power, polynomial, and rational functions chapter...

Chapter PlannerPower, Polynomial, and Rational FunctionsPower, Polynomial, and Rational Functions

84A | Chapter 2 | Power, Polynomial, and Rational Functions

All digital assets are Interactive Whiteboard ready.

Diagnostic AssessmentQuick Check, p. 85

Pacing: 2 daysLESSONLELELELELESSSSSSSSSSONONONONON

22-112222-111Pacing: 0.5 day

EXPLOREEXEXEXEXEXPLPLPLPLPLOROROROROREEEEE

22-2222-2Pacing: 2 days

LESSONLELELELELESSSSSSSSSSONONONONON

22-222222-222Pacing: 0.5 day

EXEXEXEXEXTETETETETENDNDNDNDND

22-2222-2Title Power and Radical

Functions

Graphing Technology Lab:

Behavior of GraphsPolynomial Functions

Graphing Technology Lab:

Hidden Behavior of Graphs

Objectives ■ Graph and analyze power functions.

■ Graph and analyze radical functions and solve radical equations.

■ Graph and analyze the behavior of polynomial functions.

■ Graph polynomial functions.

■ Model real-world data with polynomial functions.

■ Use TI-Nspire to explore the hidden behavior of graphs.

Key Vocabulary power function, monomial function, radical function, extraneous solutions

polynomial function, leading coefficient, leading-term test, turning point, quadratic form, repeated zero, multiplicity

NCTM Standards 2, 3, 6, 7, 8, 9, 10 3 2, 3, 4, 6, 7, 8, 9, 10 3

Multiple Representations

p. 94 p. 106

Lesson Resources connectED.mcgraw-hill.com

Worksheets

Vocabulary

Personal Tutor

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook

connectED.mcgraw-hill.com

■ TI-83/84 Plus or other graphing calculator


Worksheets

Vocabulary

Animations

Personal Tutor

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook


■ TI-Nspire technology

Resources for Every Lesson

eStudent Edition ■ eTeacher Edition

Interactive Classroom ■ Assessment

Differentiated Instruction

pp. 90, 91, 95 pp. 100, 107

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Suggested Pacing

Time Periods Instruction Review & Assess Total

45-minute 13 days 2 days 15 days

90-minute 6 days 1 day 7 days

Pacing: 2 daysLESSONLELELELELESSSSSSSSSSONONONONON

2 322-332222-333Pacing: 2 days

LESSONLELELELELESSSSSSSSSSONONONONON

22-442222-444Pacing: 2 days

SSONNLELELELELLESSSSSSSSSSONONONONON

2 52-5522-55Pacing: 1 day

SSONNLELELELELLESSSSSSSSSSONONONONON

2 62-6622-66The Remainder and Factor

TheoremsZeros of Polynomial Functions Rational Functions Nonlinear Inequalities

■ Divide polynomials using long division and synthetic division.

■ Use the Remainder and Factor Theorems.

■ Find real zeros of polynomial functions.

■ Find complex zeros of polynomial functions.

■ Analyze and graph rational functions.

■ Solve rational equations.

■ Solve polynomial inequalities.

■ Solve rational inequalities.

synthetic division, depressed polynomial, synthetic substitution

Rational Zero Theorem, Descartes’ Rule of Signs, Fundamental Theorem of Algebra, Linear Factorization Theorem, complex conjugates

rational function, asymptote, vertical asymptote, horizontal asymptote, oblique asymptote, holes

polynomial inequality, sign chart, rational inequality

2, 6, 7, 8, 9, 10 2, 3, 6, 7, 8, 9, 10 2, 3, 6, 7, 8, 9, 10 2, 6, 7, 8, 9, 10

p. 116 p. 128 p. 139 p. 146


Worksheets

Vocabulary

Personal Tutor

Virtual Manipulatives

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook


Worksheets

Vocabulary

Animations

Personal Tutor

Virtual Manipulatives

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook


Worksheets

Vocabulary

Personal Tutor

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook


Worksheets

Vocabulary

Personal Tutor

Self-Check Quiz

■ 5-Minute Check

■ Study Notebook

eStudent Edition ■ eTeacher Edition

Interactive Classroom ■ Assessment

pp. 110, 117 pp. 124, 129 pp. 132, 136 pp. 143, 144

Formative AssessmentMid-Chapter Quiz, p. 118

Study Guide and Review, pp. 148–152Practice Test, p. 153Connect to AP Calculus pp. 154–155

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SE = Student Edition, TE = Teacher Edition, CRM = Chapter Resource Masters

Assessment and Intervention

84C | Chapter 2 | Power, Polynomial, and Rational Functions

2APower, Polynomial, and Rational FunctionsPower, Polynomial, and Rational Functions

Diagnosis Prescription

DIAG

NOS

TIC

ASSE

SSM

ENT Beginning Chapter 2

Get Ready for the Chapter SE Response to Intervention TE

Beginning Every Lesson

Then, Now, Why? SE

5-Minute Check

During/After Every Lesson

FORM

ATIV

EAS

SESS

MEN

T Guided Practice SE, every exampleH.O.T. Problems SE

Spiral Review SE

Additional Examples TE

Watch Out! TE

Step 4, Assess TE

Self-Check Quizzes connectED.mcgraw-hill.com

TIER 1 Intervention Practice CRM


TIER 2 Intervention Differentiated Instruction TE; Study Guide and Intervention Masters CRM

Mid-Chapter

Mid-Chapter Quiz SE

AssessmentTIER 1 Intervention Practice CRM



Before Chapter Test

Chapter Study Guide and Review SE

Practice Test SE

Chapter Test connectED.mcgraw-hill.com

Vocabulary Review connectED.mcgraw-hill.com

Assessment

TIER 1 Intervention Practice CRM



After Chapter 2

SUM

MAT

IVE

ASSE

SSM

ENT Multiple-Choice Tests CRM

Free-Response Tests CRM

Vocabulary Test CRM

Extended Response Test CRM

Assessment

Study Guide and Intervention CRM


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Differentiated Instruction2DPower, Polynomial, and Rational FunctionsPower, Polynomial, and Rational Functions

Option 1 Reaching All LearnersAL OL

BL ELL

Logical Learners Have groups of students write and graph six functions, two for each of the following types:

f (x ) = x n, f (x ) = x -n, and f (x ) = x p _ n , where n and p are positive

integers and p _ n is in simplest form. Have students write each

function and each graph on a separate index card. Groups trade cards and try to match each graph with its function.

= g x( ) x 4

y

xO

Visual/Spatial Learners Have students use grid paper and colored pencils to help organize their work when performing synthetic division. For example, have students write 6 x 3 - 25 x 2 + 18 x + 9 divided by x - 3 with color as shown. Then have them write on grid paper with the same colors.

3 6

↓

6

-25 18 9

18 -21 -9

-7 -3 0

Option 2 Approaching Level AL

Divide students into groups of three or four. Have students take turns explaining how to determine the end behavior of the graph of a polynomial function and how to locate the zeros of the function. Have them describe how knowing this information helps them graph the polynomial.

Option 3 English Learners ELL

Provide students with a photocopy of some or all of the word problems from this chapter. Before they begin to solve the problems, have students highlight all words they do not understand. Pair each ELL student with a partner with fluent English-language skills and have them discuss the meanings of the highlighted words.

Option 4 Beyond Level BL

Have students take a picture of a nonlinear item. Ask them to superimpose a coordinate grid and sketch a curve over a section of the picture as shown. Then have students locate points on the graph and use those points and the CubicReg tool on their calculators to write a polynomial function that models that portion of the curve.

The example below is a close-up of a zebra’s pattern. The function f (x ) = 0.011 x 3 - 0.217 x 2 + 1.359 x + 0.234 comes close to this portion of the pattern.

x

y

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Focus on Mathematical ContentPower, Polynomial, and Rational FunctionsPower, Polynomial, and Rational Functions

84E | Chapter 2 | Power, Polynomial, and Rational Functions

Lesson-by-Lesson PreviewVertical Alignment

2-1 Power and Radical FunctionsFunctions of the form f (x ) = a x n , where a and n are constant real numbers, are power functions. A power function is also a type of monomial function. A monomial function is any function that can be written as f (x ) = a or f (x ) = a x n , where a and n are nonzero constant real numbers. If n is an even positive integer:

y

x

Even Degree

If n is an odd positive integer:

y

x

Odd Degree

Power functions of the form f (x ) = x p _ n can be written as radical

functions of the form f (x ) = n √ �

x p , where n and p are positive integers greater than 1 that have no common factors. Note that the domain may be restricted to nonnegative values.If n is an even positive integer:

y

x

Even Degree

If n is an odd positive integer:

y

x

Odd Degree

Before Chapter 2■ Evaluate polynomials and terms of polynomials.■ Graph quadratic functions.■ Solve quadratic equations and inequalities.■ Work with function notation and manipulate and evaluate

functions.

Chapter 2■ Graph and analyze power, radical, polynomial, and rational

functions.■ Divide polynomials using long division and synthetic

division.■ Use the Remainder and Factor Theorems.■ Find all zeros of polynomial functions.■ Solve radical and rational equations.■ Solve polynomial and rational inequalities.

After Chapter 2■ Determine the domain and range of functions using

graphs, tables, and symbols.■ Find the limit of a polynomial function, rational function,

and radical function.■ Sketch the graph of a function.■ Solve real-world applications involving absolute extrema

on a closed interval.

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2-2 Polynomial FunctionsMonomial functions are the most basic polynomial functions. Finding the sums and differences of monomial functions creates other types of polynomial functions. The end behavior of a polynomial function f (x ) = a n x n + … + a 1 x + a 0 can be determined by the degree n of the polynomial and its leading coefficient a n .

A polynomial function f (x ) = a n x n + … + a 1 x + a 0 has at most n distinct real zeros and at most n - 1 turning points. The zeros can be found by factoring. The repeated zero c occurs when the factor (x - c ) repeats itself. The number of times (x - c ) occurs is the multiplicity of c.

■ odd multiplicity: the function’s graph crosses the x-axis at c and the value of f (x ) changes signs at x = c.

■ even multiplicity: the function’s graph touches the x-axis at c and the value of f (x ) does not change signs at x = c.

2-3 The Remainder and Factor TheoremsAn algorithm similar to the long division algorithm for integers can be used to divide polynomials. Division of polynomials can result in a zero remainder or in a nonzero remainder.

Synthetic division is a shortcut for dividing a polynomial by a linear factor of the form (x - c ) using the coefficients of the dividend. So, a divisor such as (x + 2) would first be written as (x - (-2)) .

1

1

coefficients of quotient remainder

Divide

5 -4-2

Add terms.

Multiply by c.3

-2 -6

-10

x 2 + 5x 4 by x + 2.

Whether using long division or synthetic division, use zeros as placeholders for any missing term in the dividend and write the polynomial in standard form. If a polynomial function is divided by (x - c ), then the remainder equals f (c ) and (x - c ) is a factor of the polynomial if and only if f (c ) = 0.

2-4 Zeros of Polynomial FunctionsReal zeros can be either rational or irrational. The Rational Zero Theorem, on p. 119, uses the leading coefficient and the constant term of a polynomial function with integer coefficients to determine all the possible rational zeros. Direct or synthetic substitution can be used to determine which of those possible zeros are actual zeros. Narrow the search for the actual zeros by determining the interval (upper and lower bounds) within which the zeros are located or by using Descartes’ Rule of Signs.

2-5 Rational FunctionsThe quotient of two polynomial functions is a rational function. ■ Vertical asymptotes, if any, occur at real zeros (undefined

values), if any, of the denominator of the function.■ y = 0 is a horizontal asymptote if the degree n of the

numerator is less than the degree m of the denominator. ■ No horizontal asymptotes occur if n > m.■ If n = m, there is a horizontal asymptote at the ratio

of the leading coefficients of the numerator and denominator.

■ If n = m + 1, where m > 0, the graph has an oblique asymptote.

2-6 Nonlinear InequalitiesThe real zeros of a polynomial function divide the x-axis into intervals for which the values of f (x ) are either entirely positive (the graph is above the x-axis) or entirely negative (the graph is below the x-axis). A polynomial inequality can be solved using a sign chart and its end behavior. A rational inequality can be solved by first writing the inequality in general form with a single rational expression on the left side and a 0 on the right and then creating a sign chart using its real zeros and undefined points.

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Chapter 2

84 | Chapter 2 | Power, Polynomial, and Rational Functions

Chapter Project

Olé Cocoa CaféStudents use what they learned about power, radical, polynomial and rational functions and inequalities to analyze aspects of a business.

■ Have students think about selling cocoa to complete each statement.

1. My café sells on average cocoas each month at an average of $ per beverage.

2. I will sell fewer cocoas for every $ I raise the price.

3. I want total sales to equal $ .

4. For selling n cocoas, I will make p (n ) = n 2 - n hundreds of dollars in revenue. It costs me c (n ) = n + to make n cocoas.

■ Have students write a function for their total sales after raising the price of each cocoa by x dollars. Have students determine how many dollars they need to raise the price of each cocoa so that the total amount of sales equals their number from statement 3 above.

■ Have students use their answers to statement 4 to determine the minimum number of cocoas they need to sell in order to make a profit.

Key Vocabulary Introduce the key vocabulary in the chapter using the routine below.

Define: An asymptote is a line or curve that a graph approaches.

Example: The line x = -2 is a vertical asymptote of the graph of

f (x ) = 5 _ (x + 2) 2

.

Ask: Why can x never equal -2 in

f (x ) = 5 _ (x + 2) 2

? The function is

undefined for any value that makes the denominator equal to 0.

Preread/Prewrite Anticipation Guide

Study Notebook

Encourage students to begin their study of the chapter by prereading each lesson. They should think about their background knowledge and make predictions about the content. Allow time for groups to discuss the reading and generate questions. Emphasize the text features such as section headings and Key Concept and Concept Summary Boxes. Students should also complete the first column of the Anticipation Guide as well as the chapter organizer from the Study Notebook to prepare for Chapter 2.

connectED.mcgraw-hill.comcccc Your Digital Math Portal

Animation Vocabulary eGlossaryPersonal

TutorGraphingCalculator

AudioSelf-Check

PracticeWorksheets

Then Now Why?

In Chapter 1, you analyzed functions and their graphs and determined whether inverse functions existed.

In Chapter 2, you will:

Model real-world data with polynomial functions.

Use the Remainder and Factor Theorems.

Find real and complex zeros of polynomial functions.

Analyze and graph rational functions.

Solve polynomial and rational inequalities.

ARCHITECTURE Polynomial functions are often used when designing and building a new structure. Architects use functions to determine the weight and strength of the materials, analyze costs, estimate deterioration of materials, and determine the proper labor force.

PREREAD Scan the lessons of Chapter 2, and use what you already know about functions to make a prediction of the purpose of this chapter. See students’ work.

Power, Polynomial,Power, Polynomial,and Rational Functionsand Rational Functions

Mas

sim

o Pi

zzot

ti/Ph

otol

ibra

ry

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Chapter 2

85connectED.mcgraw-hill.com

Get Ready for Chapter 2

Response to Intervention (RtI)Use the Quick Check results and the Intervention Planner to help you determine your Response to Intervention. The If-Then statements in the chart below help you decide the appropriate tier of RtI and suggest intervention resources for each tier.

Essential Questions■ Why is mathematics used to model real-world situations? Sample answer: in order to study trends,

make predictions, understand phenomena in nature

■ When would a nonlinear function be used to model a real-world situation? Sample answer: When the relationship that is modeled has a rate of change that is not constant, and thus, is nonlinear.

Intervention Planner

On Level OL1If

students miss about 25% of the exercises or less,

Then choose a resource:

SE Lesson 1-1

TE Chapter Project

Practice

connectED.mcgraw-hill.comcococco Self-Check Quiz

Strategic Intervention ALapproaching grade level2

If students miss about 50% of the exercises,

Then choose a resource:

Study Guide and Intervention

connectED.mcgraw-hill.comcococco Extra Examples, Homework Help

Additional Answers

1. (x - 4)(x + 5)

2. (x - 3)(x + 8)

3. (2x - 3)(x - 7)

4. (x - 3)(3x + 4)

5. (4x - 5)(3x + 7)

6. (2x - 9)(4x - 3)

Get Ready for the Chapter

New Vocabulary

English Español

power function p. 86 función potencia

monomial function p. 86 función monomio

radical function p. 89 función radical

extraneous solutions p. 91 solución extrana

polynomial function p. 97 función polinomial

leading coefficient p. 97 coeficiente lı der

leading-term test p. 98 conducción de prueba de término

quartic function p. 99 función quartic

quadratic form p. 100 forma de ecuación cuadra tica

repeated zero p. 101 cero repetido

lower bound p. 121 más abajo ligado

upper bound p. 121 superior ligado

rational function p. 130 función racional

asymptotes p. 130 ası ntota

vertical asymptote p. 131 ası ntota vertical

horizontal asymptote p. 131 ası ntota horizontal

polynomial inequality p. 141 desigualdad de polinomio

sign chart p. 141 carta de signo

rational inequality p. 143 desigualdad racional

Review Vocabulary

complex conjugates p. X conjugados complejos a pair of complex numbers in the form a + bi and a - bi

reciprocal functions p. XX funciones recíprocas functions of the form f (x ) = a _ x

y

x

−4

−8

−4−8

8

4

4 8

f (x) = 1_x

Diagnose Readiness You have two options for checking Prerequisite Skills.

Factor each polynomial. Prerequisite Skill

1. x 2 + x - 20 2. x 2 + 5x - 24

3. 2 x 2 - 17x + 21 4. 3 x 2 - 5x - 12

5. 12 x 2 + 13x - 35 6. 8 x 2 - 42x + 27

7. GEOMETRY The area of a square can be represented by 16 x 2 + 56x + 49. Determine the expression that represents the width of the square.

Use a table to graph each function. Prerequisite Skill

8. f (x ) = 1 _ 2 x 9. f (x ) = -2

10. f (x ) = x 2 + 3 11. f (x ) = - x 2 + x - 6

12. f (x ) = 2 x 2 - 5x - 3 13. f (x ) = 3 x 2 - x - 2

14. TELEVISIONS An electronics magazine estimates that the total number of plasma televisions sold worldwide can be represented by f (x ) = 2t + 0.5 t 2 , where t is the number of days after their release date. Graph this function for 0 ≤ t ≤ 40.

Write each set of numbers in set-builder and interval notation, if

possible. (Lesson 1-1)

15. x ≤ 6 16. {-2, -1, 0, …}

17. -2 < x < 9 18. 1 < x ≤ 4

19. x < -4 or x > 5 20. x < -1 or x ≥ 7

21. MUSIC At a music store, all of the compact discs are between $9.99 and $19.99. Describe the prices in set-builder and interval notation.

1–6. See margin.

⎪4x + 7⎥

8–13. See Chapter 2 Answer Appendix.

See Chapter 2 Answer Appendix.


(9.99, 19.99); { x | 9.99 < x < 19.99, x ∈ �}

1 Textbook Option Take the Quick Check below.

Online Option Take an online self-check Chapter Readiness Quiz at connectED.mcgraw-hill.com.2

Quick Check

85

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Lesson 2-1

86 | Lesson 2-1 | Power and Radical Functions

Lesson 2-1 Resources

Resource Approaching Level AL On Level OL Beyond Level BL English Learners ELL

Teacher Edition ■ Differentiated Instruction ■ Differentiated Instruction ■ Differentiated Instruction ■ Differentiated Instruction

Chapter

Resource

Masters

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ TI-Nspire Activity

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Enrichment■ TI-Nspire Activity

■ Practice■ Word Problem Practice■ Enrichment■ TI-Nspire Activity

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ TI-Nspire Activity

Other■ Study Notebook■ 5-Minute Check

■ Study Notebook■ 5-Minute Check



1 FocusVertical Alignment

Before Lesson 2-1 Analyze parent functions and their families of graphs.

Lesson 2-1 Graph and analyze power functions. Graph and analyze radical functions, and solve radical equations.

After Lesson 2-1 Graph and analyze polynomial functions.

2 TeachPreread/PrewriteStudy Notebook

Have students complete the What You’ll Learn section in the Study Notebook.

Scaffolding QuestionsHave students read the Why? section of the lesson.

Ask:■ Can a steel cable have a diameter

less than or equal to 0 inches? Explain. No, length is always positive.

■ What will a graph of data comparing the cable diameter and the amount of weight it can support look like? Sample answer: a curve with no intercepts, increasing from left to right

86

New Vocabularypower function

monomial function

radical function

extraneous solution

ThenYou analyzed parent functions and their families of graphs. (Lesson 1-5)

Now

1Graph and analyze power functions.

2 Graph and analyze radical functions, and solve radical equations.

Why?Suspension bridges are used to span long distances by hanging, or suspending, the main deck using steel cables. The amount of weight that a steel cable can support is a function of the cable’s diameter and can be modeled by a power function.

1 Power Functions In Lesson 1-5, you studied several parent functions that can be classified as power functions. A power function is any function of the form f (x) = a x n , where a and n are

nonzero constant real numbers.

A power function is also a type of monomial function. A monomial function is any function that can be written as f (x) = a or f (x) = a x n , where a and n are nonzero constant real numbers.

Let f be the power function f (x ) = ax n , where n is a positive integer.

n Even, a Positive

y

x

f (x) = axn

Domain: (-∞, ∞) Range: [0, ∞)

x- and y-Intercept: 0

Continuity: continuous for x ∈ �

Symmetry: y-axis Minimum: (0, 0)

Decreasing: (-∞, 0) Increasing: (0, ∞)

End behavior: lim x→-∞f (x) = ∞ and

lim x→∞ f (x) = ∞

n Even, a Negative

y

x

f (x) = ax n

Domain: (-∞, ∞) Range: (-∞, 0]



Symmetry: y-axis Maximum: (0, 0)

Decreasing: (0, ∞) Increasing: (-∞, 0)

End behavior: lim x→-∞f (x) = -∞ and

lim x→∞ f (x) = -∞

n Odd, a Positive

y

x

f (x) = axn

Domain and Range: (-∞, ∞)


Continuity: continuous on (-∞, ∞)

Symmetry: origin

Extrema: none Increasing: (-∞, ∞)

End Behavior: lim x→-∞ f (x) = -∞ and

lim x→∞ f (x) = ∞

n Odd, a Negative

y

x

f (x) = ax nf (x) = ax n




Symmetry: origin

Extrema: none Decreasing: (-∞, ∞)

End Behavior: lim x→-∞ f (x) = ∞ and

lim x→∞ f (x) = -∞

Key Concept Monomial Functions

Power and Radical FunctionsPower and Radical Functions

86 | Lesson 2-1

John

Col

etti/

The

Imag

e Ba

nk/G

etty

Imag

es

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Additional Answers (Additional Example)

1a. y

8

16

24

32

x−2−4 2 4

1b. y

x−2

−20

−40

−4

20

40

2 4

■ How could you use the function that models the data to predict the weight that a cable with a diameter of 3 feet could support? Find the value of the function for an input value of 3 feet.

