponding output for F : F(t) = V(t + 2).


Given a tabular function, create a new row to represent a horizontal shift.
1. Identify the input row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each input cell.


Example 1.52
Shifting a Tabular Function Horizontally


A function   f (x)  is given in Table 1.28. Create a table for the function  g(x) = f (x − 3).


x 2 4 6 8


f(x) 1 3 7 11


Table 1.28


Solution
The formula  g(x) = f (x − 3)  tells us that the output values of  g  are the same as the output value of   f   when the
input value is 3 less than the original value. For example, we know that   f (2) = 1. To get the same output from


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the function  g,  we will need an input value that is 3 larger. We input a value that is 3 larger for  g(x)  because
the function takes 3 away before evaluating the function   f .


g(5) = f (5 − 3)
            = f (2)
            = 1


We continue with the other values to create Table 1.29.


x 5 7 9 11


x−3 2 4 6 8


f(x) 1 3 7 11


g(x) 1 3 7 11


Table 1.29


The result is that the function  g(x)  has been shifted to the right by 3. Notice the output values for  g(x)  remain
the same as the output values for   f (x),   but the corresponding input values,  x,   have shifted to the right by 3.
Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.


Analysis
Figure 1.76 represents both of the functions. We can see the horizontal shift in each point.


Chapter 1 Functions 107




Example 1.53
Identifying a Horizontal Shift of a Toolkit Function


Figure 1.77 represents a transformation of the toolkit function   f (x) = x2. Relate this new function  g(x)  to
  f (x),   and then find a formula for  g(x).


Figure 1.77


Solution
Notice that the graph is identical in shape to the   f (x) = x2   function, but the x-values are shifted to the right 2
units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function
shifted 2 units to the right, so


g(x) = f (x − 2)


Notice how we must input the value  x = 2  to get the output value  y = 0;   the x-values must be 2 units larger
because of the shift to the right by 2 units. We can then use the definition of the   f (x)  function to write a formula
for  g(x)  by evaluating   f (x − 2).


f (x) = x2


g(x) = f (x − 2)


g(x) = f (x − 2) = (x − 2)2


Analysis
To determine whether the shift is  + 2  or  − 2 , consider a single reference point on the graph. For a quadratic,
looking at the vertex point is convenient. In the original function,   f (0) = 0.  In our shifted function,  g(2) = 0. 
To obtain the output value of 0 from the function   f ,  we need to decide whether a plus or a minus sign will work
to satisfy  g(2) = f (x − 2) = f (0) = 0.  For this to work, we will need to subtract 2 units from our input values.


Example 1.54
Interpreting Horizontal versus Vertical Shifts


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The function  G(m)  gives the number of gallons of gas required to drive  m miles. Interpret  G(m) + 10  and
 G(m + 10).


Solution
G(m) + 10  can be interpreted as adding 10 to the output, gallons. This is the gas required to drive  m miles, plus
another 10 gallons of gas. The graph would indicate a vertical shift.
G(m + 10)  can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to
drive 10 miles more than  m miles. The graph would indicate a horizontal shift.


Given the function   f (x) = x,   graph the original function   f (x)  and the transformation
 g(x) = f (x + 2)  on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by
how many units?


Combining Vertical and Horizontal Shifts
Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the
output ( y- ) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input ( x- )
axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift
up or down and right or left.


Given a function and both a vertical and a horizontal shift, sketch the graph.
1. Identify the vertical and horizontal shifts from the formula.
2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant
and down for a negative constant.


3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant
and right for a negative constant.


4. Apply the shifts to the graph in either order.


Example 1.55
Graphing Combined Vertical and Horizontal Shifts


Given   f (x) = |x|,   sketch a graph of  h(x) = f (x + 1) − 3.


Solution
The function   f   is our toolkit absolute value function. We know that this graph has a V shape, with the point at the
origin. The graph of  h  has transformed   f   in two ways:   f (x + 1)  is a change on the inside of the function, giving
a horizontal shift left by 1, and the subtraction by 3 in   f (x + 1) − 3  is a change to the outside of the function,
giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 1.78.
Let us follow one point of the graph of   f (x) = |x|.


• The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0)


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1.34


• The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3)


Figure 1.78


Figure 1.79 shows the graph of  h.


Figure 1.79


Given   f (x) = |x|,   sketch a graph of  h(x) = f (x − 2) + 4.


Example 1.56
Identifying Combined Vertical and Horizontal Shifts


Write a formula for the graph shown in Figure 1.80, which is a transformation of the toolkit square root function.


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1.35


Figure 1.80


Solution
The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In
function notation, we could write that as


h(x) = f (x − 1) + 2


Using the formula for the square root function, we can write
h(x) = x − 1 + 2


Analysis
Note that this transformation has changed the domain and range of the function. This new graph has domain
 [1, ∞)  and range  [2, ∞).


Write a formula for a transformation of the toolkit reciprocal function   f (x) = 1x   that shifts the function’s
graph one unit to the right and one unit up.


Graphing Functions Using Reflections about the Axes
Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects
a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The
reflections are shown in Figure 1.81.


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Figure 1.81 Vertical and horizontal reflections of a function.


Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis.
The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.


Reflections
Given a function   f (x), a new function  g(x) = − f (x)  is a vertical reflection of the function   f (x),   sometimes called
a reflection about (or over, or through) the x-axis.
Given a function   f (x),   a new function  g(x) = f ( − x)  is a horizontal reflection of the function   f (x),   sometimes
called a reflection about the y-axis.


Given a function, reflect the graph both vertically and horizontally.
1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph
about the x-axis.


2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph
about the y-axis.


Example 1.57
Reflecting a Graph Horizontally and Vertically


Reflect the graph of  s(t) = t (a) vertically and (b) horizontally.


Solution
a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as
shown in Figure 1.82.


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Figure 1.82 Vertical reflection of the square root function


Because each output value is the opposite of the original output value, we can write
V(t) = − s(t) or V(t) = − t


Notice that this is an outside change, or vertical shift, that affects the output  s(t)  values, so the negative
sign belongs outside of the function.


b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in
Figure 1.83.


Figure 1.83 Horizontal reflection of the square root function


Because each input value is the opposite of the original input value, we can write
H(t) = s( − t) or H(t) = −t


Chapter 1 Functions 113




1.36


Notice that this is an inside change or horizontal change that affects the input values, so the negative sign
is on the inside of the function.
Note that these transformations can affect the domain and range of the functions. While the original
square root function has domain  [0, ∞)  and range  [0, ∞),   the vertical reflection gives the  V(t) 
function the range (−∞,  0] and the horizontal reflection gives the  H(t)  function the domain (−∞,  0].


Reflect the graph of   f (x) = |x − 1| (a) vertically and (b) horizontally.


Example 1.58
Reflecting a Tabular Function Horizontally and Vertically


A function   f (x)  is given as Table 1.30. Create a table for the functions below.
a.  g(x) = − f (x)
b.  h(x) = f ( − x)


x 2 4 6 8


f(x) 1 3 7 11


Table 1.30


Solution
a. For  g(x),   the negative sign outside the function indicates a vertical reflection, so the x-values stay the
same and each output value will be the opposite of the original output value. See Table 1.31.


x 2 4 6 8


 g(x)  –1 –3 –7 –11


Table 1.31
b. For  h(x),   the negative sign inside the function indicates a horizontal reflection, so each input value will
be the opposite of the original input value and the  h(x)  values stay the same as the   f (x)  values. See
Table 1.32.


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1.37


x −2 −4 −6 −8


h(x) 1 3 7 11


Table 1.32


A function   f (x)  is given as Table 1.33. Create a table for the functions below.


a. g(x) = − f (x)
b. h(x) = f ( − x)


x −2 0 2 4


f(x) 5 10 15 20


Table 1.33


Example 1.59
Applying a Learning Model Equation


A common model for learning has an equation similar to k(t) = − 2−t + 1,   where k is the percentage of
mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2t shown
in Figure 1.84. Sketch a graph of k(t).


Chapter 1 Functions 115




Figure 1.84


Solution
This equation combines three transformations into one equation.


• A horizontal reflection: f ( − t) = 2−t


• A vertical reflection:  − f ( − t) = − 2−t


• A vertical shift:  − f ( − t) + 1 = − 2−t + 1
We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two
points through each of the three transformations. We will choose the points (0, 1) and (1, 2).
1. First, we apply a horizontal reflection: (0, 1) (–1, 2).
2. Then, we apply a vertical reflection: (0, −1) (1, –2).
3. Finally, we apply a vertical shift: (0, 0) (1, 1).


This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations.
In Figure 1.85, the first graph results from a horizontal reflection. The second results from a vertical reflection.
The third results from a vertical shift up 1 unit.


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Figure 1.85


Analysis
As a model for learning, this function would be limited to a domain of  t ≥ 0,  with corresponding range  [0, 1).


Given the toolkit function   f (x) = x2,   graph  g(x) = − f (x)  and  h(x) = f ( − x). Take note of any
surprising behavior for these functions.


Determining Even and Odd Functions
Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the
toolkit functions f (x) = x2 or f (x) = |x| will result in the original graph. We say that these types of graphs are symmetric
about the y-axis. Functions whose graphs are symmetric about the y-axis are called even functions.
If the graphs of   f (x) = x3   or   f (x) = 1x  were reflected over both axes, the result would be the original graph, as shown in
Figure 1.86.


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Figure 1.86 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical
reflections reproduce the original cubic function.


We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called
an odd function.
Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example,   f (x) = 2x   is neither even
nor odd. Also, the only function that is both even and odd is the constant function   f (x) = 0.


Even and Odd Functions
A function is called an even function if for every input  x


f (x) = f ( − x)


The graph of an even function is symmetric about the y- axis.
A function is called an odd function if for every input  x


f (x) = − f ( − x)


The graph of an odd function is symmetric about the origin.


Given the formula for a function, determine if the function is even, odd, or neither.
1. Determine whether the function satisfies   f (x) = f ( − x).  If it does, it is even.
2. Determine whether the function satisfies   f (x) = − f ( − x).  If it does, it is odd.
3. If the function does not satisfy either rule, it is neither even nor odd.


Example 1.60
Determining whether a Function Is Even, Odd, or Neither


Is the function   f (x) = x3 + 2x  even, odd, or neither?


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Solution
Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the
reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.


f ( − x) = ( − x)3 + 2( − x) = − x3 − 2x


This does not return us to the original function, so this function is not even. We can now test the rule for odd
functions.


− f ( − x) = − ⎛⎝−x
3 − 2x⎞⎠ = x


3 + 2x


Because  − f ( − x) = f (x),   this is an odd function.


Analysis
Consider the graph of   f   in Figure 1.87. Notice that the graph is symmetric about the origin. For every point
 (x, y)  on the graph, the corresponding point  (−x, − y)  is also on the graph. For example, (1, 3) is on the graph
of   f ,   and the corresponding point (−1, −3) is also on the graph.


Figure 1.87


Is the function   f (s) = s4 + 3s2 + 7  even, odd, or neither?


Graphing Functions Using Stretches and Compressions
Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did
not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity.
We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each
change has a specific effect that can be seen graphically.


Chapter 1 Functions 119




Vertical Stretches and Compressions
When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically
in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is
between 0 and 1, we get a vertical compression. Figure 1.88 shows a function multiplied by constant factors 2 and 0.5
and the resulting vertical stretch and compression.


Figure 1.88 Vertical stretch and compression


Vertical Stretches and Compressions
Given a function   f (x),   a new function  g(x) = a f (x),  where  a  is a constant, is a vertical stretch or vertical
compression of the function   f (x).


• If  a > 1,   then the graph will be stretched.
• If  0 < a < 1,   then the graph will be compressed.
• If  a < 0,   then there will be combination of a vertical stretch or compression with a vertical reflection.


Given a function, graph its vertical stretch.
1. Identify the value of  a.
2. Multiply all range values by  a.
3. If   a > 1,    the graph is stretched by a factor of  a.
If  0 < a < 1,   the graph is compressed by a factor of  a.
If  a < 0,   the graph is either stretched or compressed and also reflected about the x-axis.


Example 1.61
Graphing a Vertical Stretch


A function  P(t) models the population of fruit flies. The graph is shown in Figure 1.89.


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Figure 1.89


A scientist is comparing this population to another population,  Q,  whose growth follows the same pattern, but
is twice as large. Sketch a graph of this population.


Solution
Because the population is always twice as large, the new population’s output values are always twice the original
function’s output values. Graphically, this is shown in Figure 1.90.
If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2.
The following shows where the new points for the new graph will be located.


(0, 1) → (0, 2)
(3, 3) → (3, 6)
(6, 2) → (6, 4)
(7, 0) → (7, 0)


Figure 1.90


Symbolically, the relationship is written as
Q(t) = 2P(t)


Chapter 1 Functions 121




This means that for any input  t,   the value of the function  Q  is twice the value of the function  P. Notice that the
effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal
axis. The input values,  t,   stay the same while the output values are twice as large as before.


Given a tabular function and assuming that the transformation is a vertical stretch or compression, create
a table for a vertical compression.


1. Determine the value of  a.
2. Multiply all of the output values by  a.


Example 1.62
Finding a Vertical Compression of a Tabular Function


A function   f   is given as Table 1.34. Create a table for the function  g(x) = 1
2
f (x).


x 2 4 6 8


f(x)  1 3 7 11


Table 1.34


Solution
The formula  g(x) = 1


2
f (x)  tells us that the output values of  g  are half of the output values of   f  with the same


inputs. For example, we know that   f (4) = 3. Then


g(4) = 1
2
f (4) = 1


2
(3) = 3


2


We do the same for the other values to produce Table 1.35.


x 2 4 6 8


g(x) 1
2


3
2


7
2


11
2


Table 1.35


Analysis
The result is that the function  g(x)  has been compressed vertically by  1


2
. Each output value is divided in half, so


the graph is half the original height.


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1.40 A function   f   is given as Table 1.36. Create a table for the function  g(x) = 3
4
f (x).


x 2 4 6 8


f (x) 12 16 20 0


Table 1.36


Example 1.63
Recognizing a Vertical Stretch


The graph in Figure 1.91 is a transformation of the toolkit function   f (x) = x3. Relate this new function  g(x) 
to   f (x),   and then find a formula for  g(x).


Figure 1.91


Solution
When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively
clear. In this graph, it appears that  g(2) = 2. With the basic cubic function at the same input,   f (2) = 23 = 8. 
Based on that, it appears that the outputs of  g  are  1


4
  the outputs of the function   f   because  g(2) = 1


4
f (2).  From


this we can fairly safely conclude that  g(x) = 1
4
f (x).


We can write a formula for  g  by using the definition of the function   f .


g(x) = 1
4
f (x) = 1


4
x3


Chapter 1 Functions 123




1.41 Write the formula for the function that we get when we stretch the identity toolkit function by a factor of
3, and then shift it down by 2 units.


Horizontal Stretches and Compressions
Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a
function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant
is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the
function.


Figure 1.92


Given a function  y = f (x),   the form  y = f (bx)  results in a horizontal stretch or compression. Consider the function
 y = x2. Observe Figure 1.92. The graph of  y = (0.5x)2   is a horizontal stretch of the graph of the function  y = x2   by a
factor of 2. The graph of  y = (2x)2   is a horizontal compression of the graph of the function  y = x2   by a factor of 2.


Horizontal Stretches and Compressions
Given a function   f (x),   a new function  g(x) = f (bx),  where  b  is a constant, is a horizontal stretch or horizontal
compression of the function   f (x).


• If  b > 1,   then the graph will be compressed by  1
b
.


• If  0 < b < 1,   then the graph will be stretched by  1
b
.


• If  b < 0,   then there will be combination of a horizontal stretch or compression with a horizontal reflection.


Given a description of a function, sketch a horizontal compression or stretch.
1. Write a formula to represent the function.
2. Set  g(x) = f (bx) where  b > 1  for a compression or  0 < b < 1  for a stretch.


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Example 1.64
Graphing a Horizontal Compression


Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan
twice as fast as the original population. In other words, this new population,  R,  will progress in 1 hour the
same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original
population does in 4 hours. Sketch a graph of this population.


Solution
Symbolically, we could write


R(1) = P(2),
R(2) = P(4), and in general,


 R(t) = P(2t).
See Figure 1.93 for a graphical comparison of the original population and the compressed population.


Figure 1.93 (a) Original population graph (b) Compressed population graph


Analysis
Note that the effect on the graph is a horizontal compression where all input values are half of their original
distance from the vertical axis.


Example 1.65
Finding a Horizontal Stretch for a Tabular Function


A function   f (x)  is given as Table 1.37. Create a table for the function  g(x) = f ⎛⎝12x

⎠.


Chapter 1 Functions 125




x 2 4 6 8


f(x) 1 3 7 11


Table 1.37


Solution
The formula  g(x) = f ⎛⎝12x



⎠  tells us that the output values for  g  are the same as the output values for the


function   f   at an input half the size. Notice that we do not have enough information to determine  g(2)  because
 g(2) = f ⎛⎝12 ⋅ 2



⎠ = f (1),   and we do not have a value for   f (1)  in our table. Our input values to  g will need to


be twice as large to get inputs for   f   that we can evaluate. For example, we can determine  g(4).


g(4) = f ⎛⎝
1
2
⋅ 4⎞⎠ = f (2) = 1


We do the same for the other values to produce Table 1.38.


x 4 8 12 16


g(x) 1 3 7 11


Table 1.38


Figure 1.94 shows the graphs of both of these sets of points.


Figure 1.94


Analysis
Because each input value has been doubled, the result is that the function  g(x)  has been stretched horizontally by
a factor of 2.


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1.42


Example 1.66
Recognizing a Horizontal Compression on a Graph


Relate the function  g(x)  to   f (x)  in Figure 1.95.


Figure 1.95


Solution
The graph of  g(x)  looks like the graph of   f (x)  horizontally compressed. Because   f (x)  ends at  (6, 4)  and  g(x) 
ends at  (2, 4),  we can see that the  x- values have been compressed by  1


3
,   because  6⎛⎝13



⎠ = 2. We might also


notice that  g(2) = f (6)  and  g(1) = f (3). Either way, we can describe this relationship as  g(x) = f (3x). This is
a horizontal compression by  1


3
.


Analysis
Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or
compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of  1


4
  in our


function:   f ⎛⎝14x

⎠. This means that the input values must be four times larger to produce the same result, requiring


the input to be larger, causing the horizontal stretching.


Write a formula for the toolkit square root function horizontally stretched by a factor of 3.


Performing a Sequence of Transformations
When combining transformations, it is very important to consider the order of the transformations. For example, vertically
shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then
vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the
original function gets stretched when we stretch first.
When we see an expression such as   2 f (x) + 3,  which transformation should we start with? The answer here follows
nicely from the order of operations. Given the output value of   f (x),  we first multiply by 2, causing the vertical stretch, and
then add 3, causing the vertical shift. In other words, multiplication before addition.
Horizontal transformations are a little trickier to think about. When we write  g(x) = f (2x + 3),   for example, we have to
think about how the inputs to the function  g  relate to the inputs to the function    f .  Suppose we know   f (7) = 12. What
input to  g would produce that output? In other words, what value of  x will allow  g(x) = f (2x + 3) = 12?  We would need


Chapter 1 Functions 127




 2x + 3 = 7. To solve for  x,  we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a
horizontal compression.
This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before
shifting. We can work around this by factoring inside the function.


f (bx + p) = f ⎛⎝b

⎝x +


p
b





Let’s work through an example.
f (x) = (2x + 4)2


We can factor out a 2.
f (x) = ⎛⎝2(x + 2)⎞⎠2


Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way
allows us to horizontally stretch first and then shift horizontally.


Combining Transformations
When combining vertical transformations written in the form  a f (x) + k,   first vertically stretch by  a  and then
vertically shift by  k.
When combining horizontal transformations written in the form   f (bx + h),   first horizontally shift by  h  and then
horizontally stretch by  1


b
.


When combining horizontal transformations written in the form   f (b(x + h)),   first horizontally stretch by  1
b
  and then


horizontally shift by  h.
Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical
transformations are performed first.


Example 1.67
Finding a Triple Transformation of a Tabular Function


Given Table 1.39 for the function   f (x),   create a table of values for the function  g(x) = 2 f (3x) + 1.


x 6 12 18 24


f(x) 10 14 15 17


Table 1.39


Solution
There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal
transformations,   f (3x)  is a horizontal compression by  1


3
,  which means we multiply each  x- value by  1


3
. See


Table 1.40.


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x 2 4 6 8


f(3x) 10 14 15 17


Table 1.40


Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output
values by 2. We apply this to the previous transformation. See Table 1.41.


x 2 4 6 8


2 f(3x) 20 28 30 34


Table 1.41


Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 1.42.


x 2 4 6 8


g(x) = 2 f(3x) +1 21 29 31 35


Table 1.42


Example 1.68
Finding a Triple Transformation of a Graph


Use the graph of   f (x)  in Figure 1.96 to sketch a graph of  k(x) = f ⎛⎝12x + 1

⎠− 3.


Chapter 1 Functions 129




Figure 1.96


Solution
To simplify, let’s start by factoring out the inside of the function.


f ⎛⎝
1
2
x + 1⎞⎠− 3 = f




1
2
(x + 2)⎞⎠− 3


By factoring the inside, we can first horizontally stretch by 2, as indicated by the  1
2
  on the inside of the function.


Remember that twice the size of 0 is still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch to
(4,0). See Figure 1.97.


Figure 1.97


Next, we horizontally shift left by 2 units, as indicated by  x + 2.  See Figure 1.98.


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Figure 1.98


Last, we vertically shift down by 3 to complete our sketch, as indicated by the  − 3  on the outside of the function.
See Figure 1.99.


Figure 1.99


Access this online resource for additional instruction and practice with transformation of functions.
• Function Transformations (http://openstaxcollege.org/l/functrans)


Chapter 1 Functions 131




298.


299.


300.


301.


302.


303.


304.


305.


306.


307.


308.


309.


310.


311.


312.


313.


314.


315.


316.


317.


318.


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320.


321.


322.


323.


1.5 EXERCISES
Verbal


When examining the formula of a function that is the
result of multiple transformations, how can you tell a
horizontal shift from a vertical shift?


When examining the formula of a function that is the
result of multiple transformations, how can you tell a
horizontal stretch from a vertical stretch?


When examining the formula of a function that is the
result of multiple transformations, how can you tell a
horizontal compression from a vertical compression?


When examining the formula of a function that is the
result of multiple transformations, how can you tell a
reflection with respect to the x-axis from a reflection with
respect to the y-axis?


How can you determine whether a function is odd or
even from the formula of the function?


Algebraic
Write a formula for the function obtained when the


graph of   f (x) = x  is shifted up 1 unit and to the left 2
units.


Write a formula for the function obtained when the
graph of   f (x) = |x|  is shifted down 3 units and to the right
1 unit.


Write a formula for the function obtained when the
graph of   f (x) = 1x   is shifted down 4 units and to the right 3
units.


Write a formula for the function obtained when the
graph of   f (x) = 1


x2
  is shifted up 2 units and to the left 4


units.
For the following exercises, describe how the graph of the
function is a transformation of the graph of the original
function   f .


y = f (x − 49)


y = f (x + 43)


y = f (x + 3)


y = f (x − 4)


y = f (x) + 5


y = f (x) + 8


y = f (x) − 2


y = f (x) − 7


y = f (x − 2) + 3


y = f (x + 4) − 1


For the following exercises, determine the interval(s) on
which the function is increasing and decreasing.


f (x) = 4(x + 1)2 − 5


g(x) = 5(x + 3)2 − 2


a(x) = −x + 4


k(x) = − 3 x − 1


Graphical
For the following exercises, use the graph of   f (x) = 2x
shown in Figure 1.100 to sketch a graph of each
transformation of   f (x).


Figure 1.100


g(x) = 2x + 1


h(x) = 2x − 3


w(x) = 2x − 1


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324.


325.


326.


327.


328.


329.


330.


331.


332.


For the following exercises, sketch a graph of the function
as a transformation of the graph of one of the toolkit
functions.


f (t) = (t + 1)2 − 3


h(x) = |x − 1| + 4


k(x) = (x − 2)3 − 1


m(t) = 3 + t + 2


Numeric
Tabular representations for the functions   f ,  g,   and


 h  are given below. Write  g(x)  and  h(x)  as
transformations of   f (x).


x −2 −1 0 1 2


f(x) −2 −1 −3 1 2


x −1 0 1 2 3


g(x) −2 −1 −3 1 2


x −2 −1 0 1 2


h(x) −1 0 −2 2 3


Tabular representations for the functions   f ,  g,   and
 h  are given below. Write  g(x)  and  h(x)  as
transformations of   f (x).


x −2 −1 0 1 2


f(x) −1 −3 4 2 1


x −3 −2 −1 0 1


g(x) −1 −3 4 2 1


x −2 −1 0 1 2


h(x) −2 −4 3 1 0


For the following exercises, write an equation for each
graphed function by using transformations of the graphs of
one of the toolkit functions.


Chapter 1 Functions 133




333.


334.


335.


336.


337.


338.


339.


For the following exercises, use the graphs of
transformations of the square root function to find a
formula for each of the functions.


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340.


341.


342.


343.


344.


345.


346.


347.


348.


349.


350.


351.


352.


353.


354.


355.


356.


357.


358.


359.


360.


361.


362.


For the following exercises, use the graphs of the
transformed toolkit functions to write a formula for each of
the resulting functions.


For the following exercises, determine whether the function
is odd, even, or neither.


f (x) = 3x4


g(x) = x


h(x) = 1x + 3x


f (x) = (x − 2)2


g(x) = 2x4


h(x) = 2x − x3


For the following exercises, describe how the graph of each
function is a transformation of the graph of the original
function   f .


g(x) = − f (x)


g(x) = f ( − x)


g(x) = 4 f (x)


g(x) = 6 f (x)


g(x) = f (5x)


g(x) = f (2x)


g(x) = f ⎛⎝
1
3
x⎞⎠


g(x) = f ⎛⎝
1
5
x⎞⎠


g(x) = 3 f (−x)


g(x) = − f (3x)


For the following exercises, write a formula for the function
 g  that results when the graph of a given toolkit function is
transformed as described.


The graph of   f (x) = |x|  is reflected over the  y -axis
and horizontally compressed by a factor of  1


4
.


The graph of   f (x) = x  is reflected over the  x -axis
and horizontally stretched by a factor of 2.


Chapter 1 Functions 135




363.


364.


365.


366.


367.


368.


369.


370.


371.


372.


373.


374.


375.


376.


377.


378.


The graph of   f (x) = 1
x2


  is vertically compressed by a


factor of  1
3
,   then shifted to the left 2 units and down 3


units.


The graph of   f (x) = 1x   is vertically stretched by a
factor of 8, then shifted to the right 4 units and up 2 units.


The graph of   f (x) = x2   is vertically compressed by
a factor of  1


2
,   then shifted to the right 5 units and up 1


unit.


The graph of   f (x) = x2   is horizontally stretched by
a factor of 3, then shifted to the left 4 units and down 3
units.
For the following exercises, describe how the formula is a
transformation of a toolkit function. Then sketch a graph of
the transformation.


g(x) = 4(x + 1)2 − 5


g(x) = 5(x + 3)2 − 2


h(x) = − 2|x − 4| + 3


k(x) = − 3 x − 1


m(x) = 1
2
x3


n(x) = 1
3|x − 2|


p(x) = ⎛⎝
1
3
x⎞⎠


3
− 3


q(x) = ⎛⎝
1
4
x⎞⎠


3
+ 1


a(x) = −x + 4


For the following exercises, use the graph in Figure 1.101
to sketch the given transformations.


Figure 1.101


g(x) = f (x) − 2


g(x) = − f (x)


g(x) = f (x + 1)


g(x) = f (x − 2)


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1.6 | Absolute Value Functions
Learning Objectives


In this section you will:
1.6.1 Graph an absolute value function.
1.6.2 Solve an absolute value equation.
1.6.3 Solve an absolute value inequality.


Figure 1.102 Distances in deep space can be measured in all
directions. As such, it is useful to consider distance in terms of
absolute values. (credit: "s58y"/Flickr)


Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of
thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at
distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances
in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In
this section, we will investigate absolute value functions.
Understanding Absolute Value
Recall that in its basic form   f (x) = |x|,   the absolute value function, is one of our toolkit functions. The absolute value
function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for
whatever the input value is, the output is the value without regard to sign.


Absolute Value Function
The absolute value function can be defined as a piecewise function


  f (x) = |x| = ⎧


x if x ≥ 0
−x if x < 0




Example 1.69
Determine a Number within a Prescribed Distance


Describe all values  x within or including a distance of 4 from the number 5.


Chapter 1 Functions 137




1.43


1.44


Solution
We want the distance between  x  and 5 to be less than or equal to 4. We can draw a number line, such as the one
in Figure 1.103, to represent the condition to be satisfied.


Figure 1.103


The distance from  x  to 5 can be represented using the absolute value as  |x − 5|. We want the values of  x  that
satisfy the condition |x − 5| ≤ 4.


Analysis
Note that


−4 ≤ x − 5 x − 5 ≤ 4
    1 ≤ x           x ≤ 9


So  |x − 5| ≤ 4  is equivalent to  1 ≤ x ≤ 9.
However, mathematicians generally prefer absolute value notation.


Describe all values  x within a distance of 3 from the number 2.


Example 1.70
Resistance of a Resistor


Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters:
resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters
vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can
do is to try to guarantee that the variations will stay within a specified range, often  ±1%, ± 5%,  or  ± 10%.
Suppose we have a resistor rated at 680 ohms,  ± 5%. Use the absolute value function to express the range of
possible values of the actual resistance.


Solution
5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance
should not exceed the stated variability, so, with the resistance  R  in ohms,


|R − 680| ≤ 34


Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute
value notation.


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Graphing an Absolute Value Function
The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point
is shown at the origin in Figure 1.104.


Figure 1.104


Figure 1.105 shows the graph of  y = 2|x – 3| + 4. The graph of  y = |x|  has been shifted right 3 units, vertically stretched
by a factor of 2, and shifted up 4 units. This means that the corner point is located at  (3, 4)  for this transformed function.


Figure 1.105


Example 1.71
Writing an Equation for an Absolute Value Function


Write an equation for the function graphed in Figure 1.106.


Chapter 1 Functions 139




Figure 1.106


Solution
The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units
and down 2 units from the basic toolkit function. See Figure 1.107.


Figure 1.107


We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line
is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute
value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 1.108.


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1.45


Figure 1.108


From this information we can write the equation
f (x) = 2|x − 3| − 2, treating the stretch as a vertical stretch, or
f (x) = |2(x − 3)| − 2, treating the stretch as a horizontal compression.


Analysis
Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written
interchangeably as a vertical or horizontal stretch or compression.


If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?
Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor
by putting in a known pair of values for  x  and   f (x).


f (x) = a|x − 3| − 2
Now substituting in the point (1, 2)


2 = a|1 − 3| − 2
4 = 2a
a = 2


Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically
flipped, and vertically shifted up 3 units.


Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?
Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis
when the input is zero.
No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis,
depending on how the graph has been shifted and reflected. It is possible for the absolute value function to
intersect the horizontal axis at zero, one, or two points (see Figure 1.109).


Chapter 1 Functions 141




Figure 1.109 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects
the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.


Solving an Absolute Value Equation
Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an
equation such as  8 = |2x − 6|,  we notice that the absolute value will be equal to 8 if the quantity inside the absolute value
is 8 or -8. This leads to two different equations we can solve independently.


2x − 6 = 8 or 2x − 6 = − 8
  2x = 14   2x = − 2
  x = 7   x = − 1


Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify
numbers or points on a line that are at a specified distance from a given reference point.
An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example,


|x| = 4,


|2x − 1| = 3


|5x + 2| − 4 = 9


Solutions to Absolute Value Equations
For real numbers  A  and  B,   an equation of the form  |A| = B,  with  B ≥ 0,  will have solutions when  A = B  or
A = − B.  If  B < 0,   the equation  |A| = B  has no solution.


Given the formula for an absolute value function, find the horizontal intercepts of its graph.
1. Isolate the absolute value term.
2. Use  |A| = B  to write  A = B  or  −A = B,   assuming  B > 0.
3. Solve for  x.


Example 1.72
Finding the Zeros of an Absolute Value Function


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For the function   f (x) = |4x + 1| − 7 , find the values of x such that f (x) = 0 .


Solution
0 = |4x + 1| − 7 Substitute 0 for f (x).
7 = |4x + 1| Isolate the absolute value on one side of the equation.


7 = 4x + 1 or −7 = 4x + 1 Break into two separate equations and solve.


6 = 4x −8 = 4x


x = 6
4
= 1.5 x = −8


4
= − 2


The function outputs 0 when  x = 1.5  or  x = − 2. See Figure 1.110.


Figure 1.110


For the function   f (x) = |2x − 1| − 3, find the values of  x  such that   f (x) = 0.


Should we always expect two answers when solving  |A| = B?
No. We may find one, two, or even no answers. For example, there is no solution to  2 + |3x − 5| = 1.


Given an absolute value equation, solve it.
1. Isolate the absolute value term.
2. Use  |A| = B  to write  A = B  or  A = −B.
3. Solve for  x.


Chapter 1 Functions 143




1.47


Example 1.73
Solving an Absolute Value Equation


Solve  1 = 4|x − 2| + 2.


Solution
Isolating the absolute value on one side of the equation gives the following.


 1 = 4|x − 2| + 2
  − 1 = 4|x − 2|
 − 1


4
= |x − 2|


The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative
value. At this point, we notice that this equation has no solutions.


In Example 1.73, if   f(x) = 1  and  g(x) = 4|x−2| +2 were graphed on the same set of axes, would the
graphs intersect?
No. The graphs of   f   and  g would not intersect, as shown in Figure 1.111. This confirms, graphically, that the
equation  1 = 4|x − 2|+ 2  has no solution.


Figure 1.111


Find where the graph of the function   f (x) = − |x + 2| + 3  intersects the horizontal and vertical axes.


Solving an Absolute Value Inequality
Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of
values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation
of the form


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|A| < B, |A| ≤ B, |A| > B,   or   |A| ≥ B,
where an expression  A  (and possibly but not usually  B ) depends on a variable  x.  Solving the inequality means finding the
set of all  x  that satisfy the inequality. Usually this set will be an interval or the union of two intervals.
There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the
graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic
approach is it yields solutions that may be difficult to read from the graph.
For example, we know that all numbers within 200 units of 0 may be expressed as


|x| < 200 or − 200 < x < 200
Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of
$600. We can solve algebraically for the set of values  x  such that the distance between  x  and 600 is less than 200. We
represent the distance between  x  and 600 as  |x − 600|.


|x − 600| < 200 or − 200 < x − 600 < 200
− 200 + 600 < x − 600 + 600 < 200 + 600
400 < x < 800


This means our returns would be between $400 and $800.
Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed
absolute value function, where we must determine for which values of the input the function’s output will be negative or
positive.


Given an absolute value inequality of the form  |x− A| ≤ B  for real numbers  a  and  b where  b  is positive,
solve the absolute value inequality algebraically.


1. Find boundary points by solving  |x − A| = B. 
2. Test intervals created by the boundary points to determine where  |x − A| ≤ B.
3. Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder
notation.


Example 1.74
Solving an Absolute Value Inequality


Solve  |x − 5| ≤ 4.


Solution
With both approaches, we will need to know first where the corresponding equality is true. In this case we first
will find where  |x − 5| = 4. We do this because the absolute value is a function with no breaks, so the only way
the function values can switch from being less than 4 to being greater than 4 is by passing through where the
values equal 4. Solve  |x − 5| = 4.


x − 5 = 4
       x = 9    or   


x − 5 = − 4
    x = 1  


After determining that the absolute value is equal to 4 at  x = 1  and  x = 9,  we know the graph can change only
from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:


x < 1, 1 < x < 9, and x > 9.


Chapter 1 Functions 145




To determine when the function is less than 4, we could choose a value in each interval and see if the output is
less than or greater than 4, as shown in Table 1.43.


Interval test  x f(x) < 4  or   > 4?


x < 1 0 |0 − 5| = 5 Greater than


1 < x < 9 6 |6 − 5| = 1 Less than


x > 9 11 |11 − 5| = 6 Greater than


Table 1.43


Because  1 ≤ x ≤ 9  is the only interval in which the output at the test value is less than 4, we can conclude that
the solution to  |x − 5| ≤ 4  is  1 ≤ x ≤ 9,   or  [1, 9].
To use a graph, we can sketch the function   f (x) = |x − 5|. To help us see where the outputs are 4, the line
 g(x) = 4  could also be sketched as in Figure 1.112.


Figure 1.112 Graph to find the points satisfying an absolute
value inequality.


We can see the following:
• The output values of the absolute value are equal to 4 at  x = 1  and  x = 9.
• The graph of   f   is below the graph of  g  on  1 < x < 9. This means the output values of   f (x)  are less
than the output values of  g(x).


• The absolute value is less than or equal to 4 between these two points, when  1 ≤ x ≤ 9.  In interval
notation, this would be the interval  [1, 9].


Analysis
For absolute value inequalities,


|x − A| < C, |x − A| > C,
 − C < x − A < C, x − A < − C or x − A > C.


The   <   or   >   symbol may be replaced by   ≤ or ≥ .


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1.48


So, for this example, we could use this alternative approach.
| x − 5| ≤ 4
       − 4 ≤ x − 5 ≤ 4 Rewrite by removing the absolute value bars.
−4 + 5 ≤ x − 5 + 5 ≤ 4 + 5 Isolate the x.
              1 ≤ x ≤ 9


Solve  |x + 2| ≤ 6.


Given an absolute value function, solve for the set of inputs where the output is positive (or negative).
1. Set the function equal to zero, and solve for the boundary points of the solution set.
2. Use test points or a graph to determine where the function’s output is positive or negative.


Example 1.75
Using a Graphical Approach to Solve Absolute Value Inequalities


Given the function f (x) = − 1
2|4x − 5| + 3,   determine the x- values for which the function values are


negative.


Solution
We are trying to determine where   f (x) < 0,  which is when − 1


2
|4x − 5| + 3 < 0.We begin by isolating the


absolute value.
− 1


2|4x − 5| < − 3 Multiply both sides by –2, and reverse the inequality.
          |4x − 5| > 6


Next we solve for the equality  |4x − 5| = 6.
4x − 5 = 6 4x − 5 = − 6
4x − 5 = 6 or           4x = − 1
             x = 11


4
             x = − 1


4


Now, we can examine the graph of   f   to observe where the output is negative. We will observe where the branches
are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know
that it crosses the horizontal axis at  x = − 1


4
  and  x = 11


4
  and that the graph has been reflected vertically. See


Figure 1.113.


Chapter 1 Functions 147




1.49


Figure 1.113


We observe that the graph of the function is below the x-axis left of  x = − 1
4
  and right of  x = 11


4
. This means


the function values are negative to the left of the first horizontal intercept at  x = − 1
4
,   and negative to the right


of the second intercept at  x = 11
4
. This gives us the solution to the inequality.


x < − 1
4
 or x > 11


4


In interval notation, this would be  (−∞, − 0.25) ∪ (2.75, ∞).


Solve  − 2|k − 4| ≤ − 6.


Access these online resources for additional instruction and practice with absolute value.
• Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue)
• Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2)
• Equations of Absolute Value Function (http://openstaxcollege.org/l/findeqabsval)
• Equations of Absolute Value Function 2 (http://openstaxcollege.org/l/findeqabsval2)
• Solving Absolute Value Equations (http://openstaxcollege.org/l/solveabsvalueeq)


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379.
380.


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415.


1.6 EXERCISES
Verbal


How do you solve an absolute value equation?
How can you tell whether an absolute value function


has two x-intercepts without graphing the function?
When solving an absolute value function, the isolated


absolute value term is equal to a negative number. What
does that tell you about the graph of the absolute value
function?


How can you use the graph of an absolute value
function to determine the x-values for which the function
values are negative?


How do you solve an absolute value inequality
algebraically?


Algebraic
Describe all numbers  x  that are at a distance of 4


from the number 8. Express this using absolute value
notation.


Describe all numbers  x  that are at a distance of  1
2


from the number −4. Express this using absolute value
notation.


Describe the situation in which the distance that point
 x  is from 10 is at least 15 units. Express this using
absolute value notation.


Find all function values   f (x)  such that the distance
from   f (x)  to the value 8 is less than 0.03 units. Express
this using absolute value notation.
For the following exercises, solve the equations below and
express the answer using set notation.


|x + 3| = 9


|6 − x| = 5


|5x − 2| = 11


|4x − 2| = 11


2|4 − x| = 7


3|5 − x| = 5


3|x + 1| − 4 = 5


5|x − 4| − 7 = 2


0 = − |x − 3| + 2


2|x − 3| + 1 = 2


|3x − 2| = 7


|3x − 2| = − 7


|12x − 5| = 11


|13x + 5| = 14
−|13x + 5| + 14 = 0


For the following exercises, find the x- and y-intercepts of
the graphs of each function.


f (x) = 2|x + 1| − 10


f (x) = 4|x − 3| + 4


f (x) = − 3|x − 2| − 1


f (x) = − 2|x + 1| + 6


For the following exercises, solve each inequality and write
the solution in interval notation.


|x − 2| > 10


2|v − 7| − 4 ≥ 42


|3x − 4| ≤ 8


|x − 4| ≥ 8


|3x − 5| ≥ 13


|3x − 5| ≥ − 13


|34x − 5| ≥ 7


|34x − 5| + 1 ≤ 16
Graphical
For the following exercises, graph the absolute value
function. Plot at least five points by hand for each graph.


y = |x − 1|


Chapter 1 Functions 149




416.


417.


418.


419.


420.


421.


422.


423.


424.


425.


426.


427.


428.


429.


430.


431.


432.


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435.


436.


437.


438.


439.


440.


441.


442.


y = |x + 1|


y = |x| + 1


For the following exercises, graph the given functions by
hand.


y = |x| − 2


y = − |x|


y = − |x| − 2


y = − |x − 3| − 2


f (x) = − |x − 1| − 2


f (x) = − |x + 3| + 4


f (x) = 2|x + 3| + 1


f (x) = 3|x − 2| + 3


f (x) = |2x − 4| − 3


f (x) = |3x + 9| + 2


f (x) = − |x − 1| − 3


f (x) = − |x + 4| − 3


f (x) = 1
2|x + 4| − 3


Technology
Use a graphing utility to graph f (x) = 10|x − 2| on


the viewing window [0, 4]. Identify the corresponding
range. Show the graph.


Use a graphing utility to graph
f (x) = − 100|x| + 100  on the viewing window  ⎡⎣−5, 5⎤⎦. 
Identify the corresponding range. Show the graph.
For the following exercises, graph each function using a
graphing utility. Specify the viewing window.


f (x) = − 0.1|0.1(0.2 − x)| + 0.3


f (x) = 4×109 |x − (5×109)| + 2×109


Extensions
For the following exercises, solve the inequality.


| − 2x − 23(x + 1)| + 3 > −1
If possible, find all values of a such that there are no


x- intercepts for f (x) = 2|x + 1| + a.


If possible, find all values of  a  such that there are no
 y -intercepts for   f (x) = 2|x + 1| + a.


Real-World Applications
Cities A and B are on the same east-west line.


Assume that city A is located at the origin. If the distance
from city A to city B is at least 100 miles and  x  represents
the distance from city B to city A, express this using
absolute value notation.


The true proportion  p  of people who give a favorable
rating to Congress is 8% with a margin of error of 1.5%.
Describe this statement using an absolute value equation.


Students who score within 18 points of the number 82
will pass a particular test. Write this statement using
absolute value notation and use the variable  x  for the score.


A machinist must produce a bearing that is within
0.01 inches of the correct diameter of 5.0 inches. Using  x
as the diameter of the bearing, write this statement using
absolute value notation.


The tolerance for a ball bearing is 0.01. If the true
diameter of the bearing is to be 2.0 inches and the measured
value of the diameter is  x  inches, express the tolerance
using absolute value notation.


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1.7 | Inverse Functions
Learning Objectives


In this section, you will:
1.7.1 Verify inverse functions.
1.7.2 Determine the domain and range of an inverse function, and restrict the domain of a
function to make it one-to-one.
1.7.3 Find or evaluate the inverse of a function.
1.7.4 Use the graph of a one-to-one function to graph its inverse function on the same axes.


A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one
direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the
outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional
electrical resistance heating.
If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been
studying can also run backwards. Figure 1.114 provides a visual representation of this question. In this section, we will
consider the reverse nature of functions.


Figure 1.114 Can a function “machine” operate in reverse?


Verifying That Two Functions Are Inverse Functions
Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not
familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to
convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula


C = 5
9
(F − 32)


and substitutes 75 for  F  to calculate
5
9
(75 − 32) ≈ 24°C.


Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather
forecast from Figure 1.115 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.


Chapter 1 Functions 151




Figure 1.115


At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her
algebra, and can easily solve the equation for  F  after substituting a value for  C.  For example, to convert 26 degrees
Celsius, she could write


  26 = 5
9
(F − 32)


26 ⋅ 9
5
= F − 32


  F = 26 ⋅ 9
5
+ 32 ≈ 79


After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will
be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a
different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.
The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the
input of the original function becomes the output of the inverse function and the output of the original function becomes the
input of the inverse function.
Given a function   f (x),  we represent its inverse as   f −1(x),   read as  “ f   inverse of  x.” The raised  −1  is part of the
notation. It is not an exponent; it does not imply a power of  −1  . In other words,   f −1(x)  does not mean   1


f (x)
  because


1
f (x)


  is the reciprocal of   f   and not the inverse.


The “exponent-like” notation comes from an analogy between function composition and multiplication: just as  a−1a = 1 
(1 is the identity element for multiplication) for any nonzero number  a,   so   f −1 ∘ f   equals the identity function, that is,



⎝ f


−1 ∘ f ⎞⎠(x) = f
−1 ⎛
⎝ f (x)⎞⎠ = f −1 (y) = x


This holds for all  x  in the domain of   f .  Informally, this means that inverse functions “undo” each other. However, just as
zero does not have a reciprocal, some functions do not have inverses.
Given a function   f (x),  we can verify whether some other function  g(x)  is the inverse of   f (x)  by checking whether either
 g( f (x)) = x  or   f (g(x)) = x  is true. We can test whichever equation is more convenient to work with because they are
logically equivalent (that is, if one is true, then so is the other.)
For example,  y = 4x  and  y = 1


4
x  are inverse functions.



⎝ f


−1 ∘ f ⎞⎠(x) = f
−1 (4x) = 1


4
(4x) = x


and

⎝ f ∘ f


−1⎞
⎠(x) = f




1
4
x⎞⎠ = 4




1
4
x⎞⎠ = x


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1.50


A few coordinate pairs from the graph of the function  y = 4x  are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from
the graph of the function  y = 1


4
x  are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate


pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.


Inverse Function
For any one-to-one function   f (x) = y,   a function   f −1 (x)  is an inverse function of   f   if   f −1(y) = x. This can also
be written as   f −1( f (x)) = x  for all  x  in the domain of   f .  It also follows that   f ( f −1(x)) = x  for all  x  in the domain
of   f −1   if   f −1   is the inverse of   f . 
The notation f −1 is read “ f inverse.” Like any other function, we can use any variable name as the input for f −1,
so we will often write   f −1(x), which we read as “ f inverse of x.” Keep in mind that


f −1(x) ≠ 1
f (x)


and not all functions have inverses.


Example 1.76
Identifying an Inverse Function for a Given Input-Output Pair


If for a particular one-to-one function   f (2) = 4  and   f (5) = 12,  what are the corresponding input and output
values for the inverse function?


Solution
The inverse function reverses the input and output quantities, so if


f (2) = 4, then f −1(4) = 2;


       f (5) = 12, then f−1 (12) = 5.
Alternatively, if we want to name the inverse function  g,   then  g(4) = 2  and  g(12) = 5.


Analysis
Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table
1.43.



⎝x, f(x)⎞⎠ ⎛⎝x, g(x)⎞⎠


(2, 4) (4, 2)


(5, 12) (12, 5)


Table 1.43


Given that  h−1(6) = 2,  what are the corresponding input and output values of the original function  h?  


Chapter 1 Functions 153




1.51


Given two functions    f(x)   and  g(x),   test whether the functions are inverses of each other.
1. Determine whether   f (g(x)) = x  or  g( f (x)) = x.
2. If either statement is true, then both are true, and  g = f −1   and   f = g−1.  If either statement is false, then
both are false, and  g ≠ f −1   and   f ≠ g−1.


Example 1.77
Testing Inverse Relationships Algebraically


If   f (x) = 1
x + 2


  and  g(x) = 1x − 2,   is  g = f −1?


Solution


g( f (x)) = 1⎛


1
x + 2



− 2


= x + 2 − 2
= x


so
g = f −1 and f = g−1


This is enough to answer yes to the question, but we can also verify the other formula.
f (g(x)) = 1


1
x − 2 + 2


= 1
1
x


= x


Analysis
Notice the inverse operations are in reverse order of the operations from the original function.


If   f (x) = x3 − 4  and  g(x) = x + 4 3 ,   is  g = f −1?


Example 1.78
Determining Inverse Relationships for Power Functions


If   f (x) = x3   (the cube function) and  g(x) = 1
3
x,   is  g = f −1?


Solution
f ⎛⎝g(x)⎞⎠ = x


3


27
≠ x


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1.52


No, the functions are not inverses.


Analysis


The correct inverse to the cube is, of course, the cube root   x3 = x
1
3,   that is, the one-third is an exponent, not a


multiplier.


If   f (x) = (x − 1)3  and g(x) = x3 + 1,   is  g = f −1?


Finding Domain and Range of Inverse Functions
The outputs of the function   f   are the inputs to   f −1,   so the range of   f   is also the domain of   f −1. Likewise, because the
inputs to   f   are the outputs of   f −1,   the domain of   f   is the range of   f −1. We can visualize the situation as in Figure
1.116.


Figure 1.116 Domain and range of a function and its inverse


When a function has no inverse function, it is possible to create a new function where that new function on a limited domain
does have an inverse function. For example, the inverse of   f (x) = x  is   f −1(x) = x2,   because a square “undoes” a square
root; but the square is only the inverse of the square root on the domain  [0, ∞),   since that is the range of   f (x) = x.
We can look at this problem from the other side, starting with the square (toolkit quadratic) function   f (x) = x2.  If we
want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic
function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic
function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input
corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To
put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an
inverse function. In order for a function to have an inverse, it must be a one-to-one function.
In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-
to-one. For example, we can make a restricted version of the square function   f (x) = x2  with its range limited to  [0, ∞),  
which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).
If   f (x) = (x − 1)2   on  [1, ∞),   then the inverse function is   f −1(x) = x + 1.


• The domain of   f   = range of   f −1   =  [1, ∞).
• The domain of   f −1   = range of   f   =  [0, ∞).


Chapter 1 Functions 155




Is it possible for a function to have more than one inverse?
No. If two supposedly different functions, say,  g  and  h,   both meet the definition of being inverses of another
function   f ,   then you can prove that  g = h. We have just seen that some functions only have inverses if we restrict
the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading
to different inverses. However, on any one domain, the original function still has only one unique inverse.


Domain and Range of Inverse Functions
The range of a function   f (x)  is the domain of the inverse function   f −1(x).
The domain of   f (x)  is the range of   f −1(x).


Given a function, find the domain and range of its inverse.
1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and
write the domain of the original function as the range of the inverse.


2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted
domain becomes the range of the inverse function.


Example 1.79
Finding the Inverses of Toolkit Functions


Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted
domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 1.44. We
restrict the domain in such a fashion that the function assumes all y-values exactly once.


Constant Identity Quadratic Cubic Reciprocal


f (x) = c f (x) = x f (x) = x2 f (x) = x3 f (x) = 1x


Reciprocal
squared Cube root


Square
root


Absolute
value


f (x) = 1
x2 f (x) = x


3 f (x) = x f (x) = |x|


Table 1.44


Solution
The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-
to-one, so the constant function has no meaningful inverse.
The absolute value function can be restricted to the domain  [0, ∞), where it is equal to the identity function.
The reciprocal-squared function can be restricted to the domain  (0, ∞).


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Analysis
We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure
1.117. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so
that it passes the horizontal line test, then in that restricted domain, it can have an inverse.


Figure 1.117 (a) Absolute value (b) Reciprocal squared


The domain of function   f   is  (1, ∞)  and the range of function   f   is  (−∞, −2).  Find the domain and
range of the inverse function.


Finding and Evaluating Inverse Functions
Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete
representation of the inverse function in many cases.
Inverting Tabular Functions
Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the
range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and
range.
Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or
column) of outputs becomes the row (or column) of inputs for the inverse function.


Example 1.80
Interpreting the Inverse of a Tabular Function


A function   f (t)  is given in Table 1.45, showing distance in miles that a car has traveled in  t minutes. Find and
interpret   f −1(70).


Chapter 1 Functions 157




1.54


t (minutes) 30 50 70 90


f(t) (miles) 20 40 60 70


Table 1.45


Solution
The inverse function takes an output of   f   and returns an input for   f .  So in the expression   f −1(70),   70 is an
output value of the original function, representing 70 miles. The inverse will return the corresponding input of the
original function   f ,   90 minutes, so   f −1(70) = 90. The interpretation of this is that, to drive 70 miles, it took
90 minutes.
Alternatively, recall that the definition of the inverse was that if   f (a) = b,   then   f −1(b) = a. By this definition,
if we are given   f −1(70) = a,   then we are looking for a value  a  so that   f (a) = 70.  In this case, we are looking
for a  t  so that   f (t) = 70,  which is when  t = 90.


Using Table 1.46, find and interpret (a) f (60), and (b) f −1(60).


t (minutes) 30 50 60 70 90


f (t) (miles) 20 40 50 60 70


Table 1.46


Evaluating the Inverse of a Function, Given a Graph of the Original Function
We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal
extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original
function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse
function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse
function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical
axis of the original function’s graph.


Given the graph of a function, evaluate its inverse at specific points.
1. Find the desired input on the y-axis of the given graph.
2. Read the inverse function’s output from the x-axis of the given graph.


Example 1.81
Evaluating a Function and Its Inverse from a Graph at Specific Points


A function  g(x)  is given in Figure 1.118. Find  g(3)  and  g−1(3).


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1.55


Figure 1.118


Solution
To evaluate g(3),  we find 3 on the x-axis and find the corresponding output value on the y-axis. The point
 (3, 1)  tells us that  g(3) = 1.
To evaluate  g−1(3),   recall that by definition  g−1(3) means the value of x for which  g(x) = 3. By looking for
the output value 3 on the vertical axis, we find the point  (5, 3)  on the graph, which means  g(5) = 3,   so by
definition,  g−1(3) = 5.  See Figure 1.119.


Figure 1.119


Using the graph in Figure 1.119, (a) find  g−1(1), and (b) estimate  g−1(4).


Finding Inverses of Functions Represented by Formulas
Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function
is given as a formula— for example,  y  as a function of  x— we can often find the inverse function by solving to obtain  x 
as a function of  y.


Given a function represented by a formula, find the inverse.
1. Make sure   f   is a one-to-one function.
2. Solve for  x.
3. Interchange  x  and  y.


Chapter 1 Functions 159




1.56


Example 1.82
Inverting the Fahrenheit-to-Celsius Function


Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.
C = 5


9
(F − 32)


Solution


  C = 5
9
(F − 32)


C ⋅ 9
5
= F − 32


         F = 9
5
C + 32


By solving in general, we have uncovered the inverse function. If
C = h(F) = 5


9
(F − 32),


then
F = h−1(C) = 9


5
C + 32.


In this case, we introduced a function  h  to represent the conversion because the input and output variables are
descriptive, and writing  C−1   could get confusing.


Solve for  x  in terms of  y  given  y = 1
3
(x − 5)


Example 1.83
Solving to Find an Inverse Function


Find the inverse of the function   f (x) = 2
x − 3


+ 4.


Solution
  y = 2


x − 3
+ 4 Set up an equation.


  y − 4 = 2
x − 3


Subtract 4 from both sides.


  x − 3 = 2
y − 4


Multiply both sides by x − 3 and divide by y − 4.


  x = 2
y − 4


+ 3 Add 3 to both sides.


So   f −1 (y) = 2
y − 4


+ 3  or   f −1 (x) = 2
x − 4


+ 3.


Analysis


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1.57


The domain and range of   f   exclude the values 3 and 4, respectively.   f   and   f −1   are equal at two points but are
not the same function, as we can see by creating Table 1.46.


x 1 2 5 f −1(y)


f(x) 3 2 5 y


Table 1.46


Example 1.84
Solving to Find an Inverse with Radicals


Find the inverse of the function   f (x) = 2 + x − 4.


Solution
                 y = 2 + x − 4
(y − 2)2 = x − 4


                 x = (y − 2)2 + 4
So   f −1 (x) = (x − 2)2 + 4.
The domain of   f   is  [4, ∞). Notice that the range of   f   is  [2, ∞),   so this means that the domain of the inverse
function   f −1   is also  [2, ∞).


Analysis
The formula we found for   f −1 (x)  looks like it would be valid for all real  x. However,   f −1   itself must have an
inverse (namely,   f   ) so we have to restrict the domain of   f −1   to  [2, ∞)  in order to make   f −1   a one-to-one
function. This domain of   f −1   is exactly the range of   f .


What is the inverse of the function   f (x) = 2 − x? State the domains of both the function and the
inverse function.


Finding Inverse Functions and Their Graphs
Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to
the quadratic function   f (x) = x2   restricted to the domain  [0, ∞), on which this function is one-to-one, and graph it as in
Figure 1.120.


Chapter 1 Functions 161




Figure 1.120 Quadratic function with domain restricted to [0,
∞).


Restricting the domain to  [0, ∞) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an
inverse on this restricted domain.
We already know that the inverse of the toolkit quadratic function is the square root function, that is, f −1(x) = x. What
happens if we graph both f and f −1 on the same set of axes, using the x- axis for the input to both f and f −1?


We notice a distinct relationship: The graph of   f −1(x)  is the graph of   f (x)  reflected about the diagonal line  y = x,  which
we will call the identity line, shown in Figure 1.121.


Figure 1.121 Square and square-root functions on the non-
negative domain


This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping
inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.


Example 1.85
Finding the Inverse of a Function Using Reflection about the Identity Line


Given the graph of   f (x)  in Figure 1.122, sketch a graph of   f −1(x).


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1.58


Figure 1.122


Solution
This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an
apparent domain of  (0, ∞)  and range of  (−∞, ∞),   so the inverse will have a domain of  (−∞, ∞)  and range
of  (0, ∞).
If we reflect this graph over the line  y = x,   the point  (1, 0)  reflects to  (0, 1)  and the point  (4, 2)  reflects to
 (2, 4).  Sketching the inverse on the same axes as the original graph gives Figure 1.123.


Figure 1.123 The function and its inverse, showing reflection
about the identity line


Draw graphs of the functions   f and f −1 from Example 1.83.


Chapter 1 Functions 163




Is there any function that is equal to its own inverse?
Yes. If   f = f −1,   then   f ⎛⎝ f (x)⎞⎠ = x,   and we can think of several functions that have this property. The identity
function does, and so does the reciprocal function, because


1
1
x
= x


Any function   f (x) = c − x,  where  c  is a constant, is also equal to its own inverse.


Access these online resources for additional instruction and practice with inverse functions.
• Inverse Functions (http://openstaxcollege.org/l/inversefunction)
• Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph)
• Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/
restrictdomain)


Visit this website (http://openstaxcollege.org/l/PreCalcLPC01) for additional practice questions from
Learningpod.


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443.


444.


445.
446.


447.


448.


449.


450.


451.


452.


453.


454.


455.


456.


457.


458.


459.


460.


461.


462.


463.


464.


465.


466.


1.7 EXERCISES
Verbal


Describe why the horizontal line test is an effective
way to determine whether a function is one-to-one?


Why do we restrict the domain of the function
  f (x) = x2   to find the function’s inverse?


Can a function be its own inverse? Explain.
Are one-to-one functions either always increasing or


always decreasing? Why or why not?
How do you find the inverse of a function


algebraically?


Algebraic
Show that the function   f (x) = a − x  is its own


inverse for all real numbers  a. 


For the following exercises, find   f −1(x)  for each function.


f (x) = x + 3


f (x) = x + 5


f (x) = 2 − x


f (x) = 3 − x


f (x) = x
x + 2


f (x) = 2x + 3
5x + 4


For the following exercises, find a domain on which each
function   f   is one-to-one and non-decreasing. Write the
domain in interval notation. Then find the inverse of   f  
restricted to that domain.


f (x) = (x + 7)2


f (x) = (x − 6)2


f (x) = x2 − 5


Given   f (x) = x
2
+ x  and  g(x) = 2x


1 − x
:


a. Find   f (g(x))  and  g( f (x)).


b. What does the answer tell us about the relationship
between   f (x)  and  g(x)?


For the following exercises, use function composition to
verify that   f (x)  and  g(x)  are inverse functions.


f (x) = x − 1
3   and  g(x) = x3 + 1


f (x) = − 3x + 5  and  g(x) = x − 5
−3


Graphical
For the following exercises, use a graphing utility to
determine whether each function is one-to-one.


f (x) = x


f (x) = 3x + 1
3


f (x) = −5x + 1


f (x) = x3 − 27


For the following exercises, determine whether the graph
represents a one-to-one function.


Chapter 1 Functions 165




467.


468.


469.


470.


471.


472.


473.


474.


475.


476.


477.


478.


479.


480.


481.


482.


For the following exercises, use the graph of   f   shown in
Figure 1.124.


Figure 1.124


Find   f (0).


Solve   f (x) = 0.


Find   f −1 (0).


Solve   f −1 (x) = 0.


For the following exercises, use the graph of the one-to-one
function shown in Figure 1.125.


Figure 1.125


Sketch the graph of   f −1. 


Find   f (6) and f −1(2).


If the complete graph of   f   is shown, find the domain
of   f . 


If the complete graph of   f   is shown, find the range of
  f .


Numeric
For the following exercises, evaluate or solve, assuming
that the function   f   is one-to-one.


If   f (6) = 7,   find    f −1(7).


If   f (3) = 2,   find   f −1(2).


If   f −1 (−4) = − 8,   find   f ( − 8).


If   f −1 (−2) = − 1,   find   f ( − 1).


For the following exercises, use the values listed in Table
1.47 to evaluate or solve.


x f(x)


0 8


1 0


2 7


3 4


4 2


5 6


6 5


7 3


8 9


9 1


Table 1.47


Find   f (1).


Solve   f (x) = 3.


Find   f −1 (0).


Solve   f −1 (x) = 7.


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483.


484.


485.


486.


487.


488.


489.


Use the tabular representation of   f   in Table 1.48 to
create a table for   f −1 (x).


x 3 6 9 13 14


f(x) 1 4 7 12 16


Table 1.48


Technology
For the following exercises, find the inverse function.
Then, graph the function and its inverse.


f (x) = 3
x − 2


f (x) = x3 − 1


Find the inverse function of   f (x) = 1
x − 1


. Use a
graphing utility to find its domain and range. Write the
domain and range in interval notation.


Real-World Applications
To convert from  x  degrees Celsius to  y  degrees


Fahrenheit, we use the formula   f (x) = 9
5
x + 32.  Find the


inverse function, if it exists, and explain its meaning.
The circumference  C  of a circle is a function of its


radius given by  C(r) = 2πr. Express the radius of a circle
as a function of its circumference. Call this function  r(C). 
Find  r(36π)  and interpret its meaning.


A car travels at a constant speed of 50 miles per hour.
The distance the car travels in miles is a function of time,
 t,   in hours given by  d(t) = 50t.  Find the inverse
function by expressing the time of travel in terms of the
distance traveled. Call this function  t(d).  Find  t(180)  and
interpret its meaning.


Chapter 1 Functions 167




absolute maximum
absolute minimum
absolute value equation


absolute value inequality
average rate of change


composite function


decreasing function


dependent variable
domain
even function


function
horizontal compression


horizontal line test


horizontal reflection
horizontal shift


horizontal stretch


increasing function


independent variable
input
interval notation


inverse function


local extrema
local maximum
local minimum


CHAPTER 1 REVIEW
KEY TERMS


the greatest value of a function over an interval
the lowest value of a function over an interval


an equation of the form  |A| = B,  with  B ≥ 0;   it will have solutions when  A = B  or
A = − B


a relationship in the form |A| < B,    |A| ≤ B,     |A| > B,   or   |A| ≥ B
the difference in the output values of a function found for two values of the input divided by the


difference between the inputs
the new function formed by function composition, when the output of one function is used as the


input of another
a function is decreasing in some open interval if   f (b) < f (a)  for any two input values  a  and  b  in


the given interval where  b > a
an output variable


the set of all possible input values for a relation
a function whose graph is unchanged by horizontal reflection,   f (x) = f ( − x),   and is symmetric about


the y- axis


a relation in which each input value yields a unique output value
a transformation that compresses a function’s graph horizontally, by multiplying the input by a


constant  b > 1
a method of testing whether a function is one-to-one by determining whether any horizontal line


intersects the graph more than once
a transformation that reflects a function’s graph across the y-axis by multiplying the input by  −1


a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the
input


a transformation that stretches a function’s graph horizontally by multiplying the input by a constant
 0 < b < 1


a function is increasing in some open interval if   f (b) > f (a)  for any two input values  a  and  b  in
the given interval where  b > a


an input variable
each object or value in a domain that relates to another object or value by a relationship known as a function


a method of describing a set that includes all numbers between a lower limit and an upper limit; the
lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and
a parenthesis indicating exclusion


for any one-to-one function   f (x),   the inverse is a function   f −1(x)  such that   f −1 ⎛⎝ f (x)⎞⎠ = x  for all  x
in the domain of   f ;   this also implies that   f ⎛⎝ f −1 (x)⎞⎠ = x  for all  x  in the domain of   f −1


collectively, all of a function's local maxima and minima
a value of the input where a function changes from increasing to decreasing as the input value increases.
a value of the input where a function changes from decreasing to increasing as the input value increases.


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odd function


one-to-one function
output
piecewise function
range
rate of change
relation
set-builder notation


vertical compression


vertical line test


vertical reflection
vertical shift


vertical stretch


a function whose graph is unchanged by combined horizontal and vertical reflection,   f (x) = − f ( − x),  
and is symmetric about the origin


a function for which each value of the output is associated with a unique input value
each object or value in the range that is produced when an input value is entered into a function


a function in which more than one formula is used to define the output
the set of output values that result from the input values in a relation


the change of an output quantity relative to the change of the input quantity
a set of ordered pairs


a method of describing a set by a rule that all of its members obey; it takes the form
 {x| statement about x}


a function transformation that compresses the function’s graph vertically by multiplying the
output by a constant  0 < a < 1


a method of testing whether a graph represents a function by determining whether a vertical line
intersects the graph no more than once


a transformation that reflects a function’s graph across the x-axis by multiplying the output by  −1
a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the


output
a transformation that stretches a function’s graph vertically by multiplying the output by a constant


 a > 1


KEY EQUATIONS
Constant function f (x) = c, where  c  is a constant


Identity function f (x) = x


Absolute value function f (x) = |x|


Quadratic function f (x) = x2


Cubic function f (x) = x3


Reciprocal function f (x) = 1x


Reciprocal squared function f (x) = 1x2


Square root function f (x) = x


Cube root function f (x) = x3


Chapter 1 Functions 169




Average rate of change ΔyΔx =
f (x2) − f (x1)


x2 − x1


Composite function ⎛⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠


Vertical shift g(x) = f (x) + k  (up for  k > 0 )


Horizontal shift g(x) = f (x − h) (right for  h > 0 )


Vertical reflection g(x) = − f (x)


Horizontal reflection g(x) = f ( − x)


Vertical stretch g(x) = a f (x)  ( a > 0 )


Vertical compression g(x) = a f (x)  (0 < a < 1)


Horizontal stretch g(x) = f (bx) (0 < b < 1)


Horizontal compression g(x) = f (bx)  ( b > 1 )


KEY CONCEPTS
1.1 Functions and Function Notation


• A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input,
leads to exactly one range value, or output. See Example 1.1 and Example 1.2.


• Function notation is a shorthand method for relating the input to the output in the form  y = f (x).  See Example
1.3 and Example 1.4.


• In tabular form, a function can be represented by rows or columns that relate to input and output values. See
Example 1.5.


• To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function
can be evaluated by replacing the input variable with a given value. See Example 1.6 and Example 1.7.


• To solve for a specific function value, we determine the input values that yield the specific output value. See
Example 1.8.


• An algebraic form of a function can be written from an equation. See Example 1.9 and Example 1.10.
• Input and output values of a function can be identified from a table. See Example 1.11.
• Relating input values to output values on a graph is another way to evaluate a function. See Example 1.12.
• A function is one-to-one if each output value corresponds to only one input value. See Example 1.13.
• A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point.
See Example 1.14.


• The graph of a one-to-one function passes the horizontal line test. See Example 1.15.


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1.2 Domain and Range
• The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical
operation, such as dividing by zero or taking the square root of a negative number.


• The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example
1.16.


• The domain of a function can also be determined by identifying the input values of a function written as an equation.
See Example 1.17, Example 1.18, and Example 1.19.


• Interval values represented on a number line can be described using inequality notation, set-builder notation, and
interval notation. See Example 1.20.


• For many functions, the domain and range can be determined from a graph. See Example 1.21 and Example
1.22.


• An understanding of toolkit functions can be used to find the domain and range of related functions. See Example
1.23, Example 1.24, and Example 1.25.


• A piecewise function is described by more than one formula. See Example 1.26 and Example 1.27.
• A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 1.28.


1.3 Rates of Change and Behavior of Graphs
• A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change
is determined using only the beginning and ending data. See Example 1.29.


• Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example
1.30.


• Comparing pairs of input and output values in a table can also be used to find the average rate of change. See
Example 1.31.


• An average rate of change can also be computed by determining the function values at the endpoints of an interval
described by a formula. See Example 1.32 and Example 1.33.


• The average rate of change can sometimes be determined as an expression. See Example 1.34.
• A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See
Example 1.35.


• A local maximum is where a function changes from increasing to decreasing and has an output value larger (more
positive or less negative) than output values at neighboring input values.


• A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an
output value smaller (more negative or less positive) than output values at neighboring input values.


• Minima and maxima are also called extrema.
• We can find local extrema from a graph. See Example 1.36 and Example 1.37.
• The highest and lowest points on a graph indicate the maxima and minima. See Example 1.38.


1.4 Composition of Functions
• We can perform algebraic operations on functions. See Example 1.39.
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer)
function.


• The function produced by combining two functions is a composite function. See Example 1.40 and Example
1.41.


• The order of function composition must be considered when interpreting the meaning of composite functions. See
Example 1.42.


Chapter 1 Functions 171




• A composite function can be evaluated by evaluating the inner function using the given input value and then
evaluating the outer function taking as its input the output of the inner function.


• A composite function can be evaluated from a table. See Example 1.43.
• A composite function can be evaluated from a graph. See Example 1.44.
• A composite function can be evaluated from a formula. See Example 1.45.
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to
outputs of the inner function that are in the domain of the outer function. See Example 1.46 and Example 1.47.


• Just as functions can be combined to form a composite function, composite functions can be decomposed into
simpler functions.


• Functions can often be decomposed in more than one way. See Example 1.48.


1.5 Transformation of Functions
• A function can be shifted vertically by adding a constant to the output. See Example 1.49 and Example 1.50.
• A function can be shifted horizontally by adding a constant to the input. See Example 1.51, Example 1.52, and
Example 1.53.


• Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal
shifts. See Example 1.54.


• Vertical and horizontal shifts are often combined. See Example 1.55 and Example 1.56.
• A vertical reflection reflects a graph about the  x- axis. A graph can be reflected vertically by multiplying the output
by –1.


• A horizontal reflection reflects a graph about the y- axis. A graph can be reflected horizontally by multiplying the
input by –1.


• A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not
affect the final graph. See Example 1.57.


• A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or
columns accordingly. See Example 1.58.


• A function presented as an equation can be reflected by applying transformations one at a time. See Example 1.59.
• Even functions are symmetric about the y- axis, whereas odd functions are symmetric about the origin.
• Even functions satisfy the condition   f (x) = f ( − x).
• Odd functions satisfy the condition   f (x) = − f ( − x).
• A function can be odd, even, or neither. See Example 1.60.
• A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 1.61,
Example 1.62, and Example 1.63.


• A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 1.64,
Example 1.65, and Example 1.66.


• The order in which different transformations are applied does affect the final function. Both vertical and horizontal
transformations must be applied in the order given. However, a vertical transformation may be combined with a
horizontal transformation in any order. See Example 1.67 and Example 1.68.


1.6 Absolute Value Functions
• The absolute value function is commonly used to measure distances between points. See Example 1.69.
• Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See
Example 1.70.


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• The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes
direction. See Example 1.71.


• In an absolute value equation, an unknown variable is the input of an absolute value function.
• If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown
variable. See Example 1.72.


• An absolute value equation may have one solution, two solutions, or no solutions. See Example 1.73.
• An absolute value inequality is similar to an absolute value equation but takes the form


 |A| < B,    |A| ≤ B,     |A| > B,   or   |A| ≥ B. It can be solved by determining the boundaries of the solution set and
then testing which segments are in the set. See Example 1.74.


• Absolute value inequalities can also be solved graphically. See Example 1.75.


1.7 Inverse Functions
• If  g(x)  is the inverse of   f (x),   then  g( f (x)) = f (g(x)) = x.  See Example 1.76, Example 1.77, and Example
1.78.


• Each of the toolkit functions has an inverse. See Example 1.79.
• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
• For a tabular function, exchange the input and output rows to obtain the inverse. See Example 1.80.
• The inverse of a function can be determined at specific points on its graph. See Example 1.81.
• To find the inverse of a formula, solve the equation  y = f (x)  for  x  as a function of  y. Then exchange the labels  x 
and   y.   See Example 1.82, Example 1.83, and Example 1.84.


• The graph of an inverse function is the reflection of the graph of the original function across the line  y = x.  See
Example 1.85.


CHAPTER 1 REVIEW EXERCISES
Functions and Function Notation
For the following exercises, determine whether the relation
is a function.
490. {(a, b), (c, d), (e, d)}


491. ⎧⎩⎨(5, 2), (6, 1), (6, 2), (4, 8)⎫⎭⎬


492. y2 + 4 = x,   for  x  the independent variable and  y 
the dependent variable


493. Is the graph in Figure 1.126 a function?
Figure 1.126


For the following exercises, evaluate the function at the
indicated values:
    f ( − 3);    f (2);     f ( − a);    − f (a);     f (a + h).
494. f (x) = − 2x2 + 3x


495. f (x) = 2|3x − 1|


Chapter 1 Functions 173




For the following exercises, determine whether the
functions are one-to-one.
496. f (x) = − 3x + 5


497. f (x) = |x − 3|


For the following exercises, use the vertical line test to
determine if the relation whose graph is provided is a
function.
498.


499.


500.


For the following exercises, graph the functions.
501. f (x) = |x + 1|


502. f (x) = x2 − 2


For the following exercises, use Figure 1.127 to
approximate the values.


Figure 1.127
503. f (2)


504. f (−2)


505. If   f (x) = −2,   then solve for  x.


506. If   f (x) = 1,   then solve for  x.


For the following exercises, use the function
 h(t) = − 16t2 + 80t  to find the values.


507. h(2) − h(1)
2 − 1


508. h(a) − h(1)
a − 1


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Domain and Range
For the following exercises, find the domain of each
function, expressing answers using interval notation.
509. f (x) = 2


3x + 2


510. f (x) = x − 3
x2 − 4x − 12


511. f (x) = x − 6
x − 4


512. Graph this piecewise function:
f (x) =






x + 1 x < − 2
−2x − 3 x ≥ − 2


Rates of Change and Behavior of Graphs
For the following exercises, find the average rate of change
of the functions from  x = 1 to x = 2.
513. f (x) = 4x − 3


514. f (x) = 10x2 + x


515. f (x) = − 2
x2


For the following exercises, use the graphs to determine the
intervals on which the functions are increasing, decreasing,
or constant.
516.


517.


518.


519. Find the local minimum of the function graphed in
Exercise 1.516.


520. Find the local extrema for the function graphed in
Exercise 1.517.


521. For the graph in Figure 1.128, the domain of the
function is  [−3, 3]. The range is  [−10, 10].  Find the
absolute minimum of the function on this interval.


522. Find the absolute maximum of the function graphed
in Figure 1.128.


Chapter 1 Functions 175




Figure 1.128


Composition of Functions
For the following exercises, find  ( f ∘g)(x)  and  (g ∘ f )(x) 
for each pair of functions.
523. f (x) = 4 − x,  g(x) = − 4x


524. f (x) = 3x + 2,  g(x) = 5 − 6x


525. f (x) = x2 + 2x,  g(x) = 5x + 1


526. f (x) = x + 2, g(x) = 1x


527.   f (x) = x + 3
2


, g(x) = 1 − x


For the following exercises, find  ⎛⎝ f ∘g⎞⎠  and the domain for
 ⎛⎝ f ∘g⎞⎠(x)  for each pair of functions.


528. f (x) = x + 1
x + 4


, g(x) = 1x


529. f (x) = 1
x + 3


, g(x) = 1
x − 9


530. f (x) = 1x , g(x) = x


531. f (x) = 1
x2 − 1


, g(x) = x + 1


For the following exercises, express each function  H  as
a composition of two functions   f   and  g where
 H(x) = ( f ∘g)(x).


532. H(x) = 2x − 1
3x + 4


533. H(x) = 1
(3x2 − 4)−3


Transformation of Functions
For the following exercises, sketch a graph of the given
function.
534. f (x) = (x − 3)2


535. f (x) = (x + 4)3


536. f (x) = x + 5


537. f (x) = − x3


538. f (x) = −x3


539. f (x) = 5 −x − 4


540. f (x) = 4[|x − 2| − 6]


541. f (x) = − (x + 2)2 − 1


For the following exercises, sketch the graph of the
function  g  if the graph of the function   f   is shown in
Figure 1.129.


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Figure 1.129
542. g(x) = f (x − 1)


543. g(x) = 3 f (x)


For the following exercises, write the equation for the
standard function represented by each of the graphs below.
544.


545.


For the following exercises, determine whether each
function below is even, odd, or neither.
546. f (x) = 3x4


547. g(x) = x


548. h(x) = 1x + 3x


For the following exercises, analyze the graph and
determine whether the graphed function is even, odd, or
neither.
549.


550.


551.


Absolute Value Functions
For the following exercises, write an equation for the
transformation of   f (x) = |x|.
552.


Chapter 1 Functions 177




553.


554.


For the following exercises, graph the absolute value
function.
555. f (x) = |x − 5|


556. f (x) = − |x − 3|


557. f (x) = |2x − 4|


For the following exercises, solve the absolute value
equation.
558. |x + 4| = 18


559. |13x + 5| = |34x − 2|
For the following exercises, solve the inequality and
express the solution using interval notation.
560. |3x − 2| < 7


561. |13x − 2| ≤ 7


Inverse Functions
For the following exercises, find f −1(x) for each
function.
562. f (x) = 9 + 10x


563. f (x) = x
x + 2


For the following exercise, find a domain on which the
function f is one-to-one and non-decreasing. Write the
domain in interval notation. Then find the inverse of f
restricted to that domain.
564. f (x) = x2 + 1


565. Given f (x) = x3 − 5 and g(x) = x + 53 :
a. Find f (g(x)) and g( f (x)).
b. What does the answer tell us about the
relationship between f (x) and g(x)?


For the following exercises, use a graphing utility to
determine whether each function is one-to-one.
566. f (x) = 1x


567. f (x) = − 3x2 + x


568. If f (5) = 2, find f −1(2).


569. If f (1) = 4, find f −1(4).


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CHAPTER 1 PRACTICE TEST
For the following exercises, determine whether each of the
following relations is a function.
570. y = 2x + 8


571. {(2, 1), (3, 2), ( − 1, 1), (0, − 2)}


For the following exercises, evaluate the function
  f (x) = − 3x2 + 2x  at the given input.
572. f (−2)


573.   f (a) 


574. Show that the function   f (x) = − 2(x − 1)2 + 3  is
not one-to-one.


575. Write the domain of the function   f (x) = 3 − x  in
interval notation.


576. Given   f (x) = 2x2 − 5x,   find   f (a + 1) − f (1).


577. Graph the function
f (x) =






x + 1 if −2 < x < 3


− x if x ≥ 3


578. Find the average rate of change of the function
  f (x) = 3 − 2x2 + x  by finding   f (b) − f (a)


b − a
.


For the following exercises, use the functions
  f (x) = 3 − 2x2 + x and g(x) = x  to find the composite
functions.
579. ⎛⎝g ∘ f ⎞⎠(x)


580. ⎛⎝g ∘ f ⎞⎠(1)


581. Express  H(x) = 5x2 − 3x3   as a composition of two
functions,   f   and  g,  where  ⎛⎝ f ∘g⎞⎠(x) = H(x).


For the following exercises, graph the functions by
translating, stretching, and/or compressing a toolkit
function.
582. f (x) = x + 6 − 1


583. f (x) = 1
x + 2


− 1


For the following exercises, determine whether the
functions are even, odd, or neither.
584. f (x) = − 5


x2
+ 9x6


585. f (x) = − 5
x3


+ 9x5


586. f (x) = 1x


587. Graph the absolute value function
  f (x) = − 2|x − 1| + 3.


588. Solve |2x − 3| = 17.


589. Solve  − |13x − 3| ≥ 17. Express the solution in
interval notation.


For the following exercises, find the inverse of the function.
590. f (x) = 3x − 5


591. f (x) = 4
x + 7


For the following exercises, use the graph of  g  shown in
Figure 1.130.


Figure 1.130
592. On what intervals is the function increasing?


593. On what intervals is the function decreasing?


Chapter 1 Functions 179




594. Approximate the local minimum of the function.
Express the answer as an ordered pair.


595. Approximate the local maximum of the function.
Express the answer as an ordered pair.


For the following exercises, use the graph of the piecewise
function shown in Figure 1.131.


Figure 1.131
596. Find   f (2).


597. Find   f (−2).


598. Write an equation for the piecewise function.


For the following exercises, use the values listed in Table
1.49.


x F(x)


0 1


1 3


2 5


3 7


4 9


5 11


6 13


7 15


8 17


Table 1.49
599. Find  F(6).


600. Solve the equation  F(x) = 5.


601. Is the graph increasing or decreasing on its domain?


602. Is the function represented by the graph one-to-one?


603. Find  F−1(15).


604. Given   f (x) = − 2x + 11,   find   f −1(x).


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2 | LINEAR FUNCTIONS


Figure 2.1 A bamboo forest in China (credit: “JFXie”/Flickr)


Chapter Outline
2.1 Linear Functions
2.2 Graphs of Linear Functions
2.3 Modeling with Linear Functions
2.4 Fitting Linear Models to Data


Introduction
Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although
it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the
fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.[1] In a
twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such
as the growth cycle of this bamboo plant, is a linear function.
Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain
exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world
applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate
them to data.


1. http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/


Chapter 2 Linear Functions 181




2.1 | Linear Functions
Learning Objectives


In this section, you will:
2.1.1 Represent a linear function.
2.1.2 Determine whether a linear function is increasing, decreasing, or constant.
2.1.3 Calculate and interpret slope.
2.1.4 Write the point-slope form of an equation.
2.1.5 Write and interpret a linear function.


Figure 2.2 Shanghai MagLev Train (credit: “kanegen”/Flickr)


Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for
example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 2.2). It carries passengers
comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[2]
Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second
for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a
function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate
real-world situations such as the train’s distance from the station at a given point in time.
Representing Linear Functions
The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change,
that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function
notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
Representing a Linear Function in Word Form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence
may be used to describe the function relationship.


• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus
its original distance from the station when it began moving at constant speed.


The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes
with respect to the independent variable. The rate of change for this example is constant, which means that it is the same
for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters.
The train began moving at this constant speed at a distance of 250 meters from the station.


2. http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm


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Representing a Linear Function in Function Notation
Another approach to representing linear functions is by using function notation. One example of function notation is an
equation written in the form known as the slope-intercept form of a line, where x is the input value, m is the rate of
change, and b is the initial value of the dependent variable.


(2.1)Equation form y = mx + b
Equation notation f (x) = mx + b


In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The
rate,m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250
meters. We can write a generalized equation to represent the motion of the train.


D(t) = 83t + 250


Representing a Linear Function in Tabular Form
A third method of representing a linear function is through the use of a table. The relationship between the distance from
the station and the time is represented in Figure 2.3. From the table, we can see that the distance changes by 83 meters for
every 1 second increase in time.


Figure 2.3 Tabular representation of the function D showing
selected input and output values


Can the input in the previous example be any real number?
No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real
numbers are not possible for this example. The input consists of non-negative real numbers.


Representing a Linear Function in Graphical Form
Another way to represent linear functions is visually, using a graph. We can use the function relationship from above,
D(t) = 83t + 250, to draw a graph, represented in Figure 2.4. Notice the graph is a line. When we plot a linear function,
the graph is always a line.
The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero
is the vertical intercept, or y-intercept, of the line. We can see from the graph in Figure 2.4 that the y-intercept in the
train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a
constant speed.


Figure 2.4 The graph of D(t) = 83t + 250. Graphs of linear
functions are lines because the rate of change is constant.


Chapter 2 Linear Functions 183




Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line
f (x) = 2x+1. Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The
domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.


Linear Function
A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a
line


(2.2)f (x) = mx + b
where b is the initial or starting value of the function (when input, x = 0 ), and m is the constant rate of change, or
slope of the function. The y-intercept is at (0, b).


Example 2.1
Using a Linear Function to Find the Pressure on a Diver


The pressure, P, in pounds per square inch (PSI) on the diver in Figure 2.5 depends upon her depth below the
water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate
this function in words.


Figure 2.5 (credit: Ilse Reijs and Jan-Noud Hutten)


Solution
To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth
equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.


Analysis
The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water.
The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI
for each foot her depth increases.


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Determining whether a Linear Function Is Increasing, Decreasing, or
Constant
The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear
function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values
increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope
slants upward from left to right as in Figure 2.6(a). For a decreasing function, the slope is negative. The output values
decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 2.6(b).
If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero
is horizontal as in Figure 2.6(c).


Figure 2.6


Increasing and Decreasing Functions
The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant
function.


• f (x) = mx + b is an increasing function if m > 0.
• f (x) = mx + b is an decreasing function if m < 0.
• f (x) = mx + b is a constant function if m = 0.


Example 2.2
Deciding whether a Function Is Increasing, Decreasing, or Constant


Some recent studies suggest that a teenager sends an average of 60 texts per day.[3] For each of the following
scenarios, find the linear function that describes the relationship between the input value and the output value.
Then, determine whether the graph of the function is increasing, decreasing, or constant.
a. The total number of texts a teen sends is considered a function of time in days. The input is the number
of days, and output is the total number of texts sent.


b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and
output is the total number of texts remaining for the month.


3. http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/


Chapter 2 Linear Functions 185




c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is
the number of days, and output is the total cost of texting each month.


Solution
Analyze each function.
a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive
so the function is increasing. This makes sense because the total number of texts increases with each day.


b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the
slope is negative so the function is decreasing. This makes sense because the number of texts remaining
decreases each day and this function represents the number of texts remaining in the data plan after x
days.


c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total
cost. The slope is 0 so the function is constant.


Calculating and Interpreting Slope
In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope
given input and output values. Given two values for the input, x1 and x2, and two corresponding values for the output,
y1 and y2—which can be represented by a set of points, (x1 , y1) and (x2 , y2)—we can calculate the slope m, as
follows


m =
change in output (rise)
change in input (run)


=
Δy
Δx


=
y2 − y1
x2 − x1


where Δy is the vertical displacement and Δx is the horizontal displacement. Note in function notation two corresponding
values for the output y1 and y2 for the function f , y1 = f (x1) and y2 = f (x2), so we could equivalently write


m =
f (x2) – f (x1)


x2 – x1


Figure 2.7 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the
slope measures steepness. The greater the absolute value of the slope, the steeper the line is.


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Figure 2.7 The slope of a function is calculated by the change
in y divided by the change in x. It does not matter which
coordinate is used as the (x2, y2) and which is the
(x1, y1), as long as each calculation is started with the
elements from the same coordinate pair.


Are the units for slope always units for the output
units for the input


?


Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope
could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for
the input.


Calculate Slope
The slope, or rate of change, of a function m can be calculated according to the following:


(2.3)
m =


change in output (rise)
change in input (run)


=
Δy
Δx


=
y2 − y1
x2 − x1


where x1 and x2 are input values, y1 and y2 are output values.


Given two points from a linear function, calculate and interpret the slope.
1. Determine the units for output and input values.
2. Calculate the change of output values and change of input values.
3. Interpret the slope as the change in output values per unit of the input value.


Chapter 2 Linear Functions 187




2.1


2.2


Example 2.3
Finding the Slope of a Linear Function


If f (x) is a linear function, and (3,−2) and (8,1) are points on the line, find the slope. Is this function increasing
or decreasing?


Solution
The coordinate pairs are (3,−2) and (8,1). To find the rate of change, we divide the change in output by the
change in input.


m =
change in output
change in input


= 1 − ( − 2)
8 − 3


= 3
5


We could also write the slope as m = 0.6. The function is increasing because m > 0.


Analysis
As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as
long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used.


If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function
increasing or decreasing?


Example 2.4
Finding the Population Change from a Linear Function


The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population
per year if we assume the change was constant from 2008 to 2012.


Solution
The rate of change relates the change in population to the change in time. The population increased by
27, 800 − 23, 400 = 4400 people over the four-year time interval. To find the rate of change, divide the change
in the number of people by the number of years.


4,400 people
4 years


= 1,100
people
year


So the population increased by 1,100 people per year.


Analysis
Because we are told that the population increased, we would expect the slope to be positive. This positive slope
we calculated is therefore reasonable.


The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of
population per year if we assume the change was constant from 2009 to 2012.


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Writing the Point-Slope Form of a Linear Equation
Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will
learn another way to write a linear function, the point-slope form.


y − y1 = m(x − x1)


The point-slope form is derived from the slope formula.
m =


y − y1
x − x1


assuming x ≠ x1


m(x − x1) =
y − y1
x − x1


(x − x1) Multiply both sides by (x − x1).


m(x − x1) = y − y1 Simplify.


y − y1 = m(x − x1) Rearrange.


Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can
move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form,
y − 4 = − 1


2
(x − 6) . We can convert it to the slope-intercept form as shown.


y − 4 = − 1
2
(x − 6)


y − 4 = − 1
2
x + 3 Distribute the − 1


2
.


y = − 1
2
x + 7 Add 4 to each side.


Therefore, the same line can be described in slope-intercept form as y = − 1
2
x + 7.


Point-Slope Form of a Linear Equation
The point-slope form of a linear equation takes the form


(2.4)y − y1 = m(x − x1)
where m is the slope, x1 and y1 are the x and y coordinates of a specific point through which the line passes.


Writing the Equation of a Line Using a Point and the Slope
The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told
that a line has a slope of 2 and passes through the point (4, 1). We know that m = 2 and that x1 = 4 and y1 = 1. We
can substitute these values into the general point-slope equation.


y − y1 = m(x − x1)


y − 1 = 2(x − 4)


If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.
y − 1 = 2(x − 4)
y − 1 = 2x − 8 Distribute the 2.
y = 2x − 7 Add 1 to each side.


Both equations, y − 1 = 2(x − 4) and y = 2x – 7, describe the same line. See Figure 2.8.


Chapter 2 Linear Functions 189




2.3


Figure 2.8


Example 2.5
Writing Linear Equations Using a Point and the Slope


Write the point-slope form of an equation of a line with a slope of 3 that passes through the point (6, –1). Then
rewrite it in the slope-intercept form.


Solution
Let’s figure out what we know from the given information. The slope is 3, so m = 3. We also know one point,
so we know x1 = 6 and y1 = −1. Now we can substitute these values into the general point-slope equation.


y − y1 = m(x − x1)


y − ( − 1) = 3(x − 6) Substitute known values.
y + 1 = 3(x − 6) Distribute − 1 to find point-slope orm.


Then we use algebra to find the slope-intercept form.
y + 1 = 3(x − 6)
y + 1 = 3x − 18 Distribute 3.
y = 3x − 19 Simplify to slope-intercept form.


Write the point-slope form of an equation of a line with a slope of –2 that passes through the point
(–2, 2). Then rewrite it in the slope-intercept form.


Writing the Equation of a Line Using Two Points
The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for
example, we know that a line passes through the points (0, 1) and (3, 2). We can use the coordinates of the two points
to find the slope.


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m =
y2 − y1
x2 − x1


= 2 − 1
3 − 0


= 1
3


Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1)
for our point.


y − y1 = m(x − x1)


y − 1 = 1
3
(x − 0)


As before, we can use algebra to rewrite the equation in the slope-intercept form.
y − 1 = 1


3
(x − 0)


y − 1 = 1
3
x Distribute the 1


3
.


y = 1
3
x + 1 Add 1 to each side.


Both equations describe the line shown in Figure 2.9.


Figure 2.9


Example 2.6
Writing Linear Equations Using Two Points


Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it
in the slope-intercept form.


Solution
Let’s begin by finding the slope.


Chapter 2 Linear Functions 191




2.4


m =
y2 − y1
x2 − x1


= 7 − 1
8 − 5


= 6
3


= 2


So m = 2. Next, we substitute the slope and the coordinates for one of the points into the general point-slope
equation. We can choose either point, but we will use (5, 1).


y − y1 = m(x − x1)


y − 1 = 2(x − 5)


The point-slope equation of the line is y2 – 1 = 2(x2 – 5). To rewrite the equation in slope-intercept form, we
use algebra.


y − 1 = 2(x − 5)
y − 1 = 2x − 10
y = 2x − 9


The slope-intercept equation of the line is y = 2x – 9.


Write the point-slope form of an equation of a line that passes through the points (–1, 3) and (0, 0).
Then rewrite it in the slope-intercept form.


Writing and Interpreting an Equation for a Linear Function
Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can
choose which method to use based on the information we are given. That information may be provided in the form of a
graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 2.10.


Figure 2.10


We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7)
and (4, 4). We can use these points to calculate the slope.


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m =
y2 − y1
x2 − x1


= 4 − 7
4 − 0


= − 3
4


Now we can substitute the slope and the coordinates of one of the points into the point-slope form.
y − y1 = m(x − x1)


y − 4 = − 3
4
(x − 4)


If we want to rewrite the equation in the slope-intercept form, we would find
y − 4 = − 3


4
(x − 4)


y − 4 = − 3
4
x + 3


y = − 3
4
x + 7


If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the
line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so
we can substitute m and b into the slope-intercept form of a line.


So the function is f (x) = − 3
4
x + 7, and the linear equation would be y = − 3


4
x + 7.


Given the graph of a linear function, write an equation to represent the function.
1. Identify two points on the line.
2. Use the two points to calculate the slope.
3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection.
4. Substitute the slope and y-intercept into the slope-intercept form of a line equation.


Example 2.7
Writing an Equation for a Linear Function


Write an equation for a linear function given a graph of f shown in Figure 2.11.


Chapter 2 Linear Functions 193




Figure 2.11


Solution
Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope.


m =
y2 − y1
x2 − x1


= −4 − 2
−2 − 0


= −6
−2


= 3


Substitute the slope and the coordinates of one of the points into the point-slope form.
y − y1 = m(x − x1)


y − (−4) = 3(x − (−2))


y + 4 = 3(x + 2)


We can use algebra to rewrite the equation in the slope-intercept form.
y + 4 = 3(x + 2)
y + 4 = 3x + 6
y = 3x + 2


Analysis
This makes sense because we can see from Figure 2.12 that the line crosses the y-axis at the point (0, 2) ,
which is the y-intercept, so b = 2.


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Figure 2.12


Example 2.8
Writing an Equation for a Linear Cost Function


Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which
includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the
cost for x items produced in a given month.


Solution
The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item,
which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs
is the sum of these two costs, represented by C(x) = 1250 + 37.5x.


Analysis
If Ben produces 100 items in a month, his monthly cost is represented by


C(100) = 1250 + 37.5(100)
= 5000


So his monthly cost would be $5,000.


Example 2.9
Writing an Equation for a Linear Function Given Two Points


If f is a linear function, with f (3) = −2 , and f (8) = 1 , find an equation for the function in slope-intercept
form.


Solution
We can write the given points using coordinates.


Chapter 2 Linear Functions 195




2.5


f (3) = −2 → (3, −2)
f (8) = 1 → (8, 1)


We can then use the points to calculate the slope.
m =


y2 − y1
x2 − x1


= 1 − ( − 2)
8 − 3


= 3
5


Substitute the slope and the coordinates of one of the points into the point-slope form.
y − y1 = m(x − x1)


y − ( − 2) = 3
5
(x − 3)


We can use algebra to rewrite the equation in the slope-intercept form.
y + 2 = 3


5
(x − 3)


y + 2 = 3
5
x − 9


5


y = 3
5
x − 19


5


If f (x) is a linear function, with f (2) = – 11, and f (4) = − 25, find an equation for the function in
slope-intercept form.


Modeling Real-World Problems with Linear Functions
In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately,
we can analyze the problem by first representing it as a linear function and then interpreting the components of the function.
As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many
different kinds of real-world problems.


Given a linear function f and the initial value and rate of change, evaluate f(c).
1. Determine the initial value and the rate of change (slope).
2. Substitute the values into f (x) = mx + b.
3. Evaluate the function at x = c.


Example 2.10
Using a Linear Function to Determine the Number of Songs in a Music Collection


Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for
the number of songs, N, in his collection as a function of time, t, the number of months. How many songs
will he own in a year?


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Solution
The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means
that b = 200.
The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we
know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a
line.


Figure 2.13


We can write the formula N(t) = 15t + 200.
With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words,
we can evaluate the function at t = 12.


N(12) = 15(12) + 200
= 180 + 200
= 380


Marcus will have 380 songs in 12 months.


Analysis
Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number
of songs) increases as well.


Example 2.11
Using a Linear Function to Calculate Salary Plus Commission


Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore,
Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he
sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920.
Find an equation for I(n), and interpret the meaning of the components of the equation.


Solution
The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of
change.


m = 920 − 760
5 − 3


= $160
2 policies


= $80 per policy


Chapter 2 Linear Functions 197




Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies
increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy
sold during the week.
We can then solve for the initial value.


I(n) = 80n + b
760 = 80(3) + b when n = 3, I(3) = 760
760 − 80(3) = b
520 = b


The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new
policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number
of policies sold.
We can now write the final equation.


I(n) = 80n + 520


Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for
each policy sold.


Example 2.12
Using Tabular Form to Write an Equation for a Linear Function


Table 2.1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.


w, number of weeks 0 2 4 6


P(w), number of rats 1000 1080 1160 1240


Table 2.1


Solution
We can see from the table that the initial value for the number of rats is 1000, so b = 1000.
Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2
weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per
week.


P(w) = 40w + 1000


If we did not notice the rate of change from the table we could still solve for the slope using any two points from
the table. For example, using (2, 1080) and (6, 1240)


m = 1240 − 1080
6 − 2


= 160
4


= 40


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2.6


Is the initial value always provided in a table of values like Table 2.1?
No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then
the initial value would be the corresponding output. If the initial value is not provided because there is no value
of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b,
and solve for b.


A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2.2
shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x),
where x is the number of months since the start of the experiment.


x 0 2 4 8 12


H(x) 12.5 13.5 14.5 16.5 18.5


Table 2.2


Access this online resource for additional instruction and practice with linear functions.
• Linear Functions (http://openstaxcollege.org/l/linearfunctions)


Chapter 2 Linear Functions 199




1.


2.


3.


4.


5.


6.


7.


8.


9.


10.


11.


12.


13.


14.


15.


16.


17.


18.


19.


20.


21.


22.


23.


24.


25.


26.


27.


28.


29.


30.


31.


32.


33.


34.


35.


36.


37.


2.1 EXERCISES
Verbal
Terry is skiing down a steep hill. Terry's elevation,


E(t), in feet after t seconds is given by
E(t) = 3000 − 70t. Write a complete sentence describing
Terry’s starting elevation and how it is changing over time.
Maria is climbing a mountain. Maria's elevation, E(t),


in feet after t minutes is given by E(t) = 1200 + 40t.
Write a complete sentence describing Maria’s starting
elevation and how it is changing over time.
Jessica is walking home from a friend’s house. After 2


minutes she is 1.4 miles from home. Twelve minutes after
leaving, she is 0.9 miles from home. What is her rate in
miles per hour?
Sonya is currently 10 miles from home and is walking


farther away at 2 miles per hour. Write an equation for her
distance from home t hours from now.
A boat is 100 miles away from the marina, sailing


directly toward it at 10 miles per hour. Write an equation for
the distance of the boat from the marina after t hours.
Timmy goes to the fair with $40. Each ride costs $2.


How much money will he have left after riding n rides?


Algebraic
For the following exercises, determine whether the
equation of the curve can be written as a linear function.


y = 1
4
x + 6


y = 3x − 5


y = 3x2 − 2


3x + 5y = 15


3x2 + 5y = 15


3x + 5y2 = 15


−2x2 + 3y2 = 6


− x − 3
5


= 2y


For the following exercises, determine whether each
function is increasing or decreasing.


f (x) = 4x + 3


g(x) = 5x + 6


a(x) = 5 − 2x


b(x) = 8 − 3x


h(x) = − 2x + 4


k(x) = − 4x + 1


j(x) = 1
2
x − 3


p(x) = 1
4
x − 5


n(x) = − 1
3
x − 2


m(x) = − 3
8
x + 3


For the following exercises, find the slope of the line that
passes through the two given points.


(2, 4) and (4, 10)


(1, 5) and (4, 11)


(−1, 4) and (5, 2)


(8, −2) and (4, 6)


(6, 11) and ( − 4, 3)


For the following exercises, given each set of information,
find a linear equation satisfying the conditions, if possible.


f ( − 5) = − 4, and f (5) = 2


f (−1) = 4 and f (5) = 1


(2, 4) and (4, 10)


Passes through (1, 5) and (4, 11)


Passes through (−1, 4) and (5, 2)


Passes through (−2, 8) and (4, 6)


x intercept at (−2, 0) and y intercept at (0, −3)


x intercept at (−5, 0) and y intercept at (0, 4)


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38.


39.


40.


41.


42.


Graphical
For the following exercises, find the slope of the lines
graphed.


For the following exercises, write an equation for the lines
graphed.


Chapter 2 Linear Functions 201




43.


44.


45.


46.


47.


48.


Numeric
For the following exercises, which of the tables could
represent a linear function? For each that could be linear,
find a linear equation that models the data.


x 0 5 10 15


g(x) 5 –10 –25 –40


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49.


50.


51.


52.


53.


54.


55.


56.


57.


58.


59.


60.


61.


62.


x 0 5 10 15


h(x) 5 30 105 230


x 0 5 10 15


f(x) –5 20 45 70


x 5 10 20 25


k(x) 28 13 58 73


x 0 2 4 6


g(x) 6 –19 –44 –69


x 2 4 6 8


f(x) –4 16 36 56


x 2 4 6 8


f(x) –4 16 36 56


x 0 2 6 8


k(x) 6 31 106 231


Technology
If f is a linear function,


f (0.1) = 11.5, and f (0.4) = – 5.9, find an
equation for the function.
Graph the function f on a domain of


[ – 10, 10] : f (x) = 0.02x − 0.01. Enter the function
in a graphing utility. For the viewing window, set the
minimum value of x to be −10 and the maximum value
of x to be 10.


Graph the function f on a domain of
[ – 10, 10] : f x) = 2, 500x + 4, 000


Table 2.3 shows the input, w, and output, k, for a
linear function k. a. Fill in the missing values of the table.
b. Write the linear function k, round to 3 decimal places.


w –10 5.5 67.5 b


k 30 –26 a –44


Table 2.3


Table 2.4 shows the input, p, and output, q, for a
linear function q. a. Fill in the missing values of the table.
b. Write the linear function k.


p 0.5 0.8 12 b


q 400 700 a 1,000,000


Table 2.4


Graph the linear function f on a domain of
[−10, 10] for the function whose slope is 1


8
and y-


intercept is 31
16
. Label the points for the input values of


−10 and 10.


Graph the linear function f on a domain of
[−0.1, 0.1] for the function whose slope is 75 and y-
intercept is −22.5 . Label the points for the input values of
−0.1 and 0.1.


Chapter 2 Linear Functions 203




63.


64.


65.


66.


67.


68.


69.


70.


71.


72.


73.


74.


75.


76.


Graph the linear function f where f (x) = ax + b on the
same set of axes on a domain of [−4, 4] for the following
values of a and b.
i. a = 2; b = 3
ii. a = 2; b = 4
iii. a = 2; b = – 4
iv. a = 2; b = – 5


Extensions
Find the value of x if a linear function goes through


the following points and has the following slope:
(x, 2), ( − 4, 6),  m = 3


Find the value of y if a linear function goes through the
following points and has the following slope:
(10, y), (25, 100),  m = − 5


Find the equation of the line that passes through the
following points: (a, b) and (a, b + 1)


Find the equation of the line that passes through the
following points: (2a, b) and (a, b + 1)


Find the equation of the line that passes through the
following points: (a, 0) and (c, d)


Real-World Applications
At noon, a barista notices that she has $20 in her tip jar.


If she makes an average of $0.50 from each customer, how
much will she have in her tip jar if she serves n more
customers during her shift?
A gym membership with two personal training


sessions costs $125, while gym membership with five
personal training sessions costs $260. What is cost per
session?
A clothing business finds there is a linear relationship


between the number of shirts, n, it can sell and the price,
p, it can charge per shirt. In particular, historical data
shows that 1,000 shirts can be sold at a price of $30 , while
3,000 shirts can be sold at a price of $22. Find a linear
equation in the form p(n) = mn + b that gives the price p
they can charge for n shirts.


A phone company charges for service according to the
formula: C(n) = 24 + 0.1n, where n is the number of
minutes talked, and C(n) is the monthly charge, in dollars.
Find and interpret the rate of change and initial value.


A farmer finds there is a linear relationship between the
number of bean stalks, n, she plants and the yield, y,
each plant produces. When she plants 30 stalks, each plant
yields 30 oz of beans. When she plants 34 stalks, each plant
produces 28 oz of beans. Find a linear relationships in the
form y = mn + b that gives the yield when n stalks are
planted.
A city’s population in the year 1960 was 287,500. In


1989 the population was 275,900. Compute the rate of
growth of the population and make a statement about the
population rate of change in people per year.
A town’s population has been growing linearly. In


2003, the population was 45,000, and the population has
been growing by 1,700 people each year. Write an equation,
P(t), for the population t years after 2003.


Suppose that average annual income (in dollars) for the
years 1990 through 1999 is given by the linear function:
I(x) = 1054x + 23, 286, where x is the number of
years after 1990. Which of the following interprets the
slope in the context of the problem?
a. As of 1990, average annual income was $23,286.
b. In the ten-year period from 1990–1999, average
annual income increased by a total of $1,054.


c. Each year in the decade of the 1990s, average
annual income increased by $1,054.


d. Average annual income rose to a level of $23,286
by the end of 1999.


When temperature is 0 degrees Celsius, the Fahrenheit
temperature is 32. When the Celsius temperature is 100, the
corresponding Fahrenheit temperature is 212. Express the
Fahrenheit temperature as a linear function of C, the
Celsius temperature, F(C).
a. Find the rate of change of Fahrenheit temperature
for each unit change temperature of Celsius.


b. Find and interpret F(28).
c. Find and interpret F(–40).


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2.2 | Graphs of Linear Functions
Learning Objectives


In this section, you will:
2.2.1 Graph linear functions.
2.2.2 Write the equation for a linear function from the graph of a line.
2.2.3 Given the equations of two lines, determine whether their graphs are parallel or
perpendicular.
2.2.4 Write the equation of a line parallel or perpendicular to a given line.
2.2.5 Solve a system of linear equations.


Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance
minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the
same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by
a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of
comparing functions using graphs.
Graphing Linear Functions
In Linear Functions, we saw that that the graph of a linear function is a straight line. We were also able to see the points
of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their
characteristics.
There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the
points. The second is by using the y-intercept and slope. And the third is by using transformations of the identity function
f (x) = x.


Graphing a Function by Plotting Points
To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output
values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid.
In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For
example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value
of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2
yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all
three points do not fall on the same line, we know we made an error.


Given a linear function, graph by plotting points.
1. Choose a minimum of two input values.
2. Evaluate the function at each input value.
3. Use the resulting output values to identify coordinate pairs.
4. Plot the coordinate pairs on a grid.
5. Draw a line through the points.


Example 2.13
Graphing by Plotting Points


Graph f (x) = − 2
3
x + 5 by plotting points.


Chapter 2 Linear Functions 205




2.7


Solution
Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose
multiples of 3 as input values. We will choose 0, 3, and 6.
Evaluate the function at each input value, and use the output value to identify coordinate pairs.


x = 0 f (0) = − 2
3
(0) + 5 = 5 ⇒(0, 5)


x = 3 f (3) = − 2
3
(3) + 5 = 3 ⇒(3, 3)


x = 6 f (6) = − 2
3
(6) + 5 = 1 ⇒(6, 1)


Plot the coordinate pairs and draw a line through the points. Figure 2.14 represents the graph of the function
f (x) = − 2


3
x + 5.


Figure 2.14 The graph of the linear function
f (x) = − 2


3
x + 5.


Analysis
The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant,
which indicates a negative slope. This is also expected from the negative constant rate of change in the equation
for the function.


Graph f (x) = − 3
4
x + 6 by plotting points.


Graphing a Function Using y-intercept and Slope
Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The
first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set
x = 0 in the equation.
The other characteristic of the linear function is its slope m, which is a measure of its steepness. Recall that the slope is the
rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs.
Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We
encountered both the y-intercept and the slope in Linear Functions.
Let’s consider the following function.


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f (x) = 1
2
x + 1


The slope is 1
2
. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the


point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. We
can begin graphing by plotting the point (0, 1) We know that the slope is rise over run, m = riserun. From our example, we
have m = 1


2
, which means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can rise 1 and


then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as
shown in Figure 2.15.


Figure 2.15


Graphical Interpretation of a Linear Function
In the equation f (x) = mx + b


• b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis.
• m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run)
between each successive pair of points. Recall the formula for the slope:


m =
change in output (rise)
change in input (run)


=
Δy
Δx


=
y2 − y1
x2 − x1


Do all linear functions have y-intercepts?
Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-
axis does not have a y-intercept, but it is not a function.)


Given the equation for a linear function, graph the function using the y-intercept and slope.
1. Evaluate the function at an input value of zero to find the y-intercept.
2. Identify the slope as the rate of change of the input value.
3. Plot the point represented by the y-intercept.
4. Use riserun to determine at least two more points on the line.


5. Sketch the line that passes through the points.


Chapter 2 Linear Functions 207




2.8


Example 2.14
Graphing by Using the y-intercept and Slope


Graph f (x) = − 2
3
x + 5 using the y-intercept and slope.


Solution
Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross
the y-axis at (0, 5).


According to the equation for the function, the slope of the line is −2
3
. This tells us that for each vertical decrease


in the “rise” of – 2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the
function by first plotting the y-intercept on the graph in Figure 2.16. From the initial value (0, 5) we move
down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a
line through the points.


Figure 2.16


Analysis
The graph slants downward from left to right, which means it has a negative slope as expected.


Find a point on the graph we drew in Example 2.14 that has a negative x-value.


Graphing a Function Using Transformations
Another option for graphing is to use transformations of the identity function f (x) = x . A function may be transformed by
a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.


Vertical Stretch or Compression


In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is
negative, there is also a vertical reflection of the graph. Notice in Figure 2.17 that multiplying the equation of f (x) = x
by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if
0 < m < 1. This means the larger the absolute value of m, the steeper the slope.


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Figure 2.17 Vertical stretches and compressions and reflections on the function f (x) = x.


Vertical Shift


In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line.
Notice in Figure 2.18 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if
b is positive and |b| units down if b is negative.


Chapter 2 Linear Functions 209




Figure 2.18 This graph illustrates vertical shifts of the function f (x) = x.


Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of
linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each
method.


Given the equation of a linear function, use transformations to graph the linear function in the form
f(x) =mx+ b.


1. Graph f (x) = x.
2. Vertically stretch or compress the graph by a factor m.
3. Shift the graph up or down b units.


Example 2.15
Graphing by Using Transformations


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2.9


Graph f (x) = 1
2
x − 3 using transformations.


Solution
The equation for the function shows that m = 1


2
so the identity function is vertically compressed by 1


2
. The


equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First,
graph the identity function, and show the vertical compression as in Figure 2.19.


Figure 2.19 The function, y = x, compressed by a factor of
1
2
.


Then show the vertical shift as in Figure 2.20.


Figure 2.20 The function y = 1
2
x, shifted down 3 units.


Graph f (x) = 4 + 2x, using transformations.


Chapter 2 Linear Functions 211




In Example 2.15, could we have sketched the graph by reversing the order of the transformations?
No. The order of the transformations follows the order of operations. When the function is evaluated at a given
input, the corresponding output is calculated by following the order of operations. This is why we performed the
compression first. For example, following the order: Let the input be 2.


f (2) = 1
2
(2) − 3


= 1 − 3
= − 2


Writing the Equation for a Function from the Graph of a Line
Recall that in Linear Functions, we wrote the equation for a linear function from a graph. Now we can extend what we
know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 2.21. We
can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept.