1 Power FunctionsExamples 1–3 show how to graph and analyze power functions, describing the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. Example 4 shows how to use a graphing calculator to find a power function to model a set of data.

Formative AssessmentUse the Guided Practice exercises after each example to determine students’ understanding of concepts.

Additional Example

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing.

a. f (x ) = 1 _ 2 x 6 D = (-∞, ∞);

R = [ 0, ∞); intercept: 0; lim x→-∞

f (x ) = ∞ and

lim x→∞

f (x ) = ∞; continuous

for all real numbers; decreasing: (-∞, 0); increasing: (0, ∞)

b. f (x ) = - x 5 D = (-∞, ∞); R = (-∞, ∞); intercept: 0; lim x→-∞

f (x ) = ∞ and

lim x→∞

f (x ) = -∞; continuous

for all real numbers; decreasing: (-∞, ∞)

Additional Examples also in Interactive Classroom PowerPoint® Presentations

1

Review VocabularyDegree of a Monomial The sum of the exponents of the variables of a monomial.

Review VocabularyReciprocal Functions

Reciprocal functions have the form f (x ) = a _ x . (Lesson 1-5)

Monomial functions with an even degree are also even in the sense that f (-x) = f (x). Likewise, monomial functions with an odd degree are also odd, or f (-x) = -f (x).

Example 1 Analyze Monomial Functions


a. f (x) = 1 _ 2 x 4

Evaluate the function for several x-values in its domain. Then use a smooth curve to connect each of these points to complete the graph.

x -3 -2 -1 0 1 2 3

f (x ) 40.5 8 0.5 0 0.5 8 40.5

Domain: (-∞, ∞) Range: [0, ∞)

y

x−4−8

8

12

4

4 8

Intercept: 0

End behavior: lim x→-∞ f (x) = ∞ and lim x→∞

f (x) = ∞


Decreasing: (-∞, 0) Increasing: (0, ∞)

b. f (x) = - x 7

x -3 -2 -1 0 1 2 3

f (x ) 2187 128 1 0 -1 -128 -2187

Domain: (-∞, ∞) Range: (-∞, ∞)

y

x

Intercept: 0

End behavior: lim x→-∞ f (x) = ∞ and lim x→∞

f (x) = -∞


Decreasing: (-∞, ∞)

GuidedGuidedPracticePractice

1A. f (x) = 3 x 6 1B. f (x) = - 2 _ 3 x 5

1A–B. See Chapter 2 Answer Appendix.

Recall that f (x) = 1 _ x or x -1 is undefined at x = 0. Similarly, f (x) = x -2 and f (x) = x -3 are undefined at x = 0. Because power functions can be undefined when n < 0, the graphs of these functions will contain discontinuities.

Example 2 Functions with Negative Exponents


a. f (x) = 3 x -2

x -3 -2 -1 0 1 2 3

f (x ) 0. −

3 0.75 3 undefined 3 0.75 0. −

3

Domain: (-∞, 0) ∪ (0, ∞) Range: (0, ∞)

Intercepts: none

End behavior: lim x→-∞ f (x) = 0 and lim x→∞

f (x) = 0

Continuity: infinite discontinuity at x = 0

Increasing: (-∞, 0) Decreasing: (0, ∞)

y

x−4−8 4 8

4

8

12

16


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0086_0095_AM_T_C02_L01_664293.indd 870086_0095_AM_T_C02_L01_664293.indd 87 7/31/12 10:50 AM7/31/12 10:50 AM


Focus on Mathematical ContentFunctions A power function is any function of the form f (x ) = a x n , where a and n are nonzero constant real numbers. A monomial function is a power function in which n is a positive integer. The graph of an even-degree monomial function is symmetric with respect to the y-axis. The graph of an odd-degree monomial function is symmetric with respect to the origin.

Power Functions A power function of the form f (x ) = ax n , where n is a negative integer, contains a discontinuity. A power function of the

form f (x ) = ax p _ n , where n is even and

p _ n is in

simplest form, has the domain restricted to nonnegative values.

Additional Example


a. f (x ) = 2 x -4 D = (-∞, 0) � (0, ∞); R = ( 0, ∞); no intercept; lim x→-∞

f (x ) = 0

and lim x→∞

f (x ) = 0; infinite

discontinuity at x = 0; increasing: (-∞, 0); decreasing: (0, ∞)

y

x

−20

−40

−2−4

40

20

2 4

b. f (x ) = 2 x -3 D = (-∞, 0) � (0, ∞); R = (-∞, 0) �

(0, ∞); no intercept; lim x→-∞

f (x ) = 0 and lim x→∞

f (x ) = 0;

infinite discontinuity at x = 0; decreasing: (-∞, 0) and (0, ∞)

y

x

−4

−8

−4−8

8

4

4 8

2

Teach with TechInteractive Whiteboard Have students work through the examples on the whiteboard, save their work as notes, and post them to a class Web site. This may help students focus on the lesson rather than copying notes for each type of monomial function.

b. f (x) = - 3 _ 4 x -5

x -3 -2 -1 0 1 2 3

f (x ) 0.0031 0.0234 0.75 undefined -0.75 -0.0234 -0.0031

Domain: (-∞, 0) ∪ (0, ∞) Range: (-∞, 0) ∪ (0, ∞)

Intercepts: none

End behavior: lim x→-∞

f (x) = 0 and lim x→∞

f (x) = 0


Increasing: (-∞, 0) and (0, ∞)

y

x−4−8

−4

−8

4 8

4

8

Guided GuidedPracticePractice

2A. f (x) = -

1 _ 2 x -4 2B. f (x) = 4x -3


Recall that x 1 _ n indicates the nth root of x, and x

p _ n , where

p _ n is in simplest form, indicates the nth root

of x p . If n is an even integer, then the domain must be restricted to nonnegative values .

Example 3 Rational Exponents


a. f (x) = x 5 _ 2

x 0 1 2 3 4 5 6

f (x ) 0 1 5.657 15.588 32 55.902 88.182

Domain: [0, ∞) Range: [0, ∞)

x- and y-Intercepts: 0

End behavior: lim x→∞

f (x) = ∞

Continuity: continuous on [0, ∞)

Increasing: (0, ∞)

y

x

b. f (x) = 6 x - 2 _

3

x -3 -2 -1 0 1 2 3

f (x ) 2.884 3.780 6 undefined 6 3.780 2.884

Domain: (-∞, 0) ∪ (0, ∞) Range: (0, ∞)

Intercepts: none


f (x) = 0 and lim x→∞

f (x) = 0


Increasing: (-∞, 0) Decreasing: (0, ∞)

y

x


3A. f (x) = 2 x 3 _ 4 3B. f (x) = 10 x

5 _ 3


Ro

Review VocabularyRational Exponents exponents written as fractions in simplest form. (Lesson 0-4)


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Additional Examples


a. f (x ) = x 5 _ 4 D = [0, ∞);

R = [0, ∞); intercept: 0;

lim x→∞


on [0, ∞); increasing: (0, ∞)

y

x

b. f (x ) = 4 x -

2 _ 5 D = (-∞, 0) �

(0, ∞); R = (0, ∞); no intercepts; lim x→-∞

f (x ) = 0

and lim x→∞


discontinuity at x = 0; increasing: (-∞, 0) and decreasing: (0, ∞)

ANIMALS The following data represent the body length L in centimeters and the mass M in kilograms of several African Golden cats being studied by a scientist.

L 72 72 73 74 75 76 78 79

M 11 12 13 15 15 14 15 15

L 80 83 84 85 86 88 89 90

M 14 16 16 15 17 17 18 18

a. Create a scatter plot of the data.

b. Determine a power function to model the data. y = 0.02 x 1.5

c. Use the data to predict the mass of an African Golden cat with a length of 77 centimeters. about 14.1 kg

3

4

Real-World LinkA Calorie is a unit of energy equal to the amount of heat needed to raise the temperature of one kilogram of water by 1°C. One Calorie is equivalent to 4.1868 kilojoules. The average apple contains 60 Calories.

Source: Foods & Nutrition Encyclopedia

Study Tip Regression Model A polynomial function with rounded coefficients will produce estimates different from values calculated using the unrounded regression equation. From this point forward, you can assume that when asked to use a model to estimate a value, you are to use the unrounded regression equation.

Example 4 Power Regression

BIOLOGICAL SCIENCE The following data represents the resting metabolic rate R in kilocalories per day for the mass m in kilograms of several selected animals.

m 0.3 0.4 0.7 0.8 0.85 2.4 2.6 5.5 6.4 6

R 28 35 54 66 46 135 143 331 293 292

m 7 7.9 8.41 8.5 13 29.3 29.8 39.5 83.6

R 265 327 346 363 520 956 839 1036 1948

Source: American Journal of Physical Anthropology


The scatter plot appears to resemble the square root function, which is a power function. Therefore, test a power regression model.

[0, 100] scl: 10 by [0, 2000] scl: 200

b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth, and state the correlation coefficient.

Using the PwrReg tool on a graphing calculator and rounding each coefficient to the nearest thousandth yields f (x) = 69.582 x 0.759 . The correlation coefficient r for the data, 0.995, suggests that a power regression may accurately reflect the data.

We can graph the complete (unrounded) regression by sending it to the menu. In the menu, pick up this regression equation by entering VARS , Statistics, EQ. Graph this function and the scatter plot in the same viewing window. The function appears to fit the data reasonably well.

[0, 100] scl: 10 by [0, 2000] scl: 200

c. Use the equation to predict the resting metabolic rate for a 60-kilogram animal.

Use the CALC feature on the calculator to find f (60). The value of f (60) is about 1554, so the resting metabolic rate for a 60-kilogram animal is about 1554 kilocalories.


4. CARS The table shows the braking distance in feet at several speeds in miles per hour for a specific car on a dry, well-paved roadway.

Speed 10 20 30 40 50 60 70

Distance 4.2 16.7 37.6 66.9 104.5 150.5 204.9

A. Create a scatter plot of the data.

B. Determine a power function to model the data.

C. Predict the braking distance of a car going 80 miles per hour.


y = 0.042 x 2 - 0.004x + 0.043

about 267.6 ft

2 Radical Functions An expression with rational exponents can be written in radical form.

Exponential Form Radical Form

x p

_ n = n √ � x p

Power functions with rational exponents represent the most basic of radical functions. A radical function is a function that can be written as f (x) =

n √ � x p , where n and p are positive integers greater

than 1 that have no common factors. Some examples of radical functions are shown below.

f (x) = 3 √ �� 5 x 3 f (x) = -5

3 √ �� x 4 + 3 x 2 - 1 f (x) = 4

√ �� x + 12 + 1 _ 2 x - 7


Food

colle

ctio

n/G

etty

Imag

es

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Differentiated Instruction AL OL BL

Interpersonal Learners Have students work in groups to compare solving radical equations to solving quadratic equations. Have groups write or give a short presentation about the differences and similarities in solution processes.

2 Radical FunctionsExample 5 shows how to graph and analyze radical functions, describing the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing for functions in the form f (x ) =

n √ �

x p . Example 6 shows how to solve radical equations, eliminating extraneous solutions.

Additional Example


a. f (x ) = 5 √____

2x 3 D = [ 0, ∞); R = [0, ∞); intercept: 0; lim x→∞


on [0, ∞); increasing: (0, ∞)y

x

−4

−8

−4−8

8

4

4 8

b. f (x ) = 1 _ 2

5 √ �� 3x - 4

D = (-∞, ∞);R = (-∞, ∞); x-intercept: 4 _

3 y-intercept:

about -0.6598; lim x→-∞

f (x ) = -∞ and

lim x→∞


for all real numbers; increasing: (-∞, ∞)

y

x

5


DDiifffferenttiiatteddIInnssttrruccttiioonn

Watch Out! Radical Functions Remember that when n is even, the domain and range will have restrictions.

It is important to understand the characteristics of the graphs of radical functions as well.

Let f be the radical function f (x ) = n √ � x where n is a positive integer.

n Even

y

x

f (x) = n√x

Domain and Range: [0, ∞)


Continuity: continuous on [0, ∞)

Symmetry: none Increasing: (0, ∞)

Extrema: absolute minimum at (0, 0)

End Behavior: lim x→∞ f (x) = ∞

n Odd

y

x

f (x) = n√x




Symmetry: origin Increasing: (-∞, ∞)

Extrema: none

End Behavior: lim x→-∞ f (x ) = -∞ and

lim x→∞ f (x ) = ∞

Key Concept Radical Functions

Example 5 Graph Radical Functions


a. f (x) = 2 4 √ �� 5 x 3

x 0 1 2 3 4 5

f (x ) 0 2.99 5.03 6.82 8.46 10

y

x

−4

−4

8

4

4 8

Domain and Range: [0, ∞)

x- and y-Intercepts: 0

End behavior: lim x→∞

f (x) = ∞

Continuity: continuous on [0, ∞)Increasing: (0, ∞)

b. f (x) = 1 _ 4 5

√ �� 6x - 8

x -3 -2 -1 0 1 2 3

f (x ) -0.48 -0.46 -0.42 -0.38 -0.29 0.33 0.40

y

x Domain and Range: (-∞, ∞)

x-Intercept: 4 _ 3 y-Intercept: about -0.38


f (x) = -∞ and lim x→∞

f (x) = ∞

Continuity: continuous on (-∞, ∞) Increasing: (-∞, ∞)

GuidedGuidedPractice Practice

5A. f (x) = -

3 √ ��

12 x 2 - 5 5B. f (x) = 1 _ 2

4 √ ��

2 x 3 - 16

5A–B. See margin.


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Additional Example

Solve each equation.

a. 2x = √ �� 28x + 29 - 3 -1, 5

b. 12 =3√

�� (x - 2) 2 + 8 10, -6

c. √ �� x + 1 = 1 + √ �� 2x - 12 8

6

Tips for New TeachersExtraneous Solutions Remind students that there is the possibility of extraneous solutions occuring as a result of squaring. Therefore, any possible solution must be checked.

Additional Answers (Guided Practice)

5A. y

x

−4

−8

−12

−4−8

4

4 8

f (x) = -3√12x 2 - 5

D = (-∞, ∞), R = (-∞,

3 √ � 5 ]; x-intercepts:

-

√ � 15 _

6 and -

√ � 15 _

6 ,

y-intercept: 3 √ � 5 ;

lim x→-∞

f (x ) = -∞ and

lim x→∞

f (x ) = -∞; continuous for

all real numbers; increasing: (-∞, 0); decreasing: (0, ∞)

5B. y

x

2

4

6

8

4 8 12 16

f (x) = 1_2

4√2x3 - 16

D = [ 2, ∞), R = [ 0, ∞); x-intercept: 2; lim x→∞

f (x ) = ∞;

continuous on [ 2, ∞); increasing: (2, ∞)

Differentiated Instruction AL OL ELL

Kinesthetic Learners Have students use a flow chart program or an Interactive Whiteboard to create a flow chart explaining how to solve a radical equation. Remind students that their flow charts should include a loop for the steps involved in isolating and eliminating the radicals. Then have students test their flow charts using equations from the Practice exercises.


Like radical functions, a radical equation is any equation in which a variable is in the radicand. To solve a radical equation, first isolate the radical expression. Then raise each side of the equation to a power equal to the index of the radical to eliminate the radical.

Raising each side of an equation to a power sometimes produces extraneous solutions, or solutions that do not satisfy the original equation. It is important to check for extraneous solutions.

Example 6 Solve Radical Equations


a. 2x = √ �� 100 - 12x - 2

2x = √ �� 100 - 12x - 2 Original equation

2x + 2 = √ �� 100 - 12x Isolate the radical.

4 x 2 + 8x + 4 = 100 − 12x Square each side to eliminate the radical.

4 x 2 + 20x - 96 = 0 Subtract 100 - 12x from each side.

4( x 2 + 5x - 24) = 0 Factor.

4(x + 8)(x - 3) = 0 Factor.

x + 8 = 0 or x − 3 = 0 Zero Product Property

x = -8 x = 3 Solve.

CHECK x = -8 CHECK x = 3

2x = √ �� 100 - 12 x - 2 2x = √ �� 100 - 12x - 2

-16 � √ �� 100 - 12(-8) - 2 6 � √ �� 100 - 12(3) - 2

-16 � √ �� 196 - 2 6 � √ � 64 - 2

-16 ≠ 12 � 6 = 6 �

One solution checks and the other solution does not. Therefore, the solution is 3.

b. 3 √ �� (x - 5) 2 + 14 = 50

3 √ ��

(x - 5) 2 + 14 = 50 Original equation

3 √ ��

(x - 5) 2 = 36 Isolate the radical.

(x - 5) 2 = 46,656 Raise each side to the third power. (The index is 3.)

x - 5 = ±216 Take the square root of each side.

x = 221 or -211 Add 5 to each side.

A check of the solutions in the original equation confirms that the solutions are valid.

c. √ �� x - 2 = 5 - √ �� 15 - x

√ �� x - 2 = 5 - √ �� 15 - x Original equation

x - 2 = 25 - 10 √ �� 15 - x + (15 - x) Square each side.

2x - 42 = -10 √ �� 15 - x Isolate the radical.

4 x 2 - 168x + 1764 = 100(15 - x) Square each side.

4 x 2 - 168x + 1764 = 1500 - 100x Distributive Property

4 x 2 - 68x + 264 = 0 Combine like terms.

4( x 2 - 17x + 66) = 0 Factor.

4(x - 6)(x - 11) = 0 Factor.

x - 6 = 0 or x - 11 = 0 Zero Product Property

x = 6 x = 11 Solve.

A check of the solutions in the original equation confirms that both solutions are valid.


6A. 3x = 3 + √ �� 18x - 18 6B. 3 √ �� 4x + 8 + 3 = 7 6C. √ �� x + 7 = 3 + √ �� 2 - x 1, 3 14 2

Study Tip Common Factors Remember that you can sometimes factor out a common multiple before using any other factoring methods.

Watch Out! Squaring Radical Expressions

Take extra care as you square 5 - √ �� 15 - x . While similar to using the FOIL method with binomial expressions, there are some differences. Be sure to account for every term.


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Differentiated Homework Options OL BL ELL AL

Level Assignment Two-Day Option

AL Approaching Level 1–55, 82–100 1–55 odd, 100–100 2–54 even, 82–99

OL On Level 1–61 odd, 62, 63–73 odd, 74, 75, 77, 79, 80, 82–100

1–55, 100–100 56–80, 82–99

BL Beyond Level 56–100


3 PracticeFormative AssessmentUse Exercises 1–55 to check for understanding

Then use the table below to customize your assignments for students.

Watch Out!

Common Error In Exercises 30–33, Students may forget to clear the elements stored in L1 and L2 before entering the data given for each exercise. Remind students that to clear the lists, they should press to move the cursor onto L1, and then press

ENTER ENTER to clear both

lists.

Additional Answers

17b. V(r)

r2 4 6 8

20

40

60

80

V (r) = 4_3r 3π

30a.

[0, 10] scl: 1 by [0, 1500] scl: 150

30b. y = 3.54x 2.89

30c. about 66,098.82

31a.

[0, 10] scl: 1 by [0, 150000] scl: 10000

31b. y = 0.77 x 5.75

31c. about 235,906,039

Difffferentiated Home ork OOpttiions

Exercises

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. (Examples 1 and 2)

1. f (x) = 5 x 2 2. g(x) = 8 x 5

3. h(x) = - x 3 4. f (x) = -4 x 4

5. g(x) = 1 _ 3 x 9 6. f (x) = 5 _

8 x 8

7. f (x) = -

1 _ 2 x 7 8. g(x) = -

1 _ 4 x 6

9. f (x) = 2 x -4 10. h(x) = -3 x -7

11. f (x) = -8 x -5 12. g(x) = 7 x -2

13. f (x) = -

2 _ 5 x -9 14. h(x) = 1 _

6 x -6

15. h(x) = 3 _ 4 x -3 16. f (x) = -

7 _ 10

x -8

17. GEOMETRY The volume of a sphere is given by V(r) = 4 _

3 π r 3 , where r is the radius. (Example 1)

a. State the domain and range of the function.

b. Graph the function.

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. (Example 3)

18. f (x) = 8 x 1 _ 4 19. f (x) = -6 x

1 _ 5

20. g(x) = -

1 _ 5 x

-

1 _ 3 21. f (x) = 10 x

-

1 _ 6

22. g(x) = -3 x 5 _ 8 23. h(x) = 3 _

4 x

3 _ 5

24. f (x) = -

1 _ 2 x

-

3 _ 4 25. f (x) = x

-

2 _ 3

26. h(x) = 7 x 5 _ 3 27. h(x) = -4 x

7 _ 4

28. h(x) = -5 x -

3 _ 2 29. h(x) = 2 _

3 x

-

8 _ 5

Complete each step.


b. Determine a power function to model the data.

c. Calculate the value of each model at x = 30. (Example 4)

30. 31.

x y

1 4

2 22

3 85

4 190

5 370

6 650

7 1000

8 1500

x y

1 1

2 32

3 360

4 2000

5 7800

6 25,000

7 60,000

8 130,000

32. CLIFF DIVING In the sport of cliff diving, competitors perform three dives from a height of 28 meters. Judges award divers a score from 0 to 10 points based on degree of difficulty, take-off, positions, and water entrance. The table shows the speed of a diver at various distances in the dive. (Example 4)



c. Use the function to predict the speed at which a diver would enter the water from a cliff dive of 30 meters.

33. WEATHER The wind chill temperature is the apparent temperature felt on exposed skin, taking into account the effect of the wind. The table shows the wind chill temperature produced at winds of various speeds when the actual temperature is 50°F. (Example 4)



c. Use the function to predict the wind chill temperature when the wind speed is 65 miles per hour.

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. (Example 5)

34. f (x) = 3 √ �� 6 + 3x 35. g(x) = -2 5 √ �� 1024 + 8x

36. f (x) = -

3 _ 8 6

√ �� 16x + 48 - 3 37 h(x) = 4 + √ �� 7x - 12

38. g(x) = √ ��

(1 - 4x) 3 - 16 39. f (x) = -

3 √ ��

(25x - 7) 2 - 49

40. h(x) = 1 _ 2 3

√ �� 27 - 2x - 8 41. g(x) = √ �� 22 - x - √ �� 3x - 3

42. FLUID MECHANICS The velocity of the water flowing through a hose with a nozzle can be modeled using V(P) = 12.1 √ � P , where V is the velocity in feet per second and P is the pressure in pounds per square inch. (Example 5)

a. Graph the velocity through a nozzle as a function of pressure.

b. Describe the domain, range, end behavior, and continuity of the function and determine where it is increasing or decreasing.