Figure 2.21


Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point
( − 2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the
slope must be


m = riserun =
4
2
= 2


Substituting the slope and y-intercept into the slope-intercept form of a line gives
y = 2x + 4


Given a graph of linear function, find the equation to describe the function.
1. Identify the y-intercept of an equation.
2. Choose two points to determine the slope.
3. Substitute the y-intercept and slope into the slope-intercept form of a line.


Example 2.16
Matching Linear Functions to Their Graphs


Match each equation of the linear functions with one of the lines in Figure 2.22.


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a. f (x) = 2x + 3
b. g(x) = 2x − 3
c. h(x) = − 2x + 3


d. j(x) = 1
2
x + 3


Figure 2.22


Solution
Analyze the information for each function.
a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward
from left to right. We can use two points to find the slope, or we can compare it with the other functions
listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant
because they have the same slope. Line III does not pass through (0, 3) so f must be represented by
line I.


b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, − 3) and
slant upward from left to right. It must be represented by line III.


c. This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative
slope, so it must be represented by line IV because it slants downward from left to right.


d. This function has a slope of 1
2
and a y-intercept of 3. It must pass through the point (0, 3) and slant


upward from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of
f so the line for j must be flatter. This function is represented by Line II.


Now we can re-label the lines as in Figure 2.23.


Chapter 2 Linear Functions 213




Figure 2.23


Finding the x-intercept of a Line
So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis.
A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the
x-axis. In other words, it is the input value when the output value is zero.
To find the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function
shown.


f (x) = 3x − 6


Set the function equal to 0 and solve for x.
0 = 3x − 6
6 = 3x
2 = x
x = 2


The graph of the function crosses the x-axis at the point (2, 0).


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Do all linear functions have x-intercepts?
No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of
linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function
has no x-intercepts, as shown in Figure 2.24.


Figure 2.24


x-intercept
The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b.


Example 2.17
Finding an x-intercept


Find the x-intercept of f (x) = 1
2
x − 3.


Solution
Set the function equal to zero to solve for x.


0 = 1
2
x − 3


3 = 1
2
x


6 = x
x = 6


The graph crosses the x-axis at the point (6, 0).


Analysis
A graph of the function is shown in Figure 2.25. We can see that the x-intercept is (6, 0) as we expected.


Chapter 2 Linear Functions 215




2.10


Figure 2.25 The graph of the linear function
f (x) = 1


2
x − 3.


Find the x-intercept of f (x) = 1
4
x − 4.


Describing Horizontal and Vertical Lines
There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output,
or y-value. In Figure 2.26, we see that the output has a value of 2 for every input value. The change in outputs between
any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation
f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph
represents the function f (x) = 2.


Figure 2.26 A horizontal line representing the function
f (x) = 2.


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A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the
output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a
function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator
will be zero, so the slope of a vertical line is undefined.


Notice that a vertical line, such as the one in Figure 2.27, has an x-intercept, but no y-intercept unless it’s the line x = 0.
This graph represents the line x = 2.


Figure 2.27 The vertical line, x = 2, which does not
represent a function.


Horizontal and Vertical Lines
Lines can be horizontal or vertical.
A horizontal line is a line defined by an equation in the form f (x) = b.
A vertical line is a line defined by an equation in the form x = a.


Example 2.18
Writing the Equation of a Horizontal Line


Write the equation of the line graphed in Figure 2.28.


Chapter 2 Linear Functions 217




Figure 2.28


Solution
For any x-value, the y-value is −4, so the equation is y = − 4.


Example 2.19
Writing the Equation of a Vertical Line


Write the equation of the line graphed in Figure 2.29.


Figure 2.29


Solution
The constant x-value is 7, so the equation is x = 7.


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Determining Whether Lines are Parallel or Perpendicular
The two lines in Figure 2.30 are parallel lines: they will never intersect. Notice that they have exactly the same steepness,
which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line
vertically toward the y-intercept of the other, they would become the same line.


Figure 2.30 Parallel lines.


We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same
and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.


Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines
in Figure 2.31 are perpendicular.


Figure 2.31 Perpendicular lines.


Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific
way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal
is 1. So, if m1 and  m2 are negative reciprocals of one another, they can be multiplied together to yield –1.


Chapter 2 Linear Functions 219




m1m2 = − 1


To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 1
8
, and the reciprocal of 1


8
is 8. To


find the negative reciprocal, first find the reciprocal and then change the sign.
As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the
lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the
lines are perpendicular.


f (x) = 1
4
x + 2 negative reciprocal of 1


4
is −4


f (x) = − 4x + 3 negative reciprocal of − 4 is 1
4


The product of the slopes is –1.
−4⎛⎝


1
4

⎠ = − 1


Parallel and Perpendicular Lines
Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.


f (x) = m1 x  +   b1 and g(x) = m2 x  +   b2 are parallel if m1    =   m2.
If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line.
Two lines are perpendicular lines if they intersect at right angles.


f (x) = m1 x + b1 and g(x) = m2 x + b2 are perpendicular if m1m2 = − 1, and so m2 = −
1
m1


.


Example 2.20
Identifying Parallel and Perpendicular Lines


Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of
perpendicular lines.


f (x) = 2x + 3 h(x) = − 2x + 2


g(x) = 1
2
x − 4   j(x) = 2x − 6


Solution
Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope
of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1


2
are


negative reciprocals, the equations, g(x) = 1
2
x − 4 and h(x) = − 2x + 2 represent perpendicular lines.


Analysis
A graph of the lines is shown in Figure 2.32.


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Figure 2.32


The graph shows that the lines f (x) = 2x + 3 and j(x) = 2x – 6 are parallel, and the lines g(x) = 1
2
x – 4 and


h(x) = − 2x + 2 are perpendicular.


Writing the Equation of a Line Parallel or Perpendicular to a Given
Line
If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel
or perpendicular to the given line.
Writing Equations of Parallel Lines
Suppose for example, we are given the following equation.


f (x) = 3x + 1


We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line
with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x).


g(x) = 3x + 6
h(x) = 3x + 1


p(x) = 3x + 2
3


Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). We already
know that the slope is 3. We just need to determine which value for b will give the correct line. We can begin with the
point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.


y − y1 = m(x − x1)


  y − 7 = 3(x − 1)
  y − 7 = 3x − 3
y = 3x + 4


So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7).


Chapter 2 Linear Functions 221




Given the equation of a function and a point through which its graph passes, write the equation of a line
parallel to the given line that passes through the given point.


1. Find the slope of the function.
2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for
a line.


3. Simplify.


Example 2.21
Finding a Line Parallel to a Given Line


Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0).


Solution
The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3, x = 3, and
f (x) = 0 into the slope-intercept form to find the y-intercept.


g(x) = 3x + b
0 = 3(3) + b
b = – 9


The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9.


Analysis
We can confirm that the two lines are parallel by graphing them. Figure 2.33 shows that the two lines will never
intersect.


Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.
Figure 2.33


Writing Equations of Perpendicular Lines
We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same
slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:


f (x) = 2x + 4


The slope of the line is 2, and its negative reciprocal is −1
2
. Any function with a slope of −1


2
will be perpendicular to


f (x). So the lines formed by all of the following functions will be perpendicular to f (x).


g(x) = − 1
2
x + 4


h(x) = − 1
2
x + 2


p(x) = − 1
2
x − 1


2


As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given
point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point
(4, 0). We already know that the slope is −1


2
. Now we can use the point to find the y-intercept by substituting the given


values into the slope-intercept form of a line and solving for b.


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g(x) = mx + b


        0 = − 1
2
(4) + b


        0 = − 2 + b
        2 = b
        b = 2


The equation for the function with a slope of −1
2
and a y-intercept of 2 is


g(x) = − 1
2
x + 2.


So g(x) = − 1
2
x + 2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular


lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.
A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are
perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of
perpendicular lines?
No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a
function so the definition is not contradicted.


Given the equation of a function and a point through which its graph passes, write the equation of a line
perpendicular to the given line.


1. Find the slope of the function.
2. Determine the negative reciprocal of the slope.
3. Substitute the new slope and the values for x and y from the coordinate pair provided into


g(x) = mx + b.


4. Solve for b.
5. Write the equation for the line.


Example 2.22
Finding the Equation of a Perpendicular Line


Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0).


Solution
The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or −1


3
.


Using this slope and the given point, we can find the equation for the line.
g(x) = – 1


3
x + b


0 = – 1
3
(3) + b


1 = b
b = 1


The line perpendicular to f (x) that passes through (3, 0) is g(x) = − 1
3
x + 1.


Chapter 2 Linear Functions 223




2.11


Analysis
A graph of the two lines is shown in Figure 2.34 below.


Figure 2.34


Given the function h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is


a. parallel to h(x)
b. perpendicular to h(x)


Given two points on a line and a third point, write the equation of the perpendicular line that passes through
the point.


1. Determine the slope of the line passing through the points.
2. Find the negative reciprocal of the slope.
3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.
4. Simplify.


Example 2.23
Finding the Equation of a Line Perpendicular to a Given Line Passing through a
Point


A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes
through the point (4, 5).


Solution


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2.12


From the two points of the given line, we can calculate the slope of that line.
m1 =


5 − 6
4 − ( − 2)


         = −1
6


         = − 1
6


Find the negative reciprocal of the slope.
m2 =


−1
− 1


6


          = − 1⎛⎝−61



          = 6
We can then solve for the y-intercept of the line passing through the point (4, 5).


g(x) = 6x + b


        5 = 6(4) + b
        5 = 24 + b
−19 = b
        b = −19


The equation for the line that is perpendicular to the line passing through the two given points and also passes
through point (4, 5) is


y = 6x − 19


A line passes through the points, (−2, −15) and (2,−3). Find the equation of a perpendicular line that
passes through the point, (6, 4).


Solving a System of Linear Equations Using a Graph
A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point
if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and
they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations
and therefore represent the solution to the system.
To find this point when the equations are given as functions, we can solve for an input value so that f (x) = g(x). In other
words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.


Example 2.24
Finding a Point of Intersection Algebraically


Find the point of intersection of the lines h(t) = 3t − 4 and j(t) = 5 − t.


Solution


Chapter 2 Linear Functions 225




Set h(t) = j(t).
3t − 4 = 5 − t
4t = 9


t = 9
4


This tells us the lines intersect when the input is 9
4
.


We can then find the output value of the intersection point by evaluating either function at this input.


j⎛⎝
9
4

⎠ = 5 −


9
4


= 11
4


These lines intersect at the point ⎛⎝94,
11
4

⎠.


Analysis
Looking at Figure 2.35, this result seems reasonable.


Figure 2.35


If we were asked to find the point of intersection of two distinct parallel lines, should something in the
solution process alert us to the fact that there are no solutions?
Yes. After setting the two equations equal to one another, the result would be the contradiction “0 = non-zero real
number”.


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2.13 Look at the graph in Figure 2.35 and identify the following for the function j(t) :


a. y-intercept
b. x-intercept(s)
c. slope
d. Is j(t) parallel or perpendicular to h(t) (or neither)?
e. Is j(t) an increasing or decreasing function (or neither)?
f. Write a transformation description for j(t) from the identity toolkit function f (x) = x.


Example 2.25
Finding a Break-Even Point


A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120
to produce, and sells for $140.
a. Find the cost function, C, to produce x helmets, in dollars.
b. Find the revenue function, R, from the sales of x helmets, in dollars.
c. Find the break-even point, the point of intersection of the two graphs C and R.


Solution
a. The cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.


C(x) = 120x + 250, 000


b. The revenue function is the total revenue from the sale of x helmets, R(x) = 140x.
c. The break-even point is the point of intersection of the graph of the cost and revenue functions. To find
the x-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve
for x.


C(x) = R(x)
250, 000 + 120x = 140x
250, 000 = 20x
12, 500 = x
x = 12, 500


To find y, evaluate either the revenue or the cost function at 12,500.
R(20) = 140(12, 500)


                = $1, 750, 000
The break-even point is (12,500,1,750,000).


Analysis
This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75
million dollars. See Figure 2.36


Chapter 2 Linear Functions 227




Figure 2.36


Access these online resources for additional instruction and practice with graphs of linear functions.
• Finding Input of Function from the Output and Graph (http://openstaxcollege.org/l/
findinginput)


• Graphing Functions using Tables (http://openstaxcollege.org/l/graphwithtable)


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77.


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105.


2.2 EXERCISES
Verbal
If the graphs of two linear functions are parallel,


describe the relationship between the slopes and the y-
intercepts.
If the graphs of two linear functions are perpendicular,


describe the relationship between the slopes and the y-
intercepts.
If a horizontal line has the equation f (x) = a and a


vertical line has the equation x = a, what is the point of
intersection? Explain why what you found is the point of
intersection.
Explain how to find a line parallel to a linear function


that passes through a given point.
Explain how to find a line perpendicular to a linear


function that passes through a given point.


Algebraic
For the following exercises, determine whether the lines
given by the equations below are parallel, perpendicular, or
neither parallel nor perpendicular:


4x − 7y = 10
7x + 4y = 1


3y + x = 12
−y = 8x + 1


3y + 4x = 12
−6y = 8x + 1


6x − 9y = 10
3x + 2y = 1


y = 2
3
x + 1


3x + 2y = 1


y = 3
4
x + 1


−3x + 4y = 1


For the following exercises, find the x- and y-intercepts of
each equation


f (x) = − x + 2


g(x) = 2x + 4


h(x) = 3x − 5


k(x) = − 5x + 1


−2x + 5y = 20


7x + 2y = 56


For the following exercises, use the descriptions of each
pair of lines given below to find the slopes of Line 1
and Line 2. Is each pair of lines parallel, perpendicular, or
neither?


Line 1: Passes through (0, 6) and (3, −24)
Line 2: Passes through (−1, 19) and (8, −71)


Line 1: Passes through (−8, −55) and (10,  89)
Line 2: Passes through (9, −44) and (4, −14)


Line 1: Passes through (2, 3) and (4, − 1)
Line 2: Passes through (6, 3) and (8, 5)


Line 1: Passes through (1, 7) and (5, 5)
Line 2: Passes through (−1, −3) and (1, 1)


Line 1: Passes through (0, 5) and (3, 3)
Line 2: Passes through (1, −5) and (3, −2)


Line 1: Passes through (2, 5) and (5, −1)
Line 2: Passes through (−3, 7) and (3, −5)


Write an equation for a line parallel to
f (x) = − 5x − 3 and passing through the point
(2, –12).


Write an equation for a line parallel to g(x) = 3x − 1
and passing through the point (4, 9).


Write an equation for a line perpendicular to
h(t) = − 2t + 4 and passing through the point (-4, –1).


Write an equation for a line perpendicular to
p(t) = 3t + 4 and passing through the point (3, 1).


Find the point at which the line f (x) = − 2x − 1
intersects the line g(x) = − x.


Find the point at which the line f (x) = 2x + 5
intersects the line g(x) = − 3x − 5.


Chapter 2 Linear Functions 229




106.


107.


108.


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110.


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131.


132.


133.


134.


135.


136.


Use algebra to find the point at which the line
f (x) = − 4


5
x + 274


25
intersects the line


h(x) = 9
4
x  +   73


10
.


Use algebra to find the point at which the line
f (x) = 7


4
x  +   457


60
intersects the line g(x) = 4


3
x  +   31


5
.


Graphical
For the following exercises, the given linear equation with
its graph in Figure 2.37.


Figure 2.37


f (x) = − x − 1


f (x) = − 2x − 1


f (x) = − 1
2
x − 1


f (x) = 2


f (x) = 2 + x


f (x) = 3x + 2


For the following exercises, sketch a line with the given
features.


An x-intercept of ( – 4, 0) and y-intercept of
(0, –2)


An x-intercept of ( – 2, 0) and y-intercept of
(0, 4)


A y-intercept of (0, 7) and slope −3
2


A y-intercept of (0, 3) and slope 2
5


Passing through the points ( – 6, –2) and (6, –6)


Passing through the points ( – 3, –4) and (3, 0)


For the following exercises, sketch the graph of each
equation.


f (x) = − 2x − 1


g(x) = − 3x + 2


h(x) = 1
3
x + 2


k(x) = 2
3
x − 3


f (t) = 3 + 2t


p(t) = − 2 +  3t


x = 3


x = − 2


r(x) = 4


q(x) = 3


4x = − 9y + 36


x
3


y
4
= 1


3x − 5y = 15


3x = 15


3y = 12


If g(x) is the transformation of f (x) = x after a
vertical compression by 3


4
, a shift right by 2, and a shift


down by 4
a. Write an equation for g(x).
b. What is the slope of this line?
c. Find the y-intercept of this line.


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137.


138.


139.


140.


141.


142.


143.


144.


145.


146.


If g(x) is the transformation of f (x) = x after a vertical
compression by 1


3
, a shift left by 1, and a shift up by 3


a. Write an equation for g(x).
b. What is the slope of this line?
c. Find the y-intercept of this line.


For the following exercises,, write the equation of the line
shown in the graph.


For the following exercises, find the point of intersection of
each pair of lines if it exists. If it does not exist, indicate
that there is no point of intersection.


y = 3
4
x + 1


−3x + 4y = 12


2x − 3y = 12


5y + x = 30


2x = y − 3


y + 4x = 15


x − 2y + 2 = 3
x − y = 3


5x + 3y = − 65


x − y = − 5


Extensions
Find the equation of the line parallel to the line


g(x) = − 0.01x + 2.01 through the point (1, 2).


Chapter 2 Linear Functions 231




147.


148.


149.


150.


151.


152.


Find the equation of the line perpendicular to the line
g(x) = − 0.01x+2.01 through the point (1, 2).


For the following exercises, use the functions
f (x) = − 0.1x+200 and g(x) = 20x + 0.1.


Find the point of intersection of the lines f and g.


Where is f (x) greater than g(x)? Where is g(x)
greater than f (x)?


Real-World Applications
A car rental company offers two plans for renting a


car.
Plan A: $30 per day and $0.18 per mile
Plan B: $50 per day with free unlimited mileage
Howmany miles would you need to drive for plan B to save
you money?


A cell phone company offers two plans for minutes.
Plan A: $20 per month and $1 for every one hundred texts.
Plan B: $50 per month with free unlimited texts.
How many texts would you need to send per month for plan
B to save you money?


A cell phone company offers two plans for minutes.
Plan A: $15 per month and $2 for every 300 texts.
Plan B: $25 per month and $0.50 for every 100 texts.
How many texts would you need to send per month for plan
B to save you money?


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2.3 | Modeling with Linear Functions
Learning Objectives


In this section, you will:
2.3.1 Identify steps for modeling and solving.
2.3.2 Build linear models from verbal descriptions.
2.3.3 Build systems of linear models.


Figure 2.38 (credit: EEK Photography/Flickr)


Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates
spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What
would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model
using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions
based on those relationships. In this section, we will explore examples of linear function models.
Identifying Steps to Model and Solve Problems
When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change,
we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:


1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate,
sketch a picture or define a coordinate system.


2. Carefully read the problem to identify important information. Look for information that provides values for the
variables or values for parts of the functional model, such as slope and initial value.


3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.


Chapter 2 Linear Functions 233




4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve
checking and tracking units, building a table, or even finding a formula for the function being used to model the
problem.


5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.


Building Linear Models
Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The
amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define
our variables, including units.


• Output:  M,  money remaining, in dollars
• Input:  t,   time, in weeks


So, the amount of money remaining depends on the number of weeks:  M(t)
We can also identify the initial value and the rate of change.


• Initial Value: She saved $3,500, so $3,500 is the initial value for  M.
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.


Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because
the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.
The rate of change is constant, so we can start with the linear model  M(t) = mt + b. Then we can substitute the intercept
and slope provided.


To find the  x- intercept, we set the output to zero, and solve for the input.
0 = − 400t + 3500


 t = 3500
400


= 8.75


The  x- intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that
Emily will have no money left after 8.75 weeks.
When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be
valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make
sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved
$3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent
spending starts. It is also likely that this model is not valid after the  x- intercept, unless Emily will use a credit card and
goes into debt. The domain represents the set of input values, so the reasonable domain for this function is  0 ≤ t ≤ 8.75.
In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to
analyze the information. However, the information provided may not always be the same. Sometimes we might be provided
with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information
we are given, and use it appropriately to build a linear model.
Using a Given Intercept to Build a Model
Some real-world problems provide the y- intercept, which is the constant or initial value. Once the y- intercept is known,
the x- intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents.


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Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y- intercept is the initial amount
of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the
given information to develop a linear model.


f (x) = mx + b


            = − 250x + 1000
Now we can set the function equal to 0, and solve for x to find the x- intercept.


         0 = − 250x + 1000
1000 = 250x
         4 = x
         x = 4


The  x- intercept is the number of months it takes her to reach a balance of $0. The  x- intercept is 4 months, so it will take
Hannah four months to pay off her loan.
Using a Given Input and Output to Build a Model
Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect
of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation
of the linear model equal to a specified output.


Given a word problem that includes two pairs of input and output values, use the linear function to solve a
problem.


1. Identify the input and output values.
2. Convert the data to two coordinate pairs.
3. Find the slope.
4. Write the linear model.
5. Use the model to make a prediction by evaluating the function at a given x- value.
6. Use the model to identify an x- value that results in a given y- value.
7. Answer the question posed.


Example 2.26
Using a Linear Model to Investigate a Town’s Population


A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had
grown to 8,100. Assume this trend continues.
a. Predict the population in 2013.
b. Identify the year in which the population will reach 15,000.


Solution
The two changing quantities are the population size and time. While we could use the actual year value as the
input quantity, doing so tends to lead to very cumbersome equations because the  y- intercept would correspond
to the year 0, more than 2000 years ago!
To make computation a little nicer, we will define our input as the number of years since 2004:


• Input: t, years since 2004
• Output: P(t), the town’s population


Chapter 2 Linear Functions 235




2.14


2.15


To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find
when the population would reach 15,000, we would need to solve for the input that would provide an output of
15,000. To write an equation, we need the initial value and the rate of change, or slope.
To determine the rate of change, we will use the change in output per change in input.


m =
change in output
change in input


The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004
would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable
definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5,
giving the point (5, 8100).
The two coordinate pairs are  (0, 6200)  and  (5, 8100). Recall that we encountered examples in which we were
provided two points earlier in the chapter. We can use these values to calculate the slope.


m = 8100 − 6200
5 − 0


= 1900
5


= 380 people per year


We already know the y-intercept of the line, so we can immediately write the equation:
P(t) = 380t + 6200


To predict the population in 2013, we evaluate our function at t = 9.
P(9) = 380(9) + 6, 200


= 9, 620


If the trend continues, our model predicts a population of 9,620 in 2013.
To find when the population will reach 15,000, we can set P(t) = 15000  and solve for t.


15000 = 380t + 6200
8800 = 380t
t ≈ 23.158


Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere
around the year 2027.


A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It
costs $0.25 to produce each doughnut.


a. Write a linear model to represent the cost C  of the company as a function of  x,   the number of
doughnuts produced.


b. Find and interpret the y-intercept.


A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the
population was 36,800. Assume this trend continues.


a. Predict the population in 2014.
b. Identify the year in which the population will reach 54,000.


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Using a Diagram to Model a Problem
It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and
output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine
the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances
are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched,
labeling width and height is helpful.


Example 2.27
Using a Diagram to Model Distance Walked


Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south
at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after
they start walking will they fall out of radio contact?


Solution
In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2
miles apart, which leads us to ask a new question:


“How long will it take them to be 2 miles apart?”
In this problem, our changing quantities are time and position, but ultimately we need to know how long will it
take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and
output variables.


• Input:  t,   time in hours.
• Output:  A(t),   distance in miles, and  E(t),   distance in miles


Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2.39.


Figure 2.39


Initial Value: They both start at the same intersection so when t = 0, the distance traveled by each person should
also be 0. Thus the initial value for each is 0.
Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates
of change. The slope for A is 4 and the slope for E is 3.
Using those values, we can write formulas for the distance each person has walked.


Chapter 2 Linear Functions 237




A(t) = 4t
E(t) = 3t


For this problem, the distances from the starting point are important. To notate these, we can define a coordinate
system, identifying the “starting point” at the intersection where they both started. Then we can use the variable,
A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting
point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured
from the starting point in the southward direction. Note that in defining the coordinate system, we specified both
the starting point of the measurement and the direction of measure.
We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel.
Showing the variables on the diagram is often helpful, as we can see from Figure 2.40.
Recall that we need to know how long it takes for  D,   the distance between them, to equal 2 miles. Notice that
for any given input  t,   the outputs  A(t), E(t),   and  D(t)  represent distances.


Figure 2.40


Figure 2.39 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.
Using the Pythagorean Theorem, we get:


D(t)2 = A(t)2 + E(t)2


= (4t)2 + (3t)2


= 16t2 + 9t2


= 25t2


D(t) = ± 25t2 Solve for D(t) using the square root


                = ± 5|t|
In this scenario we are considering only positive values of  t,   so our distance  D(t) will always be positive. We
can simplify this answer to  D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear
function. Because D is a linear function, we can now answer the question of when the distance between them
will reach 2 miles. We will set the output D(t) = 2  and solve for t.


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D(t) = 2
5t = 2


t = 2
5
= 0.4


They will fall out of radio contact in 0.4 hours, or 24 minutes.


Should I draw diagrams when given information based on a geometric shape?
Yes. Sketch the figure and label the quantities and unknowns on the sketch.


Example 2.28
Using a Diagram to Model Distance between Cities


There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north.
Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of
Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is
the road junction from Westborough?


Solution
It might help here to draw a picture of the situation. See Figure 2.41. It would then be helpful to introduce a
coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This
puts Agritown at coordinates (30, 10),   and Eastborough at  (20, 0).


Figure 2.41


Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:
m = 10 − 0


30 − 0
= 1


3


The equation of the road from Westborough to Agritown would be
W(x) = 1


3
x


From this, we can determine the perpendicular road to Eastborough will have slope  m = – 3. Because the town
of Eastborough is at the point (20, 0), we can find the equation:


E(x) = − 3x + b
        0 = − 3(20) + b Substitute in (20, 0)
        b = 60
E(x) = − 3x + 60


Chapter 2 Linear Functions 239




2.16


We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting
them equal,


1
3
x = − 3x + 60


10
3
x = 60


10x = 180
x = 18 Substituting this back into W(x)


y = W(18)


= 1
3
(18)


= 6


The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from
Westborough to the junction.


distance = (x2 − x1)
2 + (y2 − y1)


2


= (18 − 0)2 + (6 − 0)2


≈ 18.974 miles


Analysis
One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This
means real-world applications discussing maps need linear functions to model the distances between reference
points.


There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north.
Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of
Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road
junction from Timpson?


Building Systems of Linear Models
Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember,
when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are
three types of answers possible, as shown in Figure 2.42.


Figure 2.42


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Given a situation that represents a system of linear equations, write the system of equations and identify the
solution.


1. Identify the input and output of each linear model.
2. Identify the slope and y-intercept of each linear model.
3. Find the solution by setting the two linear functions equal to another and solving for x, or find the point
of intersection on a graph.


Example 2.29
Building a System of Linear Models to Choose a Truck Rental Company


Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front
fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a
mile[4]. When will Keep on Trucking, Inc. be the better choice for Jamal?


Solution
The two important quantities in this problem are the cost and the number of miles driven. Because we have two
companies to consider, we will define two functions.


Input d, distance driven in miles


Outputs
K(d) : cost, in dollars, for renting from Keep on Trucking
M(d) cost, in dollars, for renting from Move It Your Way


Initial Value Up-front fee: K(0) = 20 and M(0) = 16 


Rate of Change K(d) = $0.59 /mile and  P(d) = $0.63 /mile


Table 2.5


A linear function is of the form   f (x) = mx + b. Using the rates of change and initial charges, we can write the
equations


K(d) = 0.59d + 20
M(d) = 0.63d + 16


Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all
we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less,
or when  K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find the
intersection, and then see where the  K(d)  function is smaller.
These graphs are sketched in Figure 2.43, with  K(d)  in blue.


4. Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/


Chapter 2 Linear Functions 241




Figure 2.43


To find the intersection, we set the equations equal and solve:
             K(d) = M(d)
0.59d + 20 = 0.63d + 16


        4 = 0.04d
  100 = d


  d = 100
This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at
the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be
the cheaper price when more than 100 miles are driven, that is d > 100.


Access this online resource for additional instruction and practice with linear function models.
• Interpreting a Linear Function (http://openstaxcollege.org/l/interpretlinear)


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153.


154.


155.


156.


157.


158.


159.


160.


161.
162.


163.
164.


165.


166.


167.


168.


169.
170.


171.


172.


173.


174.


175.
176.


177.


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179.


180.


2.3 EXERCISES
Verbal


Explain how to find the input variable in a word
problem that uses a linear function.


Explain how to find the output variable in a word
problem that uses a linear function.


Explain how to interpret the initial value in a word
problem that uses a linear function.


Explain how to determine the slope in a word problem
that uses a linear function.


Algebraic
Find the area of a parallelogram bounded by the y


axis, the line  x = 3,   the line   f (x) = 1 + 2x, and the line
parallel to   f (x)  passing through  (2, 7).


Find the area of a triangle bounded by the x-axis, the
line   f (x) = 12 – 1


3
x,   and the line perpendicular to   f (x) 


that passes through the origin.
Find the area of a triangle bounded by the y-axis, the


line   f (x) = 9 – 6
7
x,   and the line perpendicular to   f (x) 


that passes through the origin.
Find the area of a parallelogram bounded by the x-


axis, the line  g(x) = 2,   the line   f (x) = 3x,   and the line
parallel to   f (x)  passing through  (6, 1).


For the following exercises, consider this scenario: A
town’s population has been decreasing at a constant rate. In
2010 the population was 5,900. By 2012 the population had
dropped 4,700. Assume this trend continues.


Predict the population in 2016.
Identify the year in which the population will reach 0.


For the following exercises, consider this scenario: A
town’s population has been increased at a constant rate. In
2010 the population was 46,020. By 2012 the population
had increased to 52,070. Assume this trend continues.


Predict the population in 2016.
Identify the year in which the population will reach


75,000.
For the following exercises, consider this scenario: A town
has an initial population of 75,000. It grows at a constant
rate of 2,500 per year for 5 years.


Find the linear function that models the town’s population
 P  as a function of the year,  t,  where  t  is the number of
years since the model began.


Find a reasonable domain and range for the function
 P.


If the function  P  is graphed, find and interpret the x-
and y-intercepts.


If the function  P  is graphed, find and interpret the
slope of the function.


When will the output reached 100,000?
What is the output in the year 12 years from the onset


of the model?
For the following exercises, consider this scenario: The
weight of a newborn is 7.5 pounds. The baby gained one-
half pound a month for its first year.


Find the linear function that models the baby’s weight
 W   as a function of the age of the baby, in months,  t.


Find a reasonable domain and range for the function
W .


If the function W is graphed, find and interpret the x-
and y-intercepts.


If the function W is graphed, find and interpret the
slope of the function.


When did the baby weight 10.4 pounds?
What is the output when the input is 6.2? Interpret


your answer.
For the following exercises, consider this scenario: The
number of people afflicted with the common cold in the
winter months steadily decreased by 205 each year from
2005 until 2010. In 2005, 12,025 people were afflicted.


Find the linear function that models the number of
people inflicted with the common cold  C  as a function of
the year, t.


Find a reasonable domain and range for the function
C.


If the function  C  is graphed, find and interpret the x-
and y-intercepts.


If the function  C  is graphed, find and interpret the
slope of the function.


Chapter 2 Linear Functions 243




181.
182.


183.


184.
185.
186.


187.


188.
189.


190.


191.


192.


193.


194.


195.


196.


197.


When will the output reach 0?
In what year will the number of people be 9,700?


Graphical
For the following exercises, use the graph in Figure 2.44,
which shows the profit,  y,  in thousands of dollars, of a
company in a given year,  t, where  t  represents the number
of years since 1980.


Figure 2.44


Find the linear function  y,  where  y  depends on  t,
the number of years since 1980.


Find and interpret the y-intercept.
Find and interpret the x-intercept.
Find and interpret the slope.


For the following exercises, use the graph in Figure 2.45,
which shows the profit,  y,   in thousands of dollars, of a
company in a given year,  t,  where  t  represents the
number of years since 1980.


Figure 2.45


Find the linear function  y,  where  y  depends on  t,
the number of years since 1980.


Find and interpret the y-intercept.
Find and interpret the x-intercept.


Find and interpret the slope.


Numeric
For the following exercises, use the median home values
in Mississippi and Hawaii (adjusted for inflation) shown
in Table 2.6. Assume that the house values are changing
linearly.


Year Mississippi Hawaii


1950 $25,200 $74,400


2000 $71,400 $272,700


Table 2.6


In which state have home values increased at a higher
rate?


If these trends were to continue, what would be the
median home value in Mississippi in 2010?


If we assume the linear trend existed before 1950 and
continues after 2000, the two states’ median house values
will be (or were) equal in what year? (The answer might be
absurd.)
For the following exercises, use the median home values
in Indiana and Alabama (adjusted for inflation) shown in
Table 2.7. Assume that the house values are changing
linearly.


Year Indiana Alabama


1950 $37,700 $27,100


2000 $94,300 $85,100


Table 2.7


In which state have home values increased at a higher
rate?


If these trends were to continue, what would be the
median home value in Indiana in 2010?


If we assume the linear trend existed before 1950 and
continues after 2000, the two states’ median house values
will be (or were) equal in what year? (The answer might be
absurd.)


Real-World Applications
In 2004, a school population was 1001. By 2008 the


population had grown to 1697. Assume the population is
changing linearly.


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198.


199.


200.


201.


202.


203.


204.


205.


206.


207.


a. How much did the population grow between the
year 2004 and 2008?


b. How long did it take the population to grow from
1001 students to 1697 students?


c. What is the average population growth per year?
d. What was the population in the year 2000?
e. Find an equation for the population, P, of the
school t years after 2000.


f. Using your equation, predict the population of the
school in 2011.
In 2003, a town’s population was 1431. By 2007 the


population had grown to 2134. Assume the population is
changing linearly.
a. How much did the population grow between the
year 2003 and 2007?


b. How long did it take the population to grow from
1431 people to 2134 people?


c. What is the average population growth per year?
d. What was the population in the year 2000?
e. Find an equation for the population, P of the town


t years after 2000.
f. Using your equation, predict the population of the
town in 2014.
A phone company has a monthly cellular plan where a


customer pays a flat monthly fee and then a certain amount
of money per minute used on the phone. If a customer uses
410 minutes, the monthly cost will be $71.50. If the
customer uses 720 minutes, the monthly cost will be $118.
a. Find a linear equation for the monthly cost of the
cell plan as a function of x, the number of monthly
minutes used.


b. Interpret the slope and y-intercept of the equation.
c. Use your equation to find the total monthly cost if
687 minutes are used.
A phone company has a monthly cellular data plan


where a customer pays a flat monthly fee of $10 and then a
certain amount of money per megabyte (MB) of data used
on the phone. If a customer uses 20 MB, the monthly cost
will be $11.20. If the customer uses 130 MB, the monthly
cost will be $17.80.
a. Find a linear equation for the monthly cost of the
data plan as a function of x , the number of MB
used.


b. Interpret the slope and y-intercept of the equation.
c. Use your equation to find the total monthly cost if
250 MB are used.
In 1991, the moose population in a park was


measured to be 4,360. By 1999, the population was


measured again to be 5,880. Assume the population
continues to change linearly.
a. Find a formula for the moose population, P since
1990.


b. What does your model predict the moose
population to be in 2003?
In 2003, the owl population in a park was measured to


be 340. By 2007, the population was measured again to be
285. The population changes linearly. Let the input be years
since 1990.
a. Find a formula for the owl population, P. Let the
input be years since 2003.


b. What does your model predict the owl population
to be in 2012?
The Federal Helium Reserve held about 16 billion


cubic feet of helium in 2010 and is being depleted by about
2.1 billion cubic feet each year.
a. Give a linear equation for the remaining federal
helium reserves, R, in terms of t, the number of
years since 2010.


b. In 2015, what will the helium reserves be?
c. If the rate of depletion doesn’t change, in what year
will the Federal Helium Reserve be depleted?
Suppose the world’s oil reserves in 2014 are 1,820


billion barrels. If, on average, the total reserves are
decreasing by 25 billion barrels of oil each year:
a. Give a linear equation for the remaining oil
reserves, R, in terms of t, the number of years
since now.


b. Seven years from now, what will the oil reserves
be?


c. If the rate at which the reserves are decreasing
is constant, when will the world’s oil reserves be
depleted?
You are choosing between two different prepaid cell


phone plans. The first plan charges a rate of 26 cents per
minute. The second plan charges a monthly fee of $19.95
plus 11 cents per minute. How many minutes would you
have to use in a month in order for the second plan to be
preferable?


You are choosing between two different window
washing companies. The first charges $5 per window. The
second charges a base fee of $40 plus $3 per window. How
many windows would you need to have for the second
company to be preferable?


When hired at a new job selling jewelry, you are given
two pay options:


• Option A: Base salary of $17,000 a year with a
commission of 12% of your sales


Chapter 2 Linear Functions 245




208.


209.


210.


• Option B: Base salary of $20,000 a year with a
commission of 5% of your sales


How much jewelry would you need to sell for option A to
produce a larger income?


When hired at a new job selling electronics, you are
given two pay options:


• Option A: Base salary of $14,000 a year with a
commission of 10% of your sales


• Option B: Base salary of $19,000 a year with a
commission of 4% of your sales


How much electronics would you need to sell for option A
to produce a larger income?


When hired at a new job selling electronics, you are
given two pay options:


• Option A: Base salary of $20,000 a year with a
commission of 12% of your sales


• Option B: Base salary of $26,000 a year with a
commission of 3% of your sales


How much electronics would you need to sell for option A
to produce a larger income?


When hired at a new job selling electronics, you are
given two pay options:


• Option A: Base salary of $10,000 a year with a
commission of 9% of your sales


• Option B: Base salary of $20,000 a year with a
commission of 4% of your sales


How much electronics would you need to sell for option A
to produce a larger income?


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2.4 | Fitting Linear Models to Data
Learning Objectives


In this section, you will:
2.4.1 Draw and interpret scatter plots.
2.4.2 Find the line of best fit.
2.4.3 Distinguish between linear and nonlinear relations.
2.4.4 Use a linear model to make predictions.


A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if
there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram
that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as
a scatter plot.
Drawing and Interpreting Scatter Plots
A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from
a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions.
Figure 2.46 shows a sample scatter plot.


Figure 2.46 A scatter plot of age and final exam score
variables


Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words,
there does not appear to be a relationship between the age of the student and the score on the final exam.


Example 2.30
Using a Scatter Plot to Investigate Cricket Chirps


Table 2.8 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees
Fahrenheit[5]. Plot this data, and determine whether the data appears to be linearly related.


5. Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010


Chapter 2 Linear Functions 247




Chirps 44 35 20.4 33 31 35 18.5 37 26


Temperature 80.5 70.5 57 66 68 72 52 73.5 53


Table 2.8


Solution
Plotting this data, as depicted in Figure 2.47 suggests that there may be a trend. We can see from the trend in
the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear,
though certainly not perfectly so.


Figure 2.47


Finding the Line of Best Fit
Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear
function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can
extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can
estimate the riserun.


Example 2.31
Finding a Line of Best Fit


Find a linear function that fits the data in Table 2.8 by “eyeballing” a line that seems to fit.


Solution
On a graph, we could try sketching a line.
Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of


m = 60
50


= 1.2


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and a y-intercept at 30. This gives an equation of
T(c) = 1.2c + 30


where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting
equation is represented in Figure 2.48.


Figure 2.48


Analysis
This linear equation can then be used to approximate answers to various questions we might ask about the trend.


Recognizing Interpolation or Extrapolation
While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship
will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value
inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain
and range of the data.
Figure 2.49 compares the two processes for the cricket-chirp data addressed in Example 2.31. We can see that
interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and
44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or
greater than 44.
There is a difference between making predictions inside the domain and range of values for which we have data and outside
that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer
applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period
of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to
extrapolate a cost when x = 50, that is in 50 years, the model would not apply because we could not account for factors
fifty years in the future.


Chapter 2 Linear Functions 249




Figure 2.49 Interpolation occurs within the domain and range
of the provided data whereas extrapolation occurs outside.


Interpolation and Extrapolation
Different methods of making predictions are used to analyze data.


• The method of interpolation involves predicting a value inside the domain and/or range of the data.
• The method of extrapolation involves predicting a value outside the domain and/or range of the data.
• Model breakdown occurs at the point when the model no longer applies.


Example 2.32
Understanding Interpolation and Extrapolation


Use the cricket data from Table 2.8 to answer the following questions:
a. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or
extrapolation? Make the prediction, and discuss whether it is reasonable.


b. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation?
Make the prediction, and discuss whether it is reasonable.


Solution
a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds
is inside the domain of our data, so would be interpolation. Using our model:


T(30) = 30 + 1.2(30)
= 66 degrees


Based on the data we have, this value seems reasonable.
b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is
extrapolation because 40 is outside the range of our data. Using our model:


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2.17


40 = 30 + 1.2c
10 = 1.2c
c ≈ 8.33


We can compare the regions of interpolation and extrapolation using Figure 2.50.


Figure 2.50


Analysis
Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have
no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping
altogether below around 50 degrees.


According to the data from Table 2.8, what temperature can we predict it is if we counted 20 chirps in 15
seconds?


Finding the Line of Best Fit Using a Graphing Utility
While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the
differences between the line and data values[6]. One such technique is called least squares regression and can be computed
by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[7]. Least squares
regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression.


Given data of input and corresponding outputs from a linear function, find the best fit line using linear
regression.


1. Enter the input in List 1 (L1).
2. Enter the output in List 2 (L2).
3. On a graphing utility, select Linear Regression (LinReg).


6. Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the
data values.
7. For example, http://www.shodor.org/unchem/math/lls/leastsq.html


Chapter 2 Linear Functions 251




Example 2.33
Finding a Least Squares Regression Line


Find the least squares regression line using the cricket-chirp data in Table 2.8.


Solution
1. Enter the input (chirps) in List 1 (L1).
2. Enter the output (temperature) in List 2 (L2). See Table 2.9.


L1 44 35 20.4 33 31 35 18.5 37 26


L2 80.5 70.5 57 66 68 72 52 73.5 53


Table 2.9
3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with
technology we obtain the equation:


T(c) = 30.281 + 1.143c


Analysis
Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that
using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from
66 degrees to:


T(30) = 30.281 + 1.143(30)
= 64.571
≈ 64.6 degrees


The graph of the scatter plot with the least squares regression line is shown in Figure 2.51.


Figure 2.51


Will there ever be a case where two different lines will serve as the best fit for the data?
No. There is only one best fit line.


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Distinguishing Between Linear and Non-Linear Models
As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final
exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the
correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the
user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as r. The correlation
coefficient provides an easy way to get an idea of how close to a line the data falls.
We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which
a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless.
To get a sense for the relationship between the value of r and the graph of the data, Figure 2.52 shows some large data
sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis
shows the output.


Figure 2.52 Plotted data and related correlation coefficients. (credit: “DenisBoigelot,” Wikimedia Commons)


Correlation Coefficient
The correlation coefficient is a value, r, between –1 and 1.


• r > 0 suggests a positive (increasing) relationship
• r < 0 suggests a negative (decreasing) relationship
• The closer the value is to 0, the more scattered the data.
• The closer the value is to 1 or –1, the less scattered the data is.


Example 2.34
Finding a Correlation Coefficient


Calculate the correlation coefficient for cricket-chirp data in Table 2.8.


Solution


Chapter 2 Linear Functions 253




Because the data appear to follow a linear pattern, we can use technology to calculate r. Enter the inputs and
corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation
coefficient, r = 0.9509. This value is very close to 1, which suggests a strong increasing linear relationship.
Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when
linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON.


Predicting with a Regression Line
Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make
predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that
only one such line is a best fit for the data.


Example 2.35
Using a Regression Line to Make Predictions


Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is
shown in Table 2.10[8]. Determine whether the trend is linear, and if so, find a model for the data. Use the model
to predict the consumption in 2008.


Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04


Consumption
(billions of
gallons)


113 116 118 119 123 125 126 128 131 133 136


Table 2.10


The scatter plot of the data, including the least squares regression line, is shown in Figure 2.53.


Figure 2.53


8. http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html


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2.18


Solution
We can introduce new input variable, t, representing years since 1994.
The least squares regression equation is:


C(t) = 113.318 + 2.209t


Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing
linear trend.
Using this to predict consumption in 2008 (t = 14),


C(14) = 113.318 + 2.209(14)
= 144.244


The model predicts 144.244 billion gallons of gasoline consumption in 2008.


Use the model we created using technology in Example 2.35 to predict the gas consumption in 2011. Is
this an interpolation or an extrapolation?


Access these online resources for additional instruction and practice with fitting linear models to data.
• Introduction to Regression Analysis (http://openstaxcollege.org/l/introregress)
• Linear Regression (http://openstaxcollege.org/l/linearregress)


Visit this website (http://openstaxcollege.org/l/PreCalcLPC02) for additional practice questions from
Learningpod.


Chapter 2 Linear Functions 255




211.


212.
213.
214.


215.


216.


217.


218.


219.


220.


221.


222.


223.


2.4 EXERCISES
Verbal


Describe what it means if there is a model breakdown
when using a linear model.


What is interpolation when using a linear model?
What is extrapolation when using a linear model?
Explain the difference between a positive and a


negative correlation coefficient.
Explain how to interpret the absolute value of a


correlation coefficient.


Algebraic
A regression was run to determine whether there is a


relationship between hours of TV watched per day (x) and
number of sit-ups a person can do (y). The results of the
regression are given below. Use this to predict the number
of sit-ups a person who watches 11 hours of TV can do.


y = ax + b
a = −1.341


b = 32.234
r = −0.896


A regression was run to determine whether there is a
relationship between the diameter of a tree ( x, in inches)
and the tree’s age ( y, in years). The results of the
regression are given below. Use this to predict the age of a
tree with diameter 10 inches.


y = ax + b
a = 6.301


b = −1.044
r = −0.970


For the following exercises, draw a scatter plot for the data
provided. Does the data appear to be linearly related?


0 2 4 6 8 10


–22 –19 –15 –11 –6 –2


1 2 3 4 5 6


46 50 59 75 100 136


100 250 300 450 600 750


12 12.6 13.1 14 14.5 15.2


1 3 5 7 9 11


1 9 28 65 125 216


For the following data, draw a scatter plot. If we
wanted to know when the population would reach 15,000,
would the answer involve interpolation or extrapolation?
Eyeball the line, and estimate the answer.


Year Population


1990 11,500


1995 12,100


2000 12,700


2005 13,000


2010 13,750


For the following data, draw a scatter plot. If we
wanted to know when the temperature would reach 28 °F,
would the answer involve interpolation or extrapolation?
Eyeball the line and estimate the answer.


Temperature,
°F 16 18 20 25 30


Time, seconds 46 50 54 55 62


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224.


225.


226.


227.


228.


Graphical
For the following exercises, match each scatterplot with
one of the four specified correlations in Figure 2.54 and
Figure 2.55.


Figure 2.54


Figure 2.55


r = 0.95


r = − 0.89


r = 0.26


r = − 0.39


For the following exercises, draw a best-fit line for the
plotted data.


Chapter 2 Linear Functions 257




229.


230.


231.


232.
Numeric


The U.S. Census tracks the percentage of persons 25
years or older who are college graduates. That data for
several years is given in Table 2.11[9]. Determine whether
the trend appears linear. If so, and assuming the trend
continues, in what year will the percentage exceed 35%?


9. http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/2014.


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233. 234.


Year Percent Graduates


1990 21.3


1992 21.4


1994 22.2


1996 23.6


1998 24.4


2000 25.6


2002 26.7


2004 27.7


2006 28


2008 29.4


Table 2.11


The U.S. import of wine (in hectoliters) for several
years is given in Table 2.12. Determine whether the trend
appears linear. If so, and assuming the trend continues, in
what year will imports exceed 12,000 hectoliters?


Year Imports


1992 2665


1994 2688


1996 3565


1998 4129


2000 4584


2002 5655


2004 6549


2006 7950


2008 8487


2009 9462


Table 2.12


Table 2.13 shows the year and the number of people
unemployed in a particular city for several years.
Determine whether the trend appears linear. If so, and
assuming the trend continues, in what year will the number
of unemployed reach 5?


Chapter 2 Linear Functions 259




235.


236.


237.


238.


Year Number Unemployed


1990 750


1992 670


1994 650


1996 605


1998 550


2000 510


2002 460


2004 420


2006 380


2008 320


Table 2.13


Technology
For the following exercises, use each set of data to calculate
the regression line using a calculator or other technology
tool, and determine the correlation coefficient to 3 decimal
places of accuracy.


x 8 15 26 31 56


y 23 41 53 72 103


x 5 7 10 12 15


y 4 12 17 22 24


x y x y


3 21.9 11 15.76


4 22.22 12 13.68


5 22.74 13 14.1


6 22.26 14 14.02


7 20.78 15 11.94


8 17.6 16 12.76


9 16.52 17 11.28


10 18.54 18 9.1


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239.


240.


241.


242.


243.


x y


4 44.8


5 43.1


6 38.8


7 39


8 38


9 32.7


10 30.1


11 29.3


12 27


13 25.8


x y


21 17


25 11


30 2


31 −1


40 −18


50 −40


x y


100 2000


80 1798


60 1589


55 1580


40 1390


20 1202


x y


900 70


988 80


1000 82


1010 84


1200 105


1205 108


Extensions
Graph f (x) = 0.5x + 10 . Pick a set of 5 ordered


pairs using inputs x = −2, 1, 5, 6, 9 and use linear
regression to verify that the function is a good fit for the
data.


Graph f (x) = − 2x − 10 . Pick a set of 5 ordered
pairs using inputs x = −2, 1, 5, 6, 9 and use linear
regression to verify the function.
For the following exercises, consider this scenario: The
profit of a company decreased steadily over a ten-year span.
The following ordered pairs shows dollars and the number


Chapter 2 Linear Functions 261




244.


245.
246.


247.


248.


249.


250.


251.


252.


of units sold in hundreds and the profit in thousands of over
the ten-year span, (number of units sold, profit) for specific
recorded years:
(46, 1, 600), (48, 1, 550), (50, 1, 505), (52, 1, 540), (54, 1, 495)


.
Use linear regression to determine a function P


where the profit in thousands of dollars depends on the
number of units sold in hundreds.


Find to the nearest tenth and interpret the x-intercept.
Find to the nearest tenth and interpret the y-intercept.


Real-World Applications
For the following exercises, consider this scenario: The
population of a city increased steadily over a ten-year span.
The following ordered pairs shows the population and the
year over the ten-year span, (population, year) for specific
recorded years:
(2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010)


Use linear regression to determine a function y,
where the year depends on the population. Round to three
decimal places of accuracy.


Predict when the population will hit 8,000.
For the following exercises, consider this scenario: The
profit of a company increased steadily over a ten-year span.
The following ordered pairs show the number of units sold
in hundreds and the profit in thousands of over the ten year
span, (number of units sold, profit) for specific recorded
years:
(46, 250), (48, 305), (50, 350), (52, 390), (54, 410)


.
Use linear regression to determine a function y, where


the profit in thousands of dollars depends on the number of
units sold in hundreds .


Predict when the profit will exceed one million
dollars.
For the following exercises, consider this scenario: The
profit of a company decreased steadily over a ten-year span.
The following ordered pairs show dollars and the number
of units sold in hundreds and the profit in thousands of over
the ten-year span (number of units sold, profit) for specific
recorded years:
(46, 250), (48, 225), (50, 205), (52, 180), (54, 165).


Use linear regression to determine a function y, where
the profit in thousands of dollars depends on the number of
units sold in hundreds .


Predict when the profit will dip below the $25,000
threshold.


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correlation coefficient


decreasing linear function
extrapolation
horizontal line
increasing linear function
interpolation
least squares regression


linear function


model breakdown
parallel lines
perpendicular lines
point-slope form


slope
slope-intercept form
vertical line
x-intercept


y-intercept


CHAPTER 2 REVIEW
KEY TERMS


a value, r, between –1 and 1 that indicates the degree of linear correlation of variables, or how
closely a regression line fits a data set.


a function with a negative slope: If f (x) = mx + b, then m < 0.


predicting a value outside the domain and range of the data
a line defined by f (x) = b, where b is a real number. The slope of a horizontal line is 0.


a function with a positive slope: If f (x) = mx + b, then m > 0.


predicting a value inside the domain and range of the data
a statistical technique for fitting a line to data in a way that minimizes the differences between


the line and data values
a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight


line
when a model no longer applies after a certain point


two or more lines with the same slope
two lines that intersect at right angles and have slopes that are negative reciprocals of each other


the equation for a line that represents a linear function of the form y − y1 = m(x − x1)


the ratio of the change in output values to the change in input values; a measure of the steepness of a line
the equation for a line that represents a linear function in the form f (x) = mx + b


a line defined by x = a, where a is a real number. The slope of a vertical line is undefined.
the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the


horizontal axis
the value of a function when the input value is zero; also known as initial value


KEY EQUATIONS
slope-intercept form of a line f (x) = mx + b


slope m = change in output (rise)change in input (run) =
Δy
Δx


=
y2 − y1
x2 − x1


point-slope form of a line y − y1 = m(x − x1)


KEY CONCEPTS
2.1 Linear Functions


• The ordered pairs given by a linear function represent points on a line.
• Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example
2.1.


Chapter 2 Linear Functions 263




• The rate of change of a linear function is also known as the slope.
• An equation in the slope-intercept form of a line includes the slope and the initial value of the function.
• The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the
point at which the line crosses the y-axis.


• An increasing linear function results in a graph that slants upward from left to right and has a positive slope.
• A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.
• A constant linear function results in a graph that is a horizontal line.
• Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or
constant. See Example 2.2.


• The slope of a linear function can be calculated by dividing the difference between y-values by the difference in
corresponding x-values of any two points on the line. See Example 2.3 and Example 2.4.


• The slope and initial value can be determined given a graph or any two points on the line.
• One type of function notation is the slope-intercept form of an equation.
• The point-slope form is useful for finding a linear equation when given the slope of a line and one point. See
Example 2.5.


• The point-slope form is also convenient for finding a linear equation when given two points through which a line
passes. See Example 2.6.


• The equation for a linear function can be written if the slope m and initial value b are known. See Example 2.7,
Example 2.8, and Example 2.9.


• A linear function can be used to solve real-world problems. See Example 2.10 and Example 2.11.
• A linear function can be written from tabular form. See Example 2.12.


2.2 Graphs of Linear Functions
• Linear functions may be graphed by plotting points or by using the y-intercept and slope. See Example 2.13 and
Example 2.14.


• Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches,
compressions, and reflections. See Example 2.15.


• The y-intercept and slope of a line may be used to write the equation of a line.
• The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 2.16 and
Example 2.17.


• Horizontal lines are written in the form, f (x) = b. See Example 2.18.
• Vertical lines are written in the form, x = b. See Example 2.19.
• Parallel lines have the same slope.
• Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 2.20.
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the
line and the x- and y-values of the given point into the equation, f (x) = mx + b, and using the b that results.
Similarly, the point-slope form of an equation can also be used. See Example 2.21.


• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the
exception of using the negative reciprocal slope. See Example 2.22 and Example 2.23.


• A system of linear equations may be solved setting the two equations equal to one another and solving for x. The
y-value may be found by evaluating either one of the original equations using this x-value.


• A system of linear equations may also be solved by finding the point of intersection on a graph. See Example 2.24
and Example 2.25.


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2.3 Modeling with Linear Functions
• We can use the same problem strategies that we would use for any type of function.
• When modeling and solving a problem, identify the variables and look for key values, including the slope and y-
intercept. See Example 2.26.


• Draw a diagram, where appropriate. See Example 2.27 and Example 2.28.
• Check for reasonableness of the answer.
• Linear models may be built by identifying or calculating the slope and using the y-intercept.
• The x-intercept may be found by setting  y = 0,  which is setting the expression  mx + b  equal to 0.
• The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See
Example 2.29.


• A graph of the system may be used to identify the points where one line falls below (or above) the other line.


2.4 Fitting Linear Models to Data
• Scatter plots show the relationship between two sets of data. See Example 2.30.
• Scatter plots may represent linear or non-linear models.
• The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2.31.
• Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be
used to predict values outside the domain and range of the data. See Example 2.32.


• The correlation coefficient, r, indicates the degree of linear relationship between data. See Example 2.34.
• A regression line best fits the data. See Example 2.35.
• The least squares regression line is found by minimizing the squares of the distances of points from a line passing
through the data and may be used to make predictions regarding either of the variables. See Example 2.33.


CHAPTER 2 REVIEW EXERCISES
Linear Functions
253. Determine whether the algebraic equation is linear.
2x + 3y = 7


254. Determine whether the algebraic equation is linear.
6x2 − y = 5


255. Determine whether the function is increasing or
decreasing.
f (x) = 7x − 2


256. Determine whether the function is increasing or
decreasing.
g(x) = − x + 2


257. Given each set of information, find a linear equation
that satisfies the given conditions, if possible.
Passes through (7, 5) and (3, 17)


258. Given each set of information, find a linear equation
that satisfies the given conditions, if possible.
x-intercept at (6, 0) and y-intercept at (0, 10)


259. Find the slope of the line shown in the line graph.


Chapter 2 Linear Functions 265




260. Find the slope of the line graphed.


261. Write an equation in slope-intercept form for the line
shown.


262. Does the following table represent a linear function?
If so, find the linear equation that models the data.


x –4 0 2 10


g(x) 18 –2 –12 –52


263. Does the following table represent a linear function?
If so, find the linear equation that models the data.


x 6 8 12 26


g(x) –8 –12 –18 –46


264. On June 1st, a company has $4,000,000 profit. If the
company then loses 150,000 dollars per day thereafter in
the month of June, what is the company’s profit nthday after
June 1st?


Graphs of Linear Functions
(https://cnx.org/content/m49325/latest/)
For the following exercises, determine whether the lines
given by the equations below are parallel, perpendicular, or
neither parallel nor perpendicular:


265. 2x − 6y = 12
−x + 3y = 1


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266.
y = 1


3
x − 2


3x + y = − 9


For the following exercises, find the x- and y- intercepts of
the given equation
267. 7x + 9y = −63


268. f (x) = 2x − 1


For the following exercises, use the descriptions of the pairs
of lines to find the slopes of Line 1 and Line 2. Is each pair
of lines parallel, perpendicular, or neither?
269.
Line 1: Passes through (5, 11) and (10, 1)
Line 2: Passes through (−1, 3) and (−5, 11)


270.
Line 1: Passes through (8, −10) and (0, −26)
Line 2: Passes through (2, 5) and (4, 4)


271. Write an equation for a line perpendicular to
f (x) = 5x − 1 and passing through the point (5, 20).


272. Find the equation of a line with a y- intercept of
(0, 2) and slope −1


2
.


273. Sketch a graph of the linear function f (t) = 2t − 5 .


274. Find the point of intersection for the 2 linear
functions:


x = y + 6


2x − y = 13


275. A car rental company offers two plans for renting a
car.
Plan A: 25 dollars per day and 10 cents per mile
Plan B: 50 dollars per day with free unlimited mileage
Howmany miles would you need to drive for plan B to save
you money?


Modeling with Linear Functions
276. Find the area of a triangle bounded by the y axis, the
line f (x) = 10 − 2x , and the line perpendicular to f that
passes through the origin.


277. A town’s population increases at a constant rate. In
2010 the population was 55,000. By 2012 the population


had increased to 76,000. If this trend continues, predict the
population in 2016.


278. The number of people afflicted with the common
cold in the winter months dropped steadily by 50 each year
since 2004 until 2010. In 2004, 875 people were inflicted.
Find the linear function that models the number of people
afflicted with the common cold C as a function of the year,
t. When will no one be afflicted?


For the following exercises, use the graph in Figure 2.56
showing the profit, y, in thousands of dollars, of a
company in a given year, x, where x represents years
since 1980.


Figure 2.56
279. Find the linear function y, where y depends on x,
the number of years since 1980.


280. Find and interpret the y-intercept.


For the following exercise, consider this scenario: In 2004,
a school population was 1,700. By 2012 the population had
grown to 2,500.
281. Assume the population is changing linearly.


a. How much did the population grow between
the year 2004 and 2012?
b. What is the average population growth per
year?
c. Find an equation for the population, P, of the
school t years after 2004.


For the following exercises, consider this scenario: In 2000,
the moose population in a park was measured to be 6,500.
By 2010, the population was measured to be 12,500.
Assume the population continues to change linearly.
282. Find a formula for the moose population, P.


283. What does your model predict the moose population
to be in 2020?


For the following exercises, consider this scenario: The
median home values in subdivisions Pima Central and East


Chapter 2 Linear Functions 267




Valley (adjusted for inflation) are shown in Table 2.14.
Assume that the house values are changing linearly.


Year Pima Central East Valley


1970 32,000 120,250


2010 85,000 150,000


Table 2.14
284. In which subdivision have home values increased at
a higher rate?


285. If these trends were to continue, what would be the
median home value in Pima Central in 2015?


Fitting Linear Models to Data
286. Draw a scatter plot for the data in Table 2.15. Then
determine whether the data appears to be linearly related.


0 2 4 6 8 10


–105 –50 1 55 105 160


Table 2.15


287. Draw a scatter plot for the data in Table 2.16. If we
wanted to know when the population would reach 15,000,
would the answer involve interpolation or extrapolation?


Year Population


1990 5,600


1995 5,950


2000 6,300


2005 6,600


2010 6,900


Table 2.16


288. Eight students were asked to estimate their score on
a 10-point quiz. Their estimated and actual scores are given
in Table 2.17. Plot the points, then sketch a line that fits
the data.


Predicted Actual


6 6


7 7


7 8


8 8


7 9


9 10


10 10


10 9


Table 2.17


289. Draw a best-fit line for the plotted data.


For the following exercises, consider the data in Table
2.18, which shows the percent of unemployed in a city of
people 25 years or older who are college graduates is given
below, by year.


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Year Percent Graduates


2000 6.5


2002 7.0


2005 7.4


2007 8.2


2010 9.0


Table 2.18
290. Determine whether the trend appears to be linear.
If so, and assuming the trend continues, find a linear
regression model to predict the percent of unemployed in a
given year to three decimal places.


291. In what year will the percentage exceed 12%?


292. Based on the set of data given in Table 2.19,
calculate the regression line using a calculator or other
technology tool, and determine the correlation coefficient
to three decimal places.


x 17 20 23 26 29


y 15 25 31 37 40


Table 2.19


293. Based on the set of data given in Table 2.20,
calculate the regression line using a calculator or other
technology tool, and determine the correlation coefficient
to three decimal places.


x 10 12 15 18 20


y 36 34 30 28 22


Table 2.20


For the following exercises, consider this scenario: The
population of a city increased steadily over a ten-year span.
The following ordered pairs show the population and the
year over the ten-year span (population, year) for specific
recorded years:


(3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006)


294. Use linear regression to determine a function y,
where the year depends on the population, to three decimal
places of accuracy.


295. Predict when the population will hit 12,000.


296. What is the correlation coefficient for this model to
three decimal places of accuracy?


297. According to the model, what is the population in
2014?


Chapter 2 Linear Functions 269




CHAPTER 2 PRACTICE TEST
298. Determine whether the following algebraic equation
can be written as a linear function. 2x +  3y = 7


299. Determine whether the following function is
increasing or decreasing. f (x) = − 2x +  5


300. Determine whether the following function is
increasing or decreasing. f (x) = 7x +  9


301. Given the following set of information, find a linear
equation satisfying the conditions, if possible.
Passes through (5, 1) and (3, –9)


302. Given the following set of information, find a linear
equation satisfying the conditions, if possible.
x intercept at (–4, 0) and y-intercept at (0, –6)


303. Find the slope of the line in Figure 2.57.


Figure 2.57


304. Write an equation for line in Figure 2.58.


Figure 2.58


305. Does Table 2.21 represent a linear function? If so,
find a linear equation that models the data.


x –6 0 2 4


g(x) 14 32 38 44


Table 2.21


306. Does Table 2.22 represent a linear function? If so,
find a linear equation that models the data.


x 1 3 7 11


g(x) 4 9 19 12


Table 2.22


307. At 6 am, an online company has sold 120 items
that day. If the company sells an average of 30 items per
hour for the remainder of the day, write an expression to
represent the number of items that were sold n after 6 am.


For the following exercises, determine whether the lines
given by the equations below are parallel, perpendicular, or
neither parallel nor perpendicular:


308.
y = 3


4
x − 9


−4x − 3y = 8


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309.
−2x + y = 3


3x + 3
2
y = 5


310. Find the x- and y-intercepts of the equation
2x +  7y = − 14.


311. Given below are descriptions of two lines. Find the
slopes of Line 1 and Line 2. Is the pair of lines parallel,
perpendicular, or neither?
Line 1: Passes through (−2, −6) and (3, 14)
Line 2: Passes through (2, 6) and (4, 14)


312. Write an equation for a line perpendicular to
f (x) = 4x + 3 and passing through the point (8, 10).


313. Sketch a line with a y-intercept of (0, 5) and slope
−5
2
.


314. Graph of the linear function f (x) = −x + 6 .


315. For the two linear functions, find the point of
intersection:


x = y + 2


2x − 3y = −1


316. A car rental company offers two plans for renting a
car.
Plan A: $25 per day and $0.10 per mile
Plan B: $40 per day with free unlimited mileage
Howmany miles would you need to drive for plan B to save
you money?


317. Find the area of a triangle bounded by the y axis, the
line f (x) = 12 − 4x , and the line perpendicular to f that
passes through the origin.


318. A town’s population increases at a constant rate. In
2010 the population was 65,000. By 2012 the population
had increased to 90,000. Assuming this trend continues,
predict the population in 2018.


319. The number of people afflicted with the common
cold in the winter months dropped steadily by 25 each year
since 2002 until 2012. In 2002, 8,040 people were inflicted.
Find the linear function that models the number of people
afflicted with the common cold C as a function of the year,
t. When will less than 6,000 people be afflicted?


For the following exercises, use the graph in Figure 2.59,
showing the profit, y , in thousands of dollars, of a
company in a given year, x , where x represents years
since 1980.


Figure 2.59
320. Find the linear function y , where y depends on x ,
the number of years since 1980.


321. Find and interpret the y-intercept.


322. In 2004, a school population was 1250. By 2012 the
population had dropped to 875. Assume the population is
changing linearly.


a. How much did the population drop between the
year 2004 and 2012?
b. What is the average population decline per
year?
c. Find an equation for the population, P, of the
school t years after 2004.


323. Draw a scatter plot for the data provided in Table
2.23. Then determine whether the data appears to be
linearly related.


0 2 4 6 8 10


–450 –200 10 265 500 755


Table 2.23


Chapter 2 Linear Functions 271




324. Draw a best-fit line for the plotted data.


For the following exercises, use Table 2.24, which shows
the percent of unemployed persons 25 years or older who
are college graduates in a particular city, by year.


Year Percent Graduates


2000 8.5


2002 8.0


2005 7.2


2007 6.7


2010 6.4


Table 2.24
325. Determine whether the trend appears linear. If so,
and assuming the trend continues, find a linear regression
model to predict the percent of unemployed in a given year
to three decimal places.


326. In what year will the percentage drop below 4%?


327. Based on the set of data given in Table 2.25,
calculate the regression line using a calculator or other
technology tool, and determine the correlation coefficient.
Round to three decimal places of accuracy.


x 16 18 20 24 26


y 106 110 115 120 125


Table 2.25


For the following exercises, consider this scenario: The
population of a city increased steadily over a ten-year span.
The following ordered pairs shows the population (in
hundreds) and the year over the ten-year span, (population,
year) for specific recorded years:
(4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006)


328. Use linear regression to determine a function y,
where the year depends on the population. Round to three
decimal places of accuracy.


329. Predict when the population will hit 20,000.


330. What is the correlation coefficient for this model?


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3 | POLYNOMIAL AND
RATIONAL FUNCTIONS


Figure 3.1 35-mm film, once the standard for capturing photographic images, has been made largely obsolete by digital
photography. (credit “film”: modification of work by Horia Varlan; credit “memory cards”: modification of work by Paul
Hudson)


Chapter Outline
3.1 Complex Numbers
3.2 Quadratic Functions
3.3 Power Functions and Polynomial Functions
3.4 Graphs of Polynomial Functions
3.5 Dividing Polynomials
3.6 Zeros of Polynomial Functions
3.7 Rational Functions
3.8 Inverses and Radical Functions
3.9 Modeling Using Variation


Introduction
Digital photography has dramatically changed the nature of photography. No longer is an image etched in the emulsion on a
roll of film. Instead, nearly every aspect of recording and manipulating images is now governed by mathematics. An image
becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file,
software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses
complex polynomials to transform images, allowing us to manipulate the image in order to crop details, change the color
palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter,
we will learn about these concepts and discover how mathematics can be used in such applications.


Chapter 3 Polynomial and Rational Functions 273




3.1 | Complex Numbers
Learning Objectives


In this section, you will:
3.1.1 Express square roots of negative numbers as multiples of i.
3.1.2 Plot complex numbers on the complex plane.
3.1.3 Add and subtract complex numbers.
3.1.4 Multiply and divide complex numbers.


The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of
positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a
void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still
have no solution to equations such as


x2 + 4 = 0


Our best guesses might be +2 or –2. But if we test +2 in this equation, it does not work. If we test –2, it does not work. If
we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have
described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides
solutions to problems such as these. In this section, we will explore this number system and how to work within it.
Expressing Square Roots of Negative Numbers as Multiples of i
We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative
number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary
number. The imaginary number  i  is defined as the square root of negative 1.


−1 = i


So, using properties of radicals,
i2 = ( −1)2 = − 1


We can write the square root of any negative number as a multiple of  i. Consider the square root of –25.


−25 = 25 ⋅ ( − 1)


= 25 −1
= 5i


We use  5i  and not  − 5i  because the principal root of  25  is the positive root.
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard
form when written  a + bi where  a  is the real part and  bi  is the imaginary part. For example,  5 + 2i  is a complex number.
So, too, is  3 + 4 3i.


Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real
number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number
is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.


Imaginary and Complex Numbers
A complex number is a number of the form  a + bi where


• a  is the real part of the complex number.


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3.1


• bi  is the imaginary part of the complex number.
If  b = 0,   then  a + bi  is a real number. If  a = 0  and  b  is not equal to 0, the complex number is called an imaginary
number. An imaginary number is an even root of a negative number.


Given an imaginary number, express it in standard form.
1. Write   −a  as   a −1.
2. Express   −1  as  i.
3. Write   a ⋅ i  in simplest form.


Example 3.1
Expressing an Imaginary Number in Standard Form


Express   −9  in standard form.


Solution
−9 = 9 −1 = 3i


In standard form, this is  0 + 3i.


Express   −24  in standard form.


Plotting a Complex Number on the Complex Plane
We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them
graphically. To represent a complex number we need to address the two components of the number. We use the complex
plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis
represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs  (a, b),  
where  a  represents the coordinate for the horizontal axis and  b  represents the coordinate for the vertical axis.
Let’s consider the number−2 + 3i. The real part of the complex number is−2  and the imaginary part is  3i. We plot the
ordered pair  (−2, 3)  to represent the complex number −2 + 3i  as shown in Figure 3.2.


Chapter 3 Polynomial and Rational Functions 275




Figure 3.2


Complex Plane
In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis as shown in Figure
3.3.


Figure 3.3


Given a complex number, represent its components on the complex plane.
1. Determine the real part and the imaginary part of the complex number.
2. Move along the horizontal axis to show the real part of the number.
3. Move parallel to the vertical axis to show the imaginary part of the number.
4. Plot the point.


Example 3.2
Plotting a Complex Number on the Complex Plane


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3.2


Plot the complex number  3 − 4i  on the complex plane.


Solution
The real part of the complex number is  3,   and the imaginary part is  −4i. We plot the ordered pair  (3, −4)  as
shown in Figure 3.4.


Figure 3.4


Plot the complex number  −4 − i  on the complex plane.


Adding and Subtracting Complex Numbers
Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers,
we combine the real parts and combine the imaginary parts.


Complex Numbers: Addition and Subtraction
Adding complex numbers:


(a + bi) + (c + di) = (a + c) + (b + d)i


Subtracting complex numbers:
(a + bi) − (c + di) = (a − c) + (b − d)i


Given two complex numbers, find the sum or difference.
1. Identify the real and imaginary parts of each number.
2. Add or subtract the real parts.
3. Add or subtract the imaginary parts.


Chapter 3 Polynomial and Rational Functions 277




3.3


3.4


Example 3.3
Adding Complex Numbers


Add  3 − 4i  and  2 + 5i.


Solution
We add the real parts and add the imaginary parts.


(a + bi) + (c + di) = (a + c) + (b + d)i
(3 − 4i) + (2 + 5i) = (3 + 2) + ( − 4 + 5)i


= 5 + i


Subtract  2 + 5i  from  3 – 4i.


Multiplying Complex Numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and
imaginary parts separately.
Multiplying a Complex Numbers by a Real Number
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a
binomial. So, for example,


Given a complex number and a real number, multiply to find the product.
1. Use the distributive property.
2. Simplify.


Example 3.4
Multiplying a Complex Number by a Real Number


Find the product  4(2 + 5i).


Solution
Distribute the 4.


4(2 + 5i) = (4 ⋅ 2) + (4 ⋅ 5i)
= 8 + 20i


Find the product  − 4(2 + 6i).


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3.5


Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL
is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL
method, we get


(a + bi)(c + di) = ac + adi + bci + bdi2


Because  i2 = − 1,  we have
(a + bi)(c + di) = ac + adi + bci − bd


To simplify, we combine the real parts, and we combine the imaginary parts.
(a + bi)(c + di) = (ac − bd) + (ad + bc)i


Given two complex numbers, multiply to find the product.
1. Use the distributive property or the FOIL method.
2. Simplify.


Example 3.5
Multiplying a Complex Number by a Complex Number


Multiply  (4 + 3i)(2 − 5i).


Solution
Use  (a + bi)(c + di) = (ac − bd) + (ad + bc)i


(4 + 3i)(2 − 5i) = (4 ⋅ 2 − 3 ⋅ ( − 5)) + (4 ⋅ ( − 5) + 3 ⋅ 2)i
= (8 + 15) + ( − 20 + 6)i
= 23 − 14i


Multiply  (3 − 4i)(2 + 3i).


Dividing Complex Numbers
Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot
divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term
by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator
so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator,
which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate
of  a + bi  is  a − bi.
Note that complex conjugates have a reciprocal relationship: The complex conjugate of  a + bi  is  a − bi,   and the complex
conjugate of  a − bi  is  a + bi.  Further, when a quadratic equation with real coefficients has complex solutions, the
solutions are always complex conjugates of one another.
Suppose we want to divide  c + di  by  a + bi,  where neither  a  nor  b  equals zero. We first write the division as a fraction,
then find the complex conjugate of the denominator, and multiply.


c + di
a + bi


where a ≠ 0 and b ≠ 0


Chapter 3 Polynomial and Rational Functions 279




Multiply the numerator and denominator by the complex conjugate of the denominator.
(c + di)
(a + bi)


⋅ (a − bi)
(a − bi)


= (c + di)(a − bi)
(a + bi)(a − bi)


Apply the distributive property.


= ca − cbi + adi − bdi
2


a2 − abi + abi − b2 i2


Simplify, remembering that  i2 = −1.


= ca − cbi + adi − bd( − 1)
a2 − abi + abi − b2( − 1)


= (ca + bd) + (ad − cb)i
a2 + b2


The Complex Conjugate
The complex conjugate of a complex number  a + bi  is  a − bi.  It is found by changing the sign of the imaginary part
of the complex number. The real part of the number is left unchanged.


• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.


Example 3.6
Finding Complex Conjugates


Find the complex conjugate of each number.
a. 2 + i 5


b. −1
2
i


Solution
a. The number is already in the form  a + bi. The complex conjugate is  a − bi,   or  2 − i 5.
b. We can rewrite this number in the form  a + bi  as  0 − 1


2
i. The complex conjugate is  a − bi,   or  0 + 1


2
i. 


This can be written simply as  1
2
i.


Analysis
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally
find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a
real number from an imaginary number, we can simply multiply by  i.


Given two complex numbers, divide one by the other.
1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.


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Example 3.7
Dividing Complex Numbers


Divide  (2 + 5i)  by  (4 − i).


Solution
We begin by writing the problem as a fraction.


(2 + 5i)
(4 − i)


Then we multiply the numerator and denominator by the complex conjugate of the denominator.
(2 + 5i)
(4 − i)


⋅ (4 + i)
(4 + i)


To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly
called FOIL).


(2 + 5i)
(4 − i)


⋅ (4 + i)
(4 + i)


= 8 + 2i + 20i + 5i
2


16 + 4i − 4i − i2


= 8 + 2i + 20i + 5( − 1)
16 + 4i − 4i − ( − 1)


Because i2 = − 1


= 3 + 22i
17


= 3
17


+ 22
17


i Separate real and imaginary parts.


Note that this expresses the quotient in standard form.


Example 3.8
Substituting a Complex Number into a Polynomial Function


Let   f (x) = x2 − 5x + 2. Evaluate   f (3 + i).


Solution
Substitute  x = 3 + i  into the function   f (x) = x2 − 5x + 2  and simplify.


Analysis
We write   f (3 + i) = −5 + i. Notice that the input is  3 + i  and the output is  −5 + i.


Chapter 3 Polynomial and Rational Functions 281




3.6


3.7


Let   f (x) = 2x2 − 3x. Evaluate   f (8 − i).


Example 3.9
Substituting an Imaginary Number in a Rational Function


Let   f (x) = 2 + x
x + 3


. Evaluate   f (10i).


Solution
Substitute  x = 10i  and simplify.


2 + 10i
10i + 3


Substitute 10i for x.


2 + 10i
3 + 10i


Rewrite the denominator in standard form.


2 + 10i
3 + 10i


⋅ 3 – 10i
3 – 10i


Prepare to multiply the numerator and


denominator by the complex conjugate


of the denominator.


6 – 20i + 30i – 100i2


9 – 30i + 30i – 100i2
Multiply using the distributive property or the FOIL method.


6 – 20i + 30i – 100( – 1)
9 – 30i + 30i – 100( – 1)


Substitute –1 for i2.


106 + 10i
109


Simplify.


106
109


+ 10
109


i Separate the real and imaginary parts.


Let   f (x) = x + 1
x − 4


. Evaluate   f (−i).


Simplifying Powers of i
The powers of  i  are cyclic. Let’s look at what happens when we raise  i  to increasing powers.


i1 = i


i2 = − 1


i3 = i2 ⋅ i = − 1 ⋅ i = − i


i4 = i3 ⋅ i = − i ⋅ i = − i2 = − ( − 1) = 1


i5 = i4 ⋅ i = 1 ⋅ i = i


We can see that when we get to the fifth power of  i,   it is equal to the first power. As we continue to multiply  i  by itself for
increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of  i.


i6 = i5 ⋅ i = i ⋅ i = i2 = − 1


i7 = i6 ⋅ i = i2 ⋅ i = i3 = − i


i8 = i7 ⋅ i = i3 ⋅ i = i4 = 1


i9 = i8 ⋅ i = i4 ⋅ i = i5 = i


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Example 3.10
Simplifying Powers of  i


Evaluate  i35.


Solution
Since  i4 = 1,  we can simplify the problem by factoring out as many factors of  i4   as possible. To do so, first
determine how many times 4 goes into 35:  35 = 4 ⋅ 8 + 3.


i35 = i4 ⋅ 8 + 3 = i4 ⋅ 8 ⋅ i3 = ⎛⎝i
4⎞


8
⋅ i3 = 18 ⋅ i3 = i3 = − i


Can we write  i35   in other helpful ways?
As we saw in Example 3.10, we reduced  i35   to  i3   by dividing the exponent by 4 and using the remainder to
find the simplified form. But perhaps another factorization of  i35  may be more useful. Table 3.1 shows some
other possible factorizations.


Factorization of  i35 i34 ⋅ i i33 ⋅ i2 i31 ⋅ i4 i19 ⋅ i16


Reduced form ⎛⎝i2⎞⎠
17


⋅ i i33 ⋅ (−1) i31 ⋅ 1 i19 ⋅ ⎛⎝i
4⎞


4


Simplified form (−1)17 ⋅ i −i33 i31 i19


Table 3.1
Each of these will eventually result in the answer we obtained above but may require several more steps than our
earlier method.


Access these online resources for additional instruction and practice with complex numbers.
• Adding and Subtracting Complex Numbers (http://openstaxcollege.org/l/addsubcomplex)
• Multiply Complex Numbers (http://openstaxcollege.org/l/multiplycomplex)
• Multiplying Complex Conjugates (http://openstaxcollege.org/l/multcompconj)
• Raising i to Powers (http://openstaxcollege.org/l/raisingi)


Chapter 3 Polynomial and Rational Functions 283




1.
2.


3.


4.


5.


6.


7.


8.


9.


10.


11.


12.


13.


14.


15.


16.


17.


18.


19.


20.


21.


22.


23.


24.


25.


26.


27.


28.


29.


30.


3.1 EXERCISES
Verbal
Explain how to add complex numbers.
What is the basic principle in multiplication of complex


numbers?
Give an example to show the product of two imaginary


numbers is not always imaginary.
What is a characteristic of the plot of a real number in


the complex plane?


Algebraic
For the following exercises, evaluate the algebraic
expressions.


If f (x) = x2 + x − 4,   evaluate   f (2i).


If f (x) = x3 − 2,   evaluate   f (i).


If f (x) = x2 + 3x + 5, evaluate   f (2 + i).


If f (x) = 2x2 + x − 3,   evaluate   f (2 − 3i).


If f (x) = x + 1
2 − x


,   evaluate   f (5i).


If f (x) = 1 + 2x
x + 3


,   evaluate   f (4i).


Graphical
For the following exercises, determine the number of real
and nonreal solutions for each quadratic function shown.


For the following exercises, plot the complex numbers on
the complex plane.


1 − 2i


−2 + 3i


i


−3 − 4i


Numeric
For the following exercises, perform the indicated
operation and express the result as a simplified complex
number.


(3 + 2i) + (5 − 3i)


(−2 − 4i) + (1 + 6i)


(−5 + 3i) − (6 − i)


(2 − 3i) − (3 + 2i)


( − 4 + 4i) − ( − 6 + 9i)


(2 + 3i)(4i)


(5 − 2i)(3i)


(6 − 2i)(5)


(−2 + 4i)(8)


(2 + 3i)(4 − i)


(−1 + 2i)( − 2 + 3i)


(4 − 2i)(4 + 2i)


(3 + 4i)(3 − 4i)


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31.


32.


33.


34.


35.


36.


37.


38.


39.


40.


41.


42.


43.


44.


45.


46.


47.


48.


49.


50.


51.


52.


53.


54.


55.


56.


57.


58.


3 + 4i
2


6 − 2i
3


−5 + 3i
2i


6 + 4i
i


2 − 3i
4 + 3i


3 + 4i
2 − i


2 + 3i
2 − 3i


−9 + 3 −16


− −4 − 4 −25


2 + −12
2


4 + −20
2


i8


i15


i22


Technology
For the following exercises, use a calculator to help answer
the questions.


Evaluate  (1 + i)k   for k = 4, 8, and 12. Predict the
value if  k = 16.


Evaluate  (1 − i)k   for k = 2, 6, and 10. Predict the
value if  k = 14.


Evaluate  (1 + i)k − (1 − i)k for k = 4, 8, and 12 .
Predict the value for  k = 16.


Show that a solution of  x6 + 1 = 0  is   3
2


+ 1
2
i.


Show that a solution of  x8 − 1 = 0  is   2
2


+ 2
2
i.


Extensions
For the following exercises, evaluate the expressions,
writing the result as a simplified complex number.


1
i
+ 4
i3


1
i11


− 1
i21


i7 ⎛⎝1 + i
2⎞


i−3 + 5i7


(2 + i)(4 − 2i)
(1 + i)


(1 + 3i)(2 − 4i)
(1 + 2i)


(3 + i)2


(1 + 2i)2


3 + 2i
2 + i


+ (4 + 3i)


4 + i
i


+ 3 − 4i
1 − i


3 + 2i
1 + 2i


− 2 − 3i
3 + i


Chapter 3 Polynomial and Rational Functions 285




3.2 | Quadratic Functions
Learning Objectives


In this section, you will:
3.2.1 Recognize characteristics of parabolas.
3.2.2 Understand how the graph of a parabola is related to its quadratic function.
3.2.3 Determine a quadratic function’s minimum or maximum value.
3.2.4 Solve problems involving a quadratic function’s minimum or maximum value.


Figure 3.5 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr)


Curved antennas, such as the ones shown in Figure 3.5, are commonly used to focus microwaves and radio waves to
transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna
is in the shape of a parabola, which can be described by a quadratic function.
In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile
motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide
a good opportunity for a detailed study of function behavior.
Recognizing Characteristics of Parabolas
The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has
an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the
minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph,
or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical
line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 3.6.


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Figure 3.6


The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola
crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of  x  at
which  y = 0.


Chapter 3 Polynomial and Rational Functions 287




Example 3.11
Identifying the Characteristics of a Parabola


Determine the vertex, axis of symmetry, zeros, and  y- intercept of the parabola shown in Figure 3.7.


Figure 3.7


Solution
The vertex is the turning point of the graph. We can see that the vertex is at  (3, 1). Because this parabola opens
upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry
is  x = 3. This parabola does not cross the  x- axis, so it has no zeros. It crosses the  y- axis at  (0, 7)  so this is the
y-intercept.


Understanding How the Graphs of Parabolas are Related to Their
Quadratic Functions
The general form of a quadratic function presents the function in the form


(3.1)f (x) = ax2 + bx + c
where  a, b,   and  c  are real numbers and  a ≠ 0.  If  a > 0,   the parabola opens upward. If  a < 0,   the parabola opens
downward. We can use the general form of a parabola to find the equation for the axis of symmetry.


The axis of symmetry is defined by  x = − b
2a


.  If we use the quadratic formula,  x = −b ± b2 − 4ac
2a


,   to solve
 ax2 + bx + c = 0  for the  x- intercepts, or zeros, we find the value of  x  halfway between them is always  x = − b


2a
,   the


equation for the axis of symmetry.


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Figure 3.8 represents the graph of the quadratic function written in general form as  y = x2 + 4x + 3.  In this form,
 a = 1, b = 4,   and  c = 3. Because  a > 0,   the parabola opens upward. The axis of symmetry is  x = − 4


2(1)
= − 2. 


This also makes sense because we can see from the graph that the vertical line  x = − 2  divides the graph in half. The
vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point
on the graph, in this instance,  ( − 2, − 1). The  x- intercepts, those points where the parabola crosses the  x- axis, occur at
 ( − 3, 0)  and  ( − 1, 0).


Figure 3.8


The standard form of a quadratic function presents the function in the form
(3.2)f (x) = a(x − h)2 + k


where  (h, k)  is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also
known as the vertex form of a quadratic function.
As with the general form, if  a > 0,   the parabola opens upward and the vertex is a minimum. If  a < 0,   the parabola opens
downward, and the vertex is a maximum. Figure 3.9 represents the graph of the quadratic function written in standard
form as  y = −3(x + 2)2 + 4.  Since  x – h = x + 2  in this example,  h = –2.  In this form,  a = −3, h = −2,   and  k = 4. 
Because  a < 0,   the parabola opens downward. The vertex is at  (−2, 4).


Chapter 3 Polynomial and Rational Functions 289




Figure 3.9


The standard form is useful for determining how the graph is transformed from the graph of  y = x2. Figure 3.10 is the
graph of this basic function.


Figure 3.10


If  k > 0,   the graph shifts upward, whereas if  k < 0,   the graph shifts downward. In Figure 3.9,  k > 0,   so the graph is
shifted 4 units upward. If  h > 0,   the graph shifts toward the right and if  h < 0,   the graph shifts to the left. In Figure
3.9,  h < 0,   so the graph is shifted 2 units to the left. The magnitude of  a  indicates the stretch of the graph. If |a| > 1,
the point associated with a particular  x- value shifts farther from the x-axis, so the graph appears to become narrower, and
there is a vertical stretch. But if  |a| < 1,   the point associated with a particular  x- value shifts closer to the x-axis, so the


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graph appears to become wider, but in fact there is a vertical compression. In Figure 3.9,  |a| > 1,   so the graph becomes
narrower.
The standard form and the general form are equivalent methods of describing the same function. We can see this by
expanding out the general form and setting it equal to the standard form.


                         a(x − h)2 + k = ax2 + bx + c
ax2 − 2ahx + (ah2 + k) = ax2 + bx + c


For the linear terms to be equal, the coefficients must be equal.
–2ah = b, so h = − b


2a
.


This is the axis of symmetry we defined earlier. Setting the constant terms equal:
ah2 + k = c


k = c − ah2


= c − a − ⎛⎝
b
2a



2


= c − b
2


4a


In practice, though, it is usually easier to remember that k is the output value of the function when the input is  h,   so
  f (h) = k.


Forms of Quadratic Functions
A quadratic function is a function of degree two. The graph of a quadratic function is a parabola. The general form of
a quadratic function is   f (x) = ax2 + bx + c where  a, b,   and  c  are real numbers and  a ≠ 0.
The standard form of a quadratic function is   f (x) = a(x − h)2 + k.
The vertex  (h, k)  is located at


h = – b
2a


, k = f (h) = f ⎛⎝
−b
2a

⎠.


Given a graph of a quadratic function, write the equation of the function in general form.
1. Identify the horizontal shift of the parabola; this value is  h.  Identify the vertical shift of the parabola; this
value is  k.


2. Substitute the values of the horizontal and vertical shift for  h  and  k.  in the function
  f (x) = a(x – h)2 + k.


3. Substitute the values of any point, other than the vertex, on the graph of the parabola for  x  and   f (x).
4. Solve for the stretch factor,  |a|.
5. If the parabola opens up,  a > 0.  If the parabola opens down,  a < 0  since this means the graph was
reflected about the  x- axis.


6. Expand and simplify to write in general form.


Chapter 3 Polynomial and Rational Functions 291




Example 3.12
Writing the Equation of a Quadratic Function from the Graph


Write an equation for the quadratic function  g  in Figure 3.11 as a transformation of   f (x) = x2,   and then
expand the formula, and simplify terms to write the equation in general form.


Figure 3.11


Solution
We can see the graph of g is the graph of   f (x) = x2   shifted to the left 2 and down 3, giving a formula in the form
 g(x) = a(x + 2)2 – 3.
Substituting the coordinates of a point on the curve, such as  (0, −1),  we can solve for the stretch factor.


−1 = a(0 + 2)2 − 3
2 = 4a


a = 1
2


In standard form, the algebraic model for this graph is  (g)x = 1
2
(x + 2)2 – 3.


To write this in general polynomial form, we can expand the formula and simplify terms.
g(x) = 1


2
(x + 2)2 − 3


= 1
2
(x + 2)(x + 2) − 3


= 1
2
(x2 + 4x + 4) − 3


= 1
2
x2 + 2x + 2 − 3


= 1
2
x2 + 2x − 1


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3.8


Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of
the vertex of the parabola; the vertex is unaffected by stretches and compressions.


Analysis
We can check our work using the table feature on a graphing utility. First enter  Y1 = 1


2
(x + 2)2 − 3. Next, select


 TBLSET,  then use  TblStart = – 6  and  ΔTbl = 2,  and select  TABLE.  See Table 3.1.


x –6 –4 –2 0 2


y 5 –1 –3 –1 5


Table 3.1


The ordered pairs in the table correspond to points on the graph.


A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 3.12. Find an
equation for the path of the ball. Does the shooter make the basket?


Figure 3.12 (credit: modification of work by Dan Meyer)


Given a quadratic function in general form, find the vertex of the parabola.
1. Identify  a, b, and c.
2. Find  h,   the x-coordinate of the vertex, by substituting  a  and  b  into  h = – b


2a
.


3. Find  k,   the y-coordinate of the vertex, by evaluating  k = f (h) = f ⎛⎝− b2a

⎠.


Chapter 3 Polynomial and Rational Functions 293




3.9


Example 3.13
Finding the Vertex of a Quadratic Function


Find the vertex of the quadratic function   f (x) = 2x2 – 6x + 7. Rewrite the quadratic in standard form (vertex
form).


Solution
The horizontal coordinate of the vertex will be at


h = – b
2a


= – – 6
2(2)


= 6
4


= 3
2


The vertical coordinate of the vertex will be at
k = f (h)


= f ⎛⎝
3
2



= 2⎛⎝
3
2



2
− 6⎛⎝


3
2

⎠+ 7


= 5
2


Rewriting into standard form, the stretch factor will be the same as the  a  in the original quadratic.
f (x) = ax2 + bx + c


f (x) = 2x2 − 6x + 7


Using the vertex to determine the shifts,


f (x) = 2⎛⎝x –
3
2



2
+ 5


2


Analysis
One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum
or minimum value of the output occurs,  ⎛⎝k), and where it occurs,  (x).


Given the equation  g(x) = 13 + x2 − 6x, write the equation in general form and then in standard form.


Finding the Domain and Range of a Quadratic Function
Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real
numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola
will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at
the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up
or down.


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Domain and Range of a Quadratic Function
The domain of any quadratic function is all real numbers.
The range of a quadratic function written in general form   f (x) = ax2 + bx + c with a positive  a  value is
  f (x) ≥ f ⎛⎝− b2a



⎠,   or  ⎡⎣ f ⎛⎝− b2a



⎠, ∞

⎠;   the range of a quadratic function written in general form with a negative  a 


value is   f (x) ≤ f ⎛⎝− b2a

⎠,   or  ⎛⎝−∞, f ⎛⎝− b2a





⎦.


The range of a quadratic function written in standard form   f (x) = a(x − h)2 + k with a positive  a  value is
  f (x) ≥ k;   the range of a quadratic function written in standard form with a negative  a  value is   f (x) ≤ k.


Given a quadratic function, find the domain and range.
1. Identify the domain of any quadratic function as all real numbers.
2. Determine whether  a  is positive or negative. If  a  is positive, the parabola has a minimum. If  a  is
negative, the parabola has a maximum.


3. Determine the maximum or minimum value of the parabola,  k.
4. If the parabola has a minimum, the range is given by   f (x) ≥ k,   or  ⎡⎣k, ∞).  If the parabola has a
maximum, the range is given by   f (x) ≤ k,   or  (−∞, k⎤⎦.


Example 3.14
Finding the Domain and Range of a Quadratic Function


Find the domain and range of   f (x) = − 5x2 + 9x − 1.


Solution
As with any quadratic function, the domain is all real numbers.
Because  a  is negative, the parabola opens downward and has a maximum value. We need to determine the
maximum value. We can begin by finding the  x- value of the vertex.


h = − b
2a


= − 9
2( − 5)


= 9
10


The maximum value is given by   f (h).


f ( 9
10


) = 5( 9
10


)
2
+ 9( 9


10
) − 1


= 61
20


The range is   f (x) ≤ 61
20


,   or  ⎛⎝−∞, 6120

⎦.


Chapter 3 Polynomial and Rational Functions 295




3.10 Find the domain and range of   f (x) = 2⎛⎝x − 47



2
+ 8


11
.


Determining the Maximum and Minimum Values of Quadratic
Functions
The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the
orientation of the parabola. We can see the maximum and minimum values in Figure 3.13.


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Figure 3.13


There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as
applications involving area and revenue.


Example 3.15


Chapter 3 Polynomial and Rational Functions 297




Finding the Maximum Value of a Quadratic Function


A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has
purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the
fourth side.
a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence
have length  L.


b. What dimensions should she make her garden to maximize the enclosed area?


Solution
Let’s use a diagram such as Figure 3.14 to record the given information. It is also helpful to introduce a
temporary variable,  W,   to represent the width of the garden and the length of the fence section parallel to the
backyard fence.


Figure 3.14
a. We know we have only 80 feet of fence available, and  L +W + L = 80,   or more simply,


 2L +W = 80. This allows us to represent the width,  W,   in terms of  L.
W = 80 − 2L


Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is
length multiplied by width, so


A = LW = L(80 − 2L)


A(L) = 80L − 2L2


This formula represents the area of the fence in terms of the variable length  L. The function, written in
general form, is


A(L) = − 2L2 + 80L.


b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will
be the maximum value for the area. In finding the vertex, we must be careful because the equation is
not written in standard polynomial form with decreasing powers. This is why we rewrote the function in
general form above. Since  a  is the coefficient of the squared term,  a = −2, b = 80,   and  c = 0.


To find the vertex:
h = − 80


2( − 2)
k = A(20)


= 20 and = 80(20) − 2(20)2


= 800


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The maximum value of the function is an area of 800 square feet, which occurs when  L = 20  feet. When the
shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should
enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence
has length 40 feet.


Analysis
This problem also could be solved by graphing the quadratic function. We can see where the maximum area
occurs on a graph of the quadratic function in Figure 3.15.


Figure 3.15


Given an application involving revenue, use a quadratic equation to find the maximum.
1. Write a quadratic equation for revenue.
2. Find the vertex of the quadratic equation.
3. Determine the y-value of the vertex.


Example 3.16
Finding Maximum Revenue


The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for
the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly
charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000
subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge
for a quarterly subscription to maximize their revenue?


Solution


Chapter 3 Polynomial and Rational Functions 299




Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the
price per subscription times the number of subscribers, or quantity. We can introduce variables,  p  for price per
subscription and  Q  for quantity, giving us the equation  Revenue = pQ.
Because the number of subscribers changes with the price, we need to find a relationship between the variables.
We know that currently  p = 30  and  Q = 84, 000. We also know that if the price rises to $32, the newspaper
would lose 5,000 subscribers, giving a second pair of values,  p = 32  and  Q = 79, 000.  From this we can find a
linear equation relating the two quantities. The slope will be


m = 79, 000 − 84, 000
32 − 30


= −5, 000
2


= − 2, 500


This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the
y-intercept.


Q = −2500p + b Substitute in the point Q = 84, 000 and p = 30


84, 000 = −2500(30) + b Solve for b
b = 159, 000


This gives us the linear equation  Q = −2,500p + 159, 000  relating cost and subscribers. We now return to our
revenue equation.


Revenue = pQ


Revenue = p(−2, 500p + 159, 000)


Revenue = −2, 500p2 + 159, 000p


We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will
maximize revenue for the newspaper, we can find the vertex.


h = − 159, 000
2( − 2, 500)


= 31.8


The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To
find what the maximum revenue is, we evaluate the revenue function.


maximum revenue = −2,500(31.8)2 + 159,000(31.8)
= 2,528,100


Analysis
This could also be solved by graphing the quadratic as in Figure 3.16. We can see the maximum revenue on a
graph of the quadratic function.


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Figure 3.16


Finding the x- and y-Intercepts of a Quadratic Function
Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing
parabolas. Recall that we find the  y- intercept of a quadratic by evaluating the function at an input of zero, and we find
the  x- intercepts at locations where the output is zero. Notice in Figure 3.17 that the number of  x- intercepts can vary
depending upon the location of the graph.


Figure 3.17 Number of x-intercepts of a parabola


Given a quadratic function   f(x),   find the  y- and x-intercepts.
1. Evaluate   f (0)  to find the  y- intercept.
2. Solve the quadratic equation   f (x) = 0  to find the x-intercepts.


Example 3.17
Finding the y- and x-Intercepts of a Parabola


Find the y- and x-intercepts of the quadratic   f (x) = 3x2 + 5x − 2.


Solution


Chapter 3 Polynomial and Rational Functions 301




We find the y-intercept by evaluating   f (0).
f (0) = 3(0)2 + 5(0) − 2


= − 2


So the y-intercept is at  (0, −2).
For the x-intercepts, we find all solutions of   f (x) = 0.


0 = 3x2 + 5x − 2


In this case, the quadratic can be factored easily, providing the simplest method for solution.
0 = (3x − 1)(x + 2)


0 = 3x − 1 0 = x + 2


x = 1
3


or x = − 2


So the x-intercepts are at  ⎛⎝13, 0

⎠  and  (−2, 0).


Analysis
By graphing the function, we can confirm that the graph crosses the y-axis at  (0, −2). We can also confirm that
the graph crosses the x-axis at  ⎛⎝13, 0



⎠  and  (−2, 0). See Figure 3.18


Figure 3.18


Rewriting Quadratics in Standard Form
In Example 3.17, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be
factored. We can solve these quadratics by first rewriting them in standard form.


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Given a quadratic function, find the  x- intercepts by rewriting in standard form.
1. Substitute  a  and  b  into  h = − b


2a
.


2. Substitute  x = h  into the general form of the quadratic function to find  k.
3. Rewrite the quadratic in standard form using  h  and  k.
4. Solve for when the output of the function will be zero to find the  x- intercepts.


Example 3.18
Finding the  x- Intercepts of a Parabola


Find the  x- intercepts of the quadratic function   f (x) = 2x2 + 4x − 4.


Solution
We begin by solving for when the output will be zero.


0 = 2x2 + 4x − 4


Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the
quadratic in standard form.


f (x) = a(x − h)2 + k


We know that  a = 2. Then we solve for  h  and  k.
h = − b


2a
k = f ( − 1)


= − 4
2(2)


= 2( − 1)2 + 4( − 1) − 4


= −1 = −6


So now we can rewrite in standard form.
f (x) = 2(x + 1)2 − 6


We can now solve for when the output will be zero.
0 = 2(x + 1)2 − 6


6 = 2(x + 1)2


3 = (x + 1)2


x + 1 = ± 3


x = − 1 ± 3


The graph has  x- intercepts at  (−1 − 3, 0)  and  (−1 + 3, 0).


Analysis
We can check our work by graphing the given function on a graphing utility and observing the  x- intercepts. See
Figure 3.19.


Chapter 3 Polynomial and Rational Functions 303




3.11


Figure 3.19


In a separate Try It, we found the standard and general form for the function  g(x) = 13 + x2 − 6x. Now
find the y- and  x- intercepts (if any).


Example 3.19
Solving a Quadratic Equation with the Quadratic Formula


Solve  x2 + x + 2 = 0.


Solution


Let’s begin by writing the quadratic formula:  x = −b ± b2 − 4ac
2a


.


When applying the quadratic formula, we identify the coefficients  a, b and c.  For the equation
 x2 + x + 2 = 0,  we have  a = 1, b = 1, and c = 2.  Substituting these values into the formula we have:


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x = −b ± b
2 − 4ac


2a


= −1 ± 1
2 − 4 ⋅ 1 ⋅ (2)
2 ⋅ 1


= −1 ± 1 − 8
2


= −1 ± −7
2


= −1 ± i 7
2


The solutions to the equation are  x = −1 + i 7
2


  and  x = −1 − i 7
2


  or  x = −1
2


+ i 7
2


  and  x = −1
2


− i 7
2


.


Example 3.20
Applying the Vertex and x-Intercepts of a Parabola


A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height
above ground can be modeled by the equation  H(t) = − 16t2 + 80t + 40.
a. When does the ball reach the maximum height?
b. What is the maximum height of the ball?
c. When does the ball hit the ground?


Solution
a. The ball reaches the maximum height at the vertex of the parabola.


h = − 80
2( − 16)


= 80
32


= 5
2


= 2.5
The ball reaches a maximum height after 2.5 seconds.


b. To find the maximum height, find the  y- coordinate of the vertex of the parabola.
k = H⎛⎝−


b
2a



= H(2.5)


= −16(2.5)2 + 80(2.5) + 40
= 140


The ball reaches a maximum height of 140 feet.
c. To find when the ball hits the ground, we need to determine when the height is zero,  H(t) = 0.
We use the quadratic formula.


Chapter 3 Polynomial and Rational Functions 305




3.12


t = −80 ± 80
2 − 4( − 16)(40)


2( − 16)


= −80 ± 8960
−32


Because the square root does not simplify nicely, we can use a calculator to approximate the values of the
solutions.


t = −80 − 8960
−32


≈ 5.458 or t = −80 + 8960
−32


≈ − 0.458


The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the
ground after about 5.458 seconds. See Figure 3.20


Figure 3.20


A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet
per second. The rock’s height above ocean can be modeled by the equation  H(t) = − 16t2 + 96t + 112.


a. When does the rock reach the maximum height?
b. What is the maximum height of the rock?
c. When does the rock hit the ocean?


Access these online resources for additional instruction and practice with quadratic equations.
• Graphing Quadratic Functions in General Form (http://openstaxcollege.org/l/
graphquadgen)


• Graphing Quadratic Functions in Standard Form (http://openstaxcollege.org/l/
graphquadstan)


• Quadratic Function Review (http://openstaxcollege.org/l/quadfuncrev)
• Characteristics of a Quadratic Function (http://openstaxcollege.org/l/characterquad)


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59.


60.


61.


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63.


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67.


68.


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79.


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98.


3.2 EXERCISES
Verbal
Explain the advantage of writing a quadratic function


in standard form.
How can the vertex of a parabola be used in solving


real world problems?
Explain why the condition of  a ≠ 0  is imposed in the


definition of the quadratic function.
What is another name for the standard form of a


quadratic function?
What two algebraic methods can be used to find the


horizontal intercepts of a quadratic function?


Algebraic
For the following exercises, rewrite the quadratic functions
in standard form and give the vertex.


f (x) = x2 − 12x + 32


g(x) = x2 + 2x − 3


f (x) = x2 − x


f (x) = x2 + 5x − 2


h(x) = 2x2 + 8x − 10


k(x) = 3x2 − 6x − 9


f (x) = 2x2 − 6x


f (x) = 3x2 − 5x − 1


For the following exercises, determine whether there is a
minimum or maximum value to each quadratic function.
Find the value and the axis of symmetry.


y(x) = 2x2 + 10x + 12


f (x) = 2x2 − 10x + 4


f (x) = − x2 + 4x + 3


f (x) = 4x2 + x − 1


h(t) = − 4t2 + 6t − 1


f (x) = 1
2
x2 + 3x + 1


f (x) = − 1
3
x2 − 2x + 3


For the following exercises, determine the domain and
range of the quadratic function.


f (x) = (x − 3)2 + 2


f (x) = − 2(x + 3)2 − 6


f (x) = x2 + 6x + 4


f (x) = 2x2 − 4x + 2


k(x) = 3x2 − 6x − 9


For the following exercises, solve the equations over the
complex numbers.


x2 = − 25


x2 = − 8


x2 + 36 = 0


x2 + 27 = 0


x2 + 2x + 5 = 0


x2 − 4x + 5 = 0


x2 + 8x + 25 = 0


x2 − 4x + 13 = 0


x2 + 6x + 25 = 0


x2 − 10x + 26 = 0


x2 − 6x + 10 = 0


x(x − 4) = 20


x(x − 2) = 10


2x2 + 2x + 5 = 0


5x2 − 8x + 5 = 0


Chapter 3 Polynomial and Rational Functions 307




99.


100.


101.


102.


103.


104.


105.


106.


107.


108.


109.


110.


111.


112.


113.


114.


115.


116.


117.


118.


119.


120.


5x2 + 6x + 2 = 0


2x2 − 6x + 5 = 0


x2 + x + 2 = 0


x2 − 2x + 4 = 0


For the following exercises, use the vertex  (h, k)  and a
point on the graph  (x, y)  to find the general form of the
equation of the quadratic function.


(h, k) = (2, 0), (x, y) = (4, 4)


(h, k) = (−2, −1), (x, y) = (−4, 3)


(h, k) = (0, 1), (x, y) = (2, 5)


(h, k) = (2, 3), (x, y) = (5, 12)


(h, k) = ( − 5, 3), (x, y) = (2, 9)


(h, k) = (3, 2), (x, y) = (10, 1)


(h, k) = (0, 1), (x, y) = (1, 0)


(h, k) = (1, 0), (x, y) = (0, 1)


Graphical
For the following exercises, sketch a graph of the quadratic
function and give the vertex, axis of symmetry, and
intercepts.


f (x) = x2 − 2x


f (x) = x2 − 6x − 1


f (x) = x2 − 5x − 6


f (x) = x2 − 7x + 3


f (x) = − 2x2 + 5x − 8


f (x) = 4x2 − 12x − 3


For the following exercises, write the equation for the
graphed function.


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121.


122.


123.


124.


125.


126.


127.


128.


129.


130.


Numeric
For the following exercises, use the table of values that
represent points on the graph of a quadratic function. By
determining the vertex and axis of symmetry, find the
general form of the equation of the quadratic function.


x –2 –1 0 1 2


y 5 2 1 2 5


x –2 –1 0 1 2


y 1 0 1 4 9


x –2 –1 0 1 2


y –2 1 2 1 –2


x –2 –1 0 1 2


y –8 –3 0 1 0


x –2 –1 0 1 2


y 8 2 0 2 8


Technology
For the following exercises, use a calculator to find the
answer.


Graph on the same set of axes the functions
  f (x) = x2, f (x) = 2x2, and f (x) = 1


3
x2.


What appears to be the effect of changing the coefficient?
Graph on the same set of axes


  f (x) = x2, f (x) = x2 + 2  and
  f (x) = x2, f (x) = x2 + 5  and   f (x) = x2 − 3.  What
appears to be the effect of adding a constant?


Graph on the same set of axes
  f (x) = x2, f (x) = (x − 2)2, f (x − 3)2, and f (x) = (x + 4)2.


Chapter 3 Polynomial and Rational Functions 309




131.


132.


133.


134.


135.


136.


137.


138.


139.


140.


141.


142.


143.


144.


145.


146.


147.


148.


149.


150.


151.


152.


What appears to be the effect of adding or subtracting those
numbers?


The path of an object projected at a 45 degree angle
with initial velocity of 80 feet per second is given by the
function  h(x) = −32


(80)2
x2 + x where  x  is the horizontal


distance traveled and  h(x)  is the height in feet. Use the
TRACE feature of your calculator to determine the height
of the object when it has traveled 100 feet away
horizontally.


A suspension bridge can be modeled by the quadratic
function  h(x) = .0001x2  with  − 2000 ≤ x ≤ 2000
where  |x|  is the number of feet from the center and  h(x)  is
height in feet. Use the TRACE feature of your calculator to
estimate how far from the center does the bridge have a
height of 100 feet.


Extensions
For the following exercises, use the vertex of the graph of
the quadratic function and the direction the graph opens to
find the domain and range of the function.


Vertex  (1, −2),   opens up.


Vertex  (−1, 2)  opens down.


Vertex  (−5, 11),   opens down.


Vertex  (−100, 100),   opens up.


For the following exercises, write the equation of the
quadratic function that contains the given point and has the
same shape as the given function.


Contains  (1, 1)  and has shape of   f (x) = 2x2. 
Vertex is on the  y- axis.


Contains  (−1, 4)  and has the shape of   f (x) = 2x2. 
Vertex is on the  y- axis.


Contains  (2, 3)  and has the shape of   f (x) = 3x2. 
Vertex is on the  y- axis.


Contains  (1, −3)  and has the shape of
  f (x) = − x2. Vertex is on the  y- axis.


Contains  (4, 3)  and has the shape of   f (x) = 5x2. 
Vertex is on the  y- axis.


Contains  (1, −6)  has the shape of   f (x) = 3x2. Vertex has
x-coordinate of  −1.


Real-World Applications
Find the dimensions of the rectangular corral


producing the greatest enclosed area given 200 feet of
fencing.


Find the dimensions of the rectangular corral split
into 2 pens of the same size producing the greatest possible
enclosed area given 300 feet of fencing.


Find the dimensions of the rectangular corral
producing the greatest enclosed area split into 3 pens of the
same size given 500 feet of fencing.


Among all of the pairs of numbers whose sum is 6,
find the pair with the largest product. What is the product?


Among all of the pairs of numbers whose difference is
12, find the pair with the smallest product. What is the
product?


Suppose that the price per unit in dollars of a cell
phone production is modeled by  p = $45 − 0.0125x,
where  x  is in thousands of phones produced, and the
revenue represented by thousands of dollars is  R = x ⋅ p. 
Find the production level that will maximize revenue.


A rocket is launched in the air. Its height, in meters
above sea level, as a function of time, in seconds, is given
by  h(t) = − 4.9t2 + 229t + 234.  Find the maximum
height the rocket attains.


A ball is thrown in the air from the top of a building.
Its height, in meters above ground, as a function of time, in
seconds, is given by  h(t) = − 4.9t2 + 24t + 8. How long
does it take to reach maximum height?


A soccer stadium holds 62,000 spectators. With a
ticket price of $11, the average attendance has been 26,000.
When the price dropped to $9, the average attendance rose
to 31,000. Assuming that attendance is linearly related to
ticket price, what ticket price would maximize revenue?


A farmer finds that if she plants 75 trees per acre, each
tree will yield 20 bushels of fruit. She estimates that for
each additional tree planted per acre, the yield of each tree
will decrease by 3 bushels. How many trees should she
plant per acre to maximize her harvest?


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3.3 | Power Functions and Polynomial Functions
Learning Objectives


In this section, you will:
3.3.1 Identify power functions.
3.3.2 Identify end behavior of power functions.
3.3.3 Identify polynomial functions.
3.3.4 Identify the degree and leading coefficient of polynomial functions.