1–16. See

Chapter 2

Answer

Appendix.

See margin.

18–29. See

Chapter 2

Answer

Appendix.

30–31. See margin.

Distance

(m)

Speed

(m/s)

4 8.85

8 12.52

12 15.34

16 17.71

20 19.80

24 21.69

28 23.43See margin.

f (x ) = 4.42 x 0.5

about 24.25 m/s

Wind Speed

(mph)

Wind Chill

(°F)

5 48.22

10 46.04

15 44.64

20 43.60

25 42.76

30 42.04

35 41.43

40 40.88

See margin. f (x ) = 55.14 x - 0.0797

about 39.54°F34–41. See Chapter 2 Answer Appendix.

a–b. See Chapter 2 Answer Appendix.

17a. D = (0, ∞), R = (0, ∞)


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69c. Sample answer: Yes; the problem states that the volume and pressure are inversely proportional, and in the power function, the exponent of the volume variable is -1.

Additional Answers

32a.

[0, 30] scl: 2 by [0, 30] scl: 2

33a.

[0, 45] scl: 5 by [40, 50] scl: 1

56. Yes; sample answer: The function follows the form f (x ) = a x n , where n is a positive integer. In this case,

a = 5 _ b and n = 4a.

57. Yes; sample answer: The function follows the form f (x ) = a x n , where n is a positive integer. In this case, a = -2a and n = 4.

58. No; sample answer: The function is not a power function because the variable is in the exponent.


a = 7 _ 3 and n = ab.


a = 1 _ ab

and n = 2b.

61. No; sample answer: The function is not a monomial function because the exponent for x is negative.

69a. Volume vs. Pressure

Pres

sure

(atm

osph

eres

)

0

1

2

3

4

Volume (liters)1 2 3 4

P(v)

v

43. AGRICULTURAL SCIENCE The net energy NE m required to maintain the body weight of beef cattle, in megacalories (Mcal) per day, is estimated by the formula

NE m = 0.077 4 √

�� m 3 , where m is the animal’s mass in

kilograms. One megacalorie is equal to one million calories. (Example 6)

a. Find the net energy per day required to maintain a 400-kilogram steer.

b. If 0.96 megacalorie of energy is provided per pound of whole grain corn, how much corn does a 400-kilogram steer need to consume daily to maintain its body weight?

Solve each equation. (Example 6)

44. 4 = √ �� -6 - 2x + √ �� 31 - 3x 45. 0.5x = √ �� 4 - 3x + 2

46. -3 = √ �� 22 - x - √ �� 3x - 3 47. √ ��

(2x - 5) 3 - 10 = 17

48. 4 √ ��

(4x + 164) 3 + 36 = 100 49. x = √ �� 2x - 4 + 2

50. 7 + √ ��

(-36 - 5x) 5 = 250 51. x = 5 + √ �� x + 1

52. √ �� 6x - 11 + 4 = √ �� 12x + 1 53. √ �� 4x - 40 = -20

54. √ �� x + 2 - 1 = √ �� -2 - 2x 55. 7 + 5 √ �� 1054 - 3x = 11

Determine whether each function is a monomial function given that a and b are positive integers. Explain your reasoning.

56. y = 5 _ b x 4 a 57. G(x) = -2a x 4

58. F(b) = 3a b 5x 59. y = 7 _ 3 t ab

60. H(t) = 1 _ ab

t 4b _ 2 61. y = 4 ab x -2

62. CHEMISTRY The function r = R 0 (A) 1 _ 3 can be used to

approximate the nuclear radius of an element based on its molecular mass, where r is length of the radius in meters, R 0 is a constant (about 1.2 × 10 -15 meter), and A is the molecular mass.

ElementMolecular

Mass

Carbon (C) 12.0

Helium (H) 4.0

Iodine (I) 126.9

Lead (Pb) 207.2

Sodium (Na) ?

Sulfur (S) 32.1

a. If the nuclear radius of sodium is about 3.412 × 1 0 -15 meter, what is its molecular mass?

b. The approximate nuclear radius of an element is 6.030 × 1 0 -15 meter. Identify the element.

c. The ratio of the molecular masses of two elements is 27:8. What is the ratio of their nuclear radii?

Solve each inequality.

63. 5 √ �� 1040 + 8x ≥ 4 64. 13

√ �� 41 - 7x ≥ -1

65. (1 - 4x) 3 _ 2 ≥ 125 66. √ �� 6 + 3x ≤ 9

67. (19 - 4x) 5 _ 3 - 12 ≤ -13 68. (2x - 68 )

2 _ 3 ≥ 64

69 CHEMISTRY Boyle’s Law states that, at constant temperature, the pressure of a gas is inversely proportional to its volume. The results of an experiment to explore Boyle’s Law are shown.

Volume

(liters)

Pressure

(atmospheres)

1.0 3.65

1.5 2.41

2.0 1.79

2.5 1.46

3.0 1.21

3.5 1.02

4.0 0.92


b. Determine a power function to model the pressure P as a function of volume v.

c. Based on the information provided in the problem statement, does the function you determined in part b make sense? Explain.

d. Use the model to predict the pressure of the gas if the volume is 3.25 liters.

e. Use the model to predict the pressure of the gas if the volume is 6 liters.

Without using a calculator, match each graph with the appropriate function.

70. 71.

y

x

y

x

72. 73.

y

x

y

x−4−8

4

8

−8

−4

4 8

a. f (x) = 1 _ 2

4 √

�� 3 x 5 b. g(x) = 2 _

3 x 6

c. h(x) = 4 x -3 d. p(x) = 5 3 √ �� 2x + 1

about 6.89 Mcal

about 7.18 lb

44. no solutionno solution

7

2, 4

8

no solution

46. 13 48. 23 50. -9 52. 2, 10-1

10

B


23.0

iodine

3:2

x ≥ -2 x ≤ 6

x ≤ -6 -2 ≤ x ≤ 25

x ≥ 5x ≥ 290

See margin.

P (v ) = 3.62 v -1

See margin.

about 1.12 atmospheres

about 0.60 atmosphere

b c

a d


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Additional Answers

84a. True; sample answer: f (x ) can be written

as f (x ) = b √

� x a . If b is even and a is odd,

then x ≥ 0. An even root of a negative number is undefined.

84b. False; sample answer: f (x ) can be written

as f (x ) = b √

� x a . If b is odd and a is even,

then f (x ) is defined for all x.


pp. 5–6 AL OL ELL

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Chapter 2 5 Glencoe Precalculus

2-1

Power Functions A power function is any function of the form f(x) = axn where a and n are nonzero constant real numbers. For example, f(x) = 2x2, f(x) = x

1 −

2 , or f(x) = √

� x are power functions.

Graph and analyze f(x) = 1 − 3 x3. Describe the domain,

range, intercepts, end behavior, continuity, and where the function is increasing or decreasing.Evaluate the function for several x-values in its domain. Then use a smooth curve to connect each of these points to complete the graph.

Domain: (-∞, ∞) Range: (-∞, ∞)

Intercept: 0End behavior: lim x → -∞

f(x) = -∞ and lim x → ∞

f(x) = ∞

Continuity: continuous for all real numbersIncreasing: (-∞, ∞)

Exercises


1. f(x) = 3x4 2. f(x) =

1−

5x-3

Example

Study Guide and InterventionPower and Radical Functions

x -3 -2 -1 0 1 2 3

f(x) -9 - 8 −

3 - 1 −

3 0 1 −

3 8 −

3 9

y

x4 8−4−8

4

8

−8

−4

y

x

y

x4 8−4−8

4

8

−8

−4

D = (-∞, ∞), R = [0, ∞); intercept: (0, 0); lim x → -∞

f(x) = ∞ and

lim x → ∞ f(x) = ∞; continuous for all

real numbers; decreasing: (-∞, 0), increasing: (0, ∞)

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); intercept: none;

lim x → -∞ f(x) = 0 and

lim x → ∞ f(x) = 0; infinite discontinuity

at x = 0; decreasing: (-∞, 0), decreasing: (0, ∞)

005_038_PCCRMC02_893803.indd 5 10/1/09 5:45:05 PM

Practice

p. 7 AL OL BL ELL

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NAME DATE PERIOD


2-1


1. f(x) = 2x6 2. f(x) = -

3 √ �� (2x + 5)2

3. f(x) = -

1 −

2 5 √ �� 6x - 12


4. √ �� x - 2 = 6 5. 3 √ �� 7r + 5 = -3

6. √ �� 6x + 12 - √ �� 4x + 9 = 1 7. 5 + 2x = √ �� x2 - 2x + 1

8. ENGINEERING A team of engineers must design a fuel tank in the shape of a cone. The lateral area of a cone is given by the formula L = πr √ �� r2 + h2 . Find the height of a cone with a radius of 7 inches and a lateral area of 550 square inches.

PracticePower and Radical Functions

y

x

y

x4 8−4−8

4

8

−8

−4

y

x

38

D = (-∞, ∞); R = [0, ∞);

intercept: (0, 0); lim x → -∞ f(x) = ∞ and

lim x → ∞ f(x) = ∞; continuous for all

real numbers; decreasing: (-∞, 0),

increasing: (0, ∞)

D = (-∞, ∞); R = (-∞, 0]; x-intercept: -2.5; y-intercept:

- 3 √ �� 25 ; lim x → -∞

f(x) = -∞ and

lim x → ∞ f(x) = -∞; continuous for all

real numbers; increasing: (-∞, -2.5), decreasing: (-2.5, ∞)

D = (-∞, ∞); R = (-∞, ∞);

x-intercept: 2; y-intercept: -

5 √ �� -12 −

2 ;

lim x → -∞ f(x) = ∞ and lim x → ∞

f(x) = -∞;

continuous for all real numbers; decreasing: (-∞, ∞)

- 32 −

7

- 4 − 3 4

about 24 in.

005_038_PCCRMC02_893803.indd 7 3/23/09 10:04:02 AM

Word Problem Practice

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NAME DATE PERIOD


2-1 Word Problem PracticePower and Radical Functions

1. AREA The lateral area L of a right circular cone is given by the formula L = πr √��r2

+ h2 , where r is the radius and h is the height.

a. If the height is 6 inches, use a graphing calculator to graph the lateral area as a function of the radius. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing.

D = (-∞, ∞); R = (-∞, ∞); intercept: (0, 0), lim

x → -∞f(x) =-∞,

lim x → ∞

f(x) = ∞; continuous;

increasing: (-∞, ∞)

b. How do your answers for part achange if you only graph the relevant domain?

The domain and range are each restricted to (0, ∞); the end behavior is lim

x → ∞f(x) = ∞;

it is continuous and increasing for (0, ∞).

2. WALKING SPEED A person’s walking speed can be approximated by the

function S =

√ �� 386� −

12 , where S is the

speed in feet per second and � is the walker’s leg length in inches. What is a man’s leg length if he walks at a speed of 10 feet per second? 37.3 in.

3. CHEMISTRY At a constant temperature, the volume V of a gas is inversely proportional to its pressure, V = kp−1, where k is a constant and p is the pressure. Describe the relevant domain, range, intercepts, end behavior, continuity, and where this function is increasing or decreasing for a constant of 2.86.

4. POPULATION Zipf ’s Law states that the rank of the population of cities can be expressed as a power function. The table shows the estimated population of the top ten U. S. cities.



y = 7574.504x -0.892

c. Predict the population of the city that is ranked number 11.

892,000

d. Predict the ranking of a city with a population of about 650,000.

15th or 16th

RankPopulation

(thousands)

1 8275

2 3834

3 2837

4 2208

5 1552

6 1450

7 1329

8 1267

9 1240

10 940

Source: U.S. Census Bureau

D = (0, ∞); R = (0, ∞); intercept:

none, lim x → ∞ f(x) = 0; continuous on

(0, ∞); decreasing: (0, ∞)

2000

3000

1000

0

4000

5000

6000

70008000

321 5 74 6 8 9 10

Popu

latio

n

Rank

005_038_PCCRMC02_893803.indd 8 10/1/09 5:45:19 PM

Enrichment

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NAME DATE PERIOD

2-1 EnrichmentE i h t

Nested Radical SignsIt is possible for an expression to be under more than one radical sign. Tosolve an equation with nested radical signs, you need to repeatedly raiseeach side of the equation to the appropriate power until all of the radicalsigns are removed. Study the example below.

Solve √��√√√��√√3x33 - 8 + 2(x(( + 1) = 3.

√��

√√√��√√3x - 8 + 2(x + 1) = 3 Original equation

√��√√3x - 8 + 2(x + 1) = 9 Square each side.

√��√√3x - 8 = 7 - 2x Isolate the radical.

3x - 8 = 49 - 28x + 4x44 2 Square each side.

0 = 57 - 31x + 4x44 2 Subtract 3x - 8 from each side.

0 = (-3 + x)(-19 + 4x44 ) Factor.

00 0 == -333 +++ xx 0or 0 or 0 == -191919 +++ 44 4xx444 Zero ProducZero Product Propertyt Propertyp y

33 3 == xx or 4 75or 4 75 or 4.75 == xx S lSolveSolve.

SubstituSubstituSubstitution inttion inttion into the oro the oro the original eiginal eiginal equationquationquation shows thshows thshows that 4 75at 4 75at 4.75 is an exis an exis an extraneoustraneoustraneous l tisolutionsolution ThThe so. The sol ti ilution ilution i 3s 3s 3.

i

Example

74. ELECTRICITY The voltage used by an electrical device such as a DVD player can be calculated using V = √ �� PR , where V is the voltage in volts, P is the power in watts, and R is

the resistance in ohms. The function I = √

�

P _ R

can be used to calculate the current, where I is the current in amps.

a. If a lamp uses 120 volts and has a resistance of 11 ohms, what is the power consumption of the lamp?

b. If a DVD player has a current of 10 amps and consumes 1200 watts of power, what is the resistance of the DVD player?

c. Ohm’s Law expresses voltage in terms of current and resistance. Use the equations given above to write Ohm’s Law using voltage, resistance, and amperage.

Use the points provided to determine the power function represented by the graph.

75 76.

y

x

(2, 2)

(-1, -4)

y

x

(- 1_2

, 1_24 ) (1, 2_

3 )

77. 78.

y

x

(4, 6)

(1, 3)

y

x(-3, - 1_

27 )(2, 1_

8 )

79. OPTICS A contact lens with the appropriate depth ensures proper fit and oxygen permeation. The depth of a lens can

be calculated using the formula S = r - √

��

r 2 - ( d _ 2 )

2 ,

where S is the depth, r is the radius of curvature, and d is the diameter, with all units in millimeters.

d

rs

lenseye

d

S

a. If the depth of the contact lens is 1.15 millimeters and the radius of curvature is 7.50 millimeters, what is the diameter of the contact lens?

b. If the depth of the contact lens is increased by 0.1 millimeter and the diameter of the lens is 8.2 millimeters, what radius of curvature would be required?

c. If the radius of curvature remains constant, does the depth of the contact lens increase or decrease as the diameter increases?

80. MULTIPLE REPRESENTATIONS In this problem, you will investigate the average rates of change of power functions.

a. GRAPHICAL For power functions of the form f (x) = x n , graph a function for two values of n such that 0 < n < 1, n = 1, and two values of n such that n > 1.

b. TABULAR Copy and complete the table, using your graphs from part a to analyze the average rates of change of the functions as x approaches infinity. Describe this rate as increasing, constant, or decreasing.

n f (x )Average Rate of

Change as x→∞

0 < n < 1

n = 1

n > 1

c. VERBAL Make a conjecture about the average rate of change of a power function as x approaches infinity for the intervals 0 < n < 1, n = 1, and n > 1.

H.O.T. Problems Use Higher-Order Thinking Skills

81. CHALLENGE Show that √

��

8 n · 2 7 _ 4 -n

= 2 2n + 3 √ ��

2 n + 1 .

82. REASONING Consider y = 2 x .

a. Describe the value of y if x < 0.

b. Describe the value of y if 0 < x < 1.

c. Describe the value of y if x > 1.

d. Write a conjecture about the relationship between the value of the base and the value of the power if the exponent is greater than or less than 1. Justify your answer.

83. PREWRITE Your senior project is to tutor an underclassman for four sessions on power and radical functions. Make a plan for writing that addresses purpose and audience, and has a controlling idea, logical sequence, and time frame for completion.

84. REASONING Given f (x) = x a _ b , where a and b are integers

with no common factors, determine whether each statement is true or false. Explain.

a. If the value of b is even and the value of a is odd, then the function is undefined for x < 0.

b. If the value of a is even and the value of b is odd, then the function is undefined for x < 0.

c. If the value of a is 1, then the function is defined for all x.

85. REASONING Consider f (x) = x 1 _ n

+ 5. How would you expect the graph of the function to change as n increases if n is odd and greater than or equal to 3?

86. WRITING IN MATH Use words, graphs, tables, and equations to show the relationship between functions in exponential form and in radical form.

1309 watts

12 ohms

V = IR

C

f (x ) = 4 x -1 f (x ) = 2

_ 3 x 4

f (x ) = 3 x 1

_ 2 f (x ) = x -3

7.98 mm

7.35 mm

increase

f (x ) = x 1

_ 3 decreases as x→∞

f (x ) = x 1

_ 2 decreases as x→∞

f (x ) = x constant as x→∞

f (x ) = x 2 increases as x→∞

f (x ) = x 3 increases as x→∞

80a, c. See Chapter 2 Answer Appendix.

See Chapter 2

Answer Appendix.0 < y < 1

1 < y < 2

y > 2


See students’ work.

a–c. See margin.




0086_0095_AM_S_C02_L01_664291.indd 94 4/30/12 6:05 PM



4 AssessCrystal Ball Ask students to write how they think today’s lesson on monomial functions will help them with tomorrow’s lesson on polynomial functions.

Additional Answers

84c. False; sample answer: f (x ) can

be written as f (x ) = b √

� x 1 . If b is

odd, then f (x ) is defined for all x. If b is even, then x ≥ 0 because an even root of a negative number is undefined.

88. ( f + g )(x ) = x 2 - x + 9,

D = (-∞, ∞); (f - g)(x ) =

x 2 - 3x - 9, D = (-∞, ∞); (f · g )(x ) = x 3 + 7 x 2 - 18x,

D = (-∞, ∞); ( f _ g ) (x ) = x 2 - 2x _ x + 9

,

D = (-∞, -9) � (-9, ∞)

89. ( f + g )(x ) = x 3 + x 2 - 1 _

x + 1 ,

D = (-∞, -1) � (-1, ∞);

(f - g )(x ) = - x 3 - x 2 + 2x + 1 __

x + 1 ,

D = (-∞, -1) � (-1, ∞);

(f · g )(x ) = x 2 - x ;

D = (-∞, -1) � (-1, ∞); ( f _ g )

(x ) = x __ x 3 + x 2 - x - 1

, D =

(-∞, -1) � (-1, 1) � (1, ∞)

90. ( f + g )(x ) = x 3 - 2 x 2 - 35x + 3 __

x - 7 ,

D = (-∞, 7) � (7, ∞);

( f - g )(x ) = - x 3 - 2 x 2 - 35x - 3 __ x - 7

,

D = (-∞, 7) � (7, ∞);

(f · g )(x ) = 3 x 2 + 15x _ x - 7

,

D = (-∞, 7) � (7, ∞);

( f _ g ) (x ) = 3 __ x 3 - 2 x 2 - 35x

,

D = (-∞, -5) � (-5, 0) � (0, 7) � (7, ∞)

Differentiated Instruction BL

Extension Ask students to indicate a set of values for a and b in a = √ �� x + b for which the result is always at least one real solution. When both a and b are positive integers or when a is a positive integer and b is a negative integer, the result is always at least one real solution.


Spiral Review

87. FINANCE If you deposit $1000 in a savings account with an interest rate of r compounded annually, then the balance in the account after 3 years is given by B(r) = 1000 (1 + r) 3 , where r is written as a decimal. (Lesson 1-7)

a. Find a formula for the interest rate r required to achieve a balance of B in the account after 3 years.

b. What interest rate will yield a balance of $1100 after 3 years?

Find ( f + g)(x), ( f - g)(x), ( f · g)(x), and ( f _ g ) (x) for each f (x) and g(x). State the domain of each

new function. (Lesson 1-6)

88. f (x) = x 2 - 2x 89. f (x) = x _ x + 1

90. f (x) = 3 _ x - 7

g(x) = x + 9 g(x) = x 2 - 1 g(x) = x 2 + 5x

Use the graph of f (x) to graph g(x) = | f (x)| and h(x) = f (|x|) . (Lesson 1-5)

91. f (x) = -4x + 2 92. f (x) = √ �� x + 3 - 6 93. f (x) = x 2 - 3x - 10

Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. (Lesson 1-4)

94. 95. 96. y

x−8 −4

−4

4

8

−8

4 8

f (x) = 0.5(x + 4)(x + 1)(x - 2)

y

x−8 −4

−4

4

8

−8

4 8

f (x) =

x - 3_x + 4

y

x−8 −4

−4

4

−8

4 8

_f (x) =

x 2- 1

x + 2

r = -1 + 3 √ � B _

10

3.23%



94. f is increasing on (-∞, -3), decreasing on (-3, 1), and increasing on (1, ∞).

95. f is increasing on (-∞, -4) and increasing on (-4, ∞).

96. f is increasing on (-∞, -3.5),

decreasing on (-3.5, -2), decreasing on

(-2, 0), and increasing on (0, ∞).


97. SAT/ACT If m and n are both positive, then which of

the following is equivalent to 2m √ �� 18n _

m √ � 2 ?

A 3m √ �

n D 6 √ �

n

B 6m √ �

n E 8 √ �

n

C 4 √ �

n

98. REVIEW If f (x, y) = x 2 y 3 and f (a, b) = 10, what is the value of f (2a, 2b)?

F 50 J 320

G 100 K 640

H 160

99. REVIEW The number of minutes m it takes c children to eat p pieces of pizza varies directly as the number of pieces of pizza and inversely as the number of children. If it takes 5 children 30 minutes to eat 10 pieces of pizza, how many minutes should it take 15 children to eat 50 pieces of pizza?

A 30 C 50

B 40 D 60

100. If 3 √ �� 5m + 2 = 3, then m = ?

F 3 H 5

G 4 J 6

D

J

C

H

Skills Review for Standardized Tests

0086_0095_AM_S_C02_L01_664291.indd 95 4/30/12 6:05 PM


Explore 2-2

Graphing Technology LabGraphing Technology Lab

Behavior of GraphsBehavior of Graphs

ObjectiveGraph and analyze the behavior of polynomial functions.