Figure 3.21 (credit: Jason Bay, Flickr)


Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 3.2.


Year 2009 2010 2011 2012 2013


Bird Population 800 897 992 1, 083 1, 169


Table 3.2


The population can be estimated using the function  P(t) = − 0.3t3 + 97t + 800,  where  P(t)  represents the bird
population on the island  t  years after 2009. We can use this model to estimate the maximum bird population and when it
will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we
will examine functions that we can use to estimate and predict these types of changes.
Identifying Power Functions
In order to better understand the bird problem, we need to understand a specific type of function. A power function is a
function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number.
(A number that multiplies a variable raised to an exponent is known as a coefficient.)
As an example, consider functions for area or volume. The function for the area of a circle with radius  r  is


A(r) = πr2


and the function for the volume of a sphere with radius  r  is


Chapter 3 Polynomial and Rational Functions 311




V(r) = 4
3
πr3


Both of these are examples of power functions because they consist of a coefficient,  π  or  4
3
π,  multiplied by a variable  r


raised to a power.


Power Function
A power function is a function that can be represented in the form


f (x) = kx p


where  k  and  p  are real numbers, and  k  is known as the coefficient.


Is   f(x) = 2x   a power function?
No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to
a variable power. This is called an exponential function, not a power function.


Example 3.21
Identifying Power Functions


Which of the following functions are power functions?
f (x) = 1 Constant function
f (x) = x Identify function


f (x) = x2 Quadratic function


f (x) = x3 Cubic function


f (x) = 1x Reciprocal function


f (x) = 1
x2


Reciprocal squared function


f (x) = x Square root function


f (x) = x3 Cube root function


Solution
All of the listed functions are power functions.
The constant and identity functions are power functions because they can be written as   f (x) = x0   and
  f (x) = x1   respectively.
The quadratic and cubic functions are power functions with whole number powers   f (x) = x2   and   f (x) = x3.
The reciprocal and reciprocal squared functions are power functions with negative whole number powers because
they can be written as   f (x) = x−1   and   f (x) = x−2.
The square and cube root functions are power functions with fractional powers because they can be written as
  f (x) = x1/2   or   f (x) = x1/3.


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3.13 Which functions are power functions?
f (x) = 2x2 ⋅ 4x3


g(x) = − x5 + 5x3 − 4x


h(x) = 2x
5 − 1


3x2 + 4


Identifying End Behavior of Power Functions
Figure 3.22 shows the graphs of   f (x) = x2,  g(x) = x4   and  and h(x) = x6,  which are all power functions with even,
whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the
toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the
origin.


Figure 3.22 Even-power functions


To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol  ∞  for positive
infinity and  −∞  for negative infinity. When we say that “ x  approaches infinity,” which can be symbolically written as
 x → ∞,  we are describing a behavior; we are saying that  x  is increasing without bound.
With the even-power function, as the input increases or decreases without bound, the output values become very large,
positive numbers. Equivalently, we could describe this behavior by saying that as  x  approaches positive or negative infinity,
the   f (x)  values increase without bound. In symbolic form, we could write


as x → ±∞, f (x) → ∞


Figure 3.23 shows the graphs of   f (x) = x3,  g(x) = x5,  and h(x) = x7,  which are all power functions with odd, whole-
number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the
graphs flatten near the origin and become steeper away from the origin.


Chapter 3 Polynomial and Rational Functions 313




Figure 3.23 Odd-power function


These examples illustrate that functions of the form   f (x) = xn   reveal symmetry of one kind or another. First, in Figure
3.22 we see that even functions of the form   f (x) = xn , n even,  are symmetric about the  y- axis. In Figure 3.23 we see
that odd functions of the form   f (x) = xn , n odd,  are symmetric about the origin.
For these odd power functions, as  x  approaches negative infinity,   f (x)  decreases without bound. As  x  approaches
positive infinity,   f (x)  increases without bound. In symbolic form we write


as x → −∞, f (x) → −∞
as x → ∞, f (x) → ∞


The behavior of the graph of a function as the input values get very small (  x → −∞  ) and get very large (  x → ∞  ) is
referred to as the end behavior of the function. We can use words or symbols to describe end behavior.
Figure 3.24 shows the end behavior of power functions in the form   f (x) = kxn  where  n  is a non-negative integer
depending on the power and the constant.


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Figure 3.24


Given a power function   f(x) = kxn  where  n  is a non-negative integer, identify the end behavior.
1. Determine whether the power is even or odd.
2. Determine whether the constant is positive or negative.
3. Use Figure 3.24 to identify the end behavior.


Example 3.22
Identifying the End Behavior of a Power Function


Describe the end behavior of the graph of   f (x) = x8.


Solution
The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As  x  approaches
infinity, the output (value of   f (x)  ) increases without bound. We write as  x → ∞, f (x) → ∞. As  x 
approaches negative infinity, the output increases without bound. In symbolic form, as x → −∞, f (x) → ∞.
We can graphically represent the function as shown in Figure 3.25.


Chapter 3 Polynomial and Rational Functions 315




Figure 3.25


Example 3.23
Identifying the End Behavior of a Power Function.


Describe the end behavior of the graph of   f (x) = − x9.


Solution
The exponent of the power function is 9 (an odd number). Because the coefficient is  –1  (negative), the graph
is the reflection about the  x- axis of the graph of   f (x) = x9. Figure 3.26 shows that as  x  approaches infinity,
the output decreases without bound. As  x  approaches negative infinity, the output increases without bound. In
symbolic form, we would write


as x → −∞, f (x) → ∞
as x → ∞, f (x) → −∞


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Figure 3.26


Analysis
We can check our work by using the table feature on a graphing utility.


x f(x)


–10 1,000,000,000


–5 1,953,125


0 0


5 –1,953,125


10 –1,000,000,000


Table 3.2


We can see from Table 3.2 that, when we substitute very small values for  x,   the output is very large, and when
we substitute very large values for  x,   the output is very small (meaning that it is a very large negative value).


Chapter 3 Polynomial and Rational Functions 317




3.14 Describe in words and symbols the end behavior of   f (x) = − 5x4.


Identifying Polynomial Functions
An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles
in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil
slick by combining two functions. The radius  r  of the spill depends on the number of weeks  w  that have passed. This
relationship is linear.


r(w) = 24 + 8w


We can combine this with the formula for the area  A  of a circle.
A(r) = πr2


Composing these functions gives a formula for the area in terms of weeks.
A(w) = A(r(w))


= A(24 + 8w)


= π(24 + 8w)2


Multiplying gives the formula.
A(w) = 576π + 384πw + 64πw2


This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite
number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised
to a non-negative integer power.


Polynomial Functions
Let  n  be a non-negative integer. A polynomial function is a function that can be written in the form


(3.3)f (x) = an xn + ... + a2 x2 + a1 x + a0
This is called the general form of a polynomial function. Each  ai   is a coefficient and can be any real number. Each
product  ai x i   is a term of a polynomial function.


Example 3.24
Identifying Polynomial Functions


Which of the following are polynomial functions?
f (x) = 2x3 ⋅ 3x + 4


g(x) = − x(x2 − 4)


h(x) = 5 x + 2


Solution


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The first two functions are examples of polynomial functions because they can be written in the form
  f (x) = an xn + ... + a2 x2 + a1 x + a0,   where the powers are non-negative integers and the coefficients are
real numbers.


• f (x)  can be written as   f (x) = 6x4 + 4.
• g(x)  can be written as  g(x) = − x3 + 4x.
• h(x)  cannot be written in this form and is therefore not a polynomial function.


Identifying the Degree and Leading Coefficient of a Polynomial
Function
Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the
variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically
arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of
the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading
term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient
is the coefficient of the leading term.


Terminology of Polynomial Functions
We often rearrange polynomials so that the powers are descending.


When a polynomial is written in this way, we say that it is in general form.


Given a polynomial function, identify the degree and leading coefficient.
1. Find the highest power of  x  to determine the degree function.
2. Identify the term containing the highest power of  x  to find the leading term.
3. Identify the coefficient of the leading term.


Example 3.25
Identifying the Degree and Leading Coefficient of a Polynomial Function


Identify the degree, leading term, and leading coefficient of the following polynomial functions.



      f (x) = 3 + 2x2 − 4x3
       g(t) = 5t5 − 2t3 + 7t
h(p) = 6p − p3 − 2


Chapter 3 Polynomial and Rational Functions 319




3.15


Solution
For the function   f (x),   the highest power of  x  is 3, so the degree is 3. The leading term is the term containing
that degree,  −4x3. The leading coefficient is the coefficient of that term,  −4.
For the function  g(t),   the highest power of  t  is  5,   so the degree is  5. The leading term is the term containing
that degree,  5t5. The leading coefficient is the coefficient of that term,  5.
For the function  h(p),   the highest power of  p  is  3,   so the degree is  3. The leading term is the term containing
that degree,  − p3;   the leading coefficient is the coefficient of that term,  −1.


Identify the degree, leading term, and leading coefficient of the polynomial   f (x) = 4x2 − x6 + 2x − 6.


Identifying End Behavior of Polynomial Functions
Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior,
look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will
grow significantly faster than the other terms as  x  gets very large or very small, so its behavior will dominate the graph.
For any polynomial, the end behavior of the polynomial will match the end behavior of the term of highest degree. See
Table 3.3.


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Polynomial Function LeadingTerm Graph of Polynomial Function


f (x) = 5x4 + 2x3 − x − 4


5x4


f (x) = − 2x6 − x5 + 3x4 + x3


−2x6


Table 3.3


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Polynomial Function LeadingTerm Graph of Polynomial Function


f (x) = 3x5 − 4x4 + 2x2 + 1


3x5


f (x) = − 6x3 + 7x2 + 3x + 1


−6x3


Table 3.3


Example 3.26
Identifying End Behavior and Degree of a Polynomial Function


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Describe the end behavior and determine a possible degree of the polynomial function in Figure 3.27.


Figure 3.27


Solution
As the input values  x  get very large, the output values   f (x)  increase without bound. As the input values  x  get
very small, the output values   f (x)  decrease without bound. We can describe the end behavior symbolically by
writing


as x → −∞, f (x) → −∞
as x → ∞, f (x) → ∞


In words, we could say that as  x  values approach infinity, the function values approach infinity, and as  x  values
approach negative infinity, the function values approach negative infinity.
We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of
the polynomial creating this graph must be odd and the leading coefficient must be positive.


Chapter 3 Polynomial and Rational Functions 323




3.16


3.17


Describe the end behavior, and determine a possible degree of the polynomial function in Figure 3.28.


Figure 3.28


Example 3.27
Identifying End Behavior and Degree of a Polynomial Function


Given the function   f (x) = − 3x2(x − 1)(x + 4),   express the function as a polynomial in general form, and
determine the leading term, degree, and end behavior of the function.


Solution
Obtain the general form by expanding the given expression for   f (x).


f (x) = − 3x2(x − 1)(x + 4)


= − 3x2(x2 + 3x − 4)


= − 3x4 − 9x3 + 12x2


The general form is   f (x) = − 3x4 − 9x3 + 12x2.  The leading term is  − 3x4;   therefore, the degree of the
polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is


as x → −∞, f (x) → −∞
as x → ∞, f (x) → −∞


Given the function   f (x) = 0.2(x − 2)(x + 1)(x − 5),   express the function as a polynomial in general
form and determine the leading term, degree, and end behavior of the function.


Identifying Local Behavior of Polynomial Functions
In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the
function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the
function values change from increasing to decreasing or decreasing to increasing.


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We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects
the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial
is a function, only one output value corresponds to each input value so there can be only one y-intercept  (0, a0). The
x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one
x-intercept. See Figure 3.29.


Figure 3.29


Intercepts and Turning Points of Polynomial Functions
A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing
to increasing. The y-intercept is the point at which the function has an input value of zero. The  x- intercepts are the
points at which the output value is zero.


Given a polynomial function, determine the intercepts.
1. Determine the y-intercept by setting  x = 0  and finding the corresponding output value.
2. Determine the  x- intercepts by solving for the input values that yield an output value of zero.


Example 3.28
Determining the Intercepts of a Polynomial Function


Given the polynomial function   f (x) = (x − 2)(x + 1)(x − 4),  written in factored form for your convenience,
determine the  y- and  x- intercepts.


Chapter 3 Polynomial and Rational Functions 325




Solution
The y-intercept occurs when the input is zero so substitute 0 for  x.


f (0) = (0 − 2)(0 + 1)(0 − 4)
= ( − 2)(1)( − 4)
= 8


The y-intercept is (0, 8).
The x-intercepts occur when the output is zero.


0 = (x − 2)(x + 1)(x − 4)
x − 2 = 0 or x + 1 = 0 or x − 4 = 0
x = 2 or x = − 1 or x = 4


The  x- intercepts are  (2, 0), ( – 1, 0),   and  (4, 0).
We can see these intercepts on the graph of the function shown in Figure 3.30.


Figure 3.30


Example 3.29
Determining the Intercepts of a Polynomial Function with Factoring


Given the polynomial function   f (x) = x4 − 4x2 − 45,   determine the  y- and  x- intercepts.


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3.18


Solution
The y-intercept occurs when the input is zero.


f (0) = (0)4 − 4(0)2 − 45


= − 45


The y-intercept is  (0, − 45).
The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the
polynomial.


f (x) = x4 − 4x2 − 45


             = (x2 − 9)(x2 + 5)
             = (x − 3)(x + 3)(x2 + 5)
0 = (x − 3)(x + 3)(x2 + 5)


x − 3 = 0 or x + 3 = 0 or x2 + 5 = 0
x = 3 or x = − 3 or (no real solution)


The x-intercepts are  (3, 0)  and  ( – 3, 0).
We can see these intercepts on the graph of the function shown in Figure 3.31. We can see that the function is
even because   f (x) = f (−x).


Figure 3.31


Given the polynomial function   f (x) = 2x3 − 6x2 − 20x,   determine the  y- and  x- intercepts.


Comparing Smooth and Continuous Graphs
The degree of a polynomial function helps us to determine the number of  x- intercepts and the number of turning points. A
polynomial function of  nth  degree is the product of  n  factors, so it will have at most  n  roots or zeros, or  x- intercepts. The
graph of the polynomial function of degree  n must have at most  n – 1  turning points. This means the graph has at most
one fewer turning point than the degree of the polynomial or one fewer than the number of factors.


Chapter 3 Polynomial and Rational Functions 327




3.19


A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth
curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The
graphs of polynomial functions are both continuous and smooth.


Intercepts and Turning Points of Polynomials
A polynomial of degree  n will have, at most,  n  x-intercepts and  n − 1  turning points.


Example 3.30
Determining the Number of Intercepts and Turning Points of a Polynomial


Without graphing the function, determine the local behavior of the function by finding the maximum number of
 x- intercepts and turning points for   f (x) = − 3x10 + 4x7 − x4 + 2x3.


Solution
The polynomial has a degree of  10,   so there are at most  n  x-intercepts and at most  n − 1  turning points.


Without graphing the function, determine the maximum number of  x- intercepts and turning points for
  f (x) = 108 − 13x9 − 8x4 + 14x12 + 2x3


Example 3.31
Drawing Conclusions about a Polynomial Function from the Graph


What can we conclude about the polynomial represented by the graph shown in Figure 3.32 based on its
intercepts and turning points?


Figure 3.32


Solution


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3.20


The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 3.33.


Figure 3.33


The graph has 2  x- intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4
or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4.


What can we conclude about the polynomial represented by the graph shown in Figure 3.34 based on its
intercepts and turning points?


Figure 3.34


Example 3.32
Drawing Conclusions about a Polynomial Function from the Factors


Chapter 3 Polynomial and Rational Functions 329




3.21


Given the function   f (x) = − 4x(x + 3)(x − 4),   determine the local behavior.


Solution
The  y- intercept is found by evaluating   f (0).


f (0) = − 4(0)(0 + 3)(0 − 4)
= 0


The  y- intercept is  (0, 0).
The  x- intercepts are found by determining the zeros of the function.


0 = − 4x(x + 3)(x − 4)
x = 0 or x + 3 = 0 or x − 4 = 0
x = 0 or x = − 3 or   x = 4


The  x- intercepts are  (0, 0), ( – 3, 0),   and  (4, 0).
The degree is 3 so the graph has at most 2 turning points.


Given the function   f (x) = 0.2(x − 2)(x + 1)(x − 5),   determine the local behavior.


Access these online resources for additional instruction and practice with power and polynomial functions.
• Find Key Information about a Given Polynomial Function (http://openstaxcollege.org/l/
keyinfopoly)


• End Behavior of a Polynomial Function (http://openstaxcollege.org/l/endbehavior)
• Turning Points and x-intercepts of Polynomial Functions (http://openstaxcollege.org/l/
turningpoints)


• Least Possible Degree of a Polynomial Function (http://openstaxcollege.org/l/
leastposdegree)


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153.


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184.


3.3 EXERCISES
Verbal


Explain the difference between the coefficient of a
power function and its degree.


If a polynomial function is in factored form, what
would be a good first step in order to determine the degree
of the function?


In general, explain the end behavior of a power
function with odd degree if the leading coefficient is
positive.


What is the relationship between the degree of a
polynomial function and the maximum number of turning
points in its graph?


What can we conclude if, in general, the graph of a
polynomial function exhibits the following end behavior?
As  x → −∞,   f (x) → −∞  and as
 x → ∞,   f (x) → −∞. 


Algebraic
For the following exercises, identify the function as a
power function, a polynomial function, or neither.


f (x) = x5


f (x) = ⎛⎝x
2⎞


3


f (x) = x − x4


f (x) = x
2


x2 − 1


f (x) = 2x(x + 2)(x − 1)2


f (x) = 3x + 1


For the following exercises, find the degree and leading
coefficient for the given polynomial.


−3x


7 − 2x2


−2x2 − 3x5 + x − 6


x⎛⎝4 − x
2⎞
⎠(2x + 1)


x2 (2x − 3)2


For the following exercises, determine the end behavior of
the functions.


f (x) = x4


f (x) = x3


f (x) = − x4


f (x) = − x9


f (x) = − 2x4 − 3x2 + x − 1


f (x) = 3x2 + x − 2


f (x) = x2(2x3 − x + 1)


f (x) = (2 − x)7


For the following exercises, find the intercepts of the
functions.


f (t) = 2(t − 1)(t + 2)(t − 3)


g(n) = − 2(3n − 1)(2n + 1)


f (x) = x4 − 16


f (x) = x3 + 27


f (x) = x⎛⎝x
2 − 2x − 8⎞⎠


f (x) = (x + 3)(4x2 − 1)


Graphical
For the following exercises, determine the least possible
degree of the polynomial function shown.


Chapter 3 Polynomial and Rational Functions 331




185.


186.


187.


188.


189.


190.


191.


For the following exercises, determine whether the graph of
the function provided is a graph of a polynomial function.
If so, determine the number of turning points and the least
possible degree for the function.


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192.


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202.


Numeric
For the following exercises, make a table to confirm the end
behavior of the function.


f (x) = − x3


f (x) = x4 − 5x2


f (x) = x2 (1 − x)2


f (x) = (x − 1)(x − 2)(3 − x)


f (x) = x
5


10
− x4


Chapter 3 Polynomial and Rational Functions 333




203.


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222.


Technology
For the following exercises, graph the polynomial functions
using a calculator. Based on the graph, determine the
intercepts and the end behavior.


f (x) = x3(x − 2)


f (x) = x(x − 3)(x + 3)


f (x) = x(14 − 2x)(10 − 2x)


f (x) = x(14 − 2x)(10 − 2x)2


f (x) = x3 − 16x


f (x) = x3 − 27


f (x) = x4 − 81


f (x) = − x3 + x2 + 2x


f (x) = x3 − 2x2 − 15x


f (x) = x3 − 0.01x


Extensions
For the following exercises, use the information about the
graph of a polynomial function to determine the function.
Assume the leading coefficient is 1 or –1. There may be
more than one correct answer.


The  y- intercept is  (0, − 4). The  x- intercepts are
 ( − 2, 0),  (2, 0). Degree is 2.
End behavior:
 as x → −∞,    f (x) → ∞,  as x → ∞,   f (x) → ∞.


The  y- intercept is  (0, 9). The  x- intercepts are
 ( − 3, 0),  (3, 0). Degree is 2.
End behavior:
 as x → −∞,    f (x) → −∞,  as x → ∞,   f (x) → −∞.


The  y- intercept is  (0, 0). The  x- intercepts are
 (0, 0),  (2, 0). Degree is 3.
End behavior:
 as x → −∞,    f (x) → −∞,  as x → ∞,   f (x) → ∞.


The  y- intercept is  (0, 1). The  x- intercept is
 (1, 0). Degree is 3.


End behavior:
 as x → −∞,    f (x) → ∞,  as x → ∞,   f (x) → −∞.


The  y- intercept is  (0, 1). There is no  x- intercept.
Degree is 4.
End behavior:
 as x → −∞,    f (x) → ∞,  as x → ∞,   f (x) → ∞.


Real-World Applications
For the following exercises, use the written statements to
construct a polynomial function that represents the required
information.


An oil slick is expanding as a circle. The radius of the
circle is increasing at the rate of 20 meters per day. Express
the area of the circle as a function of  d,   the number of
days elapsed.


A cube has an edge of 3 feet. The edge is increasing at
the rate of 2 feet per minute. Express the volume of the
cube as a function of  m,   the number of minutes elapsed.


A rectangle has a length of 10 inches and a width of 6
inches. If the length is increased by  x  inches and the width
increased by twice that amount, express the area of the
rectangle as a function of  x.


An open box is to be constructed by cutting out square
corners of  x- inch sides from a piece of cardboard 8 inches
by 8 inches and then folding up the sides. Express the
volume of the box as a function of  x.


A rectangle is twice as long as it is wide. Squares of
side 2 feet are cut out from each corner. Then the sides are
folded up to make an open box. Express the volume of the
box as a function of the width ( x ).


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3.4 | Graphs of Polynomial Functions
Learning Objectives


In this section, you will:
3.4.1 Recognize characteristics of graphs of polynomial functions.
3.4.2 Use factoring to find zeros of polynomial functions.
3.4.3 Identify zeros and their multiplicities.
3.4.4 Determine end behavior.
3.4.5 Understand the relationship between degree and turning points.
3.4.6 Graph polynomial functions.
3.4.7 Use the Intermediate Value Theorem.


The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 3.4.


Year 2006 2007 2008 2009 2010 2011 2012 2013


Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1


Table 3.4


The revenue can be modeled by the polynomial function
R(t) = − 0.037t4 + 1.414t3 − 19.777t2 + 118.696t − 205.332


where  R  represents the revenue in millions of dollars and  t  represents the year, with  t = 6  corresponding to 2006. Over
which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing?
These questions, along with many others, can be answered by examining the graph of the polynomial function. We have
already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local
behavior of polynomials in general.
Recognizing Characteristics of Graphs of Polynomial Functions
Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs
are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called
continuous. Figure 3.35 shows a graph that represents a polynomial function and a graph that represents a function that is
not a polynomial.


Chapter 3 Polynomial and Rational Functions 335




Figure 3.35


Example 3.33
Recognizing Polynomial Functions


Which of the graphs in Figure 3.36 represents a polynomial function?


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Figure 3.36


Solution
The graphs of   f   and  h  are graphs of polynomial functions. They are smooth and continuous.
The graphs of  g  and  k  are graphs of functions that are not polynomials. The graph of function  g  has a sharp
corner. The graph of function  k  is not continuous.


Do all polynomial functions have as their domain all real numbers?
Yes. Any real number is a valid input for a polynomial function.


Using Factoring to Find Zeros of Polynomial Functions
Recall that if   f   is a polynomial function, the values of  x  for which   f (x) = 0  are called zeros of   f .  If the equation of the
polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.
We can use this method to find  x- intercepts because at the  x- intercepts we find the input values when the output value
is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively
simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to
remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three
cases in this section:


Chapter 3 Polynomial and Rational Functions 337




1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
2. The polynomial is given in factored form.
3. Technology is used to determine the intercepts.


Given a polynomial function   f ,   find the x-intercepts by factoring.
1. Set   f (x) = 0.
2. If the polynomial function is not given in factored form:


a. Factor out any common monomial factors.
b. Factor any factorable binomials or trinomials.


3. Set each factor equal to zero and solve to find the  x- intercepts.


Example 3.34
Finding the x-Intercepts of a Polynomial Function by Factoring


Find the x-intercepts of   f (x) = x6 − 3x4 + 2x2.


Solution
We can attempt to factor this polynomial to find solutions for   f (x) = 0.


x6 − 3x4 + 2x2 = 0 Factor out the greatest


common factor.


x2(x4 − 3x2 + 2) = 0 Factor the trinomial.


x2(x2 − 1)(x2 − 2) = 0 Set each factor equal to zero.


(x2 − 1) = 0 (x2 − 2) = 0


x2 = 0 or x2 = 1 or x2 = 2
x = 0 x = ± 1 x = ± 2


This gives us five  x- intercepts:  (0, 0), (1, 0), ( − 1, 0), ( 2, 0),   and  ( − 2, 0).  See Figure 3.37. We can
see that this is an even function.


Figure 3.37


Example 3.35


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Finding the x-Intercepts of a Polynomial Function by Factoring


Find the  x- intercepts of   f (x) = x3 − 5x2 − x + 5.


Solution
Find solutions for   f (x) = 0  by factoring.


x3 − 5x2 − x + 5 = 0 Factor by grouping.


x2(x − 5) − (x − 5) = 0 Factor out the common factor.


(x2 − 1)(x − 5) = 0 Factor the diffe ence of squares.


(x + 1)(x − 1)(x − 5) = 0 Set each factor equal to zero.


x + 1 = 0 or x − 1 = 0 or x − 5 = 0
        x = − 1         x = 1          x = 5


There are three  x- intercepts:  ( − 1, 0), (1, 0),   and  (5, 0).  See Figure 3.38.


Figure 3.38


Example 3.36
Finding the y- and x-Intercepts of a Polynomial in Factored Form


Find the  y- and x-intercepts of  g(x) = (x − 2)2(2x + 3).


Solution
The y-intercept can be found by evaluating  g(0).


                                   g(0) = (0 − 2)2(2(0) + 3)
= 12


Chapter 3 Polynomial and Rational Functions 339




So the y-intercept is  (0, 12).
The x-intercepts can be found by solving  g(x) = 0.


(x − 2)2(2x + 3) = 0


(x − 2)2 = 0 (2x + 3) = 0


x − 2 = 0 or x = − 3
2


x = 2


So the  x- intercepts are  (2, 0)  and  ⎛⎝−32, 0

⎠.


Analysis
We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as
shown in Figure 3.39.


Figure 3.39


Example 3.37
Finding the x-Intercepts of a Polynomial Function Using a Graph


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3.22


Find the  x- intercepts of  h(x) = x3 + 4x2 + x − 6.


Solution
This polynomial is not in factored form, has no common factors, and does not appear to be factorable using
techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that
some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.
Looking at the graph of this function, as shown in Figure 3.40, it appears that there are x-intercepts at
 x = −3, −2,   and  1.


Figure 3.40


We can check whether these are correct by substituting these values for  x  and verifying that
h( − 3) = h( − 2) = h(1) = 0.


Since  h(x) = x3 + 4x2 + x − 6,  we have:
h( − 3) = ( − 3)3 + 4( − 3)2 + ( − 3) − 6 = − 27 + 36 − 3 − 6 = 0


h( − 2) = ( − 2)3 + 4( − 2)2 + ( − 2) − 6 = − 8 + 16 − 2 − 6 = 0


h(1) = (1)3 + 4(1)2 + (1) − 6 = 1 + 4 + 1 − 6 = 0


Each  x- intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now
write the polynomial in factored form.


h(x) = x3 + 4x2 + x − 6
= (x + 3)(x + 2)(x − 1)


Find the  y- and x-intercepts of the function   f (x) = x4 − 19x2 + 30x.


Chapter 3 Polynomial and Rational Functions 341




Identifying Zeros and Their Multiplicities
Graphs behave differently at various  x- intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept.
Other times, the graph will touch the horizontal axis and bounce off.
Suppose, for example, we graph the function


f (x) = (x + 3)(x − 2)2 (x + 1)3.


Notice in Figure 3.41 that the behavior of the function at each of the  x- intercepts is different.


Figure 3.41 Identifying the behavior of the graph at an x-
intercept by examining the multiplicity of the zero.


The  x- intercept  x = −3  is the solution of equation  (x + 3) = 0. The graph passes directly through the  x- intercept at
 x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly
through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.
The  x- intercept  x = 2  is the repeated solution of equation  (x − 2)2 = 0. The graph touches the axis at the intercept and
changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces
off of the horizontal axis at the intercept.


(x − 2)2 = (x − 2)(x − 2)


The factor is repeated, that is, the factor  (x − 2)  appears twice. The number of times a given factor appears in the factored
form of the equation of a polynomial is called themultiplicity. The zero associated with this factor,  x = 2,   has multiplicity
2 because the factor  (x − 2)  occurs twice.
The  x- intercept  x = − 1  is the repeated solution of factor  (x + 1)3 = 0. The graph passes through the axis at the
intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a
cubic—with the same S-shape near the intercept as the toolkit function   f (x) = x3. We call this a triple zero, or a zero with
multiplicity 3.
For zeros with even multiplicities, the graphs touch or are tangent to the  x- axis. For zeros with odd multiplicities, the
graphs cross or intersect the  x- axis. See Figure 3.42 for examples of graphs of polynomial functions with multiplicity 1,
2, and 3.


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Figure 3.42


For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each
increasing even power, the graph will appear flatter as it approaches and leaves the  x- axis.
For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd
power, the graph will appear flatter as it approaches and leaves the  x- axis.


Graphical Behavior of Polynomials at  x- Intercepts
If a polynomial contains a factor of the form  (x − h) p,   the behavior near the  x- intercept  h  is determined by the
power  p. We say that  x = h  is a zero of multiplicity  p.
The graph of a polynomial function will touch the  x- axis at zeros with even multiplicities. The graph will cross the
x-axis at zeros with odd multiplicities.
The sum of the multiplicities is the degree of the polynomial function.


Given a graph of a polynomial function of degree  n,   identify the zeros and their multiplicities.
1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.
2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity.
3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.
4. The sum of the multiplicities is  n.


Example 3.38
Identifying Zeros and Their Multiplicities


Use the graph of the function of degree 6 in Figure 3.43 to identify the zeros of the function and their possible
multiplicities.


Chapter 3 Polynomial and Rational Functions 343




3.23


Figure 3.43


Solution
The polynomial function is of degree  n. The sum of the multiplicities must be  n.
Starting from the left, the first zero occurs at  x = −3. The graph touches the x-axis, so the multiplicity of the zero
must be even. The zero of  −3  has multiplicity  2.
The next zero occurs at  x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity
1.
The last zero occurs at  x = 4. The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know
that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6.


Use the graph of the function of degree 5 in Figure 3.44 to identify the zeros of the function and their
multiplicities.


Figure 3.44


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Determining End Behavior
As we have already learned, the behavior of a graph of a polynomial function of the form


f (x) = an x
n + an − 1 x


n − 1 + ... + a1 x + a0


will either ultimately rise or fall as  x  increases without bound and will either rise or fall as  x  decreases without bound.
This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true
for very small inputs, say –100 or –1,000.
Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations,
when the leading term of a polynomial function,  an xn,   is an even power function, as  x  increases or decreases without
bound,   f (x)  increases without bound. When the leading term is an odd power function, as  x  decreases without bound,
  f (x)  also decreases without bound; as  x  increases without bound,   f (x)  also increases without bound. If the leading term
is negative, it will change the direction of the end behavior. Figure 3.45 summarizes all four cases.


Figure 3.45


Understanding the Relationship between Degree and Turning Points
In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning
point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).


Chapter 3 Polynomial and Rational Functions 345




Look at the graph of the polynomial function   f (x) = x4 − x3 − 4x2 + 4x  in Figure 3.46. The graph has three turning
points.


Figure 3.46


This function   f   is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a
polynomial function is always one less than the degree of the function.


Interpreting Turning Points
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or
decreasing to increasing (falling to rising).
A polynomial of degree  n will have at most  n − 1  turning points.


Example 3.39
Finding the Maximum Number of Turning Points Using the Degree of a Polynomial
Function


Find the maximum number of turning points of each polynomial function.
a. f (x) = − x3 + 4x5 − 3x2 + + 1


b. f (x) = − (x − 1)2 ⎛⎝1 + 2x2⎞⎠


Solution
a. f (x) = − x + 4x5 − 3x2 + + 1
First, rewrite the polynomial function in descending order:   f (x) = 4x5 − x3 − 3x2 + + 1
Identify the degree of the polynomial function. This polynomial function is of degree 5.
The maximum number of turning points is  5 − 1 = 4.


b. f (x) = − (x − 1)2 ⎛⎝1 + 2x2⎞⎠
First, identify the leading term of the polynomial function if the function were expanded.


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Then, identify the degree of the polynomial function. This polynomial function is of degree 4.
The maximum number of turning points is  4 − 1 = 3.


Graphing Polynomial Functions
We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial
functions. Let us put this all together and look at the steps required to graph polynomial functions.


Given a polynomial function, sketch the graph.
1. Find the intercepts.
2. Check for symmetry. If the function is an even function, its graph is symmetrical about the  y- axis, that
is,   f (−x) = f (x).  If a function is an odd function, its graph is symmetrical about the origin, that is,
  f (−x) = − f (x).


3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the  x- intercepts.
4. Determine the end behavior by examining the leading term.
5. Use the end behavior and the behavior at the intercepts to sketch a graph.
6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
7. Optionally, use technology to check the graph.


Example 3.40
Sketching the Graph of a Polynomial Function


Sketch a graph of   f (x) = −2(x + 3)2(x − 5).


Solution
This graph has two  x- intercepts. At  x = −3,   the factor is squared, indicating a multiplicity of 2. The graph
will bounce at this  x- intercept. At  x = 5,   the function has a multiplicity of one, indicating the graph will cross
through the axis at this intercept.
The y-intercept is found by evaluating   f (0).


f (0) = − 2(0 + 3)2(0 − 5)


= − 2 ⋅ 9 ⋅ ( − 5)
= 90


The  y- intercept is  (0, 90).


Chapter 3 Polynomial and Rational Functions 347




Additionally, we can see the leading term, if this polynomial were multiplied out, would be  − 2x3,   so the end
behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the
outputs increasing as the inputs approach negative infinity. See Figure 3.47.


Figure 3.47


To sketch this, we consider that:
• As  x → −∞  the function   f (x) → ∞,   so we know the graph starts in the second quadrant and is
decreasing toward the  x- axis.


• Since   f (−x) = −2(−x + 3)2 (−x – 5)  is not equal to   f (x),   the graph does not display symmetry.
• At  (−3, 0),   the graph bounces off of the  x- axis, so the function must start increasing.
At  (0, 90),   the graph crosses the  y- axis at the  y- intercept. See Figure 3.48.


Figure 3.48


Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because
the graph passes through the next intercept at  (5, 0).  See Figure 3.49.


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3.24


Figure 3.49


As  x → ∞  the function   f (x) → −∞,   so we know the graph continues to decrease, and we can stop drawing
the graph in the fourth quadrant.
Using technology, we can create the graph for the polynomial function, shown in Figure 3.50, and verify that
the resulting graph looks like our sketch in Figure 3.49.


Figure 3.50 The complete graph of the polynomial function
  f (x) = − 2(x + 3)2(x − 5)


Sketch a graph of   f (x) = 1
4
x(x − 1)4 (x + 3)3.


Using the Intermediate Value Theorem
In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of
the x-axis, we can confirm that there is a zero between them. Consider a polynomial function   f  whose graph is smooth
and continuous. The Intermediate Value Theorem states that for two numbers  a  and  b  in the domain of   f , if  a < b  and
f (a) ≠ f (b), then the function   f   takes on every value between   f (a)  and   f (b).  We can apply this theorem to a special


Chapter 3 Polynomial and Rational Functions 349




case that is useful in graphing polynomial functions. If a point on the graph of a continuous function   f   at  x = a  lies above
the  x- axis and another point at  x = b  lies below the  x- axis, there must exist a third point between  x = a  and  x = b 
where the graph crosses the  x- axis. Call this point  ⎛⎝c, f (c)⎞⎠. This means that we are assured there is a solution  c where
f (c) = 0.


In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to
a positive value, the function must cross the  x- axis. Figure 3.51 shows that there is a zero between  a  and  b.


Figure 3.51 Using the Intermediate Value Theorem to show
there exists a zero.


Intermediate Value Theorem
Let   f   be a polynomial function. The Intermediate Value Theorem states that if   f (a)  and   f (b)  have opposite
signs, then there exists at least one value  c  between  a  and  b  for which   f (c) = 0.


Example 3.41
Using the Intermediate Value Theorem


Show that the function   f (x) = x3 − 5x2 + 3x + 6  has at least two real zeros between  x = 1  and  x = 4.


Solution
As a start, evaluate   f (x)  at the integer values  x = 1, 2, 3, and4.  See Table 3.5.


x 1 2 3 4


f(x) 5 0 –3 2


Table 3.5


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3.25


We see that one zero occurs at  x = 2. Also, since   f (3)  is negative and   f (4)  is positive, by the Intermediate
Value Theorem, there must be at least one real zero between 3 and 4.
We have shown that there are at least two real zeros between  x = 1  and  x = 4.


Analysis
We can also see on the graph of the function in Figure 3.52 that there are two real zeros between  x = 1  and
 x = 4.


Figure 3.52


Show that the function   f (x) = 7x5 − 9x4 − x2   has at least one real zero between  x = 1  and  x = 2.


Writing Formulas for Polynomial Functions
Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because
a polynomial function written in factored form will have an  x- intercept where each factor is equal to zero, we can form a
function that will pass through a set of  x- intercepts by introducing a corresponding set of factors.


Factored Form of Polynomials
If a polynomial of lowest degree  p  has horizontal intercepts at  x = x1, x2, … , xn,   then the polynomial can be
written in the factored form:   f (x) = a(x − x1)


p1 (x − x2)
p2 ⋯ (x − xn)


pn   where the powers  pi   on each factor can
be determined by the behavior of the graph at the corresponding intercept, and the stretch factor  a  can be determined
given a value of the function other than the x-intercept.


Chapter 3 Polynomial and Rational Functions 351




Given a graph of a polynomial function, write a formula for the function.
1. Identify the x-intercepts of the graph to find the factors of the polynomial.
2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor.
3. Find the polynomial of least degree containing all the factors found in the previous step.
4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor.


Example 3.42
Writing a Formula for a Polynomial Function from the Graph


Write a formula for the polynomial function shown in Figure 3.53.


Figure 3.53


Solution
This graph has three  x- intercepts:  x = −3, 2,   and  5. The  y- intercept is located at  (0, 2). At  x = −3  and
 x = 5,   the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will
be linear. At  x = 2,   the graph bounces at the intercept, suggesting the corresponding factor of the polynomial
will be second degree (quadratic). Together, this gives us


f (x) = a(x + 3)(x − 2)2(x − 5)


To determine the stretch factor, we utilize another point on the graph. We will use the  y- intercept  (0, – 2),   to
solve for  a.


f (0) = a(0 + 3)(0 − 2)2(0 − 5)


− 2 = a(0 + 3)(0 − 2)2(0 − 5)
− 2 = − 60a


a = 1
30


The graphed polynomial appears to represent the function   f (x) = 1
30


(x + 3)(x − 2)2(x − 5).


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3.26 Given the graph shown in Figure 3.54, write a formula for the function shown.


Figure 3.54


Using Local and Global Extrema
With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex.
For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even
then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning
points using technology to generate a graph.
Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on
the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also
referred to as the absolute maximum and absolute minimum values of the function.


Local and Global Extrema
A local maximum or local minimum at  x = a  (sometimes called the relative maximum or minimum, respectively)
is the output at the highest or lowest point on the graph in an open interval around  x = a.  If a function has a local
maximum at  a,   then   f (a) ≥ f (x)  for all  x  in an open interval around  x = a.  If a function has a local minimum at
 a,   then   f (a) ≤ f (x)  for all  x  in an open interval around  x = a.
A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a
global maximum at  a,   then   f (a) ≥ f (x)  for all  x.  If a function has a global minimum at  a,   then   f (a) ≤ f (x)  for
all  x.
We can see the difference between local and global extrema in Figure 3.55.


Chapter 3 Polynomial and Rational Functions 353




Figure 3.55


Do all polynomial functions have a global minimum or maximum?
No. Only polynomial functions of even degree have a global minimum or maximum. For example,   f (x) = x  has
neither a global maximum nor a global minimum.


Example 3.43
Using Local Extrema to Solve Applications


An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic
then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the
box.


Solution
We will start this problem by drawing a picture like that in Figure 3.56, labeling the width of the cut-out squares
with a variable,  w.


Figure 3.56


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Notice that after a square is cut out from each end, it leaves a  (14 − 2w)  cm by  (20 − 2w)  cm rectangle for the
base of the box, and the box will be  w  cm tall. This gives the volume


V(w) = (20 − 2w)(14 − 2w)w


= 280w − 68w2 + 4w3


Notice, since the factors are  w,    20 – 2w  and  14 – 2w,   the three zeros are 10, 7, and 0, respectively. Because a
height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting
off two squares, so values  w may take on are greater than zero or less than 7. This means we will restrict the
domain of this function to  0 < w < 7. Using technology to sketch the graph of  V(w)  on this reasonable domain,
we get a graph like that in Figure 3.57. We can use this graph to estimate the maximum value for the volume,
restricted to values for  w  that are reasonable for this problem—values from 0 to 7.


Figure 3.57


From this graph, we turn our focus to only the portion on the reasonable domain,  [0, 7]. We can estimate the
maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side.
To improve this estimate, we could use advanced features of our technology, if available, or simply change our
window to zoom in on our graph to produce Figure 3.58.


Chapter 3 Polynomial and Rational Functions 355




3.27


Figure 3.58


From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the
squares measure approximately 2.7 cm on each side.


Use technology to find the maximum and minimum values on the interval  [−1, 4]  of the function
  f (x) = − 0.2(x − 2)3 (x + 1)2(x − 4).


Access the following online resource for additional instruction and practice with graphing polynomial functions.
• Intermediate Value Theorem (http://openstaxcollege.org/l/ivt)


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223.


224.


225.


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227.


228.


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234.


235.


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257.


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260.


261.


3.4 EXERCISES
Verbal


What is the difference between an  x- intercept and a
zero of a polynomial function   f ?  


If a polynomial function of degree  n  has  n  distinct
zeros, what do you know about the graph of the function?


Explain how the Intermediate Value Theorem can
assist us in finding a zero of a function.


Explain how the factored form of the polynomial
helps us in graphing it.


If the graph of a polynomial just touches the  x- axis
and then changes direction, what can we conclude about the
factored form of the polynomial?


Algebraic
For the following exercises, find the  x- or t-intercepts of
the polynomial functions.


 C(t) = 2(t − 4)(t + 1)(t − 6)


 C(t) = 3(t + 2)(t − 3)(t + 5)


 C(t) = 4t(t − 2)2(t + 1)


 C(t) = 2t(t − 3)(t + 1)2


 C(t) = 2t4 − 8t3 + 6t2


 C(t) = 4t4 + 12t3 − 40t2


  f (x) = x4 − x2


  f (x) = x3 + x2 − 20x


f (x) = x3 + 6x2 − 7x


f (x) = x3 + x2 − 4x − 4


f (x) = x3 + 2x2 − 9x − 18


f (x) = 2x3 − x2 − 8x + 4


f (x) = x6 − 7x3 − 8


f (x) = 2x4 + 6x2 − 8


f (x) = x3 − 3x2 − x + 3


f (x) = x6 − 2x4 − 3x2


f (x) = x6 − 3x4 − 4x2


f (x) = x5 − 5x3 + 4x


For the following exercises, use the Intermediate Value
Theorem to confirm that the given polynomial has at least
one zero within the given interval.


f (x) = x3 − 9x,   between  x = − 4  and  x = − 2.


f (x) = x3 − 9x,   between  x = 2  and  x = 4.


f (x) = x5 − 2x,   between  x = 1  and  x = 2.


f (x) = − x4 + 4,   between  x = 1  and  x = 3 .


f (x) = − 2x3 − x,   between  x = – 1  and  x = 1.


f (x) = x3 − 100x + 2,   between  x = 0.01  and
 x = 0.1
For the following exercises, find the zeros and give the
multiplicity of each.


f (x) = (x + 2)3 (x − 3)2


f (x) = x2 (2x + 3)5 (x − 4)2


f (x) = x3 (x − 1)3 (x + 2)


f (x) = x2 ⎛⎝x
2 + 4x + 4⎞⎠


f (x) = (2x + 1)3 ⎛⎝9x
2 − 6x + 1⎞⎠


f (x) = (3x + 2)5 ⎛⎝x
2 − 10x + 25⎞⎠


f (x) = x⎛⎝4x
2 − 12x + 9⎞⎠



⎝x


2 + 8x + 16⎞⎠


f (x) = x6 − x5 − 2x4


f (x) = 3x4 + 6x3 + 3x2


f (x) = 4x5 − 12x4 + 9x3


Chapter 3 Polynomial and Rational Functions 357




262.


263.


264.


265.


266.


267.


268.


269.


270.


271.


272.


273.


274.


f (x) = 2x4 ⎛⎝x
3 − 4x2 + 4x⎞⎠


f (x) = 4x4 ⎛⎝9x
4 − 12x3 + 4x2⎞⎠


Graphical
For the following exercises, graph the polynomial
functions. Note  x- and  y- intercepts, multiplicity, and end
behavior.


f (x) = (x + 3)2(x − 2)


g(x) = (x + 4)(x − 1)2


h(x) = (x − 1)3 (x + 3)2


k(x) = (x − 3)3 (x − 2)2


m(x) = − 2x(x − 1)(x + 3)


n(x) = − 3x(x + 2)(x − 4)


For the following exercises, use the graphs to write the
formula for a polynomial function of least degree.


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275.


276.


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279.


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283.


284.


For the following exercises, use the graph to identify zeros
and multiplicity.


For the following exercises, use the given information
about the polynomial graph to write the equation.


Degree 3. Zeros at  x = –2,  x = 1,  and  x = 3.  y-
intercept at  (0, – 4).


Degree 3. Zeros at  x = –5,  x = –2, and  x = 1.  y-
intercept at  (0, 6)


Degree 5. Roots of multiplicity 2 at  x = 3  and
 x = 1  , and a root of multiplicity 1 at  x = –3.  y-intercept
at  (0, 9)


Degree 4. Root of multiplicity 2 at  x = 4,  and a roots
of multiplicity 1 at  x = 1  and  x = –2.  y-intercept at
 (0, –3).


Degree 5. Double zero at  x = 1,   and triple zero at
 x = 3.  Passes through the point  (2, 15).


Degree 3. Zeros at  x = 4,  x = 3, and  x = 2.  y-
intercept at  (0, −24).


Chapter 3 Polynomial and Rational Functions 359




285.


286.


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288.


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292.


293.


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295.


296.


297.


298.


299.


Degree 3. Zeros at  x = −3,  x = −2  and  x = 1. 
y-intercept at  (0, 12).


Degree 5. Roots of multiplicity 2 at  x = −3  and
 x = 2  and a root of multiplicity 1 at  x = −2.
y-intercept at  (0, 4).


Degree 4. Roots of multiplicity 2 at  x = 1
2
  and roots


of multiplicity 1 at  x = 6  and  x = −2.
y-intercept at  (0,18).


Double zero at  x = −3  and triple zero at  x = 0. 
Passes through the point  (1, 32).


Technology
For the following exercises, use a calculator to approximate
local minima and maxima or the global minimum and
maximum.


f (x) = x3 − x − 1


f (x) = 2x3 − 3x − 1


f (x) = x4 + x


f (x) = − x4 + 3x − 2


f (x) = x4 − x3 + 1


Extensions
For the following exercises, use the graphs to write a
polynomial function of least degree.


Real-World Applications
For the following exercises, write the polynomial function
that models the given situation.


A rectangle has a length of 10 units and a width of 8
units. Squares of  x  by  x  units are cut out of each corner,
and then the sides are folded up to create an open box.
Express the volume of the box as a polynomial function in
terms of  x.


Consider the same rectangle of the preceding
problem. Squares of  2x  by  2x  units are cut out of each
corner. Express the volume of the box as a polynomial in
terms of  x.


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300.


301.


A square has sides of 12 units. Squares  x + 1  by  x + 1 
units are cut out of each corner, and then the sides are
folded up to create an open box. Express the volume of the
box as a function in terms of  x.


A cylinder has a radius of  x + 2  units and a height of
3 units greater. Express the volume of the cylinder as a
polynomial function.


A right circular cone has a radius of  3x + 6  and a
height 3 units less. Express the volume of the cone as a
polynomial function. The volume of a cone is  V = 1


3
πr2h 


for radius  r  and height  h.


Chapter 3 Polynomial and Rational Functions 361




3.5 | Dividing Polynomials
Learning Objectives


In this section, you will:
3.5.1 Use long division to divide polynomials.
3.5.2 Use synthetic division to divide polynomials.


Figure 3.59 Lincoln Memorial, Washington, D.C. (credit:
Ron Cogswell, Flickr)


The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width
40 m, and height 30 m.[1]We can easily find the volume using elementary geometry.


V = l ⋅ w ⋅ h
= 61.5 ⋅ 40 ⋅ 30
= 73,800


So the volume is 73,800 cubic meters  (m ³ ).  Suppose we knew the volume, length, and width. We could divide to find the
height.


h = V
l ⋅ w


= 73, 800
61.5 ⋅ 40


= 30


As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing
dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example,
suppose the volume of a rectangular solid is given by the polynomial  3x4 − 3x3 − 33x2 + 54x.  The length of the solid is
given by  3x;   the width is given by  x − 2.  To find the height of the solid, we can use polynomial division, which is the
focus of this section.
Using Long Division to Divide Polynomials
We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend
that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat.
For example, let’s divide 178 by 3 using long division.


1. National Park Service. "Lincoln Memorial Building Statistics." http://www.nps.gov/linc/historyculture/lincoln-
memorial-building-statistics.htm. Accessed 4/3/2014


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Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check
division in elementary arithmetic.


dividend = (divisor ⋅ quotient) + remainder


              178 = (3 ⋅ 59) + 1
                        = 177 + 1
                        = 178


We call this the Division Algorithm and will discuss it more formally after looking at an example.
Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write
a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial
division correspond to the digits (and place values) of the whole number division. This method allows us to divide two
polynomials. For example, if we were to divide  2x3 − 3x2 + 4x + 5  by  x + 2  using the long division algorithm, it would


look like this:
We have found


2x3 − 3x2 + 4x + 5
x + 2


= 2x2 − 7x + 18 − 31
x + 2


or
2x3 − 3x2 + 4x + 5 = (x + 2)(2x2 − 7x + 18) − 31


We can identify the dividend, the divisor, the quotient, and the remainder.


Chapter 3 Polynomial and Rational Functions 363




Writing the result in this manner illustrates the Division Algorithm.


The Division Algorithm
The Division Algorithm states that, given a polynomial dividend   f (x)  and a non-zero polynomial divisor  d(x) 
where the degree of  d(x)  is less than or equal to the degree of   f (x),   there exist unique polynomials  q(x)  and  r(x) 
such that


(3.4)f (x) = d(x)q(x) + r(x)
q(x)  is the quotient and  r(x)  is the remainder. The remainder is either equal to zero or has degree strictly less than
 d(x). 
If  r(x) = 0,   then  d(x)  divides evenly into   f (x).  This means that, in this case, both  d(x)  and  q(x)  are factors of
  f (x). 


Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.
1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term
of the divisor.


3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.


Example 3.44
Using Long Division to Divide a Second-Degree Polynomial


Divide  5x2 + 3x − 2  by  x + 1.


Solution


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The quotient is  5x − 2.  The remainder is 0. We write the result as
5x2 + 3x − 2


x + 1
= 5x − 2


or
5x2 + 3x − 2 = (x + 1)(5x − 2)


Analysis
This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and
that the divisor is a factor of the dividend.


Example 3.45
Using Long Division to Divide a Third-Degree Polynomial


Divide  6x3 + 11x2 − 31x + 15  by  3x − 2. 


Solution


Chapter 3 Polynomial and Rational Functions 365




3.28


There is a remainder of 1. We can express the result as:
6x3 + 11x2 − 31x + 15


3x − 2
= 2x2 + 5x − 7 + 1


3x − 2


Analysis
We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.


(3x − 2)(2x2 + 5x − 7) + 1 = 6x3 + 11x2 − 31x + 15


Notice, as we write our result,
• the dividend is  6x3 + 11x2 − 31x + 15 
• the divisor is  3x − 2
• the quotient is  2x2 + 5x − 7
• the remainder is  1


Divide  16x3 − 12x2 + 20x − 3  by  4x + 5. 


Using Synthetic Division to Divide Polynomials
As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a
shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.
To illustrate the process, recall the example at the beginning of the section.
Divide  2x3 − 3x2 + 4x + 5  by  x + 2  using the long division algorithm.
The final form of the process looked like this:


There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns
under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.


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Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill
any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting
the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the
leading coefficient.


We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The
bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the
quotient is  2x ² – 7x + 18  and the remainder is  –31.  The process will be made more clear in Example 3.46.


Synthetic Division
Synthetic division is a shortcut that can be used when the divisor is a binomial in the form  x − k.  In synthetic
division, only the coefficients are used in the division process.


Given two polynomials, use synthetic division to divide.
1. Write  k  for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by  k.  Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by  k.  Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has
degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so
on.


Example 3.46
Using Synthetic Division to Divide a Second-Degree Polynomial


Use synthetic division to divide  5x2 − 3x − 36  by  x − 3. 


Solution
Begin by setting up the synthetic division. Write  k  and the coefficients.


Bring down the lead coefficient. Multiply the lead coefficient by  k. 


Continue by adding the numbers in the second column. Multiply the resulting number by  k.  Write the result in
the next column. Then add the numbers in the third column.


Chapter 3 Polynomial and Rational Functions 367




The result is  5x + 12.  The remainder is 0. So  x − 3  is a factor of the original polynomial.


Analysis
Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the
remainder.
(x − 3)(5x + 12) + 0 = 5x2 − 3x − 36


Example 3.47
Using Synthetic Division to Divide a Third-Degree Polynomial


Use synthetic division to divide  4x3 + 10x2 − 6x − 20  by  x + 2. 


Solution
The binomial divisor is  x + 2  so  k = − 2.  Add each column, multiply the result by –2, and repeat until the last
column is reached.


The result is  4x2 + 2x − 10.  The remainder is 0. Thus,  x + 2  is a factor of  4x3 + 10x2 − 6x − 20. 


Analysis
The graph of the polynomial function   f (x) = 4x3 + 10x2 − 6x − 20  in Figure 3.60 shows a zero at
 x = k = −2.  This confirms that  x + 2  is a factor of  4x3 + 10x2 − 6x − 20. 


368 Chapter 3 Polynomial and Rational Functions


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Figure 3.60


Example 3.48
Using Synthetic Division to Divide a Fourth-Degree Polynomial


Use synthetic division to divide  − 9x4 + 10x3 + 7x2 − 6  by  x − 1. 


Solution
Notice there is no x-term. We will use a zero as the coefficient for that term.


Chapter 3 Polynomial and Rational Functions 369




3.29


The result is  − 9x3 + x2 + 8x + 8 + 2
x − 1


.


Use synthetic division to divide  3x4 + 18x3 − 3x + 40  by  x + 7.


Using Polynomial Division to Solve Application Problems
Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We
looked at an application at the beginning of this section. Now we will solve that problem in the following example.


Example 3.49
Using Polynomial Division in an Application Problem


The volume of a rectangular solid is given by the polynomial  3x4 − 3x3 − 33x2 + 54x.  The length of the solid
is given by  3x  and the width is given by  x − 2.  Find the height of the solid.


Solution
There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by
the expressions for the length and width. Let us create a sketch as in Figure 3.61.


Figure 3.61


We can now write an equation by substituting the known values into the formula for the volume of a rectangular
solid.


              V = l ⋅ w ⋅ h
3x4 − 3x3 − 33x2 + 54x = 3x ⋅ (x − 2) ⋅ h


To solve for  h,   first divide both sides by  3x.
3x ⋅ (x − 2) ⋅ h


3x
= 3x


4 − 3x3 − 33x2 + 54x
3x


 (x − 2)h = x3 − x2 − 11x + 18
Now solve for  h  using synthetic division.


h = x
3 − x2 − 11x + 18


x − 2


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3.30


21 −1 −11 18
2 2 −18


1 1 − 9 0


The quotient is  x2 + x − 9  and the remainder is 0. The height of the solid is  x2 + x − 9.