96 | Lesson 2-2

In Lesson 1-3, you analyzed the end behavior of functions by making a table of values and graphing

them. For a polynomial function, the behavior of the graph can be determined by analyzing specific

terms of the function.

Activity 1 Graph Polynomial Functions

Sketch each graph, and identify the end behavior of the function.

a. f (x) = x 3 + 6 x 2 - 4x + 2

Use a table of values to sketch the graph.

x -10 -5 -2 0 2 5 10

f (x ) -358 47 26 2 26 257 1562

In the graph of f (x), it appears that lim x→-∞


f (x) = ∞. [-10, 10] scl: 1 by [-40, 60] scl: 10

b. g (x) = -2 x 3 + 6 x 2 - 4x + 2

x -8 -5 -2 0 2 5 8

g (x ) 1442 422 50 2 2 -118 -670

In the graph of g(x), it appears that lim x→-∞

g(x) = ∞ and lim x→∞

g(x) = -∞.[-5, 5] scl: 1 by [-40, 60] scl: 10

c. h(x) = - x 4 + x 3 + 6 x 2 - 4x + 2

x -8 -5 -2 0 2 5 8

h (x ) -4190 -578 10 2 10 -368 -3230

In the graph of h (x), it appears that lim x→-∞

h (x) = -∞ and lim x→∞

h(x) = -∞.[-5, 5] scl: 1 by [-20, 20] scl: 4

Analyze the Results 1. Look at the terms of each function above. What differences do you see?

2. How is the end behavior of the graphs of each function affected by these differences?

3. Develop a pattern for every possible type of end behavior of a polynomial function.

4. Give an example of a polynomial function with a graph that approaches positive infinity when x approaches both negative infinity and positive infinity.

ExercisesDescribe the end behavior of each function without making a table of values or graphing.

5. f (x) = -2 x 3 + 4x 6. f (x) = 5 x 4 + 3 7. f (x) = - x 5 + 2x - 4

8. g (x) = 6 x 6 - 2 x 2 + 10x 9. g (x) = 3x - 4 x 4 10. h (x) = 6 x 2 - 3 x 3 - 2 x 6

See margin.

See margin.


Sample answer: f (x ) = x 4

5. lim x→-∞

f (x ) = ∞;

lim x→∞

f (x ) = -∞

6. lim x →-∞

f (x ) = ∞;

lim x→∞

f (x ) = ∞

7. lim x →-∞

f (x ) = ∞;

lim x→∞

f (x ) = -∞

8. lim x→-∞

g(x ) = ∞;

lim x→∞

g(x ) = ∞

9. lim x→-∞

g(x ) = -∞;

lim x→∞

g(x ) = -∞

10. lim x→-∞

h(x ) = -∞;

lim x→∞

h(x ) = -∞

Study Tip Table of Values Be sure to use enough points to get the overall shape of the graph.

0096_0096_AM_S_C02_EXP_664291.indd 96 4/30/12 6:09 PM

96 | Explore 2-2 | Graphing Technology Lab: Behavior of Graphs

From Concrete to AbstractAsk students to predict what the end behavior of the graph of each equation in the Activity would be if the leading coefficients changed signs. Then have students graph both forms of each equation to check their predictions.

Additional Answers

1. Sample answer: The first terms are all different. The first term of f (x ) has a positive coefficient and the first term of g (x ) has a negative coefficient. h (x ) has another term in front, - x 4 .

2. Sample answer: From f (x ) to g (x ), the end behavior appears reversed. h (x ) is different from the others in that as x approaches both positive and negative infinity, h (x ) approaches negative infinity.

1 FocusObjective Graph and analyze the behavior of polynomial functions.

Teaching TipIf students find that not enough of the graph is shown in the standard viewing window, they can press 3 while in the graphing window to zoom out, allowing more of the graph to be shown.

2 TeachWorking in Cooperative GroupsPair students, mixing abilities. Have students work through the Activity parts a–c.

Ask: ■ How is the equation in part a similar

to the equation in part b? They both have the terms 6 x 2 - 4x + 2.

■ How is the graph of the equation in part a the same as the graph of the equation in part b? They both pass through (0, 2).

■ Look at the equation in part c. Would you expect its graph to pass through (0, 2) also? Explain. Yes; its constant term is also 2.

Have students work through the Analyze the Results Exercises 1–4.

Practice Have students complete Exercises 5–10.

3 AssessFormative AssessmentUse Exercise 9 to assess whether students can graph a polynomial function and describe its end behavior.

0096_AM_T_C02_EXP_664293.indd 960096_AM_T_C02_EXP_664293.indd 96 7/31/12 11:01 AM7/31/12 11:01 AM

Lesson 2-2





Chapter

Resource

Masters

■ Study Guide and Intervention■ Practice■ Word Problem Practice

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Enrichment

■ Practice■ Word Problem Practice■ Enrichment







Before Lesson 2-2 Analyze graphs of functions.

Lesson 2-2 Graph polynomial functions. Model real-world data with polynomial functions.

After Lesson 2-2 Graph and solve polynomial inequalities.




Ask:■ Look at the graph. Describe where

the graph of the data changes direction. 1980, 2000, after 2000

■ Do there appear to be breaks, holes, gaps, or sharp turns in the graph? no

■ Would you expect this graph to be modeled by a linear function? No, the data tends to be a curve, not a line.

(continued on the next page)

Polynomial FunctionsPolynomial Functions

New Vocabularypolynomial function

polynomial function of

degree n

leading coefficient

leading-term test

quartic function

turning point

quadratic form

repeated zero

multiplicity

Why? The scatter plot shows total personal savings as a percent of disposable income in the United States. Often data with multiple relative extrema are best modeled by a polynomial function.

Now

1Graph polynomial functions.

2 Model real-world data with polynomial functions.

ThenYou analyzed graphs of functions. (Lesson 1-2)

1Graph Polynomial Functions In Lesson 2-1, you learned about the basic characteristics of monomial functions. Monomial functions are the most basic polynomial functions. The

sums and differences of monomial functions form other types of polynomial functions.

Let n be a nonnegative integer and let a 0 , a 1 , a 2 , …, a n - 1 , a n be real numbers with a n ≠ 0. Then the function given by

f (x) = a n x n + a n - 1 x n - 1 + � + a 2 x 2 + a 1 x + a 0

is called a polynomial function of degree n. The leading coefficient of a polynomial function is the coefficient of the variable with the greatest exponent. The leading coefficient of f (x) is a n .

You are already familiar with the following polynomial functions.

y

x

f (x) = c, c ≠ 0

y

x

f (x) = ax + c

y

x

ax 2f (x) = + bx + c

The zero function is a constant function with no degree. The graphs of polynomial functions share certain characteristics.

Graphs of Polynomial Functions

Example Nonexamples

y

x

Polynomial functions are defined and continuous for all real numbers and have smooth, rounded turns.

y

x

y

x

Graphs of polynomial functions do not have breaks, holes, gaps, or sharp corners.

Savings as a Percent of Disposable IncomeSavings as a Percent of Disposable Income

Perc

ent S

avin

gs

0

-2

2

4

6

8

10

12

Year19821966 1998 2010


0097_0107_AM_S_C02_L02_664291.indd 97 5/28/12 11:36 PM


98 | Lesson 2-2 | Polynomial Functions

■ Would you expect this graph to be modeled by a quadratic function? Explain. No, a quadratic curve has one maximum or minimum point, while this graph appears to have at least two maximum points.

1 Graph Polynomial FunctionsExample 1 shows how to graph transformations of monomial functions. Example 2 shows how to apply the leading term test to a polynomial function written in standard form to determine the end behavior of the graph of the function. Examples 3–5 show how to find zeros of a polynomial function. Example 6 shows how to use the leading term test and zeros of a polynomial to graph the polynomial.


Additional Example

Graph each function.

a. f (x ) = (x - 3) 5 y

x

−4

−8

−4−8

8

4

4 8

b. f (x ) = x 6 - 1y

x

−4

−8

−4−8

8

4

4 8


1

Additional Answers (Guided Practice)

1A. y

x

−4

−4−8

8

12

4

4 8

4 - x3f (x) =

1B. y

x

−4

−8

−4−8−12−16

8

4

g(x) = (x + 7)4

Recall that the graphs of even-degree, non-constant monomial functions resemble the graph of f (x) = x 2 , while the graphs of odd-degree monomial functions resemble the graph of f (x) = x 3 . You can use the basic shapes and characteristics of even- and odd-degree monomial functions and what you learned in Lesson 1-5 about transformations to transform graphs of monomial functions.

Example 1 Graph Transformations of Monomial Functions

Graph each function.

a. f (x) = (x - 2) 5

This is an odd-degree function, so its graph is similar to the graph of y = x 3 . The graph of f (x) = (x - 2) 5 is the graph of y = x 5 translated 2 units to the right.

y

x

f (x) = (x - 2)5

b. g(x) = - x 4 + 1

This is an even-degree function, so its graph is similar to the graph of y = x 2 . The graph of g(x) = - x 4 + 1 is the graph of y = x 4 reflected in the x-axis and translated 1 unit up.

y

x

g(x) = -x 4 + 1


1A. f (x) = 4 - x 3 1B. g(x) = (x + 7) 4

1A–B. See margin.

In Lesson 1-3, you learned that the end behavior of a function describes how the function behaves, rising or falling, at either end of its graph. As x → -∞ and x → ∞, the end behavior of any polynomial function is determined by its leading term. The leading term test uses the power and coefficient of this term to determine polynomial end behavior.

The end behavior of any non-constant polynomial function f (x ) = a n x n + � + a 1 x + a 0 can be described in one of the following four ways, as determined by the degree n of the polynomial and its leading coefficient a n .

n odd, a n positive

lim x→-∞ f (x ) = -∞ and lim x→∞

f (x ) = ∞

y = f (x)y

x

f (x) = ∞

f (x) = -∞

limx→∞

limx→-∞

n odd, a n negative

lim x→-∞ f (x ) = ∞ and lim x→∞

f (x ) = -∞

y = f (x)

y

x

f (x) = ∞

f (x) = -∞limx→∞

limx→-∞

n even, a n positive

lim x→-∞ f (x ) = ∞ and lim x→∞

f (x ) = ∞

y = f (x)

y

x

f (x) = ∞ f (x) = ∞limx→∞

limx→-∞

n even, a n negative

lim x→-∞ f (x ) = -∞ and lim x→∞

f (x ) = -∞

y = f (x)

y

x

f (x) = -∞ f (x) = -∞limx→∞

limx→-∞

Key Concept Leading Term Test for Polynomial End Behavior


0097_0107_AM_S_C02_L02_664291.indd 98 4/30/12 6:11 PM

Key Concept boxes highlight definitions, formulas, and other important ideas. Multiple representations—words, symbols, examples, models—aid students’ understanding.



Additional Example

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

a. f (x ) = 3 x 4 - x 3 + x 2 + x - 1 The degree is 4 and the leading coefficient is 3. Because the degree is even and the leading coefficient is positive, lim x→-∞

f (x ) = ∞

and lim x→∞

f (x ) = ∞.

b. f (x ) = -3 x 2 + 2 x 5 - x 3 Written in standard form, the function is f (x ) = 2 x 5 - x 3 - 3 x 2 . The degree is 5 and the leading coefficient is 2. Because the degree is odd and the leading coefficient is positive, lim x→-∞

f (x ) = -∞

and lim x→∞

f (x ) = ∞.

c. f (x ) = -2 x 5 - 1 The degree is 5 and the leading coefficient is -2. Because the degree is odd and the leading coefficient is negative, lim x→-∞

f (x ) = ∞

and lim x→∞

f (x ) = -∞.

2

Teach with TechVideo Recording Have students create a video explaining how to use the leading term test to determine the end behavior of a polynomial’s graph. Post the video to a class Web site so students can use it as an additional reference outside of class.

Example 2 Apply the Leading Term Test

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

a. f (x) = 3 x 4 - 5 x 2 - 1

The degree is 4, and the leading coefficient is 3. Because the degree is even and the leading coefficient is positive, lim x→-∞

f (x) = ∞ and lim x→∞ f (x) = ∞.

y

x

f (x) = 3x - 5x - 14 2

b. g (x) = -3 x 2 - 2 x 7 + 4 x 4

Write in standard form as g (x) = -2 x 7 + 4 x 4 - 3 x 2 . The degree is 7, and the leading coefficient is -2. Because the degree is odd and the leading coefficient is negative, lim x→-∞

f (x) = ∞ and lim x→∞ f (x) = -∞.

y

x

g(x) = -2x + 4x - 3x7 4 2

c. h(x) = x 3 - 2 x 2

The degree is 3, and the leading coefficient is 1. Because y

x

h(x) = x - 2x3 2

the degree is odd and the leading coefficient is positive, lim x→-∞

f (x) = -∞ and lim x→∞ f (x) = ∞.


2A. g(x) = 4 x 5 - 8 x 3 + 20 2B. h(x) = -2 x 6 + 11 x 4 + 2 x 2

2A. The degree is 5, and

the leading coefficient is

4. Because the degree is

odd and the leading

coefficient is positive,

lim x→-∞

f (x ) = -∞ and

lim x→∞

f (x ) = ∞.

2B. The degree is 6, and

the leading coefficient is

-2. Because the degree

is even and the leading

coefficient is negative,

lim x→-∞

f (x ) = -∞ and

lim x→∞

f (x ) = -∞.

Consider the shapes of a few typical third-degree polynomial or cubic functions and fourth-degree polynomial or quartic functions shown.

y

x

y

x

y

O x

turning point/relative maximum

y

xO

turning point/absolute minimum

Typical Cubic Functions Typical Quartic Functions

Observe the number of x-intercepts for each graph. Because an x-intercept corresponds to a real zero of the function, you can see that cubic functions have at most 3 zeros and quartic functions have at most 4 zeros.

Turning points indicate where the graph of a function changes from increasing to decreasing, and vice versa. Maxima and minima are also located at turning points. Notice that cubic functions have at most 2 turning points, and quartic functions have at most 3 turning points. These observations can be generalized as follows and shown to be true for any polynomial function.

Watch Out! Standard Form The leading term of a polynomial function is not necessarily the first term of a polynomial. However, the leading term is always the first term of a polynomial when the polynomial is written in standard form. Recall that a polynomial is written in standard form if its terms are written in descending order of exponents.


0097_0107_AM_S_C02_L02_664291.indd 99 6/5/12 11:20 PM



Interpersonal Learners Have students work together in groups to sketch the graphs of polynomial functions having a particular degree and number of real roots, for example, degree 3 and 3 real roots or degree 3 and only 1 real root. Then have students experiment with coefficients in the general form of a polynomial in order to find functions with graphs that resemble their sketches.


Additional Example

State the number of possible real zeros and turning points of f (x ) = x 3 + 5 x 2 + 4x. Then determine all of the real zeros by factoring. The degree is 3, so fhas at most 3 distinct real zeros and at most 2 turning points. f (x ) = x 3 + 5 x 2 + 4x = x (x + 1)(x + 4), so f has three zeros, 0, -1, and -4.

3

Focus on Mathematical ContentPolynomial Functions Polynomial functions are formed by adding or subtracting monomial functions and constants.

The graph of a polynomial function of the form f (x ) = a n x n + a n - 1x n - 1

+ … + a 1x + a 0 , where a n ≠ 0, n is a nonnegative integer and the coefficients of each term are real numbers, has these characteristics.

■ Degree: n

■ Maximum number of turning points: n - 1

■ At a zero of odd multiplicity, the graph crosses the x-axis.

■ At a zero of even multiplicity, the graph touches the x-axis.

■ Between zeros, the graph is either above or below the x-axis.

■ The end behavior of the graph is determined by its leading term using the leading term test.


Study Tip Look Back To review techniques for solving quadratic equations, see Lesson 0-3.

A polynomial function f of degree n ≥ 1 has at most n distinct real zeros and at most n - 1 turning points.

y

x

−80

−40

−4 −2

80

40

42

f (x) = - -3 x 6 10 x 4 15 x 2

Example Let f (x ) = 3 x 6 - 10 x 4 - 15 x 2 . Then f has at most 6 distinct real zeros and at most 5 turning points. The graph of f suggests that the function has 3 real zeros and 3 turning points.

Key Concept Zeros and Turning Points of Polynomial Functions

Recall that if f is a polynomial function and c is an x-intercept of the graph of f, then it is equivalent to say that:

• c is a zero of f,

• x = c is a solution of the equation f (x) = 0, and

• (x - c) is a factor of the polynomial f (x).

You can find the zeros of some polynomial functions using the same factoring techniques you used to solve quadratic equations.

Example 3 Zeros of a Polynomial Function

State the number of possible real zeros and turning points of f (x) = x 3 - 5 x 2 + 6x. Then determine all of the real zeros by factoring.

The degree of the function is 3, so f has at most 3 distinct real zeros and at most 3 - 1 or 2 turning points. To find the real zeros, solve the related equation f (x) = 0 by factoring.

x 3 - 5 x 2 + 6x = 0 Set f (x ) equal to 0.

x( x 2 - 5x + 6) = 0 Factor the greatest common factor, x.

x(x - 2)(x - 3) = 0 Factor completely.

So, f has three distinct real zeros, 0, 2, and 3. This is consistent with a cubic function having at most 3 distinct real zeros.

CHECK You can use a graphing calculator to graph

[-5, 5] scl: 1 by [-5, 5] scl: 1

f (x) = x 3 - 5 x 2 + 6x and confirm these zeros. Additionally, you can see that the graph has 2 turning points, which is consistent with cubic functions having at most 2 turning points.


State the number of possible real zeros and turning points of each function. Then determine all of the real zeros by factoring.

3A. f (x) = x 3 - 6 x 2 - 27x 3B. f (x) = x 4 - 8 x 2 + 15

3 zeros and 2 turning points; 0, 9, and -3

4 zeros and 3 turning points; ± √ � 5 and ±

√ � 3

In some cases, a polynomial function can be factored using quadratic techniques if it has quadratic form.

Words A polynomial expression in x is in quadratic form if it is written as au 2 + bu + c for any numbers a, b, and c, a ≠ 0, where u is some expression in x.

Symbols x 4 - 5 x 2 - 14 is in quadratic form because the expression can be written as ( x 2 ) 2 - 5 ( x 2 ) - 14. If u = x 2 , then the expression becomes u 2 - 5u - 14.

Key Concept Quadratic Form

Study Tip Look Back Recall from Lesson 1-2 that the x-intercepts of the graph of a function are also called the zeros of a function. The solutions of the corresponding equation are called the roots of the equation.


0097_0107_AM_S_C02_L02_664291.indd 100 4/30/12 6:11 PM



Additional Examples

State the number of possible real zeros and turning points for h (x ) = x 4 - 4 x 2 + 3. Then determine all of the real zeros by factoring. The degree is 4, so hhas at most 4 distinct real zeros and at most 3 turning points. h (x ) = x 4 - 4 x 2 + 3 = ( x 2 - 3)(x - 1)(x + 1), so h has four distinct real zeros, ± √ � 3 , -1, and 1.

State the number of possible real zeros and turning points of h (x ) = x 4 + 5 x 3 + 6 x 2 . Then determine all of the real zeros by factoring. The degree is 4, so h has at most 4 distinct real zeros and at most 3 turning points. h (x ) = x 4 + 5 x 3 + 6 x 2 =

x 2 (x + 2)(x + 3), so h has three zeros, 0, -2, and -3. Of the zeros, 0 is repeated.

4

5

Tips for New TeachersZeros Emphasize to students that they can confirm the zeros (where the graph crosses the x-axis) and the number of turning points by using a graphing calculator to graph the polynomial function.

Tangents Explain to students that the graph of a polynomial function can be tangent to the x-axis at a specific point and intersect the x-axis at a different point, as shown in Figure 2.2.2.

Example 4 Zeros of a Polynomial Function in Quadratic Form

State the number of possible real zeros and turning points for g (x) = x 4 - 3 x 2 - 4. Then determine all of the real zeros by factoring.

The degree of the function is 4, so g has at most 4 distinct real zeros and at most 4 - 1 or 3 turning points. This function is in quadratic form because x 4 - 3 x 2 - 4 = ( x 2 ) 2 - 3 ( x 2 ) - 4. Let u = x 2 .

( x 2 ) 2 - 3 ( x 2 ) - 4 = 0 Set g (x ) equal to 0.

u 2 - 3u - 4 = 0 Substitute u for x 2 .

(u + 1)(u - 4) = 0 Factor the quadratic expression.

( x 2 + 1)( x 2 - 4) = 0 Substitute x 2 for u.

( x 2 + 1)(x + 2)(x - 2) = 0 Factor completely.

x 2 + 1 = 0 or x + 2 = 0 or x - 2 = 0 Zero Product Property

x = ± √ �� -1 x = -2 x = 2 Solve for x.

Because ± √ �� -1 are not real zeros, g has two distinct real zeros, -2 and 2. This is consistent with a quartic function. The graph of g (x) = x 4 - 3 x 2 - 4 in Figure 2.2.1 confirms this. Notice that there are 3 turning points, which is also consistent with a quartic function.



4A. g (x) = x 4 - 9 x 2 + 18 4B. h(x) = x 5 - 6 x 3 - 16x

4A. 4 real zeros and 3 turning points; ± √ � 3 , ±

√ � 6

4B. 5 real zeros and 4 turning points; 0, ± √ � 8

If a factor (x - c) occurs more than once in the completely factored form of f (x), then its related zero c is called a repeated zero. When the zero occurs an even number of times, the graph will be tangent to the x-axis at that point. When the zero occurs an odd number of times, the graph will cross the x-axis at that point. A graph is tangent to an axis when it touches the axis at that point, but does not cross it.

Example 5 Polynomial Function with Repeated Zeros

State the number of possible real zeros and turning points of h(x) = - x 4 - x 3 + 2 x 2 . Then determine all of the real zeros by factoring.

The degree of the function is 4, so h has at most 4 distinct real zeros and at most 4 - 1 or 3 turning points. Find the real zeros.

- x 4 - x 3 + 2 x 2 = 0 Set h (x ) equal to 0.

- x 2 ( x 2 + x - 2) = 0 Factor the greatest common factor, - x 2 .

- x 2 (x - 1)(x + 2) = 0 Factor completely.

The expression above has 4 factors, but solving for x yields only 3 distinct real zeros, 0, 1, and -2. Of the zeros, 0 occurs twice.

The graph of h(x) = - x 4 - x 3 + 2 x 2 shown in Figure 2.2.2 confirms these zeros and shows that h has three turning points. Notice that at x = 1 and x = -2, the graph crosses the x-axis, but at x = 0, the graph is tangent to the x-axis.

GuidedGuided Practice Practice


5A. g (x) = -2 x 3 - 4 x 2 + 16x 5B. f (x) = 3 x 5 - 18 x 4 + 27 x 3

3 real zeros and 2 turning points; -4, 0, 2

5B. 5 real zeros and 4 turning points; 0, 3

y

x

Figure 2.2.1

y

x

Figure 2.2.2


0097_0107_AM_S_C02_L02_664291.indd 101 4/30/12 6:11 PM



Additional Example

For f (x ) = x (3x + 1) (x - 2) 2 , (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

a. The degree is 4 and leading coefficient is 3, so lim x→-∞

f (x ) = ∞ and lim x→∞

f (x ) = ∞.

b. The zeros are x = 0, x = -

1 _ 3

,

x = 2. The zero at x = 2 has multiplicity 2.

c. Sample answer: (-1, 18), (-0.1, -0.3087), (1, 4), (3, 30)

d. y

x−2−4

8

12

16

4

2 4

6

Tips for New Teachersx-value in Interval Any value for x can be chosen as long as the value falls within the interval.

2 Model DataExample 7 shows how to find the best polynomial function to model a set of data. Counting the number of turning points observed in a scatter plot of a set of data is an important step in modeling the data.

Follow-upStudents have explored modeling using polynomial functions.

Ask:■ What are the advantages of modeling

real-world situations using polynomial functions? Sample answer: They have well-known and understood properties; there are multiple models that can be considered

within the same family of functions; the computation that is done to make predictions is relatively easy to perform.

Study TipNonrepeated Zeros A nonrepeated zero can be thought of as having a multiplicity of 1 or odd multiplicity. A graph crosses the x-axis and has a sign change at every nonrepeated zero.

In h(x) = - x 2 (x - 1)(x + 2) from Example 5, the zero x = 0 y

x−4 −2

−8

8

−16

2 4

k(x) = (x - 1)3(x + 2)4

occurs 2 times. In k (x) = (x - 1) 3 (x + 2) 4 , the zero x = 1 occurs 3 times, while x = -2 occurs 4 times. Notice that in the graph of k shown, the curve crosses the x-axis at x = 1 but not at x = -2. These observations can be generalized as follows and shown to be true for all polynomial functions.

If (x - c ) m is the highest power of (x - c ) that is a factor of polynomial function f, then c is a zero of multiplicity m of f, where m is a natural number.

• If a zero c has odd multiplicity, then the graph of f crosses the x-axis at x = c and the value of f (x ) changes signs at x = c.

• If a zero c has even multiplicity, then the graph of f is tangent to the x-axis at x = c and the value of f (x ) does not change signs at x = c.

Key Concept Repeated Zeros of Polynomial Functions

You now have several tests and tools to aid you in graphing polynomial functions.

Example 6 Graph a Polynomial Function

For f (x) = x(2x + 3) (x - 1) 2 , (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

a. The product x (2x + 3) (x - 1) 2 has a leading term of x (2x) (x) 2 or 2 x 4 , so f has degree 4 and leading coefficient 2. Because the degree is even and the leading coefficient is positive, lim x→-∞

f (x) = ∞ and lim x→∞

f (x) = ∞.

b. The distinct real zeros are 0, -1.5, and 1. The zero at 1 has multiplicity 2.

c. Choose x-values that fall in the intervals determined by the zeros of the function.

Interval x-value in Interval f (x ) (x, f (x ))

(-∞, -1.5) -2 f (-2) = 18 (-2, 18)

(-1.5, 0) -1 f (-1) = -4 (-1, -4)

(0, 1) 0.5 f (0.5) = 0.5 (0.5, 0.5)

(1, ∞) 1.5 f (1.5) = 2.25 (1.5, 2.25)

d. Plot the points you found (Figure 2.2.3).The end behavior of the function tells you that the graph eventually rises to the left and to the right. You also know that the graph crosses the x-axis at nonrepeated zeros -1.5 and 0, but does not cross the x-axis at repeated zero 1, because its multiplicity is even. Draw a continuous curve through the points as shown in Figure 2.2.4. Figure 2.2.3

y

x

Figure 2.2.4

y

x


For each function, (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

6A. f (x) = -2x(x - 4) (3x - 1) 3 6B. h(x) = - x 3 + 2 x 2 + 8x



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Additional Example

POPULATION The table below shows a town’s population over an 8-year period. Year 1 refers to the year 2001, year 2 refers to the year 2002, and so on.

Year Population

1 5050

2 5510

3 5608

4 5496

5 5201

6 5089

7 5095

8 4675

a. Create a scatter plot of the data, and determine the type of polynomial function that could be used to represent the data. cubic function

[0, 10] scl: 1 by [0, 7000] scl: 1000

b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth, and state the correlation coefficient. f (x ) = 10.020 x 3 - 176.320 x 2 + 807.469x + 4454.786; r 2 = 0.89.

[0, 10] scl: 1 by [0, 7000] scl: 1000

c. Use the model to estimate the population of the town in the year 2012. 6069

d. Use the model to determine the approximate year in which the population reaches 10,712. 2015

7

Tips for New TeachersCorrelation Coefficient Some students may need to turn on the correlation coefficient feature on their calculators. To do this from the home screen, press [CATALOG], select DiagnosticOn, and then press ENTER .

2Model Data You can use a graphing calculator to model data that exhibit linear, quadratic, cubic, and quartic behavior by first examining the number of turning points suggested by a

scatter plot of the data.

SAVINGS Refer to the beginning of the lesson. The average personal savings as a percent of disposable income in the United States is given in the table.

Year 1970 1980 1990 1995 2000 2001 2002 2003 2004 2005

% Savings 9.4 10.0 7.0 4.6 2.3 1.8 2.4 2.1 2.0 -0.4

Source: U.S. Department of Commerce

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data.

Enter the data using the list feature of a graphing calculator.

[-1, 36] scl: 1 by [-1, 11] scl: 1

Let L1 be the number of years since 1970. Then create a scatter plot of the data. The curve of the scatter plot resembles the graph of a quadratic equation, so we will use a quadratic regression.

b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth, and state the correlation coefficient.

Using the QuadReg tool on a graphing calculator and rounding each coefficient to the nearest thousandth yields f (x) = -0.009 x 2 + 0.033x + 9.744. The correlation coefficient r 2 for the data is 0.96, which is close to 1, so the model is a good fit.

We can graph the complete (unrounded) regression by sending

[-1, 36] scl: 1 by [-1, 11] scl: 1

it to the menu. If you enter L 1 , L 2 , and Y 1 after QuadReg, as shown in Figure 2.2.5, the regression equation will be entered into Y 1 . Graph this function and the scatter plot in the same viewing window. The function appears to fit the data reasonably well.

c. Use the model to estimate the percent savings in 1993.

Because 1993 is 23 years after 1970, use the CALC feature on a calculator to find f (23). The value of f (23) is 5.94, so the percent savings in 1993 was about 5.94%.

d. Use the model to determine the approximate year in which the percent savings reached 6.5%.

Graph the line y = 6.5 for Y 2 . Then use 5: intersect on the

[-1, 36] scl: 1 by [-1, 11] scl: 1

CALC menu to find the point of intersection of y = 6.5 with f (x). The intersection occurs when x ≈ 21, so the approximate year in which the percent savings reached 6.5% was about 1970 + 21 or 1991.


7. POPULATION The median age of the U.S. population by year predicted through 2080 is shown.

Year 1900 1930 1960 1990 2020 2050 2080

Median Age 22.9 26.5 29.5 33.0 40.2 42.7 43.9


a. Write a polynomial function to model the data. Let L1 be the number of years since 1900.

b. Estimate the median age of the population in 2005.

c. According to your model, in what year did the median age of the population reach 30?

Sample answer: f (x ) = 0.126x + 22.732

Sample answer: 36

Sample answer: 1958

Real-World Example 7 Model Data Using Polynomial Functions

Figure 2.2.5

Real-World LinkA college graduate planning to retire at 65 needs to save an average of $10,000 per year toward retirement.

Source: Monroe Bank


Noe

l Hen

dric

kson

/Pho

todi

sc/G

etty

Imag

es

0097_0107_AM_S_C02_L02_664291.indd 103 5/31/12 6:15 PM


3 PracticeFormative AssessmentUse Exercises 1–49 to check for understanding.


Additional Answers

1. y

x

−4

−8

−4−8

8

4

4 8

(x + 5) 2f (x) =

2. y

x

−4

−8

−4−8

8

4 8

(x - 6) 3f (x) =

3. y

x

−4

−8

−4−8

8

4

4 8

x 4 - 6f (x) =

4.

x

−4

−8

−4−8

8

4

4 8

x5 + 7f (x) =

y

5.

x

−4

−8

−4−8

8

4

4 8

(2x) 4f (x) =

y




AL Approaching Level 1–49, 90–92, 94–111 1–49 odd, 110–111 2–48 even, 90–92, 94–111

OL On Level 1–67 odd, 68, 69–79 odd, 80, 81–89 odd, 90–92, 94–111

1–49, 110–111 50–92, 94–111



Exercises

Graph each function. (Example 1)

1. f (x) = (x + 5) 2 2. f (x) = (x - 6) 3

3. f (x) = x 4 - 6 4. f (x) = x 5 + 7

5. f (x) = (2x) 4 6. f (x) = (2x) 5 - 16

7. f (x) = (x - 3) 4 + 6 8. f (x) = (x + 4) 3 - 3

9. f (x) = 1 _ 3 (x - 9) 5 10. f (x) = ( 1 _

2 x)

3 + 8

11. WATER If it takes exactly one minute to drain a 10-gallon tank of water, the volume of water remaining in the tank can be approximated by v(t) = 10 (1 - t) 2 , where t is time in minutes, 0 ≤ t ≤ 1. Graph the function. (Example 1)

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test. (Example 2)

12. f (x) = -5 x 7 + 6 x 4 + 8 13. f (x) = 2 x 6 + 4 x 5 + 9 x 2

14. g (x) = 5 x 4 + 7 x 5 - 9 15. g (x) = -7 x 3 + 8 x 4 - 6 x 6

16. h (x) = 8 x 2 + 5 - 4 x 3 17. h (x) = 4 x 2 + 5 x 3 - 2 x 5

18. f (x) = x (x + 1)(x - 3) 19. g (x) = x 2 (x + 4)(-2x + 1)

20. f (x) = -x (x - 4)(x + 5) 21. g (x) = x 3 (x + 1)( x 2 - 4)

22. ORGANIC FOOD The number of acres in the United States used for organic apple production from 2000 to 2005 can be modeled by a(x) = 43.77 x 4 - 498.76 x 3 + 1310.2 x 2 + 1626.2x + 6821.5, where x = 0 is 2000. (Example 2)

a. Graph the function using a graphing calculator.

b. Describe the end behavior of the graph of the function using limits. Explain using the leading term test.

State the number of possible real zeros and turning points of each function. Then determine all of the real zeros by factoring. (Examples 3–5)

23 f (x) = x 5 + 3 x 4 + 2 x 3 24. f (x) = x 6 - 8 x 5 + 12 x 4

25. f (x) = x 4 + 4 x 2 - 21 26. f (x) = x 4 - 4 x 3 - 32 x 2

27. f (x) = x 6 - 6 x 3 - 16 28. f (x) = 4 x 8 + 16 x 4 + 12

29. f (x) = 9 x 6 - 36 x 4 30. f (x) = 6 x 5 - 150 x 3

31. f (x) = 4 x 4 - 4 x 3 - 3 x 2 32. f (x) = 3 x 5 + 11 x 4 - 20 x 3

For each function, (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function. (Example 6)

33. f (x) = x(x + 4) (x - 1) 2 34. f (x) = x 2 (x - 4)(x + 2)

35. f (x) = -x (x + 3) 2 (x - 5) 36. f (x) = 2x (x + 5) 2 (x - 3)

37. f (x) = -x(x - 3) (x + 2) 3 38. f (x) = - (x + 2) 2 (x - 4) 2

39. f (x) = 3 x 3 - 3 x 2 - 36x 40. f (x) = -2 x 3 - 4 x 2 + 6x

41. f (x) = x4 + x3 - 20 x2 42. f (x) = x5 + 3 x4 - 10 x3

43. RESERVOIRS The number of feet below the maximum water level in Wisconsin’s Rainbow Reservoir during ten months in 2007 is shown. (Example 7)

Month Level Month Level

January 4 July 9

February 5.5 August 11

March 10 September 16.5

April 9 November 11.5

May 7.5 December 8.5

Source: Wisconsin Valley Improvement Company

a. Write a model that best relates the water level as a function of the number of months since January.

b. Use the model to estimate the water level in the reservoir in October.

Use a graphing calculator to write a polynomial function to model each set of data. (Example 7)

44. x -3 -2 -1 0 1 2 3

f (x ) 8.75 7.5 6.25 5 3.75 2.5 1.25

45. x 5 7 8 10 11 12 15 16

f (x ) 2 5 6 4 -1 -3 5 9

46. x -2.53 -2 -1.5 -1 -0.5 0 0.5 1 1.5

f (x ) 23 11 7 6 6 5 3 2 4

47. x 30 35 40 45 50 55 60 65 70 75

f (x ) 52 41 32 44 61 88 72 59 66 93

48. ELECTRICITY The average retail electricity prices in the U.S. from 1970 to 2005 are shown. Projected prices for 2010 and 2020 are also shown. (Example 7)

Year Price

(¢ / kWh)

Year Price

(¢ / kWh)

1970 6.125 1995 7.5

1974 7 2000 6.625

1980 7.25 2005 6.25

1982 9.625 2010 6.25

1990 8 2020 6.375

Source: Energy Information Administration

a. Write a model that relates the price as a function of the number of years since 1970.

b. Use the model to predict the average price of electricity in 2015.

c. According to the model, during which year was the price 7¢ for the second time?

1–10. See margin.

See Chapter 2 Answer Appendix.12–21. See Chapter 2 Answer Appendix.




See margin.

14.8 ft


48a. Sample answer: f (x ) = 2.14 � 1 0 -4 x 3 -

0.018 x 2 + 0.383x + 5.976

5.93¢

1999


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43a. Sample answer: f (x ) = -0.018 x 4 + 0.355 x 3 - 2.276 x 2 + 5.722x + 3.509

54. Sample answer: f (x ) = x 3 + 3x 2 + 3x + 1

55. Sample answer: f (x ) = x 3 - 9 x 2 + 27x - 27

56. Sample answer: f (x ) = x 4 - 6 x 3 - 27 x 2 + 108x + 324

57. Sample answer: f (x ) = x 4 + 11 x 3 + 15 x 2 - 175x - 500

58. Sample answer: f (x ) = x 4 - 28 x 3 + 294 x 2 - 1372x + 2401

59. Sample answer: f (x ) = x 5 + 4 x 4

60. Sample answer: f (x ) = x 5 - 13 x 4 + 64 x 3 - 148 x 2 + 160x - 64

61. Sample answer: f (x ) = x 5 - x 4 - 6 x 3

62. Sample answer: f (x ) = x 4 + 2 x 2 + 1

63. Sample answer: f (x ) = x 6 + 3 x 4 + 3 x 2 + 1

Watch Out!

Common Error For Exercise 49, you may wish to explain to students how the quarters relate to the years 2005–2007. Some students may mistakenly equate quarter 1 with 2005, quarter 2 with 2006, and so on. Each year has four quarters, so quarters 1–4 are for 2005, quarters 5–8 are for 2006, and 9–12 are for 2007.

Additional Answers

6. y

x

−8

−16

−2−4

16

8

2 4

(2x) 5 - 16f (x) =

7. y

x

−8

−16

−8−16

16

8

8 16

f (x) = (x - 3) 4 + 6

8. y

x

−4

−8

−4−8

8

4

4 8

(x + 4) 3 - 3f (x) =

9. y

x

−8

−16

−8−16

16

8

8 16

(x - 9) 5_3

f (x) =

10. y

x

−8

−16

−8−16

16

8

8 16

( 1_2x)

3+ 8f (x) =

49. COMPUTERS The numbers of laptops sold each quarter from 2005 to 2007 are shown. Let the first quarter of 2005 be 1, and the fourth quarter of 2007 be 12.

QuartersSale

(Thousands)

1 423

2 462

3 495

4 634

5 587

6 498

7 798

8 986

9 969

10 891

11 1130

12 1347

a. Predict the end behavior of a graph of the data as x approaches infinity.

b. Use a graphing calculator to graph and model the data. Is the model a good fit? Explain your reasoning.

c. Describe the end behavior of the graph using limits. Was your prediction accurate? Explain your reasoning.

Determine whether each graph could show a polynomial function. Write yes or no. If not, explain why not.

50. 51.

y

x

−4

−4

−8

8

4

4 8

y

x

4

8

12

2-2 4-4

52. 53. y

x

x

y

4

8

4 8

−4

−8

−8 −4

Find a polynomial function of degree n with only the following real zeros. More than one answer is possible.

54. -1; n = 3 55. 3; n = 3

56. 6, -3; n = 4 57. -5, 4; n = 4

58. 7; n = 4 59. 0, -4; n = 5

60. 2, 1, 4; n = 5 61. 0, 3, -2; n = 5

62. no real zeros; n = 4 63. no real zeros; n = 6

Determine whether the degree n of the polynomial for each graph is even or odd and whether its leading coefficient a n is positive or negative.

64. 65. y

x

y

x

66. 67. y

x

y

x

68. MANUFACTURING A company manufactures aluminum containers for energy drinks.

a. Write an equation V that represents the volume of the container.

b. Write a function A in terms of r that represents the surface area of a container with a volume of 15 cubic inches.

c. Use a graphing calculator to determine the minimum possible surface area of the can.

Determine a polynomial function that has each set of zeros. More than one answer is possible.

69. 5, -3, 6 70. 4, -8, -2

71. 3, 0, 4, -1, 3 72. 1, 1, -4, 6, 0

73. 3 _ 4 , -3, -4, -

2 _ 3 74. -1, -1, 5, 0, 5 _

6

75 POPULATION The percent of the United States population living in metropolitan areas has increased.

YearPercent of

Population

1950 56.1

1960 63

1970 68.6

1980 74.8

1990 74.8

2000 79.2


a. Write a model that relates the percent as a function of the number of years since 1950.

b. Use the model to predict the percent of the population that will be living in metropolitan areas in 2015.

c. Use the model to predict the year in which 85% of the population will live in metropolitan areas.

a–c. See

Chapter 2

Answer

Appendix.

No; there is a sharp turn at x = 2.yes

yesNo; undefined at x = 0.54–63. See margin.

n is even; a n is positive. n is odd; a n is positive.

n is odd; a n is negative. n is even; a n is negative.

V = π r 2 h

b–c. See Chapter 2 Answer Appendix.


a. Sample

answer: f (x ) =

0.45x + 58.19

87.4%

c. Sample

answer: 2010

B


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Additional Answers

81a. degree = 4; lim x→-∞

= ∞ and lim x→∞

= ∞

81b. -6, -2 (multiplicity: 2), 4

81c. Sample answer: f (x ) = 0.5(x + 2 ) 2 (x + 6)(x - 4)


= -∞ and lim x→∞

= -∞

82b. -3, 2, 6 (multiplicity: 2)


pp. 11–12 AL OL ELL

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Graph Polynomial Functions A polynomial function of degree n is a function of the form

f(x) = anxn + an - 1x n - 1 + … + a2x 2 + a1x + a0,

where n is a nonnegative integer and a0, a1, a2, …, an - 1, an are real numbers with an ≠ 0. Maxima and minima are located at turning points. A polynomial function of degree n has at most n distinct real zeros and at most n - 1 turning points.

Graph f(x) = 2x4 - 3x2 - 1. Describe the end behavior of the graph of the polynomial function using limits. Explain your reasoning using the leading term test. The degree is 4 and the leading coefficient is 2. Because the degree is even and the leading coefficient is positive,

lim x → -∞

f(x) = ∞ and lim x → ∞

f(x) = ∞.

State the number of possible real zeros and turning points of f(x) = x3 + 2x2 - x - 2. Then determine all of the real zeros by factoring.The degree is 3, so f has at most 3 distinct real zeros and at most 3 - 1 or 2 turning points. To find the real zeros, solve the related equation f(x) = 0 by factoring.

x3 + 2x2 − x - 2 = 0 Set f(x) equal to 0.

x2(x + 2) - 1(x + 2) = 0 Group the terms and find the GCF.

(x2 - 1)(x + 2) = 0 Regroup using the Distributive Property.

(x + 1)(x - 1)(x + 2) = 0 Factor completely.

So, f has three distinct real zeros, x = −1, x = 1, and x = -2. The graph has two turning points.

Exercises

1. Graph the function f(x) = −3x5 + 2x2 − 1. Describe the end behavior of the graph of the polynomial function using limits. Explain your reasoning using the leading term test.

Because the degree is odd and the leading

coefficient is negative, lim x → -∞ f(x) = ∞ and

lim x → ∞ f(x) = -∞.

2. State the number of possible real zeros and turning points of f(x) = x4 − 4x3 + 5x2 − 4x + 4. Then determine all of the real zeros by factoring.

4 real zeros and 3 turning points; 2

Study Guide and InterventionPolynomial Functions

2-2

Example 1

y

x4 8−4−8

4

8

−8

−4

y

x

Example 2

005_038_PCCRMC02_893803.indd 11 10/1/09 5:45:28 PM

Practice

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Graph each function. Describe the end behavior of the graph of the polynomial function using limits. Explain your reasoning using the leading term test.

1. f(x) = -7x5 - 2x3 + 3x2 + 5 2. f(x) = x6 - 2x4 + 1

The degree is 5 and the leading coefficient is -7. Because the degree is odd and the leading coefficient is negative,

lim x → -∞ f(x) = ∞ and lim x → ∞

f(x) = -∞.


3. f(x) = x3 - 3x2 - x + 3 4. f(x) = x4 - 5x2 - 36

3 real zeros, 2 turning points; 4 real zeros, 3 turning points; -3, 3-1, 1, 3

5. The amount of food energy produced by farms increases as more energy is expended. The following table shows the amount of energy produced and the amount of energy expended to produce the food.

a. Write a polynomial function to model the set of data. Sample answer: f(x) = -0.000000039x3 + 0.00015x2 - 0.14x + 167

b. Predict the energy output when the energy input is 800 Calories. 136.5 Calories

6. For f(x) = -x(2x + 5)2(x - 3), (a) apply the leading term test, (b) determine the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function. a. Because the degree is even and the leading

coefficient is negative, lim x → -∞ f(x) = -∞ and

lim x → ∞ f(x) = -∞.

b. -2.5 (multiplicity 2), 0, 3

d.

c. (-3, -18), (-2, -10), (1, 98), (4, -676)

PracticePolynomial Functions

The degree is 6 and the leading coefficient is 1. Because the degree is even and the leading coefficient

is positive, lim x → -∞ f(x) = ∞ and

lim x → ∞ f(x) = ∞.