The area of a rectangle is given by  3x3 + 14x2 − 23x + 6.  The width of the rectangle is given by
 x + 6.  Find an expression for the length of the rectangle.


Access these online resources for additional instruction and practice with polynomial division.
• Dividing a Trinomial by a Binomial Using Long Division (http://openstaxcollege.org/l/
dividetribild)


• Dividing a Polynomial by a Binomial Using Long Division (http://openstaxcollege.org/l/
dividepolybild)


• Ex 2: Dividing a Polynomial by a Binomial Using Synthetic Division
(http://openstaxcollege.org/l/dividepolybisd2)


• Ex 4: Dividing a Polynomial by a Binomial Using Synthetic Division
(http://openstaxcollege.org/l/dividepolybisd4)


Chapter 3 Polynomial and Rational Functions 371




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339.


3.5 EXERCISES
Verbal


If division of a polynomial by a binomial results in a
remainder of zero, what can be conclude?


If a polynomial of degree  n  is divided by a binomial
of degree 1, what is the degree of the quotient?


Algebraic
For the following exercises, use long division to divide.
Specify the quotient and the remainder.



⎝x


2 + 5x − 1⎞⎠ ÷ (x − 1)



⎝2x


2 − 9x − 5⎞⎠ ÷ (x − 5)



⎝3x


2 + 23x + 14⎞⎠ ÷ (x + 7)



⎝4x


2 − 10x + 6⎞⎠ ÷ (4x + 2)



⎝6x


2 − 25x − 25⎞⎠ ÷ (6x + 5)



⎝−x


2 − 1⎞⎠ ÷ (x + 1)



⎝2x


2 − 3x + 2⎞⎠ ÷ (x + 2)



⎝x


3 − 126⎞⎠ ÷ (x − 5)



⎝3x


2 − 5x + 4⎞⎠ ÷ (3x + 1)



⎝x


3 − 3x2 + 5x − 6⎞⎠ ÷ (x − 2)



⎝2x


3 + 3x2 − 4x + 15⎞⎠ ÷ (x + 3)


For the following exercises, use synthetic division to find
the quotient.



⎝3x


3 − 2x2 + x − 4⎞⎠ ÷ (x + 3)



⎝2x


3 − 6x2 − 7x + 6⎞⎠ ÷ (x − 4)



⎝6x


3 − 10x2 − 7x − 15⎞⎠ ÷ (x + 1)



⎝4x


3 − 12x2 − 5x − 1⎞⎠ ÷ (2x + 1)



⎝9x


3 − 9x2 + 18x + 5⎞⎠ ÷ (3x − 1)



⎝3x


3 − 2x2 + x − 4⎞⎠ ÷ (x + 3)



⎝−6x


3 + x2 − 4⎞⎠ ÷ (2x − 3)



⎝2x


3 + 7x2 − 13x − 3⎞⎠ ÷ (2x − 3)



⎝3x


3 − 5x2 + 2x + 3⎞⎠ ÷ (x + 2)



⎝4x


3 − 5x2 + 13⎞⎠ ÷ (x + 4)



⎝x


3 − 3x + 2⎞⎠ ÷ (x + 2)



⎝x


3 − 21x2 + 147x − 343⎞⎠ ÷ (x − 7)



⎝x


3 − 15x2 + 75x − 125⎞⎠ ÷ (x − 5)



⎝9x


3 − x + 2⎞⎠ ÷ (3x − 1)



⎝6x


3 − x2 + 5x + 2⎞⎠ ÷ (3x + 1)



⎝x


4 + x3 − 3x2 − 2x + 1⎞⎠ ÷ (x + 1)



⎝x


4 − 3x2 + 1⎞⎠ ÷ (x − 1)



⎝x


4 + 2x3 − 3x2 + 2x + 6⎞⎠ ÷ (x + 3)



⎝x


4 − 10x3 + 37x2 − 60x + 36⎞⎠ ÷ (x − 2)



⎝x


4 − 8x3 + 24x2 − 32x + 16⎞⎠ ÷ (x − 2)



⎝x


4 + 5x3 − 3x2 − 13x + 10⎞⎠ ÷ (x + 5)



⎝x


4 − 12x3 + 54x2 − 108x + 81⎞⎠ ÷ (x − 3)



⎝4x


4 − 2x3 − 4x + 2⎞⎠ ÷ (2x − 1)



⎝4x


4 + 2x3 − 4x2 + 2x + 2⎞⎠ ÷ (2x + 1)


For the following exercises, use synthetic division to
determine whether the first expression is a factor of the
second. If it is, indicate the factorization.


x − 2,  4x3 − 3x2 − 8x + 4


372 Chapter 3 Polynomial and Rational Functions


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351.


352.


x − 2,  3x4 − 6x3 − 5x + 10


x + 3,  − 4x3 + 5x2 + 8


x − 2,  4x4 − 15x2 − 4


x − 1
2
,  2x4 − x3 + 2x − 1


x + 1
3
,  3x4 + x3 − 3x + 1


Graphical
For the following exercises, use the graph of the third-
degree polynomial and one factor to write the factored form
of the polynomial suggested by the graph. The leading
coefficient is one.


Factor is  x2 − x + 3


Factor is  (x2 + 2x + 4)


Factor is  x2 + 2x + 5


Factor is  x2 + x + 1


Factor is x2 + 2x + 2


For the following exercises, use synthetic division to find
the quotient and remainder.


4x3 − 33
x − 2


2x3 + 25
x + 3


3x3 + 2x − 5
x − 1


Chapter 3 Polynomial and Rational Functions 373




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374.


−4x3 − x2 − 12
x + 4


x4 − 22
x + 2


Technology
For the following exercises, use a calculator with CAS to
answer the questions.


Consider  xk − 1
x − 1


 with  k = 1, 2, 3. What do you
expect the result to be if  k = 4?


Consider  xk + 1
x + 1


  for  k = 1, 3, 5. What do you
expect the result to be if  k = 7?


Consider  x4 − k4
x − k


  for  k = 1, 2, 3. What do you
expect the result to be if  k = 4?


Consider   xk
x + 1


 with  k = 1, 2, 3. What do you
expect the result to be if  k = 4?


Consider   xk
x − 1


 with  k = 1, 2, 3. What do you
expect the result to be if  k = 4?


Extensions
For the following exercises, use synthetic division to
determine the quotient involving a complex number.


x + 1
x − i


x2 + 1
x − i


x + 1
x + i


x2 + 1
x + i


x3 + 1
x − i


Real-World Applications
For the following exercises, use the given length and area
of a rectangle to express the width algebraically.


Length is  x + 5,   area is  2x2 + 9x − 5.


Length is  2x + 5,   area is  4x3 + 10x2 + 6x + 15


Length is  3x – 4,   area is
 6x4 − 8x3 + 9x2 − 9x − 4


For the following exercises, use the given volume of a box
and its length and width to express the height of the box
algebraically.


Volume is  12x3 + 20x2 − 21x − 36,   length is
 2x + 3,  width is  3x − 4.


Volume is  18x3 − 21x2 − 40x + 48,   length is
 3x – 4,   width is  3x – 4.


Volume is  10x3 + 27x2 + 2x − 24,   length is
 5x – 4,   width is  2x + 3.


Volume is  10x3 + 30x2 − 8x − 24,   length is  2,
width is  x + 3.


For the following exercises, use the given volume and
radius of a cylinder to express the height of the cylinder
algebraically.


Volume is  π(25x3 − 65x2 − 29x − 3),   radius is
 5x + 1.


Volume is  π(4x3 + 12x2 − 15x − 50),   radius is
 2x + 5.


Volume is  π(3x4 + 24x3 + 46x2 − 16x − 32),
radius is  x + 4.


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3.6 | Zeros of Polynomial Functions
Learning Objectives


In this section, you will:
3.6.1 Evaluate a polynomial using the Remainder Theorem.
3.6.2 Use the Factor Theorem to solve a polynomial equation.
3.6.3 Use the Rational Zero Theorem to find rational zeros.
3.6.4 Find zeros of a polynomial function.
3.6.5 Use the Linear Factorization Theorem to find polynomials with given zeros.
3.6.6 Use Descartes’ Rule of Signs.
3.6.7 Solve real-world applications of polynomial equations


A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the
volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the
cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should
the dimensions of the cake pan be?
This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this
section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.
Evaluating a Polynomial Using the Remainder Theorem
In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials
using the Remainder Theorem. If the polynomial is divided by  x – k,   the remainder may be found quickly by evaluating
the polynomial function at  k,   that is,   f (k) Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend   f (x)  and a non-zero polynomial divisor  d(x) 
where the degree of   d(x)  is less than or equal to the degree of   f (x),   there exist unique polynomials  q(x)  and  r(x)  such
that


  f (x) = d(x)q(x) + r(x) 
If the divisor,  d(x),   is  x − k,   this takes the form


f (x) = (x − k)q(x) + r


Since the divisor  x − k  is linear, the remainder will be a constant,  r. And, if we evaluate this for  x = k,  we have
f (k) = (k − k)q(k) + r


= 0 ⋅ q(k) + r
= r


In other words,   f (k)  is the remainder obtained by dividing   f (x)  by  x − k.


The Remainder Theorem
If a polynomial   f (x)  is divided by  x − k,   then the remainder is the value   f (k). 


Given a polynomial function   f , evaluate   f(x)  at  x= k  using the Remainder Theorem.
1. Use synthetic division to divide the polynomial by  x − k.
2. The remainder is the value   f (k).


Chapter 3 Polynomial and Rational Functions 375




3.31


Example 3.50
Using the Remainder Theorem to Evaluate a Polynomial


Use the Remainder Theorem to evaluate   f (x) = 6x4 − x3 − 15x2 + 2x − 7  at  x = 2.


Solution
To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by  x − 2.


26 −1 −15 2 −7
12 22 14 32


6 11 7 16 25


The remainder is 25. Therefore,   f (2) = 25. 


Analysis
We can check our answer by evaluating   f (2). 


f (x) = 6x4 − x3 − 15x2 + 2x − 7


f (2) = 6(2)4 − (2)3 − 15(2)2 + 2(2) − 7


= 25


Use the Remainder Theorem to evaluate   f (x) = 2x5 − 3x4 − 9x3 + 8x2 + 2  at  x = − 3.


Using the Factor Theorem to Solve a Polynomial Equation
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a
polynomial are related to the factors. Recall that the Division Algorithm tells us


f (x) = (x − k)q(x) + r.


If  k  is a zero, then the remainder  r  is   f (k) = 0  and   f (x) = (x − k)q(x) + 0  or   f (x) = (x − k)q(x). 
Notice, written in this form,  x − k  is a factor of   f (x). We can conclude if  k  is a zero of   f (x),   then  x − k  is a factor of
f (x). 
Similarly, if  x − k  is a factor of   f (x),   then the remainder of the Division Algorithm   f (x) = (x − k)q(x) + r  is 0. This
tells us that  k  is a zero.
This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree  n  in the complex number
system will have  n  zeros. We can use the Factor Theorem to completely factor a polynomial into the product of  n  factors.
Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.


The Factor Theorem
According to the Factor Theorem,  k  is a zero of   f (x)  if and only if  (x − k)  is a factor of   f (x). 


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3.32


Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial.
1. Use synthetic division to divide the polynomial by  (x − k). 
2. Confirm that the remainder is 0.
3. Write the polynomial as the product of  (x − k)  and the quadratic quotient.
4. If possible, factor the quadratic.
5. Write the polynomial as the product of factors.


Example 3.51
Using the Factor Theorem to Solve a Polynomial Equation


Show that  (x + 2)  is a factor of  x3 − 6x2 − x + 30.  Find the remaining factors. Use the factors to determine the
zeros of the polynomial.


Solution
We can use synthetic division to show that  (x + 2)  is a factor of the polynomial.


−21 −6 −1 30
−2 16 −30


1 −8 15 0


The remainder is zero, so  (x + 2)  is a factor of the polynomial. We can use the Division Algorithm to write the
polynomial as the product of the divisor and the quotient:


(x + 2)(x2 − 8x + 15)


We can factor the quadratic factor to write the polynomial as
(x + 2)(x − 3)(x − 5)


By the Factor Theorem, the zeros of  x3 − 6x2 − x + 30  are –2, 3, and 5.


Use the Factor Theorem to find the zeros of   f (x) = x3 + 4x2 − 4x − 16  given that (x − 2) is a factor of
the polynomial.


Using the Rational Zero Theorem to Find Rational Zeros
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first
we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible
rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial
Consider a quadratic function with two zeros,  x = 2


5
  and  x = 3


4
 .  By the Factor Theorem, these zeros have factors


associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching
factor.


Chapter 3 Polynomial and Rational Functions 377




Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3.
Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5
and 4.
We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will
be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of
possible rational zeros.


The Rational Zero Theorem
The Rational Zero Theorem states that, if the polynomial   f (x) = an xn + an − 1 xn − 1 + ... + a1 x + a0   has integer
coefficients, then every rational zero of   f (x)  has the form  pq  where  p  is a factor of the constant term  a0   and  q  is a
factor of the leading coefficient  an.
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.


Given a polynomial function   f(x), use the Rational Zero Theorem to find rational zeros.
1. Determine all factors of the constant term and all factors of the leading coefficient.
2. Determine all possible values of  pq ,  where  p  is a factor of the constant term and  q  is a factor of the
leading coefficient. Be sure to include both positive and negative candidates.


3. Determine which possible zeros are actual zeros by evaluating each case of   f ( pq).


Example 3.52
Listing All Possible Rational Zeros


List all possible rational zeros of   f (x) = 2x4 − 5x3 + x2 − 4.


Solution
The only possible rational zeros of   f (x)  are the quotients of the factors of the last term, –4, and the factors of the
leading coefficient, 2.
The constant term is –4; the factors of –4 are  p = ±1, ±2, ±4.
The leading coefficient is 2; the factors of 2 are  q = ±1, ±2.


378 Chapter 3 Polynomial and Rational Functions


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3.33


If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by
one of the factors of 2.


p
q = ±  11,   ±  


1
2


p
q = ±  21, ±  


2
2


p
q =  ±  41, ±  


4
2


Note that  2
2
= 1  and  4


2
= 2,  which have already been listed. So we can shorten our list.


p
q =


Factors of the last
Factors of the fir t


= ±1, ±2, ±4, ± 1
2


Example 3.53
Using the Rational Zero Theorem to Find Rational Zeros


Use the Rational Zero Theorem to find the rational zeros of   f (x) = 2x3 + x2 − 4x + 1. 


Solution
The Rational Zero Theorem tells us that if  pq   is a zero of   f (x),   then  p  is a factor of 1 and  q  is a factor of 2.


p
q =


factor of constant term
factor of leading coefficie


= factor of 1
factor of 2


The factors of 1 are±1  and the factors of 2 are±1  and±2. The possible values for  pq   are±1  and  ± 12. These
are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by
substituting these values for  x  in   f (x). 


f ( − 1) = 2( − 1)3 + ( − 1)2 − 4( − 1) + 1 = 4


f (1) = 2(1)3 + (1)2 − 4(1) + 1 = 0


f ⎛⎝−
1
2

⎠ = 2

⎝−


1
2



3
+ ⎛⎝−


1
2



2
− 4⎛⎝−


1
2

⎠+ 1 = 3


f ⎛⎝
1
2

⎠ = 2


1
2



3
+ ⎛⎝


1
2



2
− 4⎛⎝


1
2

⎠+ 1 = −


1
2


Of those, −1, − 1
2
, and 1


2
  are not zeros of   f (x).  1 is the only rational zero of   f (x). 


Use the Rational Zero Theorem to find the rational zeros of   f (x) = x3 − 5x2 + 2x + 1. 


Finding the Zeros of Polynomial Functions
The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we
have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.


Chapter 3 Polynomial and Rational Functions 379




Given a polynomial function   f , use synthetic division to find its zeros.
1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the
polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.


3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is
a quadratic.


4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and
using the quadratic formula.


Example 3.54
Finding the Zeros of a Polynomial Function with Repeated Real Zeros


Find the zeros of   f (x) = 4x3 − 3x − 1. 


Solution
The Rational Zero Theorem tells us that if  pq   is a zero of   f (x),   then  p  is a factor of –1 and  q  is a factor of 4.


p
q =


factor of constant term
factor of leading coefficie


= factor of –1
factor of 4


The factors of   – 1  are±1  and the factors of  4  are±1, ±2,   and  ±4. The possible values for  pq   are
±1,  ± 1


2
,   and  ± 1


4
.  These are the possible rational zeros for the function. We will use synthetic division to


evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.
14 0 −3 −1


4 4 1


4 4 1 0


Dividing by  (x − 1)  gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as
(x − 1)(4x2 + 4x + 1).


The quadratic is a perfect square.   f (x)  can be written as
(x − 1)(2x + 1)2.


We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To
find the other zero, we can set the factor equal to 0.


2x + 1 = 0


x = − 1
2


The zeros of the function are 1 and  − 1
2
 with multiplicity 2.


Analysis


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Look at the graph of the function   f   in Figure 3.62. Notice, at  x = − 0.5,   the graph bounces off the x-axis,
indicating the even multiplicity (2,4,6…) for the zero  − 0.5.  At  x = 1,   the graph crosses the x-axis, indicating
the odd multiplicity (1,3,5…) for the zero  x = 1. 


Figure 3.62


Using the Fundamental Theorem of Algebra
Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of
complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function
has at least one complex zero. This theorem forms the foundation for solving polynomial equations.
Suppose   f   is a polynomial function of degree four, and   f (x) = 0. The Fundamental Theorem of Algebra states that
there is at least one complex solution, call it  c1. By the Factor Theorem, we can write   f (x)  as a product of  x − c1  
and a polynomial quotient. Since  x − c1   is linear, the polynomial quotient will be of degree three. Now we apply the
Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it  c2. 
So we can write the polynomial quotient as a product of  x − c2   and a new polynomial quotient of degree two. Continue to
apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will
yield a factor of   f (x). 


The Fundamental Theorem of Algebra states that, if f(x) is a polynomial of degree n > 0, then
f(x) has at least one complex zero.
We can use this theorem to argue that, if   f (x)  is a polynomial of degree  n > 0,   and  a  is a non-zero real number,
then   f (x)  has exactly  n  linear factors


f (x) = a(x − c1)(x − c2)...(x − cn)


where  c1, c2, ..., cn   are complex numbers. Therefore,   f (x)  has  n  roots if we allow for multiplicities.


Does every polynomial have at least one imaginary zero?
No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.


Example 3.55


Chapter 3 Polynomial and Rational Functions 381




Finding the Zeros of a Polynomial Function with Complex Zeros


Find the zeros of   f (x) = 3x3 + 9x2 + x + 3. 


Solution
The Rational Zero Theorem tells us that if  pq   is a zero of   f (x),   then  p  is a factor of 3 and  q  is a factor of 3.


p
q =


factor of constant term
factor of leading coefficie


= factor of 3
factor of 3


The factors of 3 are±1  and ±3. The possible values for  pq ,   and therefore the possible rational zeros for the
function, are ±3, ±1, and ± 1


3
. We will use synthetic division to evaluate each possible zero until we find one


that gives a remainder of 0. Let’s begin with –3.


−33 9 1 3
−9 0 −3


3 0 1 0


Dividing by  (x + 3)  gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as
(x + 3)(3x2 + 1)


We can then set the quadratic equal to 0 and solve to find the other zeros of the function.
3x2 + 1 = 0


x2 = − 1
3


x = ± −1
3
= ± i 3


3


The zeros of f (x) are –3 and  ± i 3
3


.


Analysis
Look at the graph of the function   f   in Figure 3.63. Notice that, at  x = − 3,   the graph crosses the x-axis,
indicating an odd multiplicity (1) for the zero  x = – 3. Also note the presence of the two turning points. This
means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So,
the end behavior of increasing without bound to the right and decreasing without bound to the left will continue.
Thus, all the x-intercepts for the function are shown. So either the multiplicity of  x = − 3  is 1 and there are
two complex solutions, which is what we found, or the multiplicity at  x = − 3  is three. Either way, our result is
correct.


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3.34


Figure 3.63


Find the zeros of   f (x) = 2x3 + 5x2 − 11x + 4.


Using the Linear Factorization Theorem to Find Polynomials with
Given Zeros
A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree
 n  will have  n  zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the
polynomial function into  n  factors. The Linear Factorization Theorem tells us that a polynomial function will have the
same number of factors as its degree, and that each factor will be in the form  (x − c),  where  c  is a complex number.
Let   f   be a polynomial function with real coefficients, and suppose  a + bi, b ≠ 0,   is a zero of   f (x).  Then, by the
Factor Theorem,  x − (a + bi)  is a factor of   f (x).  For   f   to have real coefficients,  x − (a − bi)  must also be a factor of
  f (x).  This is true because any factor other than  x − (a − bi),   when multiplied by  x − (a + bi),   will leave imaginary
components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real
coefficients. In other words, if a polynomial function   f   with real coefficients has a complex zero  a + bi,   then the
complex conjugate  a − bi  must also be a zero of   f (x). This is called the Complex Conjugate Theorem.


Complex Conjugate Theorem
According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its
degree, and each factor will be in the form  (x − c),   where  c  is a complex number.
If the polynomial function   f   has real coefficients and a complex zero in the form  a + bi,   then the complex
conjugate of the zero,  a − bi,   is also a zero.


Given the zeros of a polynomial function   f   and a point (c, f(c)) on the graph of   f ,   use the Linear
Factorization Theorem to find the polynomial function.


1. Use the zeros to construct the linear factors of the polynomial.
2. Multiply the linear factors to expand the polynomial.
3. Substitute  ⎛⎝c, f (c)⎞⎠  into the function to determine the leading coefficient.
4. Simplify.


Chapter 3 Polynomial and Rational Functions 383




3.35


Example 3.56
Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros


Find a fourth degree polynomial with real coefficients that has zeros of –3, 2,  i,   such that   f ( − 2) = 100. 


Solution
Because  x = i  is a zero, by the Complex Conjugate Theorem  x = – i  is also a zero. The polynomial must have
factors of  (x + 3),  (x − 2),  (x − i),   and  (x + i).  Since we are looking for a degree 4 polynomial, and now have
four zeros, we have all four factors. Let’s begin by multiplying these factors.


f (x) = a(x + 3)(x − 2)(x − i)(x + i)
f (x) = a(x2 + x − 6)(x2 + 1)


f (x) = a(x4 + x3 − 5x2 + x − 6)


We need to find a to ensure   f ( – 2) = 100.  Substitute  x = – 2   and   f (2) = 100  into   f (x). 
100 = a(( − 2)4 + ( − 2)3 − 5( − 2)2 + ( − 2) − 6)
100 = a( − 20)
−5 = a


So the polynomial function is
f (x) = − 5(x4 + x3 − 5x2 + x − 6)


or
f (x) = − 5x4 − 5x3 + 25x2 − 5x + 30


Analysis
We found that both  i  and  − i were zeros, but only one of these zeros needed to be given. If  i  is a zero of a
polynomial with real coefficients, then  − i must also be a zero of the polynomial because  − i  is the complex
conjugate of  i.


If  2 +3i  were given as a zero of a polynomial with real coefficients, would  2 −3i  also need to be a zero?
Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real
coefficients, the conjugate must also be a zero of the polynomial.


Find a third degree polynomial with real coefficients that has zeros of 5 and  − 2i  such that   f (1) = 10. 


Using Descartes’ Rule of Signs
There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial
function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the
number of sign changes in   f (x)  and the number of positive real zeros. For example, the polynomial function below has one
sign change.


This tells us that the function must have 1 positive real zero.


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There is a similar relationship between the number of sign changes in   f ( − x)  and the number of negative real zeros.


In this case,   f (−x)  has 3 sign changes. This tells us that   f (x)  could have 3 or 1 negative real zeros.


Descartes’ Rule of Signs
According to Descartes’ Rule of Signs, if we let   f (x) = an xn + an − 1 xn − 1 + ... + a1 x + a0   be a polynomial
function with real coefficients:


• The number of positive real zeros is either equal to the number of sign changes of   f (x)  or is less than the
number of sign changes by an even integer.


• The number of negative real zeros is either equal to the number of sign changes of   f ( − x)  or is less than the
number of sign changes by an even integer.


Example 3.57
Using Descartes’ Rule of Signs


Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for
  f (x) = − x4 − 3x3 + 6x2 − 4x − 12.


Solution
Begin by determining the number of sign changes.


Figure 3.64


There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine   f ( − x)  to determine
the number of negative real roots.
f ( − x) = − ( − x)4 − 3( − x)3 + 6( − x)2 − 4( − x) − 12


f ( − x) = − x4 + 3x3 + 6x2 + 4x − 12


Figure 3.65


Again, there are two sign changes, so there are either 2 or 0 negative real roots.
There are four possibilities, as we can see in Table 3.6.


Chapter 3 Polynomial and Rational Functions 385




3.36


Positive Real Zeros Negative Real Zeros Complex Zeros Total Zeros


2 2 0 4


2 0 2 4


0 2 2 4


0 0 4 4


Table 3.6


Analysis
We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure
3.66. We can see from the graph that the function has 0 positive real roots and 2 negative real roots.


Figure 3.66


Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real
zeros for   f (x) = 2x4 − 10x3 + 11x2 − 15x + 12.  Use a graph to verify the numbers of positive and negative
real zeros for the function.


Solving Real-World Applications
We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery
problem from the beginning of the section.


Example 3.58
Solving Polynomial Equations


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A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery
wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They
want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be
one-third of the width. What should the dimensions of the cake pan be?


Solution
Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by  V = lwh. 
We were given that the length must be four inches longer than the width, so we can express the length of the cake
as  l = w + 4. We were given that the height of the cake is one-third of the width, so we can express the height of
the cake as  h = 1


3
w. Let’s write the volume of the cake in terms of width of the cake.


V = (w + 4)(w)(1
3
w)


V = 1
3
w3 + 4


3
w2


Substitute the given volume into this equation.
351 = 1


3
w3 + 4


3
w2 Substitute 351 for V .


1053 = w3 + 4w2 Multiply both sides by 3.


0 = w3 + 4w2 − 1053 Subtract 1053 from both sides.


Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the
possible rational zeros are  ± 3,  ± 9,  ± 13,  ± 27,  ± 39,  ± 81,  ± 117,  ± 351,   and  ± 1053.  We can
use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so
we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a
small sheet cake. Use synthetic division to check  x = 1.


11 4 0 −1053
1 5 5


1 5 5 −1048


Since 1 is not a solution, we will check  x = 3.


Since 3 is not a solution either, we will test  x = 9. 


Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between
the width and the other dimensions to determine the length and height of the sheet cake pan.


l = w + 4 = 9 + 4 = 13 and h = 1
3
w = 1


3
(9) = 3


The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.


Chapter 3 Polynomial and Rational Functions 387




3.37 A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The
client tells the manufacturer that, because of the contents, the length of the container must be one meter longer
than the width, and the height must be one meter greater than twice the width. What should the dimensions of the
container be?


Access these online resources for additional instruction and practice with zeros of polynomial functions.
• Real Zeros, Factors, and Graphs of Polynomial Functions (http://openstaxcollege.org/l/
realzeros)


• Complex Factorization Theorem (http://openstaxcollege.org/l/factortheorem)
• Find the Zeros of a Polynomial Function (http://openstaxcollege.org/l/findthezeros)
• Find the Zeros of a Polynomial Function 2 (http://openstaxcollege.org/l/findthezeros2)
• Find the Zeros of a Polynomial Function 3 (http://openstaxcollege.org/l/findthezeros3)


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375.
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414.


3.6 EXERCISES
Verbal


Describe a use for the Remainder Theorem.
Explain why the Rational Zero Theorem does not


guarantee finding zeros of a polynomial function.
What is the difference between rational and real


zeros?
If Descartes’ Rule of Signs reveals a no change of


signs or one sign of changes, what specific conclusion can
be drawn?


If synthetic division reveals a zero, why should we try
that value again as a possible solution?


Algebraic
For the following exercises, use the Remainder Theorem to
find the remainder.



⎝x


4 − 9x2 + 14⎞⎠ ÷ (x − 2)



⎝3x


3 − 2x2 + x − 4⎞⎠ ÷ (x + 3)



⎝x


4 + 5x3 − 4x − 17⎞⎠ ÷ (x + 1)



⎝−3x


2 + 6x + 24⎞⎠ ÷ (x − 4)



⎝5x


5 − 4x4 + 3x3 − 2x2 + x − 1⎞⎠ ÷ (x + 6)



⎝x


4 − 1⎞⎠ ÷ (x − 4)



⎝3x


3 + 4x2 − 8x + 2⎞⎠ ÷ (x − 3)



⎝4x


3 + 5x2 − 2x + 7⎞⎠ ÷ (x + 2)


For the following exercises, use the Factor Theorem to find
all real zeros for the given polynomial function and one
factor.


f (x) = 2x3 − 9x2 + 13x − 6; x − 1


f (x) = 2x3 + x2 − 5x + 2; x + 2


f (x) = 3x3 + x2 − 20x + 12; x + 3


f (x) = 2x3 + 3x2 + x + 6;     x + 2


f (x) = − 5x3 + 16x2 − 9;     x − 3


x3 + 3x2 + 4x + 12;  x + 3


4x3 − 7x + 3;  x − 1


2x3 + 5x2 − 12x − 30,  2x + 5


For the following exercises, use the Rational Zero Theorem
to find all real zeros.


x3 − 3x2 − 10x + 24 = 0


2x3 + 7x2 − 10x − 24 = 0


x3 + 2x2 − 9x − 18 = 0


x3 + 5x2 − 16x − 80 = 0


x3 − 3x2 − 25x + 75 = 0


2x3 − 3x2 − 32x − 15 = 0


2x3 + x2 − 7x − 6 = 0


2x3 − 3x2 − x + 1 = 0


3x3 − x2 − 11x − 6 = 0


2x3 − 5x2 + 9x − 9 = 0


2x3 − 3x2 + 4x + 3 = 0


x4 − 2x3 − 7x2 + 8x + 12 = 0


x4 + 2x3 − 9x2 − 2x + 8 = 0


4x4 + 4x3 − 25x2 − x + 6 = 0


2x4 − 3x3 − 15x2 + 32x − 12 = 0


x4 + 2x3 − 4x2 − 10x − 5 = 0


4x3 − 3x + 1 = 0


8x 4 + 26x3 + 39x2 + 26x + 6


For the following exercises, find all complex solutions (real
and non-real).


x3 + x2 + x + 1 = 0


Chapter 3 Polynomial and Rational Functions 389




415.


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449.


x3 − 8x2 + 25x − 26 = 0


x3 + 13x2 + 57x + 85 = 0


3x3 − 4x2 + 11x + 10 = 0


x4 + 2x3 + 22x2 + 50x − 75 = 0


2x3 − 3x2 + 32x + 17 = 0


Graphical
For the following exercises, use Descartes’ Rule to
determine the possible number of positive and negative
solutions. Confirm with the given graph.


f (x) = x3 − 1


f (x) = x4 − x2 − 1


f (x) = x3 − 2x2 − 5x + 6


f (x) = x3 − 2x2 + x − 1


f (x) = x4 + 2x3 − 12x2 + 14x − 5


f (x) = 2x3 + 37x2 + 200x + 300


f (x) = x3 − 2x2 − 16x + 32


f (x) = 2x4 − 5x3 − 5x2 + 5x + 3


f (x) = 2x4 − 5x3 − 14x2 + 20x + 8


f (x) = 10x4 − 21x2 + 11


Numeric
For the following exercises, list all possible rational zeros
for the functions.


f (x) = x4 + 3x3 − 4x + 4


f (x) = 2x 3 + 3x2 − 8x + 5


f (x) = 3x 3 + 5x2 − 5x + 4


f (x) = 6x4 − 10x2 + 13x + 1


f (x) = 4x5 − 10x4 + 8x3 + x2 − 8


Technology
For the following exercises, use your calculator to graph the
polynomial function. Based on the graph, find the rational
zeros. All real solutions are rational.


f (x) = 6x3 − 7x2 + 1


f (x) = 4x3 − 4x2 − 13x − 5


f (x) = 8x3 − 6x2 − 23x + 6


f (x) = 12x4 + 55x3 + 12x2 − 117x + 54


f (x) = 16x4 − 24x3 + x2 − 15x + 25


Extensions
For the following exercises, construct a polynomial
function of least degree possible using the given
information.


Real roots: –1, 1, 3 and  ⎛⎝2, f (2)⎞⎠ = (2, 4)


Real roots: –1 (with multiplicity 2 and 1) and
 ⎛⎝2, f (2)⎞⎠ = (2, 4)


Real roots: –2, 1
2


(with multiplicity 2) and
 ⎛⎝−3, f (−3)⎞⎠ = (−3, 5)


Real roots:  − 1
2
  , 0,  1


2
  and  ⎛⎝−2, f (−2)⎞⎠ = (−2, 6)


Real roots: –4, –1, 1, 4 and  ⎛⎝−2, f (−2)⎞⎠ = (−2, 10)


Real-World Applications
For the following exercises, find the dimensions of the box
described.


The length is twice as long as the width. The height is
2 inches greater than the width. The volume is 192 cubic
inches.


The length, width, and height are consecutive whole
numbers. The volume is 120 cubic inches.


The length is one inch more than the width, which is
one inch more than the height. The volume is 86.625 cubic
inches.


The length is three times the height and the height is
one inch less than the width. The volume is 108 cubic
inches.


The length is 3 inches more than the width. The width
is 2 inches more than the height. The volume is 120 cubic
inches.


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450.


451.


452.


453.


454.


For the following exercises, find the dimensions of the right
circular cylinder described.


The radius is 3 inches more than the height. The
volume is  16π  cubic meters.


The height is one less than one half the radius. The
volume is  72π  cubic meters.


The radius and height differ by one meter. The radius
is larger and the volume is  48π  cubic meters.


The radius and height differ by two meters. The
height is greater and the volume is  28.125π  cubic meters.


80. The radius is  1
3
 meter greater than the height. The


volume is  98
9
π  cubic meters.


Chapter 3 Polynomial and Rational Functions 391




3.7 | Rational Functions
Learning Objectives


In this section, you will:
3.7.1 Use arrow notation.
3.7.2 Solve applied problems involving rational functions.
3.7.3 Find the domains of rational functions.
3.7.4 Identify vertical asymptotes.
3.7.5 Identify horizontal asymptotes.
3.7.6 Graph rational functions.


Suppose we know that the cost of making a product is dependent on the number of items,  x,   produced. This is given by the
equation  C(x) = 15,000x − 0.1x2 + 1000.  If we want to know the average cost for producing  x  items, we would divide
the cost function by the number of items,  x.
The average cost function, which yields the average cost per item for  x  items produced, is


f (x) = 15,000x − 0.1x
2 + 1000


x


Many other application problems require finding an average value in a similar way, giving us variables in the denominator.
Written without a variable in the denominator, this function will contain a negative integer power.
In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for
exponents. In this section, we explore rational functions, which have variables in the denominator.
Using Arrow Notation
We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit
functions. Examine these graphs, as shown in Figure 3.67, and notice some of their features.


Figure 3.67


Several things are apparent if we examine the graph of   f (x) = 1x .
1. On the left branch of the graph, the curve approaches the x-axis  (y = 0) as x → – ∞.


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2. As the graph approaches  x = 0  from the left, the curve drops, but as we approach zero from the right, the curve
rises.


3. Finally, on the right branch of the graph, the curves approaches the x-axis  (y = 0) as x → ∞.
To summarize, we use arrow notation to show that  x  or   f (x)  is approaching a particular value. See Table 3.7.


Symbol Meaning


x → a− x  approaches  a  from the left ( x < a  but close to  a )


x → a+ x  approaches  a  from the right ( x > a  but close to  a )


x → ∞ x  approaches infinity ( x  increases without bound)


x → −∞ x  approaches negative infinity ( x  decreases without bound)


f (x) → ∞ the output approaches infinity (the output increases without bound)


f (x) → −∞ the output approaches negative infinity (the output decreases without bound)


f (x) → a the output approaches  a


Table 3.7 Arrow Notation


Local Behavior of   f(x) = 1x
Let’s begin by looking at the reciprocal function,   f (x) = 1x . We cannot divide by zero, which means the function is
undefined at  x = 0;   so zero is not in the domain. As the input values approach zero from the left side (becoming very
small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We
can see this behavior in Table 3.8.


x –0.1 –0.01 –0.001 –0.0001


f(x) = 1x –10 –100 –1000 –10,000


Table 3.8


We write in arrow notation
as x → 0− , f (x) → −∞


As the input values approach zero from the right side (becoming very small, positive values), the function values increase
without bound (approaching infinity). We can see this behavior in Table 3.9.


Chapter 3 Polynomial and Rational Functions 393




x 0.1 0.01 0.001 0.0001


f(x) = 1x 10 100 1000 10,000


Table 3.9


We write in arrow notation
As x → 0+ , f (x) → ∞.


See Figure 3.68.


Figure 3.68


This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case,
the graph is approaching the vertical line  x = 0  as the input becomes close to zero. See Figure 3.69.


Figure 3.69


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Vertical Asymptote
A vertical asymptote of a graph is a vertical line  x = a where the graph tends toward positive or negative infinity as
the inputs approach  a. We write


As x → a, f (x) → ∞, or as x → a, f (x) → −∞.


End Behavior of   f(x) = 1x
As the values of  x  approach infinity, the function values approach 0. As the values of  x  approach negative infinity, the
function values approach 0. See Figure 3.70. Symbolically, using arrow notation
As x → ∞, f (x) → 0, and as x → −∞, f (x) → 0.


Figure 3.70


Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it
seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph
approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line
 y = 0.  See Figure 3.71.


Figure 3.71


Chapter 3 Polynomial and Rational Functions 395




3.38


Horizontal Asymptote
A horizontal asymptote of a graph is a horizontal line  y = b where the graph approaches the line as the inputs
increase or decrease without bound. We write


 As x → ∞ or x → −∞, f (x) → b.


Example 3.59
Using Arrow Notation


Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 3.72.


Figure 3.72


Solution
Notice that the graph is showing a vertical asymptote at  x = 2,  which tells us that the function is undefined at
 x = 2.


As x → 2− , f (x) → −∞, and as x → 2+ , f (x) → ∞.


And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a
horizontal asymptote at  y = 4. As the inputs increase without bound, the graph levels off at 4.


As x → ∞, f (x) → 4 and as x → −∞, f (x) → 4.


Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.


Example 3.60


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3.39


Using Transformations to Graph a Rational Function


Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal
and vertical asymptotes of the graph, if any.


Solution
Shifting the graph left 2 and up 3 would result in the function


f (x) = 1
x + 2


+ 3


or equivalently, by giving the terms a common denominator,
f (x) = 3x + 7


x + 2


The graph of the shifted function is displayed in Figure 3.73.


Figure 3.73


Notice that this function is undefined at  x = − 2,   and the graph also is showing a vertical asymptote at
 x = − 2.


As x → − 2− , f (x) → −∞, and as x → − 2+ , f (x) → ∞.


As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3,
indicating a horizontal asymptote at  y = 3.


As x → ±∞, f (x) → 3.


Analysis
Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.


Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that
has been shifted right 3 units and down 4 units.


Chapter 3 Polynomial and Rational Functions 397




Solving Applied Problems Involving Rational Functions
In Example 3.60, we shifted a toolkit function in a way that resulted in the function   f (x) = 3x + 7


x + 2
. This is an example of


a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many
real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations
often involve rational functions.


Rational Function
A rational function is a function that can be written as the quotient of two polynomial functions  P(x) and Q(x).


(3.5)
f (x) = P(x)


Q(x)
=


a p x
p + a p − 1 x


p − 1
+ ... + a1 x + a0


bq x
q + bq − 1 x


q − 1
+ ... + b1 x + b0


, Q(x) ≠ 0


Example 3.61
Solving an Applied Problem Involving a Rational Function


A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap
will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at
a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is
that a greater concentration than at the beginning?


Solution
Let  t  be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the
sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the
tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:


water: W(t) = 100 + 10t in gallons
sugar: S(t) = 5 + 1t in pounds


The concentration,  C,  will be the ratio of pounds of sugar to gallons of water


C(t) = 5 + t
100 + 10t


The concentration after 12 minutes is given by evaluating  C(t)  at  t = 12.


C(12) = 5 + 12
100 + 10(12)


= 17
220


This means the concentration is 17 pounds of sugar to 220 gallons of water.
At the beginning, the concentration is


C(0) = 5 + 0
100 + 10(0)


= 1
20


Since   17
220


≈ 0.08 > 1
20


= 0.05,   the concentration is greater after 12 minutes than at the beginning.


Analysis


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3.40


To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the
denominator:


1
10


= 0.1


Notice the horizontal asymptote is  y = 0.1. This means the concentration,  C,   the ratio of pounds of sugar to
gallons of water, will approach 0.1 in the long term.


There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen
arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to
sophomores at 1 p.m.


Finding the Domains of Rational Functions
A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the
function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to
find the domain of a rational function, we need to determine which inputs would cause division by zero.


Domain of a Rational Function
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.


Given a rational function, find the domain.
1. Set the denominator equal to zero.
2. Solve to find the x-values that cause the denominator to equal zero.
3. The domain is all real numbers except those found in Step 2.


Example 3.62
Finding the Domain of a Rational Function


Find the domain of   f (x) = x + 3
x2 − 9


.


Solution
Begin by setting the denominator equal to zero and solving.


x2 − 9 = 0


x2 = 9
x = ± 3


The denominator is equal to zero when  x = ± 3. The domain of the function is all real numbers except
 x = ± 3.


Analysis
A graph of this function, as shown in Figure 3.74, confirms that the function is not defined when  x = ± 3.


Chapter 3 Polynomial and Rational Functions 399




3.41


Figure 3.74


There is a vertical asymptote at  x = 3  and a hole in the graph at  x = − 3. We will discuss these types of holes
in greater detail later in this section.


Find the domain of   f (x) = 4x
5(x − 1)(x − 5)


.


Identifying Vertical Asymptotes of Rational Functions
By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are
asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine
whether a given rational function has any asymptotes, and calculate their location.
Vertical Asymptotes
The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not
common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.


Given a rational function, identify any vertical asymptotes of its graph.
1. Factor the numerator and denominator.
2. Note any restrictions in the domain of the function.
3. Reduce the expression by canceling common factors in the numerator and the denominator.
4. Note any values that cause the denominator to be zero in this simplified version. These are where the
vertical asymptotes occur.


5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.


Example 3.63
Identifying Vertical Asymptotes


Find the vertical asymptotes of the graph of  k(x) = 5 + 2x2
2 − x − x2


.


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Solution
First, factor the numerator and denominator.


k(x) = 5 + 2x
2


2 − x − x2


= 5 + 2x
2


(2 + x)(1 − x)


To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator
equal to zero:


(2 + x)(1 − x) = 0
x = − 2, 1


Neither  x = – 2  nor  x = 1  are zeros of the numerator, so the two values indicate two vertical asymptotes. The
graph in Figure 3.75 confirms the location of the two vertical asymptotes.


Figure 3.75


Removable Discontinuities
Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call
such a hole a removable discontinuity.


For example, the function   f (x) = x2 − 1
x2 − 2x − 3


 may be re-written by factoring the numerator and the denominator.


f (x) = (x + 1)(x − 1)
(x + 1)(x − 3)


Notice that  x + 1  is a common factor to the numerator and the denominator. The zero of this factor,  x = − 1,   is the
location of the removable discontinuity. Notice also that  x – 3  is not a factor in both the numerator and denominator. The
zero of this factor,  x = 3,   is the vertical asymptote. See Figure 3.76.


Chapter 3 Polynomial and Rational Functions 401




Figure 3.76


Removable Discontinuities of Rational Functions
A removable discontinuity occurs in the graph of a rational function at  x = a  if  a  is a zero for a factor in the
denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for
common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable
discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the
multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.


Example 3.64
Identifying Vertical Asymptotes and Removable Discontinuities for a Graph


Find the vertical asymptotes and removable discontinuities of the graph of  k(x) = x − 2
x2 − 4


.


Solution
Factor the numerator and the denominator.


k(x) = x − 2
(x − 2)(x + 2)


Notice that there is a common factor in the numerator and the denominator,  x – 2. The zero for this factor is
 x = 2. This is the location of the removable discontinuity.
Notice that there is a factor in the denominator that is not in the numerator,  x + 2. The zero for this factor is
 x = − 2. The vertical asymptote is  x = − 2.  See Figure 3.77.


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3.42


Figure 3.77


The graph of this function will have the vertical asymptote at  x = −2,   but at  x = 2  the graph will have a hole.


Find the vertical asymptotes and removable discontinuities of the graph of   f (x) = x2 − 25
x3 − 6x2 + 5x


.


Identifying Horizontal Asymptotes of Rational Functions
While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes
help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will
mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the leading terms
of the numerator and denominator functions.
There are three distinct outcomes when checking for horizontal asymptotes:
Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at  y = 0.


Example: f (x) = 4x + 2
x2 + 4x − 5


In this case, the end behavior is   f (x) ≈ 4x
x2


= 4x . This tells us that, as the inputs increase or decrease without bound, this


function will behave similarly to the function  g(x) = 4x ,   and the outputs will approach zero, resulting in a horizontal
asymptote at  y = 0.  See Figure 3.78. Note that this graph crosses the horizontal asymptote.


Chapter 3 Polynomial and Rational Functions 403




Figure 3.78 Horizontal Asymptote  y = 0 when   f (x) = p(x)
q(x)


,  q(x) ≠ 0 where degree of p < degree o f q.


Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.


Example: f (x) = 3x
2 − 2x + 1
x − 1


In this case, the end behavior is   f (x) ≈ 3x2x = 3x. This tells us that as the inputs increase or decrease without bound, this
function will behave similarly to the function  g(x) = 3x. As the inputs grow large, the outputs will grow and not level off,
so this graph has no horizontal asymptote. However, the graph of  g(x) = 3x  looks like a diagonal line, and since   f  will
behave similarly to  g,   it will approach a line close to  y = 3x. This line is a slant asymptote.


To find the equation of the slant asymptote, divide  3x2 − 2x + 1
x − 1


. The quotient is  3x + 1,   and the remainder is 2. The
slant asymptote is the graph of the line  g(x) = 3x + 1.  See Figure 3.79.


Figure 3.79 Slant Asymptote when   f (x) = p(x)
q(x)


,  q(x) ≠ 0 
where degree of  p > degree of q by 1. 


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Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at  y = an
bn


,  where  an  


and  bn   are the leading coefficients of  p(x)  and  q(x)  for   f (x) = p(x)q(x) , q(x) ≠ 0.


Example: f (x) = 3x
2 + 2


x2 + 4x − 5


In this case, the end behavior is   f (x) ≈ 3x2
x2


= 3. This tells us that as the inputs grow large, this function will behave
like the function  g(x) = 3,  which is a horizontal line. As  x → ±∞, f (x) → 3,   resulting in a horizontal asymptote at
 y = 3.  See Figure 3.80. Note that this graph crosses the horizontal asymptote.


Figure 3.80 Horizontal Asymptote when
  f (x) = p(x)


q(x)
,  q(x) ≠ 0 where degree of p = degree of q.


Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a
horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph
will have at most one horizontal (or slant) asymptote.
It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end
behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function


f (x) = 3x
5 − x2
x + 3


with end behavior


f (x) ≈ 3x
5


x = 3x
4,


the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.
x → ±∞, f (x) → ∞


Horizontal Asymptotes of Rational Functions
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and
denominator.


• Degree of numerator is less than degree of denominator: horizontal asymptote at  y = 0.


Chapter 3 Polynomial and Rational Functions 405




• Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
• Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.


Example 3.65
Identifying Horizontal and Slant Asymptotes


For the functions below, identify the horizontal or slant asymptote.


a. g(x) = 6x3 − 10x
2x3 + 5x2


b. h(x) = x2 − 4x + 1
x + 2


c. k(x) = x2 + 4x
x3 − 8


Solution
For these solutions, we will use   f (x) = p(x)


q(x)
, q(x) ≠ 0.


a. g(x) = 6x3 − 10x
2x3 + 5x2


:  The degree of  p = degree of q = 3,   so we can find the horizontal asymptote by


taking the ratio of the leading terms. There is a horizontal asymptote at  y = 6
2
  or  y = 3.


b. h(x) = x2 − 4x + 1
x + 2


:  The degree of  p = 2  and degree of  q = 1.  Since  p > q  by 1, there is a slant


asymptote found at  x2 − 4x + 1
x + 2


.


21 −4 1
−2 12


1 −6 13
The quotient is  x – 2  and the remainder is 13. There is a slant asymptote at  y = – x – 2.


c. k(x) = x2 + 4x
x3 − 8


:  The degree of  p = 2 <   degree of  q = 3,   so there is a horizontal asymptote  y = 0.


Example 3.66
Identifying Horizontal Asymptotes


In the sugar concentration problem earlier, we created the equation  C(t) = 5 + t
100 + 10t


.


Find the horizontal asymptote and interpret it in context of the problem.


Solution


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Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a
horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is  t,  with
coefficient 1. In the denominator, the leading term is  10t,  with coefficient 10. The horizontal asymptote will be
at the ratio of these values:


t → ∞, C(t) → 1
10


This function will have a horizontal asymptote at  y = 1
10


.


This tells us that as the values of t increase, the values of  C will approach   1
10


.  In context, this means that, as
more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon
of water or   1


10
  pounds per gallon.


Example 3.67
Identifying Horizontal and Vertical Asymptotes


Find the horizontal and vertical asymptotes of the function


f (x) = (x − 2)(x + 3)
(x − 1)(x + 2)(x − 5)


Solution
First, note that this function has no common factors, so there are no potential removable discontinuities.
The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined.
The denominator will be zero at  x = 1, – 2, and 5,   indicating vertical asymptotes at these values.
The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater
than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to
tend towards zero as the inputs get large, and so as  x → ±∞, f (x) → 0. This function will have a horizontal
asymptote at  y = 0.  See Figure 3.81.


Chapter 3 Polynomial and Rational Functions 407




3.43


Figure 3.81


Find the vertical and horizontal asymptotes of the function:
f (x) = (2x − 1)(2x + 1)


(x − 2)(x + 3)


Intercepts of Rational Functions
A rational function will have a y-intercept when the input is zero, if the function is defined at zero. A rational function
will not have a y-intercept if the function is not defined at zero.
Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is
only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function
is equal to zero.


Example 3.68
Finding the Intercepts of a Rational Function


Find the intercepts of   f (x) = (x − 2)(x + 3)
(x − 1)(x + 2)(x − 5)


.


Solution
We can find the y-intercept by evaluating the function at zero


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3.44


f (0) = (0 − 2)(0 + 3)
(0 − 1)(0 + 2)(0 − 5)


= −6
10


= − 3
5


  = − 0.6
The x-intercepts will occur when the function is equal to zero:


0 = (x − 2)(x + 3)
(x − 1)(x + 2)(x − 5)


This is zero when the numerator is zero.


0 = (x − 2)(x + 3)
x = 2, − 3


The y-intercept is  (0, –0.6),   the x-intercepts are  (2, 0)  and  (–3, 0).  See Figure 3.82.


Figure 3.82


Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational
function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes.


Graphing Rational Functions
In Example 3.67, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the
denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer
powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with
polynomials.
The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal
functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of


Chapter 3 Polynomial and Rational Functions 409




the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative
infinity. See Figure 3.83.


Figure 3.83


When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads
toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure
3.84.


Figure 3.84


For example, the graph of   f (x) = (x + 1)
2(x − 3)


(x + 3)2(x − 2)
  is shown in Figure 3.85.


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Figure 3.85


• At the x-intercept  x = − 1  corresponding to the  (x + 1)2   factor of the numerator, the graph bounces, consistent
with the quadratic nature of the factor.


• At the x-intercept  x = 3  corresponding to the  (x − 3)  factor of the numerator, the graph passes through the axis as
we would expect from a linear factor.


• At the vertical asymptote  x = − 3  corresponding to the  (x + 3)2   factor of the denominator, the graph heads
towards positive infinity on both sides of the asymptote, consistent with the behavior of the function   f (x) = 1


x2
.


• At the vertical asymptote  x = 2,   corresponding to the  (x − 2)  factor of the denominator, the graph heads towards
positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the
behavior of the function   f (x) = 1x .


Chapter 3 Polynomial and Rational Functions 411




Given a rational function, sketch a graph.
1. Evaluate the function at 0 to find the y-intercept.
2. Factor the numerator and denominator.
3. For factors in the numerator not common to the denominator, determine where each factor of the
numerator is zero to find the x-intercepts.


4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points.
5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For
those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to
zero and then solve.


6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by
setting those factors equal to 0 and then solve.


7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant
asymptotes.


8. Sketch the graph.


Example 3.69
Graphing a Rational Function


Sketch a graph of   f (x) = (x + 2)(x − 3)
(x + 1)2(x − 2)


.


Solution
We can start by noting that the function is already factored, saving us a step.
Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept:


f (0) = (0 + 2)(0 − 3)
(0 + 1)2(0 − 2)


= 3


To find the x-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to
zero, we find x-intercepts at  x = –2  and  x = 3. At each, the behavior will be linear (multiplicity 1), with the
graph passing through the intercept.
We have a y-intercept at  (0, 3)  and x-intercepts at  (–2, 0)  and  (3, 0).
To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when
 x + 1 = 0  and when  x – 2 = 0,   giving us vertical asymptotes at  x = –1  and  x = 2.
There are no common factors in the numerator and denominator. This means there are no removable
discontinuities.
Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal
asymptote at  y = 0.
To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between
the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the
asymptotes, letting us fill in the middle portion of the graph as shown in Figure 3.86.


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3.45


Figure 3.86


The factor associated with the vertical asymptote at  x = −1 was squared, so we know the behavior will be
the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the
asymptote on the right, so the graph will head toward positive infinity on the left as well.
For the vertical asymptote at  x = 2,   the factor was not squared, so the graph will have opposite behavior on
either side of the asymptote. See Figure 3.87. After passing through the x-intercepts, the graph will then level
off toward an output of zero, as indicated by the horizontal asymptote.


Figure 3.87


Given the function   f (x) = (x + 2)
2(x − 2)


2(x − 1)2(x − 3)
,   use the characteristics of polynomials and rational


functions to describe its behavior and sketch the function.


Chapter 3 Polynomial and Rational Functions 413




Writing Rational Functions
Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use
information given by a graph to write the function. A rational function written in factored form will have an x-intercept
where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.)
As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing
a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the
denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a
corresponding set of factors.


Writing Rational Functions from Intercepts and Asymptotes
If a rational function has x-intercepts at  x = x1, x2, ..., xn,   vertical asymptotes at  x = v1, v2, … , vm,   and
no  xi = any v j,   then the function can be written in the form:


f (x) = a
(x − x1)


p1 (x − x2)
p2 ⋯ (x − xn)


pn


(x − v1)
q1 (x − v2)


q2 ⋯ (x − vm)
qn


where the powers  pi   or  qi   on each factor can be determined by the behavior of the graph at the corresponding
intercept or asymptote, and the stretch factor  a  can be determined given a value of the function other than the x-
intercept or by the horizontal asymptote if it is nonzero.


Given a graph of a rational function, write the function.
1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine
the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small
multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)


2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote
to determine the factors and their powers.


3. Use any clear point on the graph to find the stretch factor.


Example 3.70
Writing a Rational Function from Intercepts and Asymptotes


Write an equation for the rational function shown in Figure 3.88.


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Figure 3.88


Solution
The graph appears to have x-intercepts at  x = – 2  and  x = 3. At both, the graph passes through the intercept,
suggesting linear factors. The graph has two vertical asymptotes. The one at  x = – 1  seems to exhibit the basic
behavior similar to  1x ,  with the graph heading toward positive infinity on one side and heading toward negative
infinity on the other. The asymptote at  x = 2  is exhibiting a behavior similar to   1


x2
,  with the graph heading


toward negative infinity on both sides of the asymptote. See Figure 3.89.


Figure 3.89


Chapter 3 Polynomial and Rational Functions 415




We can use this information to write a function of the form


f (x) = a (x + 2)(x − 3)
(x + 1)(x − 2)2


.


To find the stretch factor, we can use another clear point on the graph, such as the y-intercept  (0, –2).


−2 = a (0 + 2)(0 − 3)
(0 + 1)(0 − 2)2


−2 = a−6
4


a = −8
−6


= 4
3


This gives us a final function of   f (x) = 4(x + 2)(x − 3)
3(x + 1)(x − 2)2


.


Access these online resources for additional instruction and practice with rational functions.
• Graphing Rational Functions (http://openstaxcollege.org/l/graphrational)
• Find the Equation of a Rational Function (http://openstaxcollege.org/l/equatrational)
• Determining Vertical and Horizontal Asymptotes (http://openstaxcollege.org/l/asymptote)
• Find the Intercepts, Asymptotes, and Hole of a Rational Function
(http://openstaxcollege.org/l/interasymptote)


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455.