Energy Input

(Calories)606 970 1121 1227 1318 1455 1636 2030 2182 2242

Energy Output

(Calories)133 144 148 157 171 175 187 193 198 198

2-2

y

x4 8−4−8

4

−8

−4

y

x4 8−4−8

4

8

−8

−4

[-10, 10] scl: 1 by [-750, 200] scl: 50

005_038_PCCRMC02_893803.indd 13 10/1/09 5:45:36 PM


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NAME DATE PERIOD


1. VOLUME The volume V of a right circular cone is given by the function V =

1−

3πr2h, where r is the radius of the

base and h is the height.

a. If the height of a cone is 8 meters, describe the end behavior of the graph of the function using limits. Explain your reasoning using the leading term test.

The degree is 2, and the

leading coefficient is8π−3

.

Because the degree is even and the leading coefficient is positive, lim

x → -∞f(x) = ∞ and

lim x → ∞

f(x) = ∞.

b. How many turning points does the graph have?

1 turning point

2. BOX A box is being constructed from a piece of cardboard that is 24 inches wide by 16 inches long. A square with side length x is removed from each corner of the cardboard.

a. Write a polynomial function to model the volume of the box. Describe the end behavior of the graph of the model using limits.

V(x) = 4x3 - 80x2 + 384x; The degree is 3, and the leading coefficient is 4. Because the degree is odd and the leading coefficient is positive,

limx → -∞

f(x) = -∞ and

lim x → ∞

f(x) = ∞.

b. Find the real zeros of the function. 0, 8, and 12

3. ENROLLMENT The freshmen fall enrollment, in thousands, in degree-granting institutions from 2000 –2005 is shown in the table.

Source: National Center for Education Statistics

a. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth. State the correlation coefficient.

Sample answer: f(x) = -58.875x2 + 728.832x +15,304.607, r ≈ 0.998

b. Use the model to estimate the enrollment in 2012.Sample answer: 15,573,000

4. WATER QUALITY Water quality varies with the season. This table shows the average hardness (amount of dissolved minerals) of water in the Missouri River measured at Kansas City, Missouri.

a. Write a polynomial function to model the data set. Sample answer: f(x) = 0.053x4 - 1.565x3 +19.701x2 - 110.033x + 397.677

b. According to the model, what will the hardness be during the following February? 468

Word Problem PracticePolynomial Functions

Month J A S O N D

Hardness

(CACO3 ppm)

170 180 210 230 295 300

Month J F M A M J

Hardness

(CACO3 ppm)

310 250 180 175 230 175

Year Enrollment

2000 15,312

2001 15,928

2002 16,612

2003 16,911

2004 17,272

2005 17,487

2-2

005_038_PCCRMC02_893803.indd 14 10/1/09 5:45:42 PM

Enrichment

p. 15 OL BL

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on

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NAME DATE PERIOD

Graphic AdditionOne way to sketch the graphs of some polynomialfunctions is to use addition of ordinates. This method is useful when a polynomial function f (x) can be written as the sum of two other functions, g(x) and h(x), that are easier to graph. Then, each f (x) can be found by mentally adding the corresponding g(x) and h(x). Thegraph at the right shows how to construct the graph off (x) = -

1−

2x3

+

1−

2x2

- 8 from the graphs of g(x) = -

1−

2x3

and h(x) = 1−

2x2

- 8.

In each problem, the graphs of g(x(( ) and h(x(( ) are shown. Use addition of ordinates to graph a new polynomial function f(ff x(( ), such thatf(ff x(( ) = g(x(( ) + h(x(( ). Then write the equation for f(ff x(( ).

1.

xx

y

-88

-1616

16

--22--44--66 22 44 66-4 2 4

gg (((xx))) = -= - xx33xx11122

hh (((xx))) == - xx22xxxx +++ 881222

ff (((xx )))

2.

EnrichmentEnrichmentEnrichmentEnrichmentE i h2 22-22 22 2

f (x)x

f (x)x = h (x)x + g (x)x

g(x )

h(x )

x

xx

y

ggg (((xx ))) = xx33111222

ff (((xx )))

hhh (((xx ))) == xx22 -- 121212112

-888

-16166

888

1616

--22--44--66 2 22 44 66--4 4

Create a function with the following characteristics. Graph the function.

76. degree = 5, 3 real zeros, lim x→∞

= ∞

77. degree = 6, 4 real zeros, lim x→∞

= -∞

78. degree = 5, 2 distinct real zeros, 1 of which has a multiplicity of 2, lim x→∞

= ∞

79. degree = 6, 3 distinct real zeros, 1 of which has a multiplicity of 2, lim x→∞

= -∞

80. WEATHER The temperatures in degrees Celsius from 10 a.m. to 7 p.m. during one day for a city are shown where x is the number of hours since 10 a.m.

Time Temp. Time Temp.

0 4.1 5 10

1 5.7 6 7

2 7.2 7 4.6

3 7.3 8 2.3

4 9.4 9 -0.4

a. Graph the data.

b. Use a graphing calculator to model the data using a polynomial function with a degree of 3.

c. Repeat part b using a function with a degree of 4.

d. Which function is a better model? Explain.

For each of the following graphs:

a. Determine the degree and end behavior.

b. Locate the zeros and their multiplicity. Assume all of the zeros are integral values.

c. Use the given point to determine a function that fits the graph.

81. 82.

x

y

60

84

−60

−120

−8 −4

120

(2, −128)

x

y

40

84

−40

−80

−8 −4

80(−2, 64)

83. 84.

x

y

4

−8

−4

−8 −4

8

4

(−3, −9)

8

x

y

4

−8

−4

−8 −4

8

4(0, 3.6)

8

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

85. f (x) = 16 x 4 + 72 x 2 + 80 86. f (x) = -12 x 3 - 44 x 2 - 40x

87 f (x) = -24 x 4 + 24 x 3 - 6 x 2 88. f (x) = x 3 + 6 x 2 - 4x - 24

89. MULTIPLE REPRESENTATIONS In this problem, you will investigate the behavior of combinations of polynomial functions.

a. GRAPHICAL Graph f (x), g(x), and h(x) in each row on the same graphing calculator screen. For each graph, modify the window to observe the behavior both on a large scale and very close to the origin.

f (x ) = g (x ) = h (x ) =

x 2 + x x 2 x

x 3 - x x 3 -x

x 3 + x 2 x 3 x 2

b. ANALYTICAL Describe the behavior of each graph of f (x) in terms of g(x) or h(x) near the origin.

c. ANALYTICAL Describe the behavior of each graph of f (x) in terms of g(x) or h(x) as x approaches ∞ and -∞.

d. VERBAL Predict the behavior of a function that is a combination of two functions a and b such that f (x) = a + b, where a is the term of higher degree.


90. ERROR ANALYSIS Colleen and Martin are modeling the data shown. Colleen thinks the model should be f (x) = 5.754 x 3 + 2.912 x 2 - 7.516x + 0.349. Martin thinks it should be f (x) = 3.697 x 2 + 11.734x - 2.476. Is either of them correct? Explain your reasoning.

x f (x ) x f(x)

-2 -19 0.5 -2

-1 5 1 1.5

0 0.4 2 43

91. REASONING Can a polynomial function have both an absolute maximum and an absolute minimum? Explain your reasoning.

92. REASONING Explain why the constant function f (x) = c, c ≠ 0, has degree 0, but the zero function f (x) = 0 has no degree.

93. CHALLENGE Use factoring by grouping to determine the zeros of f (x) = x 3 + 5 x 2 - x 2 - 5x - 12x - 60. Explain each step.

94. REASONING How is it possible for more than one function to be represented by the same degree, end behavior, and distinct real zeros? Provide an example to explain your reasoning.

95. REASONING What is the minimum degree of a polynomial function that has an absolute maximum, a relative maximum, and a relative minimum? Explain your reasoning.

96. WRITING IN MATH Explain how you determine the best polynomial function to use when modeling data.

76–79. See Chapter 2

Answer Appendix.

a–d. See Chapter 2 Answer Appendix.



a–d. See

Chapter 2

Answer

Appendix.








C


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Watch Out!

Error Analysis For Exercise 90, students should see that Colleen’s model has an odd degree, while Martin’s has an even degree.

4 AssessName the Math Ask students to describe the graph of a polynomial function. Sample answer: a smooth curve that is defined and continuous for all real numbers

Additional Answers

82c. Sample answer: f (x ) = -0.25(x - 6 ) 2 (x - 2)(x + 3)


= ∞ and lim x→∞

= ∞

83b. -4, 3 (multiplicity: 2), -1

83c. Sample answer: f (x ) = 1 _ 8 (x - 3) 2

(x + 1)(x + 4)


= -∞ and

lim x→∞

= ∞

84b. -3 (multiplicity: 2), -1, 2 (multiplicity: 2)

84c. Sample answer: f (x ) = 1 _ 10

(x - 2 ) 2 (x + 1)(x + 3 ) 2

85. 4 real zeros and 3 turning points; no real zeros

86. 3 real zeros and 2 turning points; 0,

-2, and -

5 _ 3

87. 4 real zeros and 3 turning points; 0, and 1 _

2

88. 3 real zeros and 2 turning points; -6, -2, and 2Differentiated Instruction BL

Extension Pose the following question to students. Is it always appropriate to use a polynomial model for a scatter plot in order to predict beyond the range of the data given? Ask students to give examples to support their responses. No, because realistic conditions may not exist beyond the data given; Sample answer: a scatter plot of sports records may show that the record time for completing a 100-meter race has decreased over the years. The model for the scatter plot will decrease indefinitely, becoming zero, then negative. However, the race will never be run in zero or negative time.


Spiral Review

Solve each equation. (Lesson 2-1)

97. √ �� z + 3 = 7 98. d + √ �� d 2 - 8 = 4 99. √ �� x - 8 = √ �� 13 + x

100. REMODELING An installer is replacing the carpet in a 12-foot by 15-foot living room. The new carpet costs $13.99 per square yard. The formula f (x) = 9x converts square yards to square feet. (Lesson 1-7)

a. Find the inverse f -1 (x). What is the significance of f -1 (x)?

b. How much will the new carpet cost?

Given f (x) = 2 x 2 - 5x + 3 and g (x) = 6x + 4, find each function. (Lesson 1-6)

101. ( f + g)(x) 102. [ f ◦ g](x) 103. [ g ◦ f ](x)

Describe how the graphs of f (x) = x 2 and g (x) are related. Then write an equation for g (x). (Lesson 1-5)

104. 105. 106.

x−4

4

8

4 8 12

y

12g(x)

yx

−8 −4

−12

−8

−4

4 8

g(x)

y

x−4

−8

−4

4

8

12

g(x)

107. BUSINESS A company creates a new product that costs $25 per item to produce. They hire a marketing analyst to help determine a selling price. After collecting and analyzing data relating selling price s to yearly consumer demand d, the analyst estimates demand for the product using d = -200s + 15,000. (Lesson 1-4)

a. If yearly profit is the difference between total revenue and production costs, determine a selling price s ≥ 25, that will maximize the company’s yearly profit P. (Hint: P = sd - 25d)

b. What are the risks of determining a selling price using this method?

46 3 no solution

f -1 (x) = x _ 9 ; it will allow the installer to convert the

square footage of the living room to square yards, then

the cost of the new carpet can be calculated.

$279.80

2 x 2 + x + 7 72 x 2 + 66x + 15 12 x 2 - 30x + 22

104–106. See Chapter

2 Answer

Appendix.

$50 Sample answer: The company’s competition

might offer a similar product at a lower cost.

108. SAT/ACT The figure shows theintersection of three lines. Thefigure is not drawn to scale.

x = 3x°

2x°

x°A 16 D 60

B 20 E 90

C 30

109. Over the domain 2 < x ≤ 3, which of the following functions contains the greatest values of y?

F y = x + 3 _ x - 2

H y = x 2 - 3

G y = x - 5 _ x + 1

J y = 2x

110. MULTIPLE CHOICE Which of the following equations represents the result of shifting the parent function y = x 3 up 4 units and right 5 units?

A y + 4 = (x + 5) 3 C y + 4 = (x - 5) 3

B y - 4 = (x + 5) 3 D y - 4 = (x - 5) 3

111. REVIEW Which of the following describes the numbers

in the domain of h(x) = √ �� 2x - 3

_ x - 5

?

F x ≠ 5 H x ≥ 3 _ 2 , x ≠ 5

G x ≥ 3 _ 2 J x ≠ 3 _

2

C

F

D

H



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Extend 2-2

ObjectiveUse TI-Nspire technology to explore the hidden behavior of graphs.

Using graphing technologies such as computers and calculators is an efficient way to be able to graph

and evaluate functions. It is important, however, to consider the limitations of graphing technology

when interpreting graphs.

Activity 1 Hidden Behavior of Graphs

Determine the zeros of f (x) = x 3 - x 2 - 60.7x + 204 graphically.

Step 1 Open a new Graphs and Geometry page, and graph the function.

In the default window, it appears that the function has two zeroes, one between -10 and -8 and one between 4 and 6.

Step 2 From the Window menu, choose Window Settings. Change the dimensions of the window as shown.

The behavior of the graph is much clearer in the larger window. It still appears that the function has two zeros, one between -8 and -10 and one between 4 and 6.

Step 3 From the Window menu, choose Window Settings. Change the window to [2, 8] by [-2, 2].

By enlarging the graph in the area where it appears that the zero occurs, it is clear that there is no zero between the values of 4 and 6. Therefore, the graph only has one zero.

Analyze the Results 1. In addition to the limitation discovered in the previous steps, how can graphing calculators limit

your ability to interpret graphs?

2. What are some ways to avoid these limitations?

ExercisesDetermine the zeros of each polynomial graphically. Watch for hidden behavior.

3. x 3 + 6.5 x 2 - 46.5x + 60 4. x 4 - 3 x 3 + 12 x 2 + 6x - 7

5. x 5 + 7 x 3 + 4 x 2 - x + 10.9 6. x 4 - 19 x 3 + 107.2 x 2 - 162x + 73

1. Sample answer: The

window settings may

not allow for the

complete viewing of

the graph. It may also

skew the graph

depending on the

intervals and scales

of the axes.

2. Sample answer: The

zoom tools and

window settings can

be used to display the

necessary intervals.

-11.2, 2.2, 2.5 -0.89, 0.58

-1.3 8.1, 8.9

Graphing Technology LabGraphing Technology Lab

Hidden Behavior of Graphs Hidden Behavior of Graphs

Study Tip Window Settings You can choose values for the window based on observation of your graph, or you can use one of the zoom tools such as the box zoom that allows you to zoom in on a certain area of a graph.

108 | Lesson 2-2

0108_AM_S_C02_EXT_664291.indd 108 4/30/12 6:11 PM

108 | Extend 2-2 | Graphing Technology Lab: Hidden Behavior of Graphs

1 FocusObjective Use TI-Nspire technology to explore the hidden behavior of graphs.

Teaching TipThe graphing calculator opens on the same screen as when it was turned off. Have students press c to begin the lab.

2 TeachWorking in Cooperative GroupsPair students with different abilities. Have students work through Activity Steps 1–3.

Ask:■ How many possible real zeros can a

cubic function have? 1, 2, or 3

■ If this function has only one real zero, what type of zeros are the other two? imaginary zeros

■ How many possible real zeros can a quartic function have? 0, 1, 2, 3, or 4

Practice Have students work through Exercises 3–6.

3 AssessFormative AssessmentUse Exercise 6 to assess whether students can use TI-Nspire technology to graph functions and find zeros.

From Concrete to AbstractAsk students to describe when they are certain that a value is a zero of a graphed function.

Extending the ConceptAsk students how they could keep the graph about the same in Step 1 but have either two or three zeros in Step 3.

108_AM_T_C02_EXT_664293.indd 108108_AM_T_C02_EXT_664293.indd 108 7/26/12 11:58 AM7/26/12 11:58 AM

Lesson 2-3


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Before Lesson 2-3 Factor quadratic expressions to solve equations.

Lesson 2-3 Divide polynomials using long division and synthetic division. Use the Remainder and Factor Theorems.

After Lesson 2-3 Find all zeros of a polynomial.




Ask:■ How could you find a polynomial

function to represent the average height of redwood trees as they age? Create a scatter plot of the data on a graphing calculator, count the number of turns that appear, or use a graphing calculator to get a regression function for the graph.




Teacher Edition ■ Differentiated Instruction ■ Differentiated Instruction ■ Differentiated Instruction

Chapter

Resource

Masters


■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Enrichment







54p8.625

1Divide Polynomials Consider the polynomial function f (x) = 6 x 3 - 25 x 2 + 18x + 9. If you know that f has a zero at x = 3, then you also know that (x - 3) is a factor of f (x). Because f(x)

is a third-degree polynomial, you know that there exists a second-degree polynomial q(x) such that

f (x) = (x - 3) · q (x).

This implies that q(x) can be found by dividing 6 x 3 - 25 x 2 + 18x + 9 by (x - 3) because

q (x) = f (x)

_ x - 3

, if x ≠ 3.

To divide polynomials, we can use an algorithm similar to that of long division with integers.

Example 1 Use Long Division to Factor Polynomials

Factor 6 x 3 - 25 x 2 + 18x + 9 completely using long division if (x - 3) is a factor.

6 x 2 - 7x - 3

x - 3 � �� 6 x 3 - 25 x 2 + 18x + 9

(-) __________ 6 x 3 - 18 x 2 Multiply divisor by 6 x 2 because 6 x 3

_ x = 6 x 2 .

-7 x 2 + 18x Subtract and bring down next term.

(-) ___________ -7 x 2 + 21x Multiply divisor by -7x because -

7 x 2

_ x = -7x.

-3x + 9 Subtract and bring down next term.

(-) ________ -3x + 9 Multiply divisor by -3 because -3x _ x = -3.

0 Subtract. Notice that the remainder is 0.

From this division, you can write 6 x 3 - 25 x 2 + 18x + 9 = (x - 3)(6 x 2 - 7x - 3).

Factoring the quadratic expression yields 6 x 3 - 25 x 2 + 18x + 9 = (x - 3)(2x - 3)(3x + 1).

So, the zeros of the polynomial function y

x

−4

−8

−2−4

8

12

4

2 4

6 3 - 25 2 + 18 + 9f (x) =

f (x) = 6 x 3 - 25 x 2 + 18x + 9 are 3, 3 _ 2 , and

- 1 _ 3 . The x-intercepts of the graph of f (x)

shown support this conclusion.


Factor each polynomial completely using the given factor and long division.

1A. x 3 + 7 x 2 + 4x - 12; x + 6

1B. 6 x 3 - 2 x 2 - 16x - 8; 2x - 4

(x + 2)(x - 1)(x + 6)

(x + 1)(2x - 4)(3x + 2)

Why?The redwood trees of Redwood National Park in California are the oldest living species in the world. The trees can grow up to 350 feet and can live up to 2000 years. Synthetic division can be used to determine the height of one of the trees during a particular year.

New Vocabularysynthetic division

depressed polynomial

synthetic substitution

Now

1 Divide polynomials using long division and synthetic division.

2Use the Remainder and Factor Theorems.

ThenYou factored quadratic expressions to solve equations. (Lesson 0–3)

The Remainder and Factor TheoremsThe Remainder and Factor Theorems


Dig

ital V

isio

n/Pu

nchS

tock

0109_0117_AM_S_C02_L03_664291.indd 109 5/31/12 10:27 PM


■ Suppose the polynomial function that represents the average height of redwood trees as they age is f (x ) = -0.001 x 2 + 1.15x + 8. How could you estimate the height of a redwood tree that is 1000 years old? Substitute 1000 for x and evaluate.

■ How many possible real zeros does this function have? Explain how you know. 2; The degree of the polynomial is 2.

1 Divide PolynomialsExamples 1–4 show how to divide polynomials using long division and synthetic division. This includes the need for the polynomials to be in standard form and to use zero coefficients for missing powers as place holders.


Additional Examples

Factor 6 x 3 + 17 x 2 - 104x + 60 completely using long division if (2x - 5) is a factor. (2x - 5)(3x - 2)(x + 6)

Divide 6 x 3 - 5 x 2 + 9x + 6 by 2x - 1. 3 x 2 - x + 4 + 10 _

2x - 1 ,

x ≠ 1 _ 2


1

2

Differentiated Instruction AL OL

Logical Learners To help students understand polynomial division, show some examples of division from arithmetic, such as 235 ÷ 22 = 10 R15 and 235 = 22 × 10 + 15. Work through a few examples, first with numbers and then with polynomials, to make the concepts clearer.

GuidedPractice

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Study TipGraphical Check You can also check the result in Example 2 using a graphing calculator. The graphs of Y 1 = 9 x 3 - x - 3 and Y 2 = (3 x 2 - 2x + 1) ·(3x + 2) - 5 are identical.

[-5, 5] scl: 1 by [-8, 2] scl: 1

Long division of polynomials can result in a zero remainder, as in Example 1, or a nonzero remainder, as in the example below. Notice that just as with integer long division, the result of polynomial division is expressed using the quotient, remainder, and divisor.

x + 3

x + 2 � �� x 2 + 5x - 4

(-) _______ x 2 + 2x 3x - 4

__________ (-) 3x + 6

-10

x 2 + 5x - 4 _ x + 2

= x + 3 + -10 _

x + 2 , x ≠ -2

Quotient

Divisor

Remainder

Excluded value

Divisor

Remainder

Quotient

Dividend

Dividend

Divisor

Recall that a dividend can be expressed in terms of the divisor, quotient, and remainder.

divisor · quotient + remainder = dividend

(x + 2) · (x + 3) + (-10) = x 2 + 5x - 4

This leads to a definition for polynomial division.

Let f (x ) and d (x ) be polynomials such that the degree of d (x ) is less than or equal to the degree of f (x ) and d (x ) ≠ 0. Then there exist unique polynomials q (x) and r (x) such that

f (x )

_ d (x )

= q(x ) + r (x )

_ d (x )

or f (x ) = d (x ) · q (x ) + r (x ),

where r (x ) = 0 or the degree of r (x ) is less than the degree of d (x ). If r (x ) = 0, then d(x ) divides evenly into f (x ).

Key Concept Polynomial Division

Before dividing, be sure that each polynomial is written in standard form and that placeholders with zero coefficients are inserted where needed for missing powers of the variable.

Example 2 Long Division with Nonzero Remainder

Divide 9 x 3 - x - 3 by 3x + 2.

First rewrite 9 x 3 - x - 3 as 9 x 2 + 0 x 2 - x - 3. Then divide.