456.


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484.


485.


3.7 EXERCISES
Verbal


What is the fundamental difference in the algebraic
representation of a polynomial function and a rational
function?


What is the fundamental difference in the graphs of
polynomial functions and rational functions?


If the graph of a rational function has a removable
discontinuity, what must be true of the functional rule?


Can a graph of a rational function have no vertical
asymptote? If so, how?


Can a graph of a rational function have no x-
intercepts? If so, how?


Algebraic
For the following exercises, find the domain of the rational
functions.


f (x) = x − 1
x + 2


f (x) = x + 1
x2 − 1


f (x) = x
2 + 4


x2 − 2x − 8


f (x) = x
2 + 4x − 3


x4 − 5x2 + 4


For the following exercises, find the domain, vertical
asymptotes, and horizontal asymptotes of the functions.


f (x) = 4
x − 1


f (x) = 2
5x + 2


f (x) = x
x2 − 9


f (x) = x
x2 + 5x − 36


f (x) = 3 + x
x3 − 27


f (x) = 3x − 4
x3 − 16x


f (x) = x
2 − 1


x3 + 9x2 + 14x


f (x) = x + 5
x2 − 25


f (x) = x − 4
x − 6


f (x) = 4 − 2x
3x − 1


For the following exercises, find the x- and y-intercepts for
the functions.


f (x) = x + 5
x2 + 4


f (x) = x
x2 − x


f (x) = x
2 + 8x + 7


x2 + 11x + 30


f (x) = x
2 + x + 6


x2 − 10x + 24


f (x) = 94 − 2x
2


3x2 − 12


For the following exercises, describe the local and end
behavior of the functions.


f (x) = x
2x + 1


f (x) = 2x
x − 6


f (x) = −2x
x − 6


f (x) = x
2 − 4x + 3


x2 − 4x − 5


f (x) = 2x
2 − 32


6x2 + 13x − 5


For the following exercises, find the slant asymptote of the
functions.


f (x) = 24x
2 + 6x


2x + 1


Chapter 3 Polynomial and Rational Functions 417




486.


487.


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490.


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507.


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509.


510.


511.


512.


f (x) = 4x
2 − 10


2x − 4


f (x) = 81x
2 − 18


3x − 2


f (x) = 6x
3 − 5x


3x2 + 4


f (x) = x
2 + 5x + 4
x − 1


Graphical
For the following exercises, use the given transformation
to graph the function. Note the vertical and horizontal
asymptotes.


The reciprocal function shifted up two units.
The reciprocal function shifted down one unit and left


three units.
The reciprocal squared function shifted to the right 2


units.
The reciprocal squared function shifted down 2 units


and right 1 unit.
For the following exercises, find the horizontal intercepts,
the vertical intercept, the vertical asymptotes, and the
horizontal or slant asymptote of the functions. Use that
information to sketch a graph.


p(x) = 2x − 3
x + 4


q(x) = x − 5
3x − 1


s(x) = 4
(x − 2)2


r(x) = 5
(x + 1)2


f (x) = 3x
2 − 14x − 5


3x2 + 8x − 16


g(x) = 2x
2 + 7x − 15


3x2 − 14 + 15


a(x) = x
2 + 2x − 3
x2 − 1


b(x) = x
2 − x − 6
x2 − 4


h(x) = 2x
2 + x − 1
x − 4


k(x) = 2x
2 − 3x − 20
x − 5


w(x) = (x − 1)(x + 3)(x − 5)
(x + 2)2(x − 4)


z(x) = (x + 2)
2 (x − 5)


(x − 3)(x + 1)(x + 4)


For the following exercises, write an equation for a rational
function with the given characteristics.


Vertical asymptotes at  x = 5  and  x = − 5,   x-
intercepts at  (2, 0)  and  ( − 1, 0),   y-intercept at  (0, 4)


Vertical asymptotes at  x = − 4  and  x = − 1,
x-intercepts at  (1, 0)  and  (5, 0),   y-intercept at  (0, 7)


Vertical asymptotes at  x = − 4  and  x = − 5,   x-
intercepts at  (4, 0)  and  (−6, 0),  Horizontal asymptote at
 y = 7


Vertical asymptotes at  x = − 3  and  x = 6,   x-
intercepts at  (−2, 0)  and  (1, 0),  Horizontal asymptote at
 y = − 2


Vertical asymptote at  x = − 1,  Double zero at
 x = 2,   y-intercept at  (0, 2)


Vertical asymptote at  x = 3,  Double zero at
 x = 1,   y-intercept at  (0, 4)


For the following exercises, use the graphs to write an
equation for the function.


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513.


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521.


522.


Numeric
For the following exercises, make tables to show the
behavior of the function near the vertical asymptote and
reflecting the horizontal asymptote


f (x) = 1
x − 2


f (x) = x
x − 3


f (x) = 2x
x + 4


f (x) = 2x
(x − 3)2


Chapter 3 Polynomial and Rational Functions 419




523.


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542.


f (x) = x
2


x2 + 2x + 1


Technology
For the following exercises, use a calculator to graph   f (x). 
Use the graph to solve   f (x) > 0.


f (x) = 2
x + 1


f (x) = 4
2x − 3


f (x) = 2
(x − 1)(x + 2)


f (x) = x + 2
(x − 1)(x − 4)


f (x) = (x + 3)
2


(x − 1)2 (x + 1)


Extensions
For the following exercises, identify the removable
discontinuity.


f (x) = x
2 − 4
x − 2


f (x) = x
3 + 1
x + 1


f (x) = x
2 + x − 6
x − 2


f (x) = 2x
2 + 5x − 3
x + 3


f (x) = x
3 + x2
x + 1


Real-World Applications
For the following exercises, express a rational function that
describes the situation.


A large mixing tank currently contains 200 gallons of
water, into which 10 pounds of sugar have been mixed. A
tap will open, pouring 10 gallons of water per minute into
the tank at the same time sugar is poured into the tank at a
rate of 3 pounds per minute. Find the concentration (pounds
per gallon) of sugar in the tank after  t minutes.


A large mixing tank currently contains 300 gallons of
water, into which 8 pounds of sugar have been mixed. A tap
will open, pouring 20 gallons of water per minute into the
tank at the same time sugar is poured into the tank at a rate


of 2 pounds per minute. Find the concentration (pounds per
gallon) of sugar in the tank after  t minutes.
For the following exercises, use the given rational function
to answer the question.


The concentration  C  of a drug in a patient’s
bloodstream  t  hours after injection in given by
 C(t) = 2t


3 + t2
. What happens to the concentration of the


drug as  t  increases?
The concentration  C  of a drug in a patient’s


bloodstream  t  hours after injection is given by
 C(t) = 100t


2t2 + 75
. Use a calculator to approximate the


time when the concentration is highest.
For the following exercises, construct a rational function
that will help solve the problem. Then, use a calculator to
answer the question.


An open box with a square base is to have a volume of
108 cubic inches. Find the dimensions of the box that will
have minimum surface area. Let  x  = length of the side of
the base.


A rectangular box with a square base is to have a
volume of 20 cubic feet. The material for the base costs 30
cents/ square foot. The material for the sides costs 10 cents/
square foot. The material for the top costs 20 cents/square
foot. Determine the dimensions that will yield minimum
cost. Let  x  = length of the side of the base.


A right circular cylinder has volume of 100 cubic
inches. Find the radius and height that will yield minimum
surface area. Let  x  = radius.


A right circular cylinder with no top has a volume of
50 cubic meters. Find the radius that will yield minimum
surface area. Let  x  = radius.


A right circular cylinder is to have a volume of 40
cubic inches. It costs 4 cents/square inch to construct the
top and bottom and 1 cent/square inch to construct the rest
of the cylinder. Find the radius to yield minimum cost. Let
 x  = radius.


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3.8 | Inverses and Radical Functions
Learning Objectives


In this section, you will:
3.8.1 Find the inverse of a polynomial function.
3.8.2 Restrict the domain to find the inverse of a polynomial function.


A mound of gravel is in the shape of a cone with the height equal to twice the radius.


Figure 3.90


The volume is found using a formula from elementary geometry.
V = 1


3
πr2h


= 1
3
πr2(2r)


= 2
3
πr3


We have written the volume  V   in terms of the radius  r. However, in some cases, we may start out with the volume and
want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with
a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula


r = 3V


3


This function is the inverse of the formula for  V   in terms of  r.
In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we
encounter in the process.
Finding the Inverse of a Polynomial Function
Two functions   f   and  g  are inverse functions if for every coordinate pair in   f , (a, b),   there exists a corresponding
coordinate pair in the inverse function,  g, (b,  a).  In other words, the coordinate pairs of the inverse functions have the
input and output interchanged.
For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse
function.


Chapter 3 Polynomial and Rational Functions 421




For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure 3.91. We can
use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.


Figure 3.91


Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system
at the cross section, with  x measured horizontally and  y measured vertically, with the origin at the vertex of the parabola.
See Figure 3.92.


Figure 3.92


From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the
equation will have form  y(x) = ax2. Our equation will need to pass through the point (6, 18), from which we can solve for
the stretch factor  a.


18 = a62


a = 18
36


= 1
2


Our parabolic cross section has the equation
y(x) = 1


2
x2


We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of
the water depth. For any depth  y  the width will be given by  2x,   so we need to solve the equation above for  x  and find the


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inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two
inputs that produce the same output, one positive and one negative.
To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes
sense to restrict ourselves to positive  x  values. On this domain, we can find an inverse by solving for the input variable:


   y = 1
2
x2


2y = x2


x = ± 2y


This is not a function as written. We are limiting ourselves to positive  x  values, so we eliminate the negative solution,
giving us the inverse function we’re looking for.


y = x
2


2
, x > 0


Because  x  is the distance from the center of the parabola to either side, the entire width of the water at the top will be  2x. 
The trough is 3 feet (36 inches) long, so the surface area will then be:


Area = l ⋅ w
= 36 ⋅ 2x
= 72x
= 72 2y


This example illustrates two important points:
1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power
functions.


Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial
functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation
  f −1(x).
Warning:   f −1(x)  is not the same as the reciprocal of the function   f (x). This use of “–1” is reserved to denote inverse
functions. To denote the reciprocal of a function   f (x),  we would need to write  ⎛⎝ f (x)⎞⎠−1 = 1f (x).


An important relationship between inverse functions is that they “undo” each other. If   f −1   is the inverse of a function   f ,  
then   f   is the inverse of the function   f −1.  In other words, whatever the function   f   does to  x,   f −1   undoes it—and vice-
versa. More formally, we write


f −1 ⎛⎝ f (x)⎞⎠ = x,  for all x in the domain of f
and


f ⎛⎝ f
−1 (x)⎞⎠ = x,  for all x in the domain of f −1


Verifying Two Functions Are Inverses of One Another
Two functions,   f   and  g,   are inverses of one another if for all  x  in the domain of   f   and  g.


g⎛⎝ f (x)⎞⎠ = f ⎛⎝g(x)⎞⎠ = x


Chapter 3 Polynomial and Rational Functions 423




3.46


Given a polynomial function, find the inverse of the function by restricting the domain in such a way that
the new function is one-to-one.


1. Replace   f (x) with  y.
2. Interchange  x  and  y.
3. Solve for  y,   and rename the function   f −1(x).


Example 3.71
Verifying Inverse Functions


Show that   f (x) = 1
x + 1


  and   f −1 (x) = 1x − 1  are inverses, for  x ≠ 0, − 1 .


Solution
We must show that   f −1 ⎛⎝ f (x)⎞⎠ = x  and   f ⎛⎝ f −1 (x)⎞⎠ = x.


f −1( f (x)) = f −1 ⎛⎝
1


x + 1



= 1
1


x + 1


− 1


= (x + 1) − 1
= x


f ( f −1(x)) = f ⎛⎝
1
x − 1





= 1⎛

1
x − 1



⎠+ 1


= 1
1
x


= x


Therefore,   f (x) = 1
x + 1


  and   f −1 (x) = 1x − 1  are inverses.


Show that   f (x) = x + 5
3


  and   f −1 (x) = 3x − 5  are inverses.


Example 3.72
Finding the Inverse of a Cubic Function


Find the inverse of the function   f (x) = 5x3 + 1.


Solution


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3.47


This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know
it is one-to-one. Solving for the inverse by solving for  x.


y = 5x3 + 1


x = 5y3 + 1


x − 1 = 5y3


x − 1
5


= y3


f −1(x) = x − 1
5


3


Analysis
Look at the graph of   f   and   f – 1. Notice that the two graphs are symmetrical about the line  y = x. This is
always the case when graphing a function and its inverse function.
Also, since the method involved interchanging  x  and  y,   notice corresponding points. If  (a, b)  is on the graph
of   f , then  (b, a)  is on the graph of   f – 1.  Since  (0, 1)  is on the graph of   f ,   then  (1, 0)  is on the graph of
  f – 1.  Similarly, since  (1, 6)  is on the graph of   f , then  (6, 1)  is on the graph of   f – 1.  See Figure 3.93.


Figure 3.93


Find the inverse function of   f (x) = x + 43 .


Restricting the Domain to Find the Inverse of a Polynomial Function
So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However,
as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain


Chapter 3 Polynomial and Rational Functions 425




restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an
inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.


Restricting the Domain
If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes
one-to-one, thus creating a new function, this new function will have an inverse.


Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the
inverse.


1. Restrict the domain by determining a domain on which the original function is one-to-one.
2. Replace   f (x) with y.
3. Interchange  x and y.
4. Solve for  y,   and rename the function or pair of function   f −1(x).
5. Revise the formula for   f −1(x)  by ensuring that the outputs of the inverse function correspond to the
restricted domain of the original function.


Example 3.73
Restricting the Domain to Find the Inverse of a Polynomial Function


Find the inverse function of   f :
a. f (x) = (x − 4)2, x ≥ 4


b. f (x) = (x − 4)2, x ≤ 4


Solution
The original function   f (x) = (x − 4)2   is not one-to-one, but the function is restricted to a domain of  x ≥ 4  or
 x ≤ 4  on which it is one-to-one. See Figure 3.94.


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Figure 3.94


To find the inverse, start by replacing   f (x) with the simple variable  y.
             y = (x − 4)2 Interchange x and y.
             x = (y − 4)2 Take the square root.
    ± x = y − 4 Add 4 to both sides.
4 ± x = y


This is not a function as written. We need to examine the restrictions on the domain of the original function to
determine the inverse. Since we reversed the roles of  x  and  y  for the original   f (x),  we looked at the domain:
the values  x  could assume. When we reversed the roles of  x  and  y,   this gave us the values  y  could assume.
For this function,  x ≥ 4,   so for the inverse, we should have  y ≥ 4,  which is what our inverse function gives.
a. The domain of the original function was restricted to  x ≥ 4,   so the outputs of the inverse need to be the
same,   f (x) ≥ 4,   and we must use the + case:


f −1(x) = 4 + x


Chapter 3 Polynomial and Rational Functions 427




b. The domain of the original function was restricted to  x ≤ 4,   so the outputs of the inverse need to be the
same,   f (x) ≤ 4,   and we must use the – case:


f −1(x) = 4 − x


Analysis
On the graphs in Figure 3.95, we see the original function graphed on the same set of axes as its inverse function.
Notice that together the graphs show symmetry about the line  y = x. The coordinate pair  (4, 0)  is on the graph
of   f   and the coordinate pair  (0, 4)  is on the graph of   f −1.  For any coordinate pair, if  (a, b)  is on the graph
of   f ,   then  (b, a)  is on the graph of   f −1.  Finally, observe that the graph of   f   intersects the graph of   f −1 on
the line  y = x.  Points of intersection for the graphs of   f   and   f −1  will always lie on the line  y = x.


Figure 3.95


Example 3.74
Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified


Restrict the domain and then find the inverse of
f (x) = (x − 2)2 − 3.


Solution
We can see this is a parabola with vertex at  (2, – 3)  that opens upward. Because the graph will be decreasing
on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it
will be one-to-one by limiting the domain to  x ≥ 2.
To find the inverse, we will use the vertex form of the quadratic. We start by replacing   f (x) with a simple
variable,  y,   then solve for  x.


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                         y = (x − 2)2 − 3 Interchange x and y.
                         x = (y − 2)2 − 3 Add 3 to both sides.
               x + 3 = (y − 2)2 Take the square root.
   ± x + 3 = y − 2 Add 2 to both sides.
2 ± x + 3 = y Rename the function.


           f −1(x) = 2 ± x + 3
Now we need to determine which case to use. Because we restricted our original function to a domain of  x ≥ 2,  
the outputs of the inverse should be the same, telling us to utilize the + case


f −1(x) = 2 + x + 3


If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way
we may easily observe the coordinates of the vertex to help us restrict the domain.


Analysis
Notice that we arbitrarily decided to restrict the domain on  x ≥ 2. We could just have easily opted to restrict the
domain on  x ≤ 2,   in which case   f −1(x) = 2 − x + 3. Observe the original function graphed on the same set
of axes as its inverse function in Figure 3.96. Notice that both graphs show symmetry about the line  y = x. The
coordinate pair  (2, − 3)  is on the graph of   f   and the coordinate pair  (−3, 2)  is on the graph of   f −1. Observe
from the graph of both functions on the same set of axes that


domain of f = range of f – 1 = [2, ∞)


and
domain of f – 1 = range of f = [ – 3, ∞)


Finally, observe that the graph of   f   intersects the graph of   f −1   along the line  y = x.


Figure 3.96


Chapter 3 Polynomial and Rational Functions 429




3.48 Find the inverse of the function   f (x) = x2 + 1,   on the domain  x ≥ 0.


Solving Applications of Radical Functions
Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we
want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the
original function is limited.


Given a radical function, find the inverse.
1. Determine the range of the original function.
2. Replace   f (x)  with  y,   then solve for  x.
3. If necessary, restrict the domain of the inverse function to the range of the original function.


Example 3.75
Finding the Inverse of a Radical Function


Restrict the domain and then find the inverse of the function   f (x) = x − 4.


Solution
Note that the original function has range   f (x) ≥ 0. Replace   f (x) with  y,   then solve for  x.


y = x − 4 Replace f (x)withy.


x = y − 4 Interchangexandy.


x = y − 4 Square each side.


x2 = y − 4 Add 4.


x2 + 4 = y Rename the function f −1(x).


f −1(x) = x2 + 4


Recall that the domain of this function must be limited to the range of the original function.
f −1(x) = x2 + 4, x ≥ 0


Analysis
Notice in Figure 3.97 that the inverse is a reflection of the original function over the line  y = x. Because the
original function has only positive outputs, the inverse function has only positive inputs.


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3.49


Figure 3.97


Restrict the domain and then find the inverse of the function   f (x) = 2x + 3.


Solving Applications of Radical Functions
Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to
solve the problem posed at the start of the section.


Example 3.76
Solving an Application with a Cubic Function


A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in
terms of the radius is given by


V = 2
3
πr3


Find the inverse of the function  V = 2
3
πr3   that determines the volume  V   of a cone and is a function of the radius


 r. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use
 π = 3.14.


Solution
Start with the given function for  V . Notice that the meaningful domain for the function is  r ≥ 0  since negative
radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse
function) is  V ≥ 0.  Solve for  r  in terms of  V ,   using the method outlined previously.


Chapter 3 Polynomial and Rational Functions 431




  V = 2
3
πr3


r3 = 3V


Solve for r3.


  r = 3V


3
Solve for r.


This is the result stated in the section opener. Now evaluate this for  V = 100  and  π = 3.14.


r = 3V


3


= 3 ⋅ 100
2 ⋅ 3.14


3


≈ 47.7707
3


≈ 3.63


Therefore, the radius is about 3.63 ft.


Determining the Domain of a Radical Function Composed with Other Functions
When radical functions are composed with other functions, determining domain can become more complicated.


Example 3.77
Finding the Domain of a Radical Function Composed with a Rational Function


Find the domain of the function   f (x) = (x + 2)(x − 3)
(x − 1)


.


Solution
Because a square root is only defined when the quantity under the radical is non-negative, we need to determine
where  (x + 2)(x − 3)


(x − 1)
≥ 0. The output of a rational function can change signs (change from positive to negative


or vice versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at x =
–2, 1, and 3.
To determine the intervals on which the rational expression is positive, we could test some values in the
expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as
shown in Figure 3.98.


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Figure 3.98


This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one
vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function,
and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the
denominator. There is a y-intercept at  (0, 6).
From the y-intercept and x-intercept at  x = − 2,  we can sketch the left side of the graph. From the behavior at
the asymptote, we can sketch the right side of the graph.
From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure
that the original function   f (x) will be defined.   f (x)  has domain  − 2 ≤ x < 1 or x ≥ 3,   or in interval notation,
 [ − 2, 1) ∪ [3, ∞).


Finding Inverses of Rational Functions
As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function,
particularly of rational functions that are the ratio of linear functions, such as in concentration applications.


Example 3.78
Finding the Inverse of a Rational Function


The function  C = 20 + 0.4n
100 + n


  represents the concentration  C  of an acid solution after  n mL of 40% solution has
been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for  n 
in terms of  C. Then use your result to determine how much of the 40% solution should be added so that the final
mixture is a 35% solution.


Solution


Chapter 3 Polynomial and Rational Functions 433




3.50


We first want the inverse of the function. We will solve for  n  in terms of  C.
  C = 20 + 0.4n


100 + n
  C(100 + n) = 20 + 0.4n
  100C + Cn = 20 + 0.4n
  100C − 20 = 0.4n − Cn
  100C − 20 = (0.4 − C)n


  n = 100C − 20
0.4 − C


Now evaluate this function for  C = 0.35 (35%).


 n = 100(0.35) − 20
0.4 − 0.35


= 15
0.05


= 300


We can conclude that 300 mL of the 40% solution should be added.


Find the inverse of the function   f (x) = x + 3
x − 2


.


Access these online resources for additional instruction and practice with inverses and radical functions.
• Graphing the Basic Square Root Function (http://openstaxcollege.org/l/graphsquareroot)
• Find the Inverse of a Square Root Function (http://openstaxcollege.org/l/inversesquare)
• Find the Inverse of a Rational Function (http://openstaxcollege.org/l/inverserational)
• Find the Inverse of a Rational Function and an Inverse Function Value
(http://openstaxcollege.org/l/rationalinverse)


• Inverse Functions (http://openstaxcollege.org/l/inversefunction)


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3.8 EXERCISES
Verbal


Explain why we cannot find inverse functions for all
polynomial functions.


Why must we restrict the domain of a quadratic
function when finding its inverse?


When finding the inverse of a radical function, what
restriction will we need to make?


The inverse of a quadratic function will always take
what form?


Algebraic
For the following exercises, find the inverse of the function
on the given domain.


f (x) = (x − 4)2, [4, ∞)


f (x) = (x + 2)2, [ − 2, ∞)


f (x) = (x + 1)2 − 3, [ − 1, ∞)


f (x) = 2 − 3 + x


f (x) = 3x2 + 5,   (−∞, 0⎤⎦, [0, ∞)


f (x) = 12 − x2, [0, ∞)


f (x) = 9 − x2, [0, ∞)


f (x) = 2x2 + 4, [0, ∞)


For the following exercises, find the inverse of the
functions.


f (x) = x3 + 5


f (x) = 3x3 + 1


f (x) = 4 − x3


f (x) = 4 − 2x3


For the following exercises, find the inverse of the
functions.


f (x) = 2x + 1


f (x) = 3 − 4x


f (x) = 9 + 4x − 4


f (x) = 6x − 8 + 5


f (x) = 9 + 2 x3


f (x) = 3 − x3


f (x) = 2
x + 8


f (x) = 3
x − 4


f (x) = x + 3
x + 7


f (x) = x − 2
x + 7


f (x) = 3x + 4
5 − 4x


f (x) = 5x + 1
2 − 5x


f (x) = x2 + 2x, [ − 1, ∞)


f (x) = x2 + 4x + 1, [ − 2, ∞)


f (x) = x2 − 6x + 3, [3, ∞)


Graphical
For the following exercises, find the inverse of the function
and graph both the function and its inverse.


f (x) = x2 + 2,  x ≥ 0


f (x) = 4 − x2,  x ≥ 0


f (x) = (x + 3)2,  x ≥ − 3


f (x) = (x − 4)2,  x ≥ 4


f (x) = x3 + 3


f (x) = 1 − x3


f (x) = x2 + 4x,  x ≥ − 2


f (x) = x2 − 6x + 1,  x ≥ 3


Chapter 3 Polynomial and Rational Functions 435




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f (x) = 2x


f (x) = 1
x2


,  x ≥ 0


For the following exercises, use a graph to help determine
the domain of the functions.


f (x) = (x + 1)(x − 1)x


f (x) = (x + 2)(x − 3)
x − 1


f (x) = x(x + 3)
x − 4


f (x) = x
2 − x − 20
x − 2


f (x) = 9 − x
2


x + 4


Technology
For the following exercises, use a calculator to graph the
function. Then, using the graph, give three points on the
graph of the inverse with y-coordinates given.


f (x) = x3 − x − 2,  y = 1, 2, 3


f (x) = x3 + x − 2, y = 0, 1, 2


f (x) = x3 + 3x − 4, y = 0, 1, 2


f (x) = x3 + 8x − 4, y = − 1, 0, 1


f (x) = x4 + 5x + 1, y = − 1, 0, 1


Extensions
For the following exercises, find the inverse of the
functions with  a, b, c  positive real numbers.


f (x) = ax3 + b


f (x) = x2 + bx


f (x) = ax2 + b


f (x) = ax + b
3


f (x) = ax + bx + c


Real-World Applications
For the following exercises, determine the function
described and then use it to answer the question.


An object dropped from a height of 200 meters has a
height,  h(t),   in meters after  t  seconds have lapsed, such
that  h(t) = 200 − 4.9t2. Express  t  as a function of height,
 h,   and find the time to reach a height of 50 meters.


An object dropped from a height of 600 feet has a
height,  h(t),   in feet after  t  seconds have elapsed, such
that  h(t) = 600 − 16t2. Express  t  as a function of height
 h,   and find the time to reach a height of 400 feet.


The volume,  V ,   of a sphere in terms of its radius,
 r,   is given by  V(r) = 4


3
πr3. Express  r  as a function of


 V ,   and find the radius of a sphere with volume of 200
cubic feet.


The surface area,  A,   of a sphere in terms of its
radius,  r,   is given by  A(r) = 4πr2. Express  r  as a
function of  V ,   and find the radius of a sphere with a
surface area of 1000 square inches.


A container holds 100 ml of a solution that is 25 ml
acid. If  n ml of a solution that is 60% acid is added, the
function  C(n) = 25 + .6n


100 + n
  gives the concentration,  C,   as


a function of the number of ml added,  n. Express  n  as a
function of  C  and determine the number of mL that need to
be added to have a solution that is 50% acid.


The period  T ,   in seconds, of a simple pendulum as a
function of its length  l,   in feet, is given by
 T(l) = 2π l


32.2
  . Express  l  as a function of  T   and


determine the length of a pendulum with period of 2
seconds.


The volume of a cylinder ,  V ,   in terms of radius,
 r,   and height,  h,   is given by  V = πr2h.  If a cylinder
has a height of 6 meters, express the radius as a function of
 V   and find the radius of a cylinder with volume of 300
cubic meters.


The surface area,  A,   of a cylinder in terms of its
radius,  r,   and height,  h,   is given by  A = 2πr2 + 2πrh. 
If the height of the cylinder is 4 feet, express the radius as a
function of  V   and find the radius if the surface area is 200
square feet.


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607.


608.


The volume of a right circular cone,  V ,   in terms of
its radius,  r,   and its height,  h,   is given by  V = 1


3
πr2h. 


Express  r  in terms of  h  if the height of the cone is 12 feet
and find the radius of a cone with volume of 50 cubic
inches.


Consider a cone with height of 30 feet. Express the
radius,  r,   in terms of the volume,  V ,   and find the radius
of a cone with volume of 1000 cubic feet.


Chapter 3 Polynomial and Rational Functions 437




3.9 | Modeling Using Variation
Learning Objectives


In this section, you will:
3.9.1 Solve direct variation problems.
3.9.2 Solve inverse variation problems.
3.9.3 Solve problems involving joint variation.


A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission
on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn
$736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one,
between earnings, sales, and commission rate.
Solving Direct Variation Problems
In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula  e = 0.16s 
tells us her earnings,  e,   come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a
table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 3.10.


 s  , sales
prices e = 0.16s Interpretation


$4,600 e = 0.16(4,600) = 736 A sale of a $4,600 vehicle results in $736 earnings.


$9,200 e = 0.16(9,200) = 1,472 A sale of a $9,200 vehicle results in $1472 earnings.


$18,400 e = 0.16(18,400) = 2,944 A sale of a $18,400 vehicle results in $2944earnings.


Table 3.10


Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the
vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases
as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct
variation. Each variable in this type of relationship varies directly with the other.
Figure 3.99 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of
the car. The formula  y = kxn   is used for direct variation. The value  k  is a nonzero constant greater than zero and is called
the constant of variation. In this case,  k = 0.16  and  n = 1. 


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Figure 3.99


Direct Variation
If  x and y  are related by an equation of the form


(3.6) y = kxn  
then we say that the relationship is direct variation and  y  varies directly with the  nth  power of  x.  In direct
variation relationships, there is a nonzero constant ratio  k = y


xn
,  where  k  is called the constant of variation, which


help defines the relationship between the variables.


Given a description of a direct variation problem, solve for an unknown.
1. Identify the input,  x, and the output,  y. 
2. Determine the constant of variation. You may need to divide  y  by the specified power of  x  to determine
the constant of variation.


3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.


Example 3.79
Solving a Direct Variation Problem


The quantity  y  varies directly with the cube of  x.  If  y = 25 when  x = 2,   find  y when  x  is 6.


Solution
The general formula for direct variation with a cube is  y = kx3. The constant can be found by dividing  y  by the
cube of  x. 


Chapter 3 Polynomial and Rational Functions 439




3.51


k =
y


x3


= 25
23


= 25
8


Now use the constant to write an equation that represents this relationship.
y = 25


8
x3


Substitute  x = 6  and solve for  y.


y = 25
8
(6)3


= 675


Analysis
The graph of this equation is a simple cubic, as shown in Figure 3.100.


Figure 3.100


Do the graphs of all direct variation equations look like Example 3.79?
No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc.
But all of the graphs pass through  (0,0).


The quantity  y  varies directly with the square of  x.  If  y = 24 when  x = 3,   find  y when  x  is 4.


Solving Inverse Variation Problems
Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula
 T = 14,000


d
  gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic


Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.
If we create Table 3.11, we observe that, as the depth increases, the water temperature decreases.


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d,  depth T = 14,000
d Interpretation


500 ft 14,000500 = 28 At a depth of 500 ft, the water temperature is 28° F.


350 ft 14,000350 = 40 At a depth of 350 ft, the water temperature is 40° F.


250 ft 14,000250 = 56 At a depth of 250 ft, the water temperature is 56° F.


Table 3.11


We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities
are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships
are also called inverse variations.
For our example, Figure 3.101 depicts the inverse variation. We say the water temperature varies inversely with the depth
of the water because, as the depth increases, the temperature decreases. The formula  y = kx   for inverse variation in this case
uses  k = 14,000. 


Figure 3.101


Inverse Variation
If  x  and  y  are related by an equation of the form


(3.7)y = k
xn


where  k  is a nonzero constant, then we say that  y varies inversely with the  nth  power of  x.  In inversely
proportional relationships, or inverse variations, there is a constant multiple  k = xn y. 


Example 3.80
Writing a Formula for an Inversely Proportional Relationship


Chapter 3 Polynomial and Rational Functions 441




A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the
tourist drives.


Solution
Recall that multiplying speed by time gives distance. If we let  t  represent the drive time in hours, and  v  represent
the velocity (speed or rate) at which the tourist drives, then  vt = distance. Because the distance is fixed at 100
miles,  vt = 100.  Solving this relationship for the time gives us our function.


t(v) = 100v


= 100v−1


We can see that the constant of variation is 100 and, although we can write the relationship using the negative
exponent, it is more common to see it written as a fraction.


Given a description of an indirect variation problem, solve for an unknown.
1. Identify the input,  x,   and the output,  y.
2. Determine the constant of variation. You may need to multiply  y  by the specified power of  x  to determine
the constant of variation.


3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.


Example 3.81
Solving an Inverse Variation Problem


A quantity  y  varies inversely with the cube of  x.  If  y = 25 when  x = 2,   find  y when  x  is 6.


Solution
The general formula for inverse variation with a cube is  y = k


x3
. The constant can be found by multiplying  y  by


the cube of  x.
k = x3 y


= 23 ⋅ 25
= 200


Now we use the constant to write an equation that represents this relationship.
y = k


x3
,   k = 200


y = 200
x3


Substitute  x = 6  and solve for  y.


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3.52


y = 200
63


= 25
27


Analysis
The graph of this equation is a rational function, as shown in Figure 3.102.


Figure 3.102


A quantity  y  varies inversely with the square of  x.  If  y = 8 when  x = 3,   find  y when  x  is 4.


Solving Problems Involving Joint Variation
Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends
on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called
joint variation. For example, the cost of busing students for each school trip varies with the number of students attending
and the distance from the school. The variable  c, cost, varies jointly with the number of students,  n, and the distance,
 d. 


Joint Variation
Joint variation occurs when a variable varies directly or inversely with multiple variables.
For instance, if  x  varies directly with both  y  and  z,   we have  x = kyz.  If  x  varies directly with  y  and inversely with
z, we have  x = kyz . Notice that we only use one constant in a joint variation equation.


Example 3.82
Solving Problems Involving Joint Variation


A quantity  x  varies directly with the square of  y  and inversely with the cube root of  z.  If  x = 6 when  y = 2 
and  z = 8,   find  x when  y = 1  and  z = 27. 


Chapter 3 Polynomial and Rational Functions 443




3.53


Solution
Begin by writing an equation to show the relationship between the variables.


x =
ky2


z3


Substitute  x = 6,   y = 2,   and  z = 8  to find the value of the constant  k.


6 = k2
2


8
3


6 = 4k
2


3 = k


Now we can substitute the value of the constant into the equation for the relationship.


x =
3y2


z3


To find  x when  y = 1  and  z = 27,  we will substitute values for  y  and  z  into our equation.


x = 3(1)
2


27
3


= 1


 x  varies directly with the square of  y  and inversely with  z.  If  x = 40 when  y = 4  and  z = 2,   find  x
when  y = 10  and  z = 25.


Access these online resources for additional instruction and practice with direct and inverse variation.
• Direct Variation (http://openstaxcollege.org/l/directvariation)
• Inverse Variation (http://openstaxcollege.org/l/inversevariatio)
• Direct and Inverse Variation (http://openstaxcollege.org/l/directinverse)


Visit this website (http://openstaxcollege.org/l/PreCalcLPC03) for additional practice questions from
Learningpod.


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3.9 EXERCISES
Verbal


What is true of the appearance of graphs that reflect a
direct variation between two variables?


If two variables vary inversely, what will an equation
representing their relationship look like?


Is there a limit to the number of variables that can
jointly vary? Explain.


Algebraic
For the following exercises, write an equation describing
the relationship of the given variables.


 y  varies directly as  x  and when  x = 6, y = 12.


 y  varies directly as the square of  x  and when
 x = 4,   y = 80. 


 y  varies directly as the square root of  x  and when
 x = 36,  y = 24.


 y  varies directly as the cube of  x  and when
 x = 36,  y = 24.


 y  varies directly as the cube root of  x  and when
 x = 27,   y = 15.


 y  varies directly as the fourth power of  x  and when
 x = 1,   y = 6.


 y  varies inversely as  x  and when  x = 4,   y = 2.


 y  varies inversely as the square of  x  and when
 x = 3,   y = 2.


 y  varies inversely as the cube of  x  and when
 x = 2,   y = 5.


 y  varies inversely as the fourth power of  x  and
when  x = 3,   y = 1. 


 y  varies inversely as the square root of  x  and when
 x = 25,  y = 3.


 y  varies inversely as the cube root of  x  and when
 x = 64,  y = 5.


 y  varies jointly with  x  and  z  and when
 x = 2 and z = 3,   y = 36.


 y  varies jointly as  x, z, and w  and when
 x = 1,   z = 2,   w = 5,  then y = 100.


 y  varies jointly as the square of  x  and the square of
 z  and when  x = 3 and z = 4,  then y = 72.


 y  varies jointly as  x  and the square root of  z  and
when  x = 2 and z = 25,  then y = 100.


 y varies jointly as the square of  x the cube of  z and
the square root of  W. When
 x = 1, z = 2, and w = 36, then y = 48.


 y  varies jointly as  x and z  and inversely as  w. 
When  x = 3,  z = 5,   and w = 6,  then y = 10.


 y  varies jointly as the square of  x  and the square
root of  z  and inversely as the cube of  w. When
 x = 3, z = 4, and w = 3, then y = 6.


 y  varies jointly as  x  and  z  and inversely as the
square root of  w  and the square of  t .When
 x = 3, z = 1, w = 25, and t = 2, then y = 6.


Numeric
For the following exercises, use the given information to
find the unknown value.


 y  varies directly as  x. When
 x = 3, then y = 12. Find y when x = 20.


 y  varies directly as the square of  x. When
 x = 2, then y = 16. Find y when x = 8.


 y  varies directly as the cube of  x. When
 x = 3, then y = 5. Find y when x = 4.


 y  varies directly as the square root of  x. When
 x = 16, then y = 4. Find y when x = 36.


 y  varies directly as the cube root of  x. When
 x = 125, then y = 15. Find y when x = 1, 000.


 y  varies inversely with  x. When
 x = 3, then y = 2. Find y when x = 1.


Chapter 3 Polynomial and Rational Functions 445




638.


639.


640.


641.


642.


643.


644.


645.


646.


647.


648.


649.


650.


651.


652.


653.


654.


655.


656.


657.


658.


659.


660.


661.


662.


 y  varies inversely with the square of  x. When
 x = 4, then y = 3. Find y when x = 2.


 y  varies inversely with the cube of  x. When
 x = 3, then y = 1. Find y when x = 1.


 y  varies inversely with the square root of  x. When
 x = 64,   then  y = 12.  Find  y when  x = 36.


 y  varies inversely with the cube root of  x. When
 x = 27,   then  y = 5.  Find  y when  x = 125.


 y  varies jointly as  x and z.When  x = 4  and
 z = 2,   then  y = 16.  Find  y when  x = 3  and  z = 3.


 y  varies jointly as  x, z, and w.When  x = 2,
z = 1,   and  w = 12,   then  y = 72.  Find  y when
 x = 1,   z = 2,   and  w = 3.


 y  varies jointly as  x  and the square of  z. When
 x = 2  and  z = 4,   then  y = 144.  Find  y when  x = 4
and  z = 5.


 y  varies jointly as the square of  x  and the square
root of  z. When  x = 2  and  z = 9,   then  y = 24.  Find  y 
when  x = 3  and  z = 25.


 y  varies jointly as  x  and  z  and inversely as  w. 
When  x = 5,   z = 2,   and  w = 20,   then y = 4. 
Find  y when  x = 3  and  z = 8,   and   w = 48.


 y  varies jointly as the square of  x  and the cube of  z 
and inversely as the square root of  w. When  x = 2,
z = 2,   and  w = 64,   then  y = 12.  Find  y when
 x = 1,   z = 3,   and  w = 4.


 y  varies jointly as the square of  x  and of  z  and
inversely as the square root of  w  and of  t .When  x = 2,
z = 3,   w = 16,   and  t = 3,   then  y = 1.  Find  y when
 x = 3,   z = 2,   w = 36,   and  t = 5.


Technology
For the following exercises, use a calculator to graph the
equation implied by the given variation.


 y  varies directly with the square of  x  and when
 x = 2, y = 3.


 y  varies directly as the cube of  x  and when
 x = 2, y = 4.


 y  varies directly as the square root of  x  and when
 x = 36, y = 2.


 y  varies inversely with  x  and when
 x = 6, y = 2.


 y  varies inversely as the square of  x  and when
 x = 1, y = 4.


Extensions
For the following exercises, use Kepler’s Law, which states
that the square of the time,  T ,   required for a planet to orbit
the Sun varies directly with the cube of the mean distance,
 a,   that the planet is from the Sun.


Using the Earth’s time of 1 year and mean distance of
93 million miles, find the equation relating  T   and  a. 


Use the result from the previous exercise to determine
the time required for Mars to orbit the Sun if its mean
distance is 142 million miles.


Using Earth’s distance of 150 million kilometers, find
the equation relating  T   and  a. 


Use the result from the previous exercise to determine
the time required for Venus to orbit the Sun if its mean
distance is 108 million kilometers.


Using Earth’s distance of 1 astronomical unit (A.U.),
determine the time for Saturn to orbit the Sun if its mean
distance is 9.54 A.U.


Real-World Applications
For the following exercises, use the given information to
answer the questions.


The distance  s  that an object falls varies directly with
the square of the time,  t,   of the fall. If an object falls 16
feet in one second, how long for it to fall 144 feet?


The velocity  v  of a falling object varies directly to
the time,  t,   of the fall. If after 2 seconds, the velocity of
the object is 64 feet per second, what is the velocity after 5
seconds?


The rate of vibration of a string under constant tension
varies inversely with the length of the string. If a string is
24 inches long and vibrates 128 times per second, what is
the length of a string that vibrates 64 times per second?


The volume of a gas held at constant temperature
varies indirectly as the pressure of the gas. If the volume of


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663.


664.


665.


666.


667.


668.


a gas is 1200 cubic centimeters when the pressure is 200
millimeters of mercury, what is the volume when the
pressure is 300 millimeters of mercury?


The weight of an object above the surface of the Earth
varies inversely with the square of the distance from the
center of the Earth. If a body weighs 50 pounds when it is
3960 miles from Earth’s center, what would it weigh it were
3970 miles from Earth’s center?


The intensity of light measured in foot-candles varies
inversely with the square of the distance from the light
source. Suppose the intensity of a light bulb is 0.08 foot-
candles at a distance of 3 meters. Find the intensity level at
8 meters.


The current in a circuit varies inversely with its
resistance measured in ohms. When the current in a circuit
is 40 amperes, the resistance is 10 ohms. Find the current if
the resistance is 12 ohms.


The force exerted by the wind on a plane surface
varies jointly with the square of the velocity of the wind and
with the area of the plane surface. If the area of the surface
is 40 square feet surface and the wind velocity is 20 miles
per hour, the resulting force is 15 pounds. Find the force on
a surface of 65 square feet with a velocity of 30 miles per
hour.


The horsepower (hp) that a shaft can safely transmit
varies jointly with its speed (in revolutions per minute
(rpm) and the cube of the diameter. If the shaft of a certain
material 3 inches in diameter can transmit 45 hp at 100 rpm,
what must the diameter be in order to transmit 60 hp at 150
rpm?


The kinetic energy  K  of a moving object varies
jointly with its mass  m  and the square of its velocity  v.  If
an object weighing 40 kilograms with a velocity of 15
meters per second has a kinetic energy of 1000 joules, find
the kinetic energy if the velocity is increased to 20 meters
per second.


Chapter 3 Polynomial and Rational Functions 447




arrow notation


axis of symmetry


coefficient
complex conjugate


complex number


complex plane


constant of variation


continuous function


degree
Descartes’ Rule of Signs


direct variation


Division Algorithm


end behavior
Factor Theorem
Fundamental Theorem of Algebra
general form of a quadratic function


global maximum
global minimum
horizontal asymptote


imaginary number
Intermediate Value Theorem


CHAPTER 3 REVIEW
KEY TERMS


a way to symbolically represent the local and end behavior of a function by using arrows to indicate that
an input or output approaches a value


a vertical line drawn through the vertex of a parabola around which the parabola is symmetric; it is
defined by  x = − b


2a
.


a nonzero real number multiplied by a variable raised to an exponent
the complex number in which the sign of the imaginary part is changed and the real part of the


number is left unchanged; when added to or multiplied by the original complex number, the result is a real number
the sum of a real number and an imaginary number, written in the standard form  a + bi,  where  a  is


the real part, and  bi  is the imaginary part
a coordinate system in which the horizontal axis is used to represent the real part of a complex number


and the vertical axis is used to represent the imaginary part of a complex number
the non-zero value  k  that helps define the relationship between variables in direct or inverse


variation
a function whose graph can be drawn without lifting the pen from the paper because there are no


breaks in the graph
the highest power of the variable that occurs in a polynomial


a rule that determines the maximum possible numbers of positive and negative real zeros
based on the number of sign changes of   f (x)  and   f ( − x) 


the relationship between two variables that are a constant multiple of each other; as one quantity
increases, so does the other


given a polynomial dividend   f (x)  and a non-zero polynomial divisor  d(x)  where the degree of
 d(x)  is less than or equal to the degree of   f (x),   there exist unique polynomials  q(x)  and  r(x)  such that
  f (x) = d(x)q(x) + r(x)  where  q(x)  is the quotient and  r(x)  is the remainder. The remainder is either equal to zero
or has degree strictly less than  d(x). 


the behavior of the graph of a function as the input decreases without bound and increases without bound
 k  is a zero of polynomial function   f (x)  if and only if  (x − k)  is a factor of   f (x)


a polynomial function with degree greater than 0 has at least one complex zero
the function that describes a parabola, written in the form


  f (x) = ax2 + bx + c,  where  a, b,   and  c  are real numbers and  a ≠ 0.
highest turning point on a graph;   f (a)  where   f (a) ≥ f (x)  for all  x.
lowest turning point on a graph;   f (a)  where   f (a) ≤ f (x)  for all  x.


a horizontal line  y = b where the graph approaches the line as the inputs increase or decrease
without bound.


a number in the form  bi where  i = −1
for two numbers  a  and  b  in the domain of   f ,   if  a < b  and   f (a) ≠ f (b),   then the


function   f   takes on every value between   f (a)  and   f (b);   specifically, when a polynomial function changes from a
negative value to a positive value, the function must cross the  x- axis


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inverse variation
inversely proportional


invertible function
joint variation
leading coefficient
leading term
Linear Factorization Theorem


multiplicity


polynomial function


power function


rational function
Rational Zero Theorem


Remainder Theorem
removable discontinuity


smooth curve
standard form of a quadratic function


synthetic division
term of a polynomial function


turning point
varies directly
varies inversely
vertex


vertex form of a quadratic function
vertical asymptote


zeros


the relationship between two variables in which the product of the variables is a constant
a relationship where one quantity is a constant divided by the other quantity; as one quantity


increases, the other decreases
any function that has an inverse function


a relationship where a variable varies directly or inversely with multiple variables
the coefficient of the leading term


the term containing the highest power of the variable
allowing for multiplicities, a polynomial function will have the same number of factors


as its degree, and each factor will be in the form  (x − c),   where  c  is a complex number
the number of times a given factor appears in the factored form of the equation of a polynomial; if a


polynomial contains a factor of the form  (x − h) p,  x = h  is a zero of multiplicity  p.
a function that consists of either zero or the sum of a finite number of non-zero terms, each of


which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer
power.


a function that can be represented in the form   f (x) = kx p   where  k  is a constant, the base is a variable,
and the exponent,  p,   is a constant


a function that can be written as the ratio of two polynomials
the possible rational zeros of a polynomial function have the form  pq  where  p  is a factor of the


constant term and  q  is a factor of the leading coefficient.
if a polynomial   f (x)  is divided by  x − k,   then the remainder is equal to the value   f (k) 
a single point at which a function is undefined that, if filled in, would make the function


continuous; it appears as a hole on the graph of a function
a graph with no sharp corners


the function that describes a parabola, written in the form
  f (x) = a(x − h)2 + k,  where  (h, k)  is the vertex.


a shortcut method that can be used to divide a polynomial by a binomial of the form  x − k 
any  ai x i   of a polynomial function in the form


  f (x) = an xn + ... + a2 x2 + a1 x + a0
the location at which the graph of a function changes direction
a relationship where one quantity is a constant multiplied by the other quantity
a relationship where one quantity is a constant divided by the other quantity


the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic
function


another name for the standard form of a quadratic function
a vertical line  x = a where the graph tends toward positive or negative infinity as the inputs


approach  a
in a given function, the values of  x  at which  y = 0,   also called roots


KEY EQUATIONS


Chapter 3 Polynomial and Rational Functions 449




general form of a quadratic function f (x) = ax2 + bx + c


the quadratic formula x = −b ± b2 − 4ac
2a


standard form of a quadratic function f (x) = a(x − h)2 + k


general form of a polynomial function f (x) = an xn + ... + a2 x2 + a1 x + a0


Division Algorithm f (x) = d(x)q(x) + r(x) where q(x) ≠ 0


Rational Function f (x) = P(x)Q(x) =
a p x


p + a p − 1 x
p − 1


+ ... + a1 x + a0


bq x
q + bq − 1 x


q − 1
+ ... + b1 x + b0


, Q(x) ≠ 0


Direct variation y = kxn, k is a nonzero constant.


Inverse variation y = kxn, k is a nonzero constant.


KEY CONCEPTS
3.1 Complex Numbers


• The square root of any negative number can be written as a multiple of  i.  See Example 3.1.
• To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the
real axis, and the vertical axis is the imaginary axis. See Example 3.2.


• Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See
Example 3.3.


• Complex numbers can be multiplied and divided.
• To multiply complex numbers, distribute just as with polynomials. See Example 3.4, Example 3.5, and
Example 3.8.


• To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the
denominator to eliminate the complex number from the denominator. See Example 3.6, Example 3.7, and
Example 3.9.


• The powers of  i  are cyclic, repeating every fourth one. See Example 3.10.


3.2 Quadratic Functions
• A polynomial function of degree two is called a quadratic function.
• The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down.
• The axis of symmetry is the vertical line passing through the vertex. The zeros, or  x- intercepts, are the points at
which the parabola crosses the  x- axis. The  y- intercept is the point at which the parabola crosses the  y- axis. See
Example 3.11, Example 3.17, and Example 3.18.


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• Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex
of a parabola. Either form can be written from a graph. See Example 3.12.


• The vertex can be found from an equation representing a quadratic function. See Example 3.13.
• The domain of a quadratic function is all real numbers. The range varies with the function. See Example 3.14.
• A quadratic function’s minimum or maximum value is given by the  y- value of the vertex.
• The minimum or maximum value of a quadratic function can be used to determine the range of the function and to
solve many kinds of real-world problems, including problems involving area and revenue. See Example 3.15 and
Example 3.16.


• Some quadratic equations must be solved by using the quadratic formula. See Example 3.19.
• The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example 3.20.


3.3 Power Functions and Polynomial Functions
• A power function is a variable base raised to a number power. See Example 3.21.
• The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior.
• The end behavior depends on whether the power is even or odd. See Example 3.22 and Example 3.23.
• A polynomial function is the sum of terms, each of which consists of a transformed power function with positive
whole number power. See Example 3.24.


• The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term
containing the highest power of the variable is called the leading term. The coefficient of the leading term is called
the leading coefficient. See Example 3.25.


• The end behavior of a polynomial function is the same as the end behavior of the power function represented by the
leading term of the function. See Example 3.26 and Example 3.27.


• A polynomial of degree  n  will have at most  n  x-intercepts and at most  n − 1  turning points. See Example 3.28,
Example 3.29, Example 3.30, Example 3.31, and Example 3.32.


3.4 Graphs of Polynomial Functions
• Polynomial functions of degree 2 or more are smooth, continuous functions. See Example 3.33.
• To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
See Example 3.34, Example 3.35, and Example 3.36.


• Another way to find the  x- intercepts of a polynomial function is to graph the function and identify the points at
which the graph crosses the  x- axis. See Example 3.37.


• The multiplicity of a zero determines how the graph behaves at the  x- intercepts. See Example 3.38.
• The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
• The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
• The end behavior of a polynomial function depends on the leading term.
• The graph of a polynomial function changes direction at its turning points.
• A polynomial function of degree  n  has at most  n − 1  turning points. See Example 3.39.
• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that
the final graph has at most  n − 1  turning points. See Example 3.40 and Example 3.42.


• Graphing a polynomial function helps to estimate local and global extremas. See Example 3.43.
• The Intermediate Value Theorem tells us that if   f (a) and f (b)  have opposite signs, then there exists at least one
value  c  between  a  and  b  for which   f (c) = 0.  See Example 3.41.


Chapter 3 Polynomial and Rational Functions 451




3.5 Dividing Polynomials
• Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See
Example 3.44 and Example 3.45.


• The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the
quotient added to the remainder.


• Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form  x − k.  See
Example 3.46, Example 3.47, and Example 3.48.


• Polynomial division can be used to solve application problems, including area and volume. See Example 3.49.


3.6 Zeros of Polynomial Functions
• To find   f (k),   determine the remainder of the polynomial   f (x) when it is divided by  x − k.  See Example 3.50.
•  k  is a zero of   f (x)  if and only if  (x − k)  is a factor of   f (x). See Example 3.51.
• Each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term
divided by a factor of the leading coefficient. See Example 3.52 and Example 3.53.


• When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
• Synthetic division can be used to find the zeros of a polynomial function. See Example 3.54.
• According to the Fundamental Theorem, every polynomial function has at least one complex zero. See Example
3.55.


• Every polynomial function with degree greater than 0 has at least one complex zero.
• Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor
will be in the form  (x − c),  where  c  is a complex number. See Example 3.56.


• The number of positive real zeros of a polynomial function is either the number of sign changes of the function or
less than the number of sign changes by an even integer.


• The number of negative real zeros of a polynomial function is either the number of sign changes of   f ( − x)  or less
than the number of sign changes by an even integer. See Example 3.57.


• Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division.
See Example 3.58.


3.7 Rational Functions


• We can use arrow notation to describe local behavior and end behavior of the toolkit functions   f (x) = 1x   and
  f (x) = 1


x2
.  See Example 3.59.


• A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one
vertical asymptote. See Example 3.60.


• Application problems involving rates and concentrations often involve rational functions. See Example 3.61.
• The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.
See Example 3.62.


• The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and
the numerator is not zero. See Example 3.63.


• A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and
denominator to be zero. See Example 3.64.


• A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator
functions. See Example 3.65, Example 3.66, Example 3.67, and Example 3.68.


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• Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See
Example 3.69.


• If a rational function has x-intercepts at  x = x1, x2, … , xn,   vertical asymptotes at  x = v1, v2, … , vm,  
and no  xi = any v j,   then the function can be written in the form


f (x) = a
(x − x1)


p1 (x − x2)
p2 ⋯ (x − xn)


pn


(x − v1)
q1 (x − v2)


q2 ⋯ (x − vm)
qn


See Example 3.70.


3.8 Inverses and Radical Functions
• The inverse of a quadratic function is a square root function.
• If   f −1   is the inverse of a function   f ,   then   f   is the inverse of the function   f −1.  See Example 3.71.
• While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See
Example 3.72.


• To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.
See Example 3.73 and Example 3.74.


• When finding the inverse of a radical function, we need a restriction on the domain of the answer. See Example
3.75 and Example 3.77.


• Inverse and radical and functions can be used to solve application problems. See Example 3.76 and Example
3.78.


3.9 Modeling Using Variation
• A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See
Example 3.79.


• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation. See
Example 3.80.


• Two variables that are inversely proportional to one another will have a constant multiple. See Example 3.81.
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship
joint variation. See Example 3.82.


CHAPTER 3 REVIEW EXERCISES
You have reached the end of Chapter 3: Polynomial and
Rational Functions. Let’s review some of the Key Terms,
Concepts and Equations you have learned.
Complex Numbers
Perform the indicated operation with complex numbers.
669. (4 + 3i) + (−2 − 5i)


670. (6 − 5i) − (10 + 3i)


671. (2 − 3i)(3 + 6i)


672. 2 − i
2 + i


Solve the following equations over the complex number
system.
673. x2 − 4x + 5 = 0


674. x2 + 2x + 10 = 0


Quadratic Functions
For the following exercises, write the quadratic function in
standard form. Then, give the vertex and axes intercepts.
Finally, graph the function.
675. f (x) = x2 − 4x − 5


676. f (x) = − 2x2 − 4x


Chapter 3 Polynomial and Rational Functions 453




For the following problems, find the equation of the
quadratic function using the given information.
677. The vertex is ( – 2, 3) and a point on the graph is
 (3, 6).


678. The vertex is  ( – 3, 6.5)  and a point on the graph is
 (2, 6).


Answer the following questions.
679. A rectangular plot of land is to be enclosed by
fencing. One side is along a river and so needs no fence.
If the total fencing available is 600 meters, find the
dimensions of the plot to have maximum area.


680. An object projected from the ground at a 45 degree
angle with initial velocity of 120 feet per second has height,
 h,   in terms of horizontal distance traveled,  x,   given by
 h(x) = −32


(120)2
x2 + x.  Find the maximum height the


object attains.