3 x 2 - 2x + 1

3x + 2 � �� 9 x 3 + 0 x 2 - x - 3

(-) _________ 9 x 3 + 6 x 2

-6 x 2 - x

(-) __________ -6 x 2 - 4x

3x - 3

(-) ______ 3x + 2

-5

You can write this result as

9 x 3 - x - 3 _ 3x + 2

= 3 x 2 - 2x + 1 + -5 _ 3x + 2

, x ≠ -

2 _ 3

= 3 x 2 - 2x + 1 - 5 _ 3x + 2

, x ≠ -

2 _ 3 .

CHECK Multiply to check this result.

(3x + 2)(3 x 2 - 2x + 1) + (-5) � 9 x 3 - x - 3

9 x 3 - 6 x 2 + 3x + 6 x 2 - 4x + 2 - 5 � 9 x 3 - x - 3

9 x 3 - x - 3 = 9 x 3 - x - 3 �


Divide using long division.

2A. (8 x 3 - 18 x 2 + 21x - 20) ÷ (2x - 3) 2B. (-3 x 3 + x 2 + 4x - 66) ÷ (x - 5)

2A. 4 x 2 - 3x + 6 - 2

_ 2x - 3

, x ≠ 3

_ 2

-3 x 2 - 14x - 66 - 396

_ x - 5

, x ≠ 5

When dividing polynomials, the divisor can have a degree higher than 1. This can sometimes result in a quotient with missing terms.

Study TipProper vs. Improper A rational expression is considered improper if the degree of the numerator is greater than or equal to the degree of the denominator. So in

the division algorithm, f (x )

_ d (x )

is an

improper rational expression,

while r (x )

_ d (x )

is a proper rational

expression.

110 | Lesson 2-3 | The Remainder and Factor Theorems

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Additional Example

Divide x 3 - x 2 - 14x + 4 by x 2 - 5x + 6.

x + 4 - 20 __ x 2 - 5x + 6

3

Focus on Mathematical ContentSynthetic Division Synthetic Division is a method of dividing a polynomial by a factor in the form (x - c ). For synthetic division to work, the polynomial must be written in standard form with zeros as placeholders for any missing powers. The divisor must be in the form (x - c). For polynomials being divided by divisors that have a degree 2 or higher, long division is used.

Example 3 Division by Polynomial of Degree 2 or Higher

Divide 2 x 4 - 4 x 3 + 13 x 2 + 3x - 11 by x 2 - 2x + 7.

2 x 2 -1

x 2 - 2x + 7 � �� 2 x 4 - 4 x 3 + 13 x 2 + 3x - 11

(-) _______________ 2 x 4 - 4 x 3 + 14 x 2

- x 2 + 3x - 11

(-) ____________ - x 2 + 2x - 7

x - 4

You can write this result as 2 x 4 - 4 x 3 + 13 x 2 + 3x - 11 ___ x 2 - 2x + 7

= 2 x 2 - 1 + x - 4 _ x 2 - 2x + 7

.

Guided GuidedPracticePractice


3A. (2 x 3 + 5 x 2 - 7x + 6) ÷ ( x 2 + 3x - 4) 3B. (6 x 5 - x 4 + 12 x 2 + 15x) ÷ (3 x 3 - 2 x 2 + x)

3A. 2x - 1 + 4x + 2

_ x 2 + 3x - 4

3B. 2 x 2 + x + 11 x 2 + 15x

__ 3 x 3 - 2 x 2 + x

Synthetic division is a shortcut for dividing a polynomial by a linear factor of the form x - c. Consider the long division from Example 1.

We can use the synthetic division shown in the example above to outline a procedure for synthetic division of any polynomial by a binomial.

To divide a polynomial by the factor x - c, complete each step.

Step 1 Write the coefficients of the dividend in standard form. Write the related zero c of the divisor x - c in the box. Bring down the first coefficient.

Step 2 Multiply the first coefficient by c. Write the product under the second coefficient.

Step 3 Add the product and the second coefficient.

Step 4 Repeat Steps 2 and 3 until you reach a sum in the last column. The numbers along the bottom row are the coefficients of the quotient. The power of the first term is one less than the degree of the dividend. The final number is the remainder.

Example

Divide 6 x 3 - 25 x 2 + 18x + 9 by x - 3.

3 6 -25 18 9

18 -21 -9

6 -7 -3 0

coefficients

of quotient

remainder

= Add terms. = Multiply by c, and write

the product.

Key Concept Synthetic Division Algorithm

Long Division Suppress Variables Collapse Vertically Synthetic Division

Notice the coefficients highlighted in colored text.

Suppress x and powers of x. Collapse the long division vertically, eliminating duplications.

Change the signs of the divisor and the numbers on the second line.

6 x 2 - 7x - 3

x - 3 � �� 6 x 3 - 25 x 2 + 18x + 9

(-) __________ 6 x 3 - 18 x 2

-7 x 2 + 18x

(-) ___________ -7 x 2 + 21x

-3x + 9 (-) ________ -3x + 9

0

6 - 7 - 3

- 3 � �� 6 - 25 + 18 + 9

(-) 6 - 18

-7 + 18

(-) -7 + 21

-3 + 9

(-) -3 + 9

0

-3 6 -25 18 9

-18 21 9

6 -7 -3 0

3 6 -25 18 9

18 -21 -9

6 -7 -3 0

The number now representing the divisor is the related zero of the binomial x - c. Also, by changing the signs on the second line, we are now adding instead of subtracting.

Study TipDivision by Zero In Example 3, this division is not defined for x 2 - 2 x + 7 = 0. From this point forward in this lesson, you can assume that x cannot take on values for which the indicated division is undefined.


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Tips for New TeachersZero Coefficients In Examples 2 and 4, emphasize the importance of writing each polynomial in standard form. Have students leave space in a long division problem or insert a zero in a synthetic division problem if any powers of x in the dividend have coefficients of zero. Have each student do a few synthetic division problems independently when the concept is first presented so that the necessary steps will be learned.

Additional Example

Divide using synthetic division.

a. (2 x 5 - 4 x 4 - 3 x 3 - 6 x 2 - 5x - 8) ÷ (x - 3) 2 x 4 + 2 x 3 + 3 x 2 + 3x + 4

+ 4 _ x - 3

b. (8 x 4 + 38 x 3 + 5 x 2 + 3x + 3) ÷ (4x + 1) 2 x 3 + 9 x 2 - x + 1 + 2 _

4x + 1

4

Teach with TechWiki Have students create a wiki page explaining how to set up the synthetic division of a division problem involving polynomials. Be sure that they explain how they suppress the variables as well as how they change the signs of the divisor and the numbers on the second line.

2 The Remainder and Factor

TheoremsExample 5 shows how to use the Remainder Theorem. Example 6 shows how to use the Factor Theorem to determine if x - c is a factor of a polynomial f (x ).

GuidedPractice

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As with division of polynomials by long division, remember to use zeros as placeholders for any missing terms in the dividend. When a polynomial is divided by one of its binomial factors x - c, the quotient is called a depressed polynomial.

Example 4 Synthetic Division


a. (2 x 4 - 5 x 2 + 5x - 2) ÷ (x + 2)

Because x + 2 = x - (-2), c = -2. Set up the synthetic division as follows, using zero as a placeholder for the missing x 3 -term in the dividend. Then follow the synthetic division procedure.

-2 2 0 -5 5 -2 -2 2 0 -5 5 -2

-4 8 -6 2

2 -4 3 -1 0

coefficients of

depressed polynomial

remainder

= Add terms.

= Multiply by c, and

write the product.

The quotient has degree one less than that of the dividend, so

2 x 4 - 5 x 2 + 5x - 2 __ x + 2

= 2 x 3 - 4 x 2 + 3x - 1. Check this result.

b. (10 x 3 - 13 x 2 + 5x - 14) ÷ (2x - 3)

Rewrite the division expression so that the divisor is of the form x - c.

10 x 3 - 13 x 2 + 5x - 14 __ 2x - 3

= (10 x 3 - 13 x 2 + 5x - 14) ÷ 2

___ (2x - 3) ÷ 2

or 5 x 3 - 13 _

2 x 2 + 5 _

2 x - 7 __

x - 3 _ 2 .

So, c = 3 _ 2 . Perform the synthetic division.

3 _ 2 5 - 13 _

2 5 _

2 -7 3 _

2 5 - 13 _

2 5 _

2 -7

15 _ 2 3 _

2 6

5 1 4 -1

So, 10 x 3 - 13 x 2 + 5x - 14 __ 2x - 3

= 5 x 2 + x + 4 - 1 _ x - 3 _

2 or 5 x 2 + x + 4 - 2 _

2x - 3 . Check this result.


4A. (4 x 3 + 3 x 2 - x + 8) ÷ (x - 3) 4B. (6 x 4 + 11 x 3 - 15 x 2 - 12x + 7) ÷ (3x + 1)4 x 2 + 15x + 44 +

140

_ x - 3

2 x 3 + 3 x 2 - 6x - 2 +

9

_ 3x + 1

2The Remainder and Factor Theorems When d(x) is the divisor (x - c) with degree 1, the remainder is the real number r. So, the division algorithm simplifies to

f (x) = (x - c) · q(x) + r.

Evaluating f (x) for x = c, we find that

f (c) = (c - c) · q(c) + r = 0 · q(c) + r or r.

So, f (c) = r, which is the remainder. This leads us to the following theorem.

If a polynomial f (x ) is divided by x - c, the remainder is r = f (c ).

Key Concept Remainder Theorem

Technology TipUsing Graphs To check your division, you can graph the polynomial division expression and the depressed polynomial with the remainder. The graphs should coincide.


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Additional Examples

REAL ESTATE Suppose 800 units of beachfront property have tenants paying $600 per week. Research indicates that for each $10 decrease in rent, 15 more units would be rented. The weekly revenue from the rentals is given by R (x ) = -150 x 2 + 1000x + 480,000, where x is the number of $10 decreases the property manager is willing to take. Use the Remainder Theorem to find the revenue from the properties if the property manager decreases the rent by $50. $481,250

Use the Factor Theorem to determine if the binomials given are factors of f (x ). Use the binomials that are factors to write a factored form of f (x ).

a. f (x ) = x 3 - 18 x 2 + 60x + 25; (x - 5), (x + 5) f (x ) = (x - 5)( x 2 - 13x - 5)

b. f (x ) = x 3 - 2 x 2 - 13x - 10; (x - 5), (x + 2) f (x ) = (x - 5)(x + 2)(x + 1)

5

6

Tips for New TeachersRemainder Theorem Explain to students that synthetic substitution is performed exactly like synthetic division. However, the last line of numbers is interpreted differently. Synthetic division uses the entire last line of numbers for the quotient and remainder. For synthetic substitution, only the last number in the last line is of interest, since that is the number that gives the value of the function for x = c.

The Remainder Theorem indicates that to evaluate a polynomial function f(x) for x = c, you can divide f (x) by x - c using synthetic division. The remainder will be f (c). Using synthetic division to evaluate a function is called synthetic substitution.

FOOTBALL The number of tickets sold during the Northside High School football season can be modeled by t (x) = x 3 - 12 x 2 + 48x + 74, where x is the number of games played. Use the Remainder Theorem to find the number of tickets sold during the twelfth game of the Northside High School football season.

To find the number of tickets sold during the twelfth game, use synthetic substitution to evaluate t (x) for x = 12.

12 1 -12 48 74

12 0 576

1 0 48 650

The remainder is 650, so t(12) = 650. Therefore, 650 tickets were sold during the twelfth game of the season.

CHECK You can check your answer using direct substitution.

t(x) = x 3 - 12 x 2 + 48x + 74 Original function

t(12) = (12 ) 3 - 12(12 ) 2 + 48(12) + 74 or 650 � Substitute 12 for x and simplify.


5. FOOTBALL Use the Remainder Theorem to determine the number of tickets sold during the thirteenth game of the season. 867

Real-World Example 5 Use the Remainder Theorem

If you use the Remainder Theorem to evaluate f (x) at x = c and the result is f (c) = 0, then you know that c is a zero of the function and (x - c) is a factor. This leads us to another useful theorem that provides a test to determine whether (x - c) is a factor of f (x).

A polynomial f (x ) has a factor (x - c ) if and only if f (c ) = 0.

Key Concept Factor Theorem

You can use synthetic division to perform this test.

Example 6 Use the Factor Theorem

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

a. f (x) = 4 x 4 + 21 x 3 + 25 x 2 - 5x + 3; (x - 1), (x + 3)

Use synthetic division to test each factor, (x - 1) and (x + 3).

1 4 21 25 -5 3

4 25 50 45

4 25 50 45 48

Because the remainder when f (x) is divided by (x - 1) is 48, f (1) = 48 and (x - 1) is not a factor.

-3 4 21 25 -5 3

-12 -27 6 -3

4 9 -2 1 0

Because the remainder when f (x) is divided by (x + 3) is 0, f (-3) = 0 and (x + 3) is a factor.

Because (x + 3) is a factor of f (x), we can use the quotient of f (x) ÷ (x + 3) to write a factored form of f (x).

f (x) = (x + 3)(4 x 3 + 9 x 2 - 2x + 1)

Real-World LinkHigh school football rules are similar to most college and professional football rules. Two major differences are that the quarters are 12 minutes as opposed to 15 minutes and kick-offs take place at the 40-yard line instead of the 30-yard line .

Source: National Federation of State High School Associations


Chu

ck E

cker

t/A

lam

y

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14. 36 x 4 - 36 x 3 + 24 x 2 + 9x + 6 + 12 _ 3x + 2

15. x + 4

16. 2 x 2 - 8x + 9 + 5x + 24 _

2 x 2 + x − 12

17. 2 x 2 − 4x + 2 + 2 _ 3 x 3 + 2x + 3

18. 4 x 2 - x − 3 − x + 10 __

3 x 3 + 2 x 2 - x + 6

19. x 3 + x 2 + 5x + 4 + 2 _ x − 2

20. 2 x 3 - 2 x 2 + 4x - 4 + 8 _ x + 3

21. 3 x 3 + 3 x 2 + 12x + 24 + 48 _ x - 4

22. x 4 - 2 x 3 + x 2 + 4x + 1 + 4 _ x + 2

Additional Answers

9. 5 x 3 + 17 x 2 + 74x + 295 + 1192 _ x − 4

10. x 5 - 4 x 4 + 9 x 3 - 19 x 2 + 41x -

83 + 190 _ x + 2

11. 2 x 3 - 8 x 2 + 22x - 47 + 100 _ x + 2

12. 2 x 3 - x 2 - 41x - 20

13. 3 x 5 + 3 x 3 - 6 x 2 - 2x + 4 -

2 _ 2x - 1

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CHECK If (x + 3) is a factor of f (x) = 4 x 4 + 21 x 3 +

[-10, 10] scl: 1 by [-10, 30] scl: 2

25 x 2 - 5x + 3, then -3 is a zero of the function and (-3, 0) is an x-intercept of the graph. Graph f (x) using a graphing calculator and confirm that (-3, 0) is a point on the graph. �

b. f (x) = 2 x 3 - x 2 - 41x - 20; (x + 4), (x - 5)

Use synthetic division to test the factor (x + 4).

-4 2 -1 -41 -20

-8 36 20

2 -9 -5 0

Because the remainder when f(x) is divided by (x + 4) is 0, f (-4) = 0 and (x + 4) is a factor of f (x).

Next, test the second factor, (x - 5), with the depressed polynomial 2 x 2 - 9x - 5.

5 2 -9 -5

10 5

2 1 0

Because the remainder when the quotient of f(x) ÷ (x + 4) is divided by (x - 5) is 0, f (5) = 0 and (x - 5) is a factor of f (x).

Because (x + 4) and (x - 5) are factors of f(x), we can use the final quotient to write a factored form of f (x).

f (x) = (x + 4)(x - 5)(2x + 1)

CHECK The graph of f (x) = 2 x 3 - x 2 - 41x - 20 confirms that x = -4, x = 5, and x = -

1 _ 2 are

zeros of the function. �

[-10, 10] scl: 1 by [-120, 80] scl: 20


Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

6A. f (x) = 3 x 3 - x 2 - 22x + 24; (x - 2), (x + 5)

6B. f (x) = 4 x 3 - 34 x 2 + 54x + 36; (x - 6), (x - 3)

yes, no; f (x ) = (x - 2)(3x - 4)(x + 3)

yes, yes; f (x ) = 2(x - 6)(x - 3)(2x + 1)

You can see that synthetic division is a useful tool for factoring and finding the zeros of polynomial functions.

If r is the remainder obtained after a synthetic division of f (x ) by (x - c ), then the following statements are true.

• r is the value of f (c ).

• If r = 0, then (x - c ) is a factor of f (x ).

• If r = 0, then c is an x-intercept of the graph of f.

• If r = 0, then x = c is a solution of f (x ) = 0.

Concept Summary Synthetic Division and Remainders

Technology Tip Zeros You can confirm the zeros on the graph of a function by using the zero feature on the CALC menu of a graphing calculator.


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Watch Out!

Synthetic Division For Exercises 19–28, watch for students who might use the wrong sign for the number in the synthetic division box. Remind students that the divisor must be written in the form (x - c ). For example, a divisor such as (x + 3) must be written as (x - (-3)) to see that the number to use in the synthetic division box is -3 and not 3.

Additional Answers

23. 6 x 4 + 14 x 3 + 12 x 2 + 12x + 18

+ 46 _ 2x - 3

24. 12 x 3 - 6 x 2 + 6x - 12

25. 15 x 4 + 12 x 3 + 9 x 2 + 6x + 20 _ 3

+ 76 _ 3(3x - 2)

26. 12 x 4 + 4 x 3 + 16 x 2 - 4x + 15 _ 4

+ 9 _ 4(4x + 1)

27. 12 x 5 + 6 x 4 - 3 x 3 - 5

28. 8 x 5 - 12 x 3 + 6 x 2 - 15

38. yes, no; f (x ) = (x + 2)( x 3 - 4 x 2 - x + 3)

39. no, no

40. no, no

41. yes, yes; f (x ) = (3x - 1)(x - 5)(x - 4)(x + 2)

42. yes, no; f (x ) = (4x - 1)( x 3 - 9x - 30)

43. no, no

44. no, no

45. yes, no; f (x ) = (4x + 3)( x 4 - 3 x 3 + 12 x 2 - 3x + 21)



AL Approaching Level 1–47, 62, 63, 68–77 1–47 odd 2–46 even, 62, 63, 68–77

OL On Level 1–51 odd, 52, 53–59 odd, 60, 62, 63, 68–77

1–47 48–60, 62, 63, 68–77



Exercises

Factor each polynomial completely using the given factor and long division. (Example 1)

1. x 3 + 2 x 2 - 23x - 60; x + 4

2. x 3 + 2 x 2 - 21x + 18; x - 3

3. x 3 + 3 x 2 - 18x - 40; x - 4

4. 4 x 3 + 20 x 2 - 8x - 96; x + 3

5. -3 x 3 + 15 x 2 + 108x - 540; x - 6

6. 6 x 3 - 7 x 2 - 29x - 12; 3x + 4

7. x 4 + 12 x 3 + 38 x 2 + 12x - 63; x 2 + 6x + 9

8. x 4 - 3 x 3 - 36 x 2 + 68x + 240; x 2 - 4x - 12

Divide using long division. (Examples 2 and 3)

9. (5 x 4 - 3 x 3 + 6 x 2 - x + 12) ÷ (x - 4)

10. ( x 6 - 2 x 5 + x 4 - x 3 + 3 x 2 - x + 24) ÷ (x + 2)

11. (4 x 4 - 8 x 3 + 12 x 2 - 6x + 12) ÷ (2x + 4)

12. (2 x 4 - 7 x 3 - 38 x 2 + 103x + 60) ÷ (x - 3)

13. (6 x 6 - 3 x 5 + 6 x 4 - 15 x 3 + 2 x 2 + 10x - 6) ÷ (2x - 1)

14. (108 x 5 - 36 x 4 + 75 x 2 + 36x + 24) ÷ (3x + 2)

15. ( x 4 + x 3 + 6 x 2 + 18x - 216) ÷ ( x 3 - 3 x 2 + 18x - 54)

16. (4 x 4 - 14 x 3 - 14 x 2 + 110x - 84) ÷ (2 x 2 + x - 12)

17. 6 x 5 - 12 x 4 + 10 x 3 - 2 x 2 - 8x + 8 ___ 3 x 3 + 2x + 3

18. 12 x 5 + 5 x 4 - 15 x 3 + 19 x 2 - 4x - 28 ___ 3 x 3 + 2 x 2 - x + 6

Divide using synthetic division. (Example 4)

19. ( x 4 - x 3 + 3 x 2 - 6x - 6) ÷ (x - 2)

20. (2 x 4 + 4 x 3 - 2 x 2 + 8x - 4) ÷ (x + 3)

21 (3 x 4 - 9 x 3 - 24x - 48) ÷ (x - 4)

22. ( x 5 - 3 x 3 + 6 x 2 + 9x + 6) ÷ (x + 2)

23. (12 x 5 + 10 x 4 - 18 x 3 - 12 x 2 - 8) ÷ (2x - 3)

24. (36 x 4 - 6 x 3 + 12 x 2 - 30x - 12) ÷ (3x + 1)

25. (45 x 5 + 6 x 4 + 3 x 3 + 8x + 12) ÷ (3x - 2)

26. (48 x 5 + 28 x 4 + 68 x 3 + 11x + 6) ÷ (4x + 1)

27. (60 x 6 + 78 x 5 + 9 x 4 - 12 x 3 - 25x - 20) ÷ (5x + 4)

28. 16 x 6 - 56 x 5 - 24 x 4 + 96 x 3 - 42 x 2 - 30x + 105 ____ 2x - 7

29. EDUCATION The number of U.S. students, in thousands, that graduated with a bachelor’s degree from 1970 to 2006 can be modeled by g(x) = 0.0002 x 5 - 0.016 x 4 + 0.512 x 3 - 7.15 x 2 + 47.52x + 800.27, where x is the number of years since 1970. Use synthetic substitution to find the number of students that graduated in 2005. Round to the nearest thousand. (Example 5)

30. SKIING The distance in meters that a person travels on skis can be modeled by d(t) = 0.2 t 2 + 3t, where t is the time in seconds. Use the Remainder Theorem to find the distance traveled after 45 seconds. (Example 5)

Find each f (c) using synthetic substitution. (Example 5)

31. f (x) = 4 x 5 - 3 x 4 + x 3 - 6 x 2 + 8x - 15; c = 3

32. f (x) = 3 x 6 - 2 x 5 + 4 x 4 - 2 x 3 + 8x - 3; c = 4

33. f (x) = 2 x 6 + 5 x 5 - 3 x 4 + 6 x 3 - 9 x 2 + 3x - 4; c = 5

34. f (x) = 4 x 6 + 8 x 5 - 6 x 3 - 5 x 2 + 6x - 4; c = 6

35. f (x) = 10 x 5 + 6 x 4 - 8 x 3 + 7 x 2 - 3x + 8; c = -6

36. f (x) = -6 x 7 + 4 x 5 - 8 x 4 + 12 x 3 - 15 x 2 - 9x + 64; c = 2

37. f (x) = -2 x 8 + 6 x 5 - 4 x 4 + 12 x 3 - 6x + 24; c = 4

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x). (Example 6)

38. f (x) = x 4 - 2 x 3 - 9 x 2 + x + 6; (x + 2), (x - 1)

39. f (x) = x 4 + 2 x 3 - 5 x 2 + 8x + 12; (x - 1), (x + 3)

40. f (x) = x 4 - 2 x 3 + 24 x 2 + 18x + 135; (x - 5), (x + 5)

41. f (x) = 3 x 4 - 22 x 3 + 13 x 2 + 118x - 40; (3x - 1), (x - 5)

42. f (x) = 4 x 4 - x 3 - 36 x 2 - 111x + 30; (4x - 1), (x - 6)

43. f (x) = 3 x 4 - 35 x 3 + 38 x 2 + 56x + 64; (3x - 2), (x + 2)

44. f (x) = 5 x 5 + 38 x 4 - 68 x 2 + 59x + 30; (5x - 2), (x + 8)

45. f(x) = 4 x 5 - 9 x 4 + 39 x 3 + 24 x 2 + 75x + 63; (4x + 3), (x - 1)

46. TREES The height of a tree in feet at various ages in years is given in the table.

Age Height Age Height

2 3.3 24 73.8

6 13.8 26 82.0

10 23.0 28 91.9

14 42.7 30 101.7

20 60.7 36 111.5

a. Use a graphing calculator to write a quadratic equation to model the growth of the tree.

b. Use synthetic division to evaluate the height of the tree at 15 years.