Power Functions and Polynomial Functions
For the following exercises, determine if the function is a
polynomial function and, if so, give the degree and leading
coefficient.
681. f (x) = 4x5 − 3x3 + 2x − 1


682. f (x) = 5x + 1 − x2


683. f (x) = x2 ⎛⎝3 − 6x + x2⎞⎠


For the following exercises, determine end behavior of the
polynomial function.
684. f (x) = 2x4 + 3x3 − 5x2 + 7


685. f (x) = 4x3 − 6x2 + 2


686. f (x) = 2x2(1 + 3x − x2)


Graphs of Polynomial Functions
For the following exercises, find all zeros of the polynomial
function, noting multiplicities.
687. f (x) = (x + 3)2(2x − 1)(x + 1)3


688. f (x) = x5 + 4x4 + 4x3


689. f (x) = x3 − 4x2 + x − 4


For the following exercises, based on the given graph,
determine the zeros of the function and note multiplicity.
690.


691.


692. Use the Intermediate Value Theorem to show that
at least one zero lies between 2 and 3 for the function
  f (x) = x3 − 5x + 1


Dividing Polynomials
For the following exercises, use long division to find the
quotient and remainder.


693. x3 − 2x2 + 4x + 4
x − 2


694. 3x4 − 4x2 + 4x + 8
x + 1


For the following exercises, use synthetic division to find
the quotient. If the divisor is a factor, then write the factored
form.


695. x3 − 2x2 + 5x − 1
x + 3


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696. x3 + 4x + 10
x − 3


697. 2x3 + 6x2 − 11x − 12
x + 4


698. 3x4 + 3x3 + 2x + 2
x + 1


Zeros of Polynomial Functions
For the following exercises, use the Rational Zero Theorem
to help you solve the polynomial equation.
699. 2x3 − 3x2 − 18x − 8 = 0


700. 3x3 + 11x2 + 8x − 4 = 0


701. 2x4 − 17x3 + 46x2 − 43x + 12 = 0


702. 4x4 + 8x3 + 19x2 + 32x + 12 = 0


For the following exercises, use Descartes’ Rule of Signs to
find the possible number of positive and negative solutions.
703. x3 − 3x2 − 2x + 4 = 0


704. 2x4 − x3 + 4x2 − 5x + 1 = 0


Rational Functions
For the following rational functions, find the intercepts and
the vertical and horizontal asymptotes, and then use them
to sketch a graph.
705. f (x) = x + 2


x − 5


706. f (x) = x2 + 1
x2 − 4


707. f (x) = 3x2 − 27
x2 + x − 2


708. f (x) = x + 2
x2 − 9


For the following exercises, find the slant asymptote.


709. f (x) = x2 − 1
x + 2


710. f (x) = 2x3 − x2 + 4
x2 + 1


Inverses and Radical Functions
For the following exercises, find the inverse of the function
with the domain given.
711. f (x) = (x − 2)2,  x ≥ 2


712. f (x) = (x + 4)2 − 3,  x ≥ − 4


713. f (x) = x2 + 6x − 2,  x ≥ − 3


714. f (x) = 2x3 − 3


715. f (x) = 4x + 5 − 3


716. f (x) = x − 3
2x + 1


Modeling Using Variation
For the following exercises, find the unknown value.
717.  y  varies directly as the square of  x.  If when
 x = 3, y = 36,   find  y  if  x = 4.


718.  y  varies inversely as the square root of  x  If when
 x = 25, y = 2,   find  y  if  x = 4.


719.  y  varies jointly as the cube of  x  and as  z.  If when
 x = 1  and  z = 2,   y = 6,   find  y  if  x = 2  and  z = 3.


720.  y  varies jointly as  x  and the square of  z  and
inversely as the cube of  w.  If when  x = 3,   z = 4,   and
 w = 2,   y = 48,   find  y  if  x = 4,   z = 5,   and  w = 3.


For the following exercises, solve the application problem.
721. The weight of an object above the surface of the
earth varies inversely with the distance from the center of
the earth. If a person weighs 150 pounds when he is on
the surface of the earth (3,960 miles from center), find the
weight of the person if he is 20 miles above the surface.


722. The volume  V   of an ideal gas varies directly with
the temperature  T   and inversely with the pressure P. A
cylinder contains oxygen at a temperature of 310 degrees K
and a pressure of 18 atmospheres in a volume of 120 liters.
Find the pressure if the volume is decreased to 100 liters
and the temperature is increased to 320 degrees K.


Chapter 3 Polynomial and Rational Functions 455




CHAPTER 3 PRACTICE TEST
Perform the indicated operation or solve the equation.
723. (3 − 4i)(4 + 2i)


724. 1 − 4i
3 + 4i


725. x2 − 4x + 13 = 0


Give the degree and leading coefficient of the following
polynomial function.
726. f (x) = x3 ⎛⎝3 − 6x2 − 2x2⎞⎠


Determine the end behavior of the polynomial function.
727. f (x) = 8x3 − 3x2 + 2x − 4


728. f (x) = − 2x2(4 − 3x − 5x2)


Write the quadratic function in standard form. Determine
the vertex and axes intercepts and graph the function.
729. f (x) = x2 + 2x − 8


Given information about the graph of a quadratic function,
find its equation.
730. Vertex  (2, 0)  and point on graph  (4, 12).


Solve the following application problem.
731. A rectangular field is to be enclosed by fencing. In
addition to the enclosing fence, another fence is to divide
the field into two parts, running parallel to two sides. If
1,200 feet of fencing is available, find the maximum area
that can be enclosed.


Find all zeros of the following polynomial functions, noting
multiplicities.
732. f (x) = (x − 3)3(3x − 1)(x − 1)2


733. f (x) = 2x6 − 6x5 + 18x4


Based on the graph, determine the zeros of the function and
multiplicities.
734.


Use long division to find the quotient.


735. 2x3 + 3x − 4
x + 2


Use synthetic division to find the quotient. If the divisor is
a factor, write the factored form.


736. x4 + 3x2 − 4
x − 2


737. 2x3 + 5x2 − 7x − 12
x + 3


Use the Rational Zero Theorem to help you find the zeros
of the polynomial functions.
738. f (x) = 2x3 + 5x2 − 6x − 9


739. f (x) = 4x4 + 8x3 + 21x2 + 17x + 4


740. f (x) = 4x4 + 16x3 + 13x2 − 15x − 18


741. f (x) = x5 + 6x4 + 13x3 + 14x2 + 12x + 8


Given the following information about a polynomial
function, find the function.
742. It has a double zero at  x = 3  and zeroes at  x = 1 
and  x = − 2  . It’s y-intercept is  (0, 12). 


743. It has a zero of multiplicity 3 at  x = 1
2
  and another


zero at  x = − 3  . It contains the point  (1, 8).


Use Descartes’ Rule of Signs to determine the possible
number of positive and negative solutions.
744. 8x3 − 21x2 + 6 = 0


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For the following rational functions, find the intercepts and
horizontal and vertical asymptotes, and sketch a graph.
745. f (x) = x + 4


x2 − 2x − 3


746. f (x) = x2 + 2x − 3
x2 − 4


Find the slant asymptote of the rational function.


747. f (x) = x2 + 3x − 3
x − 1


Find the inverse of the function.
748. f (x) = x − 2 + 4


749. f (x) = 3x3 − 4


750. f (x) = 2x + 3
3x − 1


Find the unknown value.
751.  y  varies inversely as the square of  x  and when
 x = 3,   y = 2.  Find  y  if  x = 1.


752.  y  varies jointly with  x  and the cube root of  z.  If
when  x = 2  and  z = 27,   y = 12,   find  y  if  x = 5  and
 z = 8.


Solve the following application problem.
753. The distance a body falls varies directly as the square
of the time it falls. If an object falls 64 feet in 2 seconds,
how long will it take to fall 256 feet?


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4 | EXPONENTIAL AND
LOGARITHMIC FUNCTIONS


Figure 4.1 Electron micrograph of E.Coli bacteria (credit: “Mattosaurus,” Wikimedia Commons)


Chapter Outline
4.1 Exponential Functions
4.2 Graphs of Exponential Functions
4.3 Logarithmic Functions
4.4 Graphs of Logarithmic Functions
4.5 Logarithmic Properties
4.6 Exponential and Logarithmic Equations
4.7 Exponential and Logarithmic Models
4.8 Fitting Exponential Models to Data


Introduction
Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see
hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth,


Chapter 4 Exponential and Logarithmic Functions 459




nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But
that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to
the body.
Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When
conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to
form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.[1]
For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. Table 4.1 shows
the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one
thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over
16 million!


Hour 0 1 2 3 4 5 6 7 8 9 10


Bacteria 1 2 4 8 16 32 64 128 256 512 1024


Table 4.1


In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns
such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential
functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data.


4.1 | Exponential Functions
Learning Objectives


In this section, you will:
4.1.1 Evaluate exponential functions.
4.1.2 Find the equation of an exponential function.
4.1.3 Use compound interest formulas.
4.1.4 Evaluate exponential functions with base e.


India is the second most populous country in the world with a population of about  1.25  billion people in 2013. The
population is growing at a rate of about  1.2%  each year[2]. If this rate continues, the population of India will exceed China’s
population by the year  2031.When populations grow rapidly, we often say that the growth is “exponential,” meaning that
something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning.
In this section, we will take a look at exponential functions, which model this kind of rapid growth.
Identifying Exponential Functions
When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for
each unit increase in input. For example, in the equation   f (x) = 3x + 4, the slope tells us the output increases by 3 each
time the input increases by 1. The scenario in the India population example is different because we have a percent change
per unit time (rather than a constant change) in the number of people.
Defining an Exponential Function
A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011,
2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish,
dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in
2050.


1. Todar, PhD, Kenneth. Todar's Online Textbook of Bacteriology. http://textbookofbacteriology.net/growth_3.html.
2. http://www.worldometers.info/world-population/. Accessed February 24, 2014.


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What exactly does it mean to grow exponentially? What does the word double have in common with percent increase?
People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the
media.


• Percent change refers to a change based on a percent of the original amount.
• Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments
of time, that is, a percent increase of the original amount over time.


• Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of
time, that is, a percent decrease of the original amount over time.


For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will
construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1.
We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and
increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 4.2.


x f(x) = 2x g(x) = 2x


0 1 0


1 2 2


2 4 4


3 8 6


4 16 8


5 32 10


6 64 12


Table 4.2


From Table 4.2 we can infer that for these two functions, exponential growth dwarfs linear growth.
• Exponential growth refers to the original value from the range increases by the same percentage over equal
increments found in the domain.


• Linear growth refers to the original value from the range increases by the same amount over equal increments
found in the domain.


Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential
growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input
increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the
output whenever the input was increased by one.
The general form of the exponential function is   f (x) = abx,  where  a  is any nonzero number,  b  is a positive real number
not equal to 1.


• If  b > 1, the function grows at a rate proportional to its size.
• If  0 < b < 1, the function decays at a rate proportional to its size.


Let’s look at the function   f (x) = 2x   from our example. We will create a table (Table 4.3) to determine the corresponding
outputs over an interval in the domain from  −3  to  3.


Chapter 4 Exponential and Logarithmic Functions 461




x −3 −2 −1 0 1 2 3


f(x) = 2x 2−3 = 18
2−2 = 1


4
2−1 = 1


2 2
0 = 1 21 = 2 22 = 4 23 = 8


Table 4.3


Let us examine the graph of   f   by plotting the ordered pairs we observe on the table in Figure 4.2, and then make a few
observations.


Figure 4.2


Let’s define the behavior of the graph of the exponential function   f (x) = 2x   and highlight some its key characteristics.
• the domain is  (−∞, ∞),
• the range is  (0, ∞),
• as  x → ∞, f (x) → ∞,
• as  x → −∞, f (x) → 0,
• f (x)  is always increasing,
• the graph of   f (x) will never touch the x-axis because base two raised to any exponent never has the result of zero.
•  y = 0  is the horizontal asymptote.
• the y-intercept is 1.


Exponential Function
For any real number  x, an exponential function is a function with the form


(4.1)f (x) = abx


where


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4.1


•  a  is the a non-zero real number called the initial value and
•  b  is any positive real number such that  b ≠ 1.
• The domain of   f   is all real numbers.
• The range of   f   is all positive real numbers if  a > 0.
• The range of   f   is all negative real numbers if  a < 0.
• The y-intercept is  (0, a), and the horizontal asymptote is  y = 0.


Example 4.1
Identifying Exponential Functions


Which of the following equations are not exponential functions?
• f (x) = 43(x − 2)


• g(x) = x3


• h(x) = ⎛⎝13



x


• j(x) = (−2)x


Solution
By definition, an exponential function has a constant as a base and an independent variable as an exponent.
Thus,  g(x) = x3   does not represent an exponential function because the base is an independent variable. In fact,
 g(x) = x3   is a power function.
Recall that the base b of an exponential function is always a positive constant, and  b ≠ 1. Thus,   j(x) = (−2)x  
does not represent an exponential function because the base,  −2, is less than  0.


Which of the following equations represent exponential functions?


• f (x) = 2x2 − 3x + 1
• g(x) = 0.875x


• h(x) = 1.75x + 2


• j(x) = 1095.6−2x


Evaluating Exponential Functions
Recall that the base of an exponential function must be a positive real number other than  1.Why do we limit the base b  to
positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:


• Let  b = − 9  and  x = 1
2
. Then   f (x) = f ⎛⎝12



⎠ = (−9)


1
2 = −9, which is not a real number.


Chapter 4 Exponential and Logarithmic Functions 463




4.2


Why do we limit the base to positive values other than 1? Because base 1  results in the constant function. Observe what
happens if the base is 1 :


• Let  b = 1. Then   f (x) = 1x = 1  for any value of  x.
To evaluate an exponential function with the form   f (x) = bx, we simply substitute x with the given value, and calculate
the resulting power. For example:
Let   f (x) = 2x. What is f (3)?


f (x) = 2x


f (3) = 23 Substitute x = 3.
= 8 Evaluate the power.


To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.
For example:
Let   f (x) = 30(2)x. What is   f (3)?


f (x) = 30(2)x


f (3) = 30(2)3 Substitute x = 3.
= 30(8) Simplify the power fir t.


= 240 Multiply.


Note that if the order of operations were not followed, the result would be incorrect:
f (3) = 30(2)3 ≠ 603 = 216,000


Example 4.2
Evaluating Exponential Functions


Let   f (x) = 5(3)x + 1. Evaluate   f (2) without using a calculator.


Solution
Follow the order of operations. Be sure to pay attention to the parentheses.


f (x) = 5(3)x + 1


f (2) = 5(3)2 + 1 Substitute x = 2.


= 5(3)3 Add the exponents.


= 5(27) Simplify the power.


= 135 Multiply.


Let f (x) = 8(1.2)x − 5. Evaluate   f (3)  using a calculator. Round to four decimal places.


Defining Exponential Growth
Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday
language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely
in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.


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Exponential Growth
A function that models exponential growth grows by a rate proportional to the amount present. For any real number
 x  and any positive real numbers  a   and  b  such that  b ≠ 1, an exponential growth function has the form


f (x) = abx


where
• a  is the initial or starting value of the function.
• b  is the growth factor or growth multiplier per unit  x .


In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To
differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores
and expands by opening 50 new stores a year, so its growth can be represented by the function  A(x) = 100 + 50x. 
Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be
represented by the function  B(x) = 100(1 + 0.5)x.
A few years of growth for these companies are illustrated in Table 4.4.


Year, x Stores, Company A Stores, Company B


0 100 + 50(0) = 100 100(1 + 0.5)0 = 100


1 100 + 50(1) = 150 100(1 + 0.5)1 = 150


2 100 + 50(2) = 200 100(1 + 0.5)2 = 225


3 100 + 50(3) = 250 100(1 + 0.5)3 = 337.5


x A(x) = 100 + 50x B(x) = 100(1 + 0.5)x


Table 4.4


The graphs comparing the number of stores for each company over a five-year period are shown in Figure 4.3.We can see
that, with exponential growth, the number of stores increases much more rapidly than with linear growth.


Chapter 4 Exponential and Logarithmic Functions 465




Figure 4.3 The graph shows the numbers of stores Companies
A and B opened over a five-year period.


Notice that the domain for both functions is  [0, ∞), and the range for both functions is  [100, ∞). After year 1, Company
B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company B,  B(x) = 100(1 + 0.5)x. 
In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and  1 + 0.5 = 1.5
represents the growth factor. Generalizing further, we can write this function as  B(x) = 100(1.5)x, where 100 is the initial
value,  1.5  is called the base, and  x  is called the exponent.


Example 4.3
Evaluating a Real-World Exponential Model


At the beginning of this section, we learned that the population of India was about  1.25  billion in the year
2013, with an annual growth rate of about  1.2%. This situation is represented by the growth function


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4.3


 P(t) = 1.25(1.012)t, where  t  is the number of years since  2013. To the nearest thousandth, what will the
population of India be in  2031?


Solution
To estimate the population in 2031, we evaluate the models for  t = 18, because 2031 is  18 years after 2013.
Rounding to the nearest thousandth,


P(18) = 1.25(1.012)18 ≈ 1.549


There will be about 1.549 billion people in India in the year 2031.


The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about
 0.6%. This situation is represented by the growth function  P(t) = 1.39(1.006)t, where  t  is the number of
years since  2013. To the nearest thousandth, what will the population of China be for the year 2031? How does
this compare to the population prediction we made for India in Example 4.3?


Finding Equations of Exponential Functions
In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we
are given information about an exponential function without knowing the function explicitly. We must use the information
to first write the form of the function, then determine the constants  a  and  b, and evaluate the function.


Given two data points, write an exponential model.
1. If one of the data points has the form  (0, a), then  a  is the initial value. Using  a, substitute the second
point into the equation   f (x) = a(b)x, and solve for  b.


2. If neither of the data points have the form  (0, a), substitute both points into two equations with the form
  f (x) = a(b)x.  Solve the resulting system of two equations in two unknowns to find  a  and  b.


3. Using the  a  and  b  found in the steps above, write the exponential function in the form   f (x) = a(b)x.


Example 4.4
Writing an Exponential Model When the Initial Value Is Known


In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The
population was growing exponentially. Write an algebraic function  N(t)  representing the population  (N)  of deer
over time  t.


Solution
We let our independent variable  t  be the number of years after 2006. Thus, the information given in the problem
can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be
measured as years after 2006, we have given ourselves the initial value for the function,  a = 80. We can now
substitute the second point into the equation  N(t) = 80bt   to find  b :


Chapter 4 Exponential and Logarithmic Functions 467




N(t) = 80bt


  180 = 80b6 Substitute using point (6, 180).
9
4


= b6 Divide and write in lowest terms.


  b = ⎛⎝94



1
6


Isolate b using properties of exponents.


  b ≈ 1.1447 Round to 4 decimal places.
NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four
places for the remainder of this section.
The exponential model for the population of deer is  N(t) = 80(1.1447)t.  (Note that this exponential function
models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the
model may not be useful in the long term.)
We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in
Figure 4.4 passes through the initial points given in the problem,  (0, 80)  and  (6, 180). We can also see that
the domain for the function is  [0, ∞), and the range for the function is  [80, ∞).


Figure 4.4 Graph showing the population of deer over time,
 N(t) = 80(1.1447)t, t  years after 2006


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4.4 A wolf population is growing exponentially. In 2011,  129 wolves were counted. By  2013,  the population
had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation?
Write the equation representing the population  N   of wolves over time  t.


Example 4.5
Writing an Exponential Model When the Initial Value is Not Known


Find an exponential function that passes through the points  (−2, 6)  and  (2, 1).


Solution
Because we don’t have the initial value, we substitute both points into an equation of the form   f (x) = abx, and
then solve the system for  a  and  b.


• Substituting  (−2, 6)  gives  6 = ab−2  
• Substituting  (2, 1)  gives  1 = ab2  


Use the first equation to solve for  a  in terms of  b :


Substitute  a  in the second equation, and solve for  b :


Use the value of  b  in the first equation to solve for the value of  a :


Thus, the equation is   f (x) = 2.4492(0.6389)x.
We can graph our model to check our work. Notice that the graph in Figure 4.5 passes through the initial points
given in the problem,  (−2, 6)  and  (2, 1). The graph is an example of an exponential decay function.


Chapter 4 Exponential and Logarithmic Functions 469




4.5


Figure 4.5 The graph of   f (x) = 2.4492(0.6389)x  models
exponential decay.


Given the two points  (1, 3)  and  (2, 4.5), find the equation of the exponential function that passes
through these two points.


Do two points always determine a unique exponential function?
Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-
coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not
every graph that looks exponential really is exponential. We need to know the graph is based on a model that
shows the same percent growth with each unit increase in  x, which in many real world cases involves time.


Given the graph of an exponential function, write its equation.
1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible.
Try to choose points that are as far apart as possible to reduce round-off error.


2. If one of the data points is the y-intercept  (0, a) , then  a  is the initial value. Using  a, substitute the
second point into the equation   f (x) = a(b)x, and solve for  b.


3. If neither of the data points have the form  (0, a), substitute both points into two equations with the form
f (x) = a(b)x.  Solve the resulting system of two equations in two unknowns to find  a  and  b.


4. Write the exponential function,   f (x) = a(b)x.


Example 4.6
Writing an Exponential Function Given Its Graph


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4.6


Find an equation for the exponential function graphed in Figure 4.6.


Figure 4.6


Solution
We can choose the y-intercept of the graph,  (0, 3), as our first point. This gives us the initial value,  a = 3. 
Next, choose a point on the curve some distance away from  (0, 3)  that has integer coordinates. One such point
is  (2, 12).


y = abx Write the general form of an exponential equation.


y = 3bx Substitute the initial value 3 for a.


12 = 3b2 Substitute in 12 for y and 2 for x.


4 = b2 Divide by 3.


b = ± 2 Take the square root.


Because we restrict ourselves to positive values of  b, we will use  b = 2.  Substitute  a  and  b  into the standard
form to yield the equation   f (x) = 3(2)x.


Find an equation for the exponential function graphed in Figure 4.7.


Figure 4.7


Chapter 4 Exponential and Logarithmic Functions 471




4.7


Given two points on the curve of an exponential function, use a graphing calculator to find the equation.
1. Press [STAT].
2. Clear any existing entries in columns L1 or L2.
3. In L1, enter the x-coordinates given.
4. In L2, enter the corresponding y-coordinates.
5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press


[ENTER].
6. The screen displays the values of a and b in the exponential equation  y = a ⋅ bx .


Example 4.7
Using a Graphing Calculator to Find an Exponential Function


Use a graphing calculator to find the exponential equation that includes the points  (2, 24.8)  and  (5, 198.4).


Solution
Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the
L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4.
Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values  a = 6.2  and  b = 2 will be
displayed. The exponential equation is  y = 6.2 ⋅ 2x.


Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6,
481.07).


Applying the Compound-Interest Formula
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use
compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated
value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an
investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In
fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal
rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula, which is an exponential function of the
variables time  t, principal  P, APR  r, and number of compounding periods in a year  n :


A(t) = P⎛⎝1 + rn



nt


For example, observe Table 4.5, which shows the result of investing $1,000 at 10% for one year. Notice how the value of
the account increases as the compounding frequency increases.


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Frequency Value after 1 year


Annually $1100


Semiannually $1102.50


Quarterly $1103.81


Monthly $1104.71


Daily $1105.16


Table 4.5


The Compound Interest Formula
Compound interest can be calculated using the formula


(4.2)A(t) = P⎛⎝1 + rn⎞⎠
nt


where
• A(t)  is the account value,
• t  is measured in years,
• P  is the starting amount of the account, often called the principal, or more generally present value,
• r  is the annual percentage rate (APR) expressed as a decimal, and
• n  is the number of compounding periods in one year.


Example 4.8
Calculating Compound Interest


If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the
account be worth in 10 years?


Solution
Because we are starting with $3,000,  P = 3000. Our interest rate is 3%, so  r = 0.03. Because we are
compounding quarterly, we are compounding 4 times per year, so  n = 4. We want to know the value of the
account in 10 years, so we are looking for  A(10), the value when  t = 10.


A(t) = P⎛⎝1 + rn



nt
Use the compound interest formula.


A(10) = 3000⎛⎝1 +
0.03
4



4⋅10
Substitute using given values.


≈ $4045.05 Round to two decimal places.


Chapter 4 Exponential and Logarithmic Functions 473




4.8


4.9


The account will be worth about $4,045.05 in 10 years.


An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What
will the investment be worth in 30 years?


Example 4.9
Using the Compound Interest Formula to Solve for the Principal


A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college
tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the
account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually
(twice a year). To the nearest dollar, how much will Lily need to invest in the account now?


Solution
The nominal interest rate is 6%, so  r = 0.06.  Interest is compounded twice a year, so  k = 2.
We want to find the initial investment,  P, needed so that the value of the account will be worth $40,000 in  18 
years. Substitute the given values into the compound interest formula, and solve for  P.


  A(t) = P⎛⎝1 + rn⎞⎠
nt


Use the compound interest formula.


40,000 = P⎛⎝1 +
0.06
2



2(18)


Substitute using given values A, r, n, and t.


40,000 = P(1.03)36 Simplify.


40,000
(1.03)36


= P Isolate P.


  P ≈ $13, 801 Divide and round to the nearest dollar.
Lily will need to invest $13,801 to have $40,000 in 18 years.


Refer to Example 4.9. To the nearest dollar, how much would Lily need to invest if the account is
compounded quarterly?


Evaluating Functions with Base e
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 4.6 shows
that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding.
This might lead us to ask whether this pattern will continue.
Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 4.6.


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Frequency A(t) = ⎛⎝1 + 1n⎞⎠
n


Value


Annually ⎛⎝1 + 11



1
$2


Semiannually ⎛⎝1 + 12



2
$2.25


Quarterly ⎛⎝1 + 14



4
$2.441406


Monthly ⎛⎝1 + 112



12
$2.613035


Daily ⎛⎝1 + 1365



365
$2.714567


Hourly ⎛⎝1 + 18766



8766
$2.718127


Once per minute ⎛⎝1 + 1525960



525960
$2.718279


Once per second ⎛⎝1 + 131557600



31557600
$2.718282


Table 4.6


These values appear to be approaching a limit as  n  increases without bound. In fact, as  n  gets larger and larger, the
expression  ⎛⎝1 + 1n⎞⎠


n   approaches a number used so frequently in mathematics that it has its own name: the letter  e. This
value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation
to six decimal places is shown below.


The Number e
The letter e represents the irrational number



⎝1 +


1
n



n
, as n increases without bound


The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation,
 e ≈ 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first
investigated and discovered many of its properties.


Chapter 4 Exponential and Logarithmic Functions 475




4.10


Example 4.10
Using a Calculator to Find Powers of e


Calculate  e3.14. Round to five decimal places.


Solution
On a calculator, press the button labeled  [ex]. The window shows  ⎡⎣e^( ⎤⎦. Type  3.14  and then close parenthesis,
 ⎡⎣)⎤⎦.  Press [ENTER]. Rounding to  5  decimal places,  e3.14 ≈ 23.10387. Caution: Many scientific calculators
have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of  e.


Use a calculator to find  e−0.5. Round to five decimal places.


Investigating Continuous Growth
So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used
as the base for exponential functions. Exponential models that use  e  as the base are called continuous growth or decay
models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid
dynamics.


The Continuous Growth/Decay Formula
For all real numbers  t, and all positive numbers  a  and  r, continuous growth or decay is represented by the formula


(4.3)A(t) = aert


where
• a  is the initial value,
• r  is the continuous growth rate per unit time,
• and  t  is the elapsed time.


If  r > 0  , then the formula represents continuous growth. If  r < 0  , then the formula represents continuous decay.
For business applications, the continuous growth formula is called the continuous compounding formula and takes the
form


A(t) = Pert


where
• P  is the principal or the initial invested,
• r  is the growth or interest rate per unit time,
• and t  is the period or term of the investment.


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4.11


Given the initial value, rate of growth or decay, and time  t, solve a continuous growth or decay function.
1. Use the information in the problem to determine  a , the initial value of the function.
2. Use the information in the problem to determine the growth rate  r.


a. If the problem refers to continuous growth, then  r > 0.
b. If the problem refers to continuous decay, then  r < 0.


3. Use the information in the problem to determine the time  t.
4. Substitute the given information into the continuous growth formula and solve for  A(t).


Example 4.11
Calculating Continuous Growth


A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much
was in the account at the end of one year?


Solution
Since the account is growing in value, this is a continuous compounding problem with growth rate  r = 0.10. The
initial investment was $1,000, so  P = 1000. We use the continuous compounding formula to find the value after
 t = 1  year:


A(t) = Pert Use the continuous compounding formula.


= 1000(e)0.1 Substitute known values for P, r, and t.


≈ 1105.17 Use a calculator to approximate.


The account is worth $1,105.17 after one year.


A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be
the value of the investment in 30 years?


Example 4.12
Calculating Continuous Decay


Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3
days?


Solution
Since the substance is decaying, the rate,  17.3% , is negative. So,  r = − 0.173. The initial amount of
radon-222 was  100 mg, so  a = 100. We use the continuous decay formula to find the value after  t = 3  days:


A(t) = aert Use the continuous growth formula.


= 100e−0.173(3) Substitute known values for a, r, and t.


≈ 59.5115 Use a calculator to approximate.


Chapter 4 Exponential and Logarithmic Functions 477




4.12


So 59.5115 mg of radon-222 will remain.


Using the data in Example 4.12, how much radon-222 will remain after one year?


Access these online resources for additional instruction and practice with exponential functions.
• Exponential Growth Function (http://openstaxcollege.org/l/expgrowth)
• Compound Interest (http://openstaxcollege.org/l/compoundint)


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4.1 EXERCISES
Verbal
Explain why the values of an increasing exponential


function will eventually overtake the values of an
increasing linear function.
Given a formula for an exponential function, is it


possible to determine whether the function grows or decays
exponentially just by looking at the formula? Explain.
The Oxford Dictionary defines the word nominal as a


value that is “stated or expressed but not necessarily
corresponding exactly to the real value.”[3] Develop a
reasonable argument for why the term nominal rate is used
to describe the annual percentage rate of an investment
account that compounds interest.


Algebraic
For the following exercises, identify whether the statement
represents an exponential function. Explain.
The average annual population increase of a pack of


wolves is 25.


A population of bacteria decreases by a factor of  1
8


every  24  hours.
The value of a coin collection has increased by  3.25% 


annually over the last  20  years.
For each training session, a personal trainer charges his


clients  $5  less than the previous training session.


The height of a projectile at time  t  is represented by the
function  h(t) = − 4.9t2 + 18t + 40.


For the following exercises, consider this scenario: For
each year  t, the population of a forest of trees is
represented by the function  A(t) = 115(1.025)t.  In a
neighboring forest, the population of the same type of tree
is represented by the function  B(t) = 82(1.029)t.  (Round
answers to the nearest whole number.)
Which forest’s population is growing at a faster rate?
Which forest had a greater number of trees initially?


By how many?
Assuming the population growth models continue to


represent the growth of the forests, which forest will have a
greater number of trees after  20  years? By how many?


Assuming the population growth models continue to
represent the growth of the forests, which forest will have a
greater number of trees after  100  years? By how many?
Discuss the above results from the previous four


exercises. Assuming the population growth models
continue to represent the growth of the forests, which forest
will have the greater number of trees in the long run? Why?
What are some factors that might influence the long-term
validity of the exponential growth model?
For the following exercises, determine whether the
equation represents exponential growth, exponential decay,
or neither. Explain.


y = 300(1 − t)5


y = 220(1.06)x


y = 16.5(1.025)
1
x


y = 11, 701(0.97)t


For the following exercises, find the formula for an
exponential function that passes through the two points
given.


(0, 6)  and  (3, 750)


(0, 2000)  and  (2, 20)



⎝−1,


3
2

⎠  and  (3, 24)


(−2, 6)  and  (3, 1)


(3, 1)  and  (5, 4)


For the following exercises, determine whether the table
could represent a function that is linear, exponential, or
neither. If it appears to be exponential, find a function that
passes through the points.


x 1 2 3 4


f(x) 70 40 10 -20


3. Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina.


Chapter 4 Exponential and Logarithmic Functions 479




25.


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48.


x 1 2 3 4


h(x) 70 49 34.3 24.01


x 1 2 3 4


m(x) 80 61 42.9 25.61


x 1 2 3 4


f(x) 10 20 40 80


x 1 2 3 4


g(x) -3.25 2 7.25 12.5


For the following exercises, use the compound interest
formula,  A(t) = P⎛⎝1 + rn⎞⎠


nt
.


After a certain number of years, the value of an
investment account is represented by the equation
 10, 250⎛⎝1 + 0.0412





120
. What is the value of the account?


What was the initial deposit made to the account in the
previous exercise?
How many years had the account from the previous


exercise been accumulating interest?
An account is opened with an initial deposit of $6,500


and earns  3.6%  interest compounded semi-annually. What
will the account be worth in  20  years?
How much more would the account in the previous


exercise have been worth if the interest were compounding
weekly?
Solve the compound interest formula for the principal,


 P .


Use the formula found in the previous exercise to
calculate the initial deposit of an account that is worth
 $14, 472.74  after earning  5.5%  interest compounded
monthly for  5  years. (Round to the nearest dollar.)
How much more would the account in the previous two


exercises be worth if it were earning interest for  5 more
years?
Use properties of rational exponents to solve the


compound interest formula for the interest rate,  r.
Use the formula found in the previous exercise to


calculate the interest rate for an account that was
compounded semi-annually, had an initial deposit of $9,000
and was worth $13,373.53 after 10 years.
Use the formula found in the previous exercise to


calculate the interest rate for an account that was
compounded monthly, had an initial deposit of $5,500, and
was worth $38,455 after 30 years.
For the following exercises, determine whether the
equation represents continuous growth, continuous decay,
or neither. Explain.


y = 3742(e)0.75t


y = 150(e)
3.25
t


y = 2.25(e)−2t


Suppose an investment account is opened with an
initial deposit of  $12, 000  earning  7.2%  interest
compounded continuously. How much will the account be
worth after  30  years?
How much less would the account from Exercise 42 be


worth after  30  years if it were compounded monthly
instead?


Numeric
For the following exercises, evaluate each function. Round
answers to four decimal places, if necessary.


f (x) = 2(5)x, for   f (−3)


f (x) = − 42x + 3, for   f (−1)


f (x) = ex, for   f (3)


f (x) = − 2ex − 1, for   f (−1)


f (x) = 2.7(4)−x + 1 + 1.5, for f (−2)


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68.


f (x) = 1.2e2x − 0.3, for   f (3)


f (x) = − 3
2
(3)−x + 3


2
, for   f (2)


Technology
For the following exercises, use a graphing calculator to
find the equation of an exponential function given the
points on the curve.


(0, 3)  and  (3, 375)


(3, 222.62)  and  (10, 77.456)


(20, 29.495)  and  (150, 730.89)


(5, 2.909)  and  (13, 0.005)


(11,310.035)  and (25,356.3652)


Extensions
The annual percentage yield (APY) of an investment


account is a representation of the actual interest rate earned
on a compounding account. It is based on a compounding
period of one year. Show that the APY of an account that
compounds monthly can be found with the formula
 APY = ⎛⎝1 + r12





12
− 1.


Repeat the previous exercise to find the formula for the
APY of an account that compounds daily. Use the results
from this and the previous exercise to develop a function
 I(n)  for the APY of any account that compounds  n  times
per year.
Recall that an exponential function is any equation


written in the form   f (x) = a ⋅ bx   such that a and
b are positive numbers and b ≠ 1. Any positive
number b can be written as b = en for some value
of n . Use this fact to rewrite the formula for an
exponential function that uses the number e as a base.


In an exponential decay function, the base of the
exponent is a value between 0 and 1. Thus, for some
number  b > 1, the exponential decay function can be
written as   f (x) = a ⋅ ⎛⎝1b





x
. Use this formula, along with


the fact that  b = en, to show that an exponential decay
function takes the form   f (x) = a(e)−nx   for some positive
number  n  .
The formula for the amount  A  in an investment


account with a nominal interest rate  r  at any time  t  is


given by  A(t) = a(e)rt, where  a  is the amount of
principal initially deposited into an account that compounds
continuously. Prove that the percentage of interest earned to
principal at any time  t  can be calculated with the formula
 I(t) = ert − 1.


Real-World Applications
The fox population in a certain region has an annual


growth rate of 9% per year. In the year 2012, there were
23,900 fox counted in the area. What is the fox population
predicted to be in the year 2020?
A scientist begins with 100 milligrams of a radioactive


substance that decays exponentially. After 35 hours, 50mg
of the substance remains. How many milligrams will
remain after 54 hours?
In the year 1985, a house was valued at $110,000. By


the year 2005, the value had appreciated to $145,000. What
was the annual growth rate between 1985 and 2005?
Assume that the value continued to grow by the same
percentage. What was the value of the house in the year
2010?
A car was valued at $38,000 in the year 2007. By 2013,


the value had depreciated to $11,000 If the car’s value
continues to drop by the same percentage, what will it be
worth by 2017?
Jamal wants to save $54,000 for a down payment on a


home. How much will he need to invest in an account with
8.2% APR, compounding daily, in order to reach his goal in
5 years?
Kyoko has $10,000 that she wants to invest. Her bank


has several investment accounts to choose from, all
compounding daily. Her goal is to have $15,000 by the time
she finishes graduate school in 6 years. To the nearest
hundredth of a percent, what should her minimum annual
interest rate be in order to reach her goal? (Hint: solve the
compound interest formula for the interest rate.)
Alyssa opened a retirement account with 7.25% APR


in the year 2000. Her initial deposit was $13,500. How
much will the account be worth in 2025 if interest
compounds monthly? How much more would she make if
interest compounded continuously?
An investment account with an annual interest rate of


7% was opened with an initial deposit of $4,000 Compare
the values of the account after 9 years when the interest is
compounded annually, quarterly, monthly, and
continuously.


Chapter 4 Exponential and Logarithmic Functions 481




4.2 | Graphs of Exponential Functions
Learning Objectives


4.2.1 Graph exponential functions.
4.2.2 Graph exponential functions using transformations.


As we discussed in the previous section, exponential functions are used for many real-world applications such as finance,
forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation
gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about
things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It
gives us another layer of insight for predicting future events.
Graphing Exponential Functions
Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function
of the form   f (x) = bx  whose base is greater than one. We’ll use the function   f (x) = 2x. Observe how the output values
in Table 4.7 change as the input increases by  1.


x −3 −2 −1 0 1 2 3


f(x) = 2x 18
1
4


1
2 1 2 4 8


Table 4.7


Each output value is the product of the previous output and the base,  2. We call the base  2  the constant ratio. In fact, for
any exponential function with the form   f (x) = abx,  b  is the constant ratio of the function. This means that as the input
increases by 1, the output value will be the product of the base and the previous output, regardless of the value of  a.
Notice from the table that


• the output values are positive for all values of x;
• as  x  increases, the output values increase without bound; and
• as  x  decreases, the output values grow smaller, approaching zero.


Figure 4.8 shows the exponential growth function   f (x) = 2x.


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Figure 4.8 Notice that the graph gets close to the x-axis, but
never touches it.


The domain of   f (x) = 2x   is all real numbers, the range is  (0, ∞), and the horizontal asymptote is  y = 0.
To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form   f (x) = bx  
whose base is between zero and one. We’ll use the function  g(x) = ⎛⎝12





x
. Observe how the output values in Table 4.8


change as the input increases by  1.


x -3 -2 -1 0 1 2 3


g(x) = ⎛⎝
1
2



x
8 4 2 1 12


1
4


1
8


Table 4.8


Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant
ratio  1


2
.


Notice from the table that
• the output values are positive for all values of  x;
• as  x  increases, the output values grow smaller, approaching zero; and
• as  x  decreases, the output values grow without bound.


Figure 4.9 shows the exponential decay function,  g(x) = ⎛⎝12



x
.


Chapter 4 Exponential and Logarithmic Functions 483




Figure 4.9


The domain of  g(x) = ⎛⎝12



x
  is all real numbers, the range is  (0, ∞), and the horizontal asymptote is  y = 0.


Characteristics of the Graph of the Parent Function f(x) = bx
An exponential function with the form   f (x) = bx,  b > 0,  b ≠ 1, has these characteristics:


• one-to-one function
• horizontal asymptote:  y = 0
• domain:  ( – ∞, ∞)
• range:  (0, ∞)
• x-intercept: none
• y-intercept:  (0, 1) 
• increasing if  b > 1
• decreasing if  b < 1


Figure 4.10 compares the graphs of exponential growth and decay functions.


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Figure 4.10


Given an exponential function of the form   f(x) = bx, graph the function.
1. Create a table of points.
2. Plot at least  3  point from the table, including the y-intercept  (0, 1).
3. Draw a smooth curve through the points.
4. State the domain,  (−∞, ∞), the range,  (0, ∞), and the horizontal asymptote,  y = 0.


Example 4.13


Sketching the Graph of an Exponential Function of the Form f(x) = bx


Sketch a graph of   f (x) = 0.25x.  State the domain, range, and asymptote.


Solution
Before graphing, identify the behavior and create a table of points for the graph.


• Since  b = 0.25  is between zero and one, we know the function is decreasing. The left tail of the graph
will increase without bound, and the right tail will approach the asymptote  y = 0.


• Create a table of points as in Table 4.9.


x −3 −2 −1 0 1 2 3


f(x) = 0.25x 64 16 4 1 0.25 0.0625 0.015625


Table 4.9
• Plot the y-intercept,  (0, 1), along with two other points. We can use  (−1, 4)  and  (1, 0.25).


Chapter 4 Exponential and Logarithmic Functions 485




4.13


Draw a smooth curve connecting the points as in Figure 4.11.


Figure 4.11


The domain is  (−∞, ∞);   the range is  (0, ∞);   the horizontal asymptote is  y = 0.


Sketch the graph of   f (x) = 4x.  State the domain, range, and asymptote.


Graphing Transformations of Exponential Functions
Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions,
we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function
f (x) = bx  without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted,
reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the
transformations applied.
Graphing a Vertical Shift
The first transformation occurs when we add a constant  d  to the parent function   f (x) = bx, giving us a vertical shift
 d  units in the same direction as the sign. For example, if we begin by graphing a parent function,   f (x) = 2x, we
can then graph two vertical shifts alongside it, using  d = 3 :   the upward shift,  g(x) = 2x + 3  and the downward shift,
 h(x) = 2x − 3. Both vertical shifts are shown in Figure 4.12.


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Figure 4.12


Observe the results of shifting   f (x) = 2x   vertically:
• The domain,  (−∞, ∞)  remains unchanged.
• When the function is shifted up  3  units to  g(x) = 2x + 3 :


◦ The y-intercept shifts up  3  units to  (0, 4).
◦ The asymptote shifts up  3  units to  y = 3.
◦ The range becomes  (3, ∞).


• When the function is shifted down  3  units to  h(x) = 2x − 3 :
◦ The y-intercept shifts down  3  units to  (0, − 2).
◦ The asymptote also shifts down  3  units to  y = − 3.
◦ The range becomes  (−3, ∞).


Graphing a Horizontal Shift
The next transformation occurs when we add a constant  c  to the input of the parent function   f (x) = bx, giving us a
horizontal shift  c  units in the opposite direction of the sign. For example, if we begin by graphing the parent function
  f (x) = 2x, we can then graph two horizontal shifts alongside it, using  c = 3 :   the shift left,  g(x) = 2x + 3, and the shift
right,  h(x) = 2x − 3. Both horizontal shifts are shown in Figure 4.13.


Chapter 4 Exponential and Logarithmic Functions 487




Figure 4.13


Observe the results of shifting   f (x) = 2x   horizontally:
• The domain,  (−∞, ∞), remains unchanged.
• The asymptote,  y = 0, remains unchanged.
• The y-intercept shifts such that:


◦ When the function is shifted left  3  units to  g(x) = 2x + 3, the y-intercept becomes  (0, 8). This is because
 2x + 3 = (8)2x, so the initial value of the function is  8.


◦ When the function is shifted right  3  units to  h(x) = 2x − 3, the y-intercept becomes  ⎛⎝0, 18

⎠. Again, see that


 2x − 3 = ⎛⎝18

⎠2


x, so the initial value of the function is  1
8
.


Shifts of the Parent Function f(x) = bx


For any constants  c  and  d, the function   f (x) = bx + c + d  shifts the parent function   f (x) = bx


• vertically  d  units, in the same direction of the sign of  d.
• horizontally  c  units, in the opposite direction of the sign of  c.
• The y-intercept becomes  ⎛⎝0, bc + d⎞⎠.
• The horizontal asymptote becomes  y = d.
• The range becomes  (d, ∞).
• The domain,  (−∞, ∞), remains unchanged.


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Given an exponential function with the form   f(x) = bx+ c+ d, graph the translation.
1. Draw the horizontal asymptote  y = d.
2. Identify the shift as  (−c, d).  Shift the graph of   f (x) = bx   left  c  units if  c  is positive, and right  c  units
if c  is negative.


3. Shift the graph of   f (x) = bx   up  d  units if  d  is positive, and down  d  units if  d  is negative.
4. State the domain,  (−∞, ∞), the range,  (d, ∞), and the horizontal asymptote  y = d.


Example 4.14
Graphing a Shift of an Exponential Function


Graph   f (x) = 2x + 1 − 3.  State the domain, range, and asymptote.


Solution
We have an exponential equation of the form   f (x) = bx + c + d, with  b = 2,  c = 1, and  d = −3.
Draw the horizontal asymptote  y = d , so draw  y = −3.
Identify the shift as  (−c, d), so the shift is  (−1, −3).
Shift the graph of   f (x) = bx   left 1 units and down 3 units.


Figure 4.14


The domain is  (−∞, ∞);   the range is  (−3, ∞);   the horizontal asymptote is  y = −3.


Chapter 4 Exponential and Logarithmic Functions 489




4.14


4.15


Graph   f (x) = 2x − 1 + 3.  State domain, range, and asymptote.


Given an equation of the form   f(x) = bx+ c+ d  for  x, use a graphing calculator to approximate the
solution.


• Press [Y=]. Enter the given exponential equation in the line headed “Y1=”.
• Enter the given value for   f (x)  in the line headed “Y2=”.
• Press [WINDOW]. Adjust the y-axis so that it includes the value entered for “Y2=”.
• Press [GRAPH] to observe the graph of the exponential function along with the line for the specified
value of   f (x).


• To find the value of  x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect”
and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of
the function.


Example 4.15
Approximating the Solution of an Exponential Equation


Solve  42 = 1.2(5)x + 2.8  graphically. Round to the nearest thousandth.


Solution
Press [Y=] and enter  1.2(5)x + 2.8  next to Y1=. Then enter 42 next to Y2=. For a window, use the values –3 to
3 for  x  and –5 to 55 for  y.  Press [GRAPH]. The graphs should intersect somewhere near  x = 2.
For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The
x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a
different window or use a different value for Guess?) To the nearest thousandth,  x ≈ 2.166.


Solve  4 = 7.85(1.15)x − 2.27  graphically. Round to the nearest thousandth.


Graphing a Stretch or Compression
While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression
occurs when we multiply the parent function   f (x) = bx   by a constant  |a| > 0.  For example, if we begin by graphing
the parent function   f (x) = 2x, we can then graph the stretch, using  a = 3, to get  g(x) = 3(2)x   as shown on the left in
Figure 4.15, and the compression, using  a = 1


3
, to get  h(x) = 1


3
(2)x   as shown on the right in Figure 4.15.


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Figure 4.15 (a)  g(x) = 3(2)x   stretches the graph of   f (x) = 2x   vertically by a factor of  3.  (b)  h(x) = 1
3
(2)x   compresses


the graph of   f (x) = 2x   vertically by a factor of  1
3
.


Stretches and Compressions of the Parent Function f(x) = bx
For any factor  a > 0, the function   f (x) = a(b)x


• is stretched vertically by a factor of  a  if  |a| > 1.
• is compressed vertically by a factor of  a  if  |a| < 1.
• has a y-intercept of  (0, a).
• has a horizontal asymptote at  y = 0, a range of  (0, ∞), and a domain of  (−∞, ∞), which are unchanged
from the parent function.


Example 4.16


Graphing the Stretch of an Exponential Function


Sketch a graph of   f (x) = 4⎛⎝12



x
.  State the domain, range, and asymptote.


Solution
Before graphing, identify the behavior and key points on the graph.


• Since  b = 1
2
  is between zero and one, the left tail of the graph will increase without bound as  x 


decreases, and the right tail will approach the x-axis as  x  increases.


• Since  a = 4, the graph of   f (x) = ⎛⎝12



x
 will be stretched by a factor of  4.


Chapter 4 Exponential and Logarithmic Functions 491




4.16


• Create a table of points as shown in Table 4.10.


x −3 −2 −1 0 1 2 3


f(x) = 4⎛⎝
1
2



x
32 16 8 4 2 1 0.5


Table 4.10
• Plot the y-intercept,  (0, 4), along with two other points. We can use  (−1, 8)  and  (1, 2).


Draw a smooth curve connecting the points, as shown in Figure 4.16.


Figure 4.16


The domain is  (−∞, ∞);   the range is  (0, ∞);   the horizontal asymptote is  y = 0.


Sketch the graph of   f (x) = 1
2
(4)x.  State the domain, range, and asymptote.


Graphing Reflections
In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we
multiply the parent function   f (x) = bx   by  −1, we get a reflection about the x-axis. When we multiply the input by  −1,
we get a reflection about the y-axis. For example, if we begin by graphing the parent function   f (x) = 2x, we can then
graph the two reflections alongside it. The reflection about the x-axis,  g(x) = −2x, is shown on the left side of Figure
4.17, and the reflection about the y-axis  h(x) = 2−x, is shown on the right side of Figure 4.17.


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Figure 4.17 (a)  g(x) = − 2x   reflects the graph of   f (x) = 2x   about the x-axis. (b)  g(x) = 2−x   reflects the graph of
  f (x) = 2x   about the y-axis.


Reflections of the Parent Function f(x) = bx
The function   f (x) = − bx


• reflects the parent function   f (x) = bx   about the x-axis.
• has a y-intercept of  (0, − 1).
• has a range of  (−∞, 0)
• has a horizontal asymptote at  y = 0  and domain of  (−∞, ∞), which are unchanged from the parent function.


The function   f (x) = b−x


• reflects the parent function   f (x) = bx   about the y-axis.
• has a y-intercept of  (0, 1), a horizontal asymptote at  y = 0, a range of  (0, ∞), and a domain of


 (−∞, ∞), which are unchanged from the parent function.


Example 4.17
Writing and Graphing the Reflection of an Exponential Function


Chapter 4 Exponential and Logarithmic Functions 493




4.17


Find and graph the equation for a function,  g(x), that reflects   f (x) = ⎛⎝14



x
about the x-axis. State its domain,


range, and asymptote.


Solution
Since we want to reflect the parent function   f (x) = ⎛⎝14





x
  about the x-axis, we multiply   f (x)  by  − 1  to get,


 g(x) = − ⎛⎝14



x
. Next we create a table of points as in Table 4.11.


x −3 −2 −1 0 1 2 3


g(x) = − ⎛⎝
1
4



x
−64 −16 −4 −1 −0.25 −0.0625 −0.0156


Table 4.11


Plot the y-intercept,  (0, −1), along with two other points. We can use  (−1, −4)  and  (1, −0.25).
Draw a smooth curve connecting the points:


Figure 4.18


The domain is  (−∞, ∞);   the range is  (−∞, 0);   the horizontal asymptote is  y = 0.


Find and graph the equation for a function,  g(x), that reflects   f (x) = 1.25x   about the y-axis. State its
domain, range, and asymptote.


Summarizing Translations of the Exponential Function
Now that we have worked with each type of translation for the exponential function, we can summarize them in Table 4.12
to arrive at the general equation for translating exponential functions.


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Translations of the Parent Function   f(x) = bx


Translation Form


Shift
• Horizontally  c  units to the left
• Vertically  d  units up


f (x) = bx + c + d


Stretch and Compress
• Stretch if  |a| > 1
• Compression if  0 < |a| < 1


f (x) = abx


Reflect about the x-axis f (x) = − bx


Reflect about the y-axis f (x) = b−x = ⎛⎝1b



x


General equation for all translations f (x) = abx + c + d


Table 4.12


Translations of Exponential Functions
A translation of an exponential function has the form


(4.4) f (x) = abx + c + d
Where the parent function,  y = bx,  b > 1, is


• shifted horizontally  c  units to the left.
• stretched vertically by a factor of  |a|  if  |a| > 0.
• compressed vertically by a factor of  |a|  if  0 < |a| < 1.
• shifted vertically  d  units.
• reflected about the x-axis when  a < 0.


Note the order of the shifts, transformations, and reflections follow the order of operations.


Example 4.18


Writing a Function from a Description


Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.
• f (x) = ex   is vertically stretched by a factor of  2  , reflected across the y-axis, and then shifted up  4 
units.


Chapter 4 Exponential and Logarithmic Functions 495




4.18


Solution
We want to find an equation of the general form   f (x) = abx + c + d. We use the description provided to find
 a, b, c, and  d.
• We are given the parent function   f (x) = ex, so  b = e.
• The function is stretched by a factor of  2 , so  a = 2.
• The function is reflected about the y-axis. We replace  x with  − x  to get:  e−x.
• The graph is shifted vertically 4 units, so  d = 4.


Substituting in the general form we get,
f (x) = abx + c + d


= 2e−x + 0 + 4
= 2e−x + 4


The domain is  (−∞, ∞);   the range is  (4, ∞);   the horizontal asymptote is  y = 4.


Write the equation for function described below. Give the horizontal asymptote, the domain, and the
range.


• f (x) = ex   is compressed vertically by a factor of  1
3
, reflected across the x-axis and then shifted


down  2 units.


Access this online resource for additional instruction and practice with graphing exponential functions.
• Graph Exponential Functions (http://openstaxcollege.org/l/graphexpfunc)


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69.


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84.


85.


86.


4.2 EXERCISES
Verbal
What role does the horizontal asymptote of an


exponential function play in telling us about the end
behavior of the graph?
What is the advantage of knowing how to recognize


transformations of the graph of a parent function
algebraically?


Algebraic
The graph of   f (x) = 3x   is reflected about the y-axis


and stretched vertically by a factor of  4. What is the
equation of the new function,  g(x)?   State its y-intercept,
domain, and range.


The graph of   f (x) = ⎛⎝12



−x
  is reflected about the y-


axis and compressed vertically by a factor of  1
5
. What is


the equation of the new function,  g(x)?   State its y-
intercept, domain, and range.


The graph of   f (x) = 10x   is reflected about the x-axis
and shifted upward  7  units. What is the equation of the
new function,  g(x)?   State its y-intercept, domain, and
range.


The graph of   f (x) = (1.68)x   is shifted right  3  units,
stretched vertically by a factor of  2, reflected about the x-
axis, and then shifted downward  3  units. What is the
equation of the new function,  g(x)?   State its y-intercept
(to the nearest thousandth), domain, and range.


The graph of   f (x) = − 1
2


1
4



x − 2
+ 4  is shifted left


 2  units, stretched vertically by a factor of  4, reflected
about the x-axis, and then shifted downward  4  units. What
is the equation of the new function,  g(x)?   State its y-
intercept, domain, and range.


Graphical
For the following exercises, graph the function and its
reflection about the y-axis on the same axes, and give the
y-intercept.


f (x) = 3⎛⎝
1
2



x


g(x) = − 2(0.25)x


h(x) = 6(1.75)−x


For the following exercises, graph each set of functions on
the same axes.


f (x) = 3⎛⎝
1
4



x
, g(x) = 3(2)x, and  h(x) = 3(4)x


f (x) = 1
4
(3)x, g(x) = 2(3)x, and  h(x) = 4(3)x


For the following exercises, match each function with one
of the graphs in Figure 4.19.


Figure 4.19


f (x) = 2(0.69)x


f (x) = 2(1.28)x


f (x) = 2(0.81)x


f (x) = 4(1.28)x


f (x) = 2(1.59)x


f (x) = 4(0.69)x


For the following exercises, use the graphs shown in
Figure 4.20. All have the form   f (x) = abx.


Chapter 4 Exponential and Logarithmic Functions 497




87.


88.


89.


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93.


94.


95.


96.


97.


98.


99.


100.


101.


102.


103.


104.


105.


106.


107.


Figure 4.20


Which graph has the largest value for  b?
Which graph has the smallest value for  b?
Which graph has the largest value for  a?
Which graph has the smallest value for  a?


For the following exercises, graph the function and its
reflection about the x-axis on the same axes.


f (x) = 1
2
(4)x


f (x) = 3(0.75)x − 1


f (x) = − 4(2)x + 2


For the following exercises, graph the transformation of
f (x) = 2x. Give the horizontal asymptote, the domain,