47. BICYCLING Patrick is cycling at an initial speed v 0 of 4 meters per second. When he rides downhill, the bike accelerates at a rate a of 0.4 meter per second squared. The vertical distance from the top of the hill to the bottom

of the hill is 25 meters. Use d(t) = v 0 t + 1 _ 2 a t 2 to find how

long it will take Patrick to ride down the hill, where d(t) is distance traveled and t is given in seconds.

(x + 3)(x - 5)(x + 4)

(x - 1)(x + 6)(x - 3)

(x + 5)(x + 2)(x - 4)

4(x + 4)(x - 2)(x + 3)

-3(x - 5)(x + 6)(x - 6)(2x + 1)(x - 3)(3x + 4)

(x + 7)(x - 1)(x + 3 ) 2

(x + 5)(x - 4)(x + 2)(x - 6)

9–18. See margin.


2,151,000 students

540 m

711

11,165

45,536

247,388

-67,978

-686

-125,184


B

a. f (x ) = -0.001 x 2 +

3.44x - 6.39about 44.985 ft

5 seconds


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Additional Answer

52b.

Volume of a Box

Volu

me

( in3 )

0

60

120

180

v (x)

x

Height (in.)2 4 6

( ) = 3 3 - 47 2 + 180



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3 x 2 - 3x + 1 x + 1 � ��

3x 3 + 0x 2 - 2x - 4 Insert a placeholder for x 2.

(-) 3 x

3 + 3 x 2

- 3x 2 - 2x (-) - 3x 2 - 3x x - 4 (-) x + 1 -5

Study Guide and InterventionThe Remainder and Factor Theorems

Divide Polynomials You can divide polynomials by using either long or synthetic division. The process of dividing polynomials by using long division is similar to that in arithmetic, with the alternating of dividing, multiplying, and subtracting. To set up long division for polynomials, be sure the dividend is in standard form, with placeholders for missing terms. Synthetic division is a shortcut for long division. It only works when the divisor is in the form x - c.

a. Divide 3x3 - 2x - 4 by x + 1 using long division.

b. Divide 3x 3 - 2x - 4 by x + 1 using synthetic division.

Exercises


1. ( 2x 3 + 3x 2 - 8x + 3) ÷ (x - 1) 2. ( 3x 2 + 4x - 12) ÷ (x - 2)

2x 2 + 5x - 3 3x + 10 + 8 −

x - 2


3. ( x 3 - 28x - 48) ÷ (x + 4) 4. ( x 4 - 3x 2 + 12) ÷ (x - 1)

x 2 - 4x - 12 x 3 + x 2 - 2x - 2 + 10 −

x - 1

Example

Divide x into each term, starting on the left. Write the

quotient above the term with the same exponent.

Multiply the quotient by the divisor, and subtract the

polynomial from the dividend. Repeat.

The quotient is 3x 2 - 3x + 1 - 5 −

x + 1 .

Write the value of c in the box and the coefficients of the

dividend in the first row. Insert a placeholder for x2. Bring down

the first coefficient. Multiply it by c and write the product under

the second coefficient. Add the numbers in this column. Repeat.

The coefficients of the terms in the quotient appear in the

bottom row, where the powers of the variables decrease by one

and the first power is one less than the degree of the dividend.

The last number is the remainder.

2-3

-1 3 0 -2 -4 -3 3 -1

3 -3 1 -5

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1. ( 2x 4 + 14x 3 - 2x 2 - 14x) ÷ (x + 7) 2. ( 3t 3 - 10t 2 + t - 5) ÷ (t - 4)

2x 3 -2x 3t 2 + 2t + 9 + 31 −

t - 4


3. ( y 3 + y 2 - 10) ÷ (y + 3) 4. ( n 4 - n 3 - 10n 2 + 4n + 24) ÷ (n + 2)

y 2 - 2y + 6 - 28 −

y + 3 n 3 - 3n 2 - 4n + 12

5. ( x 4 - 3x 3 - 15x 2 + 19x + 30) ÷ (x - 5) 6. (x 3 - 8x 2 - 29x + 180) ÷ (x - 10)

x 3 + 2x 2 - 5x - 6 x 2 + 2x - 9 + 90 −

x - 10

Find each f(c) using synthetic substitution.

7. f(x) = x 3 + 6x 2 - 9x - 54 ; c = 3 8. f(x) = 3x 4 - 6x 2 - 30 ; c = 2

0 -6

9. f(x) = -x 4 + 2x 3 - x 2 + 7x + 5 ; c = -1 10. f(x) = x 5 + 6x 3 + 9x - 3 ; c = 4

-6 1441

Use the Factor Theorem to determine if the binomials given are factors of f(x). Use the binomials that are factors to write a factored form of f(x).

11. f(x) = x 3 - 7x − 6; (x + 2), (x - 1)

yes; no; f(x) = (x + 2) (x + 1) (x - 3)

12. f(x) = 2x3 + x2 − 50x − 25; (x + 5), (x - 5)

yes; yes; f(x) = (x + 5) (x - 5) (2x + 1)

13. f(x) = x4 - x3 - 7x2 + x + 6; (x + 2), (x + 4)

yes; no; f(x) = (x + 2) (x + 1) (x - 3) (x - 1)

14. f(x) = 3x 4 - 4x3 - 61x2 + 22x + 40; (3x + 2), (x - 1)

yes; yes; f(x) = (3x + 2) (x + 4) (x - 5) (x - 1)

15. ADVERTISING An advertising manager uses the function f(x) = 0.0003x 4 - 0.02x 3 - 3x 2 + 522x - 4600 to predict the profit a commercial will earn a company, based on the number of seconds the commercial lasts, x. Use synthetic substitution to find the profit of a 35-second commercial. Round to the nearest dollar.

$9588

PracticeThe Remainder and Factor Theorems

2-3

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1. ENROLLMENT According to the National Center for Education Statistics, the projected public school enrollment for grades 9 through 12, in thousands, from 2006–2016 can be approximated by f(x) = - 0.99x 4 + 46.44x 3 - 766.31x 2 +

5209.01x + 2555.91, where x is the number of years since 2000. Use synthetic substitution to find the projected enrollment in 2013.

14,519,940

2. STOCK The value of a share of stock, rounded to the nearest dollar, a given number of weeks after a business restructuring is shown in the table.

a. Use a graphing calculator to write a cubic function to model the data. Round each coefficient to the nearest thousandth.

f(x) = 0.017x 3 - 0.428x 2 +4.429x + 14.293

b. Use synthetic substitution to find the value of a share of stock 12 weeks after the restructuring.

$35.19

3. PHYSICS A rock is dropped from the top of a cliff that is 80 meters high. The height of the rock in t seconds is given by the formula h(t) = −4.9t2

+ 80. Show how to use synthetic substitution to find the height of the rock 0.5 second after it is dropped.

0.5 -4.9 0 80

-2.45 -1.225

-4.9 -2.45 78.775

The answer is the remainder: 78.775 m.

4. ART The size, in square feet, and cost of paintings in an art store are shown in the table.

a. Use a graphing calculator to write a quadratic function to model the data. Round each coefficient to the nearest thousandth.

f(x) = -49.356x 2 + 473.928x -456.931

b. Use synthetic substitution to predict the cost of a 2-foot by 3-foot painting.

$609.82

Word Problem PracticeThe Remainder and Factor Theorems

Week Value ($)

1 18

3 25

4 26

6 29

8 32

9 29

10 35

15 42

Size (ft2) Cost ($)

1.2 55

3.5 620

8 225

3 500

2.75 430

4.5 715

5 695

7.3 295

2-3

005_038_PCCRMC02_893803.indd 19 10/1/09 5:46:03 PM

Enrichment

p. 20 OL BL

NAME DATE PERIOD

EnrichmentEnrichmentEnrichmentEnrichmentE i h

The Bisection Method for Approximating Real ZerosThe bisection method can be used to approximate zeros of polynomial functions like f(ff x) = x 3

+ x 2- 3x - 3. Since f(1)ff = -4 and f(2)ff = 3, there

is at least one real zero between 1 and 2. The midpoint of this interval

is 1 + 2−

2= 1.5. Since f(1.5) ff = -1.875, the zero is between 1.5 and 2. The

midpoint of this interval is 1.5 + 2−

2= 1.75. Since f(1.75) ff = 0.172, the

zero is between 1.5 and 1.75. 1.5 + 1.75−

2 = 1.625 and f(1.625) ff = -0.94.

The zero is between 1.625 and 1.75. The midpoint of this interval is

1.625 + 1.75−

2 = 1.6875. Since f(1.6875)ff = -0.41, the zero is between 1.6875

and 1.75. Therefore, the zero is 1.7 to the nearest tenth. The diagram belowsummarizes the bisection method.

U i g thUsing th bi te bisecti thion meth dod, appr i toximate t thto the n t tearest t th thenth the

sign of f (x )

1 1 51 1.5

1 6251.6251.625

1 68751.6875

1 751.75

+ +----

22xx ::

2 32-32 32 3

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Factor each polynomial using the given factor and long division. Assume n > 0.

48. x 3n + x 2n - 14 x n - 24; x n + 2

49. x 3n + x 2n - 12 x n + 10; x n - 1

50. 4x 3n + 2 x 2n - 10 x n + 4; 2 x n + 4

51. 9 x 3n + 24 x 2n - 171 x n + 54; 3 x n - 1

52. MANUFACTURING An 18-inch by 20-inch sheet of cardboard is cut and folded into a bakery box.

18 in.

20 in.

x x

a. Write a polynomial function to model the volume of the box.


c. The company wants the box to have a volume of 196 cubic inches. Write an equation to model this situation.

d. Find a positive integer for x that satisfies the equation found in part c.

Find the value of k so that each remainder is zero.

53. x 3 - k x 2 + 2x - 4 __ x - 2

54. x 3 + 18 x 2 + kx + 4 __ x + 2

55. x 3 + 4 x 2 - kx + 1 __ x + 1

56. 2 x 3 - x 2 + x + k __ x - 1

57. SCULPTING Esteban will use a block of clay that is 3 feet by 4 feet by 5 feet to make a sculpture. He wants to reduce the volume of the clay by removing the same amount from the length, the width, and the height.

a. Write a polynomial function to model the situation.


c. He wants to reduce the volume of the clay to 3 _ 5 of

the original volume. Write an equation to model the situation.

d. How much should he take from each dimension?

Use the graphs and synthetic division to completely factor each polynomial.

58. f (x) = 8 x 4 + 26 x 3 - 103 x 2 - 156x + 45 (Figure 2.3.1)

59. f (x) = 6 x 5 + 13 x 4 - 153 x 3 + 54 x 2 + 724x - 840 (Figure 2.3.2)

y

x

−400

−800

800

400

−4−8 4 8

f (x)

y

x

−1600

4800

1600

3200

−4−8 4 8

f (x)

Figure 2.3.1 Figure 2.3.2

60. MULTIPLE REPRESENTATIONS In this problem, you will explore the upper and lower bounds of a function.

a. GRAPHICAL Graph each related polynomial function, and determine the greatest and least zeros. Then copy and complete the table.

PolynomialGreatest

Zero

Least

Zero

x 3 - 2 x 2 - 11x + 12

x 4 + 6 x 3 + 3 x 2 - 10x

x 5 - x 4 - 2 x 3

b. NUMERICAL Use synthetic division to evaluate each function in part a for three integer values greater than the greatest zero.

c. VERBAL Make a conjecture about the characteristics of the last row when synthetic division is used to evaluate a function for an integer greater than its greatest zero.

d. NUMERICAL Use synthetic division to evaluate each function in part a for three integer values less than the least zero.

e. VERBAL Make a conjecture about the characteristics of the last row when synthetic division is used to evaluate a function for a number less than its least zero.


61 CHALLENGE Is (x - 1) a factor of 18 x 165 - 15 x 135 + 8 x 105 - 15 x 55 + 4? Explain your reasoning.

62. WRITING IN MATH Explain how you can use a graphing calculator, synthetic division, and factoring to completely factor a fifth-degree polynomial with rational coefficients, three integral zeros, and two non-integral, rational zeros.

63. REASONING Determine whether the statement below is true or false. Explain.

If h(y) = (y + 2)(3 y 2 + 11y - 4) - 1, then the remainder

of h(y)

_ y + 2

is -1.

CHALLENGE Find k so that the quotient has a 0 remainder.

64. x 3 + kx 2 - 34x + 56 __ x + 7

65. x 6 + k x 4 - 8 x 3 + 173 x 2 - 16x - 120 ___ x - 1

66. k x 3 + 2 x 2 - 22x - 4 __ x - 2

67. CHALLENGE If 2 x 2 - dx + (31 - d 2 )x + 5 has a factor x - d, what is the value of d if d is an integer?

68. WRITING IN MATH Compare and contrast polynomial division using long division and using synthetic division.

( x n + 2)( x n - 4)( x n + 3)

( x 2n + 2 x n - 10)( x n - 1)

(2 x n - 1)(2 x n + 4)( x n - 1)

3( x n - 3)( x n + 6)(3 x n - 1)

v (x ) = 3 x 3 - 47 x 2 + 180x

See margin.

196 = 3 x 3 - 47 x 2 + 180x

2 in.

C

2 34

-4 -2

v (x ) = - x 3 + 12 x 2 - 47x + 60See margin.

36 = - x 3 + 12 x 2 - 47x + 60

about 0.60 ft

f (x ) = (x + 5)(x - 3)(4x - 1)(2x + 3)

f (x ) = (x + 6)(x - 2 ) 2 (3x - 7)(2x + 5)

a–d. See margin.

Sample answer: The elements in the last

row alternate between nonnegative and nonpositive.

See margin.


True; sample answer: The Remainder

Theorem states that if h (y ) is divided by

y - (-2), then the remainder is r = h (-2), which is -1.

1

-30

5

-5



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54p8.625

4 AssessTicket Out the Door Ask students to write the remainder of x 3 - x 2 -

5x - 3 when it is divided by x - 3. 0

Additional Answers

57b.

Volume of Clay

Volu

me

( ft3 )

0

20

40

60

Removed Amount (ft)1 2 3

= - 3 + 12 2 - 47 + 60( )

x

v(x)

60a.

PolynomialGreatest

ZeroLeast Zero

x 3 - 2 x 2 - 11x + 12 4 -3

x 4 + 6 x 3 + 3 x 2 - 10x 1 -5

x 5 - x 4 - 2 x 3 2 -1

60b. Sample answer: f (x ) = x 3 - 2 x 2 - 11x + 12: f (5) = 32, f (7) = 180, f (8) = 308; f (x ) = x 4 + 6 x 3 + 3 x 2 - 10x: f (2) = 56, f (3) = 240, f (4) = 648; f (x ) = x 5 - x 4 - 2 x 3 : f (3) = 108, f (5) = 2250, f (7) = 13,720

60c. Sample answer: All of the elements in the last row of the synthetic division are positive.

60d. Sample answer: f (x ) = x 3 - 2 x 2 - 11x + 12: f (-4) = -40, f (-5) = -108, f (-6) = -210; f (x ) = x 4 + 6 x 3 + 3 x 2 - 10x : f (-6) = 168, f (-7) = 560, f (-9) = 2520; f (x ) = x 5 - x 4 - 2 x 3 : f (-2) = -32, f (-3) = -270, f (-4) = -1152

61. Yes; sample answer: Let f (x ) be the related polynomial function. Using the Factor Theorem, since f (1) = 0, (x - 1) is a factor of the polynomial.


Extension Ask students to find the values of a, b, and c so that when x 6 - 2 x 4 + a x 2 + bx + c is divided by ( x - 1), (x - 2), and (x + 4), the remainder is 0. -125, 342, -216

DDiifffferentiiateddIInsttructtiion

Spiral Review

Determine whether the degree n of the polynomial for each graph is even or odd and whether its leading coefficient a n is positive or negative. (Lesson 2–2)

69. 70. 71. y

x

y

x

y

x

72. SKYDIVING The approximate time t in seconds that it takes an object to fall a distance of

d feet is given by t = √

�

d _ 16

. Suppose a skydiver falls 11 seconds before the parachute opens.

How far does the skydiver fall during this time period? (Lesson 2–1)

73. FIRE FIGHTING The velocity v and maximum height h of water being pumped into the air are related by v = √ �� 2gh , where g is the acceleration due to gravity (32 feet/secon d 2 ). (Lesson 1–7)

a. Determine an equation that will give the maximum height of the water as a function of its velocity.

b. The Mayfield Fire Department must purchase a pump that is powerful enough to propel water 80 feet into the air. Will a pump that is advertised to project water with a velocity of 75 feet/second meet the fire department’s needs? Explain.

even; negative odd; positive odd; negative

1936 ft

h = v 2

_ 64

Yes; the pump can propel water to a height of about 88 ft.

74. SAT/ACT In the figure, an equilateral triangle is drawn with an altitude that is also the diameter of the circle. If the perimeter of the triangle is 36, what is the circumference of the circle?

A 6 √ � 2 π C 12 √ � 2 π E 36π

B 6 √ � 3 π D 12 √ � 3 π

75. REVIEW If (3, -7) is the center of a circle and (8, 5) is on the circle, what is the circumference of the circle?

F 13π H 18π K 26π

G 15π J 25π

76. REVIEW The first term in a sequence is x. Each subsequent term is three less than twice the preceding term. What is the 5th term in the sequence?

A 8x - 21 C 16x - 39 E 32x - 43

B 8x - 15 D 16x - 45

77. Use the graph of the polynomial function. Which is not a factor of x 5 + x 4 - 3 x 3 - 3 x 2 - 4x - 4?

F (x - 2)

G (x + 2)

H (x - 1)

J (x + 1)

x

y

−12

−2−4

−4

−8

2 4

f (x) = x 5 + x 4 - 3x 3 - 3x 2 - 4x - 4

B

K

D

H



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Chapter 2 Mid-Chapter QuizChapter 2 Mid-Chapter Quiz

118 | Chapter 2 | Mid-Chapter Quiz

Graph and analyze each function. Describe its domain, range,

intercepts, end behavior, continuity, and where the function is

increasing or decreasing. (Lesson 2-1)

1. f (x ) = 2 x 3 2. f (x ) = -

2 _ 3 x 4

3. f (x ) = 3 x -8 4. f (x ) = 4 x 2 _ 5

5. TREES The heights of several fir trees and the areas under their branches are shown in the table. (Lesson 2-1)

Height (m) Area ( m 2 )

4.2 37.95

2.1 7.44

3.4 23.54

1.7 4.75

4.6 46.48



c. Predict the area under the branches of a fir tree that is 7.6 meters high.

Solve each equation. (Lesson 2-1)

6. √ �� 5x + 7 = 13

7. √ �� 2x - 2 + 1 = x

8. √ �� 3x + 10 + 1 = √ �� x + 11

9. -5 = 4 √ ��

(6x + 3 ) 3 - 32

State the number of possible real zeros and turning points of each

function. Then find all of the real zeros by factoring. (Lesson 2-2)

10. f (x ) = x 2 - 11x - 26

11. f (x ) = 3 x 5 + 2 x 4 - x 3

12. f (x ) = x 4 + 9 x 2 - 10

13. MULTIPLE CHOICE Which of the following describes the possible end behavior of a polynomial of odd degree? (Lesson 2-2)

A lim x→∞

f (x ) = 5; lim x→-∞

f (x ) = 5

B lim x→∞

f (x ) = -∞; lim x→-∞

f (x ) = -∞

C lim x→∞

f (x ) = ∞; lim x→-∞

f (x ) = ∞

D lim x→∞

f (x ) = -∞; lim x→-∞

f (x ) = ∞

Describe the end behavior of the graph of each polynomial function

using limits. Explain your reasoning using the leading term

test. (Lesson 2-2)

14. f (x ) = -7 x 4 - 3 x 3 - 8 x 2 + 23x + 7

15. f (x ) = -5 x 5 + 4 x 4 + 12 x 2 - 8

16. ENERGY Crystal’s electricity consumption measured in kilowatt hours (kWh) for the past 12 months is shown below. (Lesson 2-2)

Month Consumption

(kWh)

Month Consumption

(kWh)

January 240 July 300

February 135 August 335

March 98 September 390

April 110 October 345

May 160 November 230

June 230 December 100

a. Determine a model for the number of kilowatt hours Crystal used as a function of the number of months since January.

b. Use the model to predict how many kilowatt hours Crystal will use the following January. Does this answer make sense? Explain your reasoning.

Divide using synthetic division. (Lesson 2-3)

17. (5 x 3 - 7 x 2 + 8x - 13) ÷ (x - 1)

18. ( x 4 - x 3 - 9x + 18) ÷ (x - 2)

19. (2 x 3 - 11 x 2 + 9x - 6) ÷ (2x - 1)

Determine each f (c ) using synthetic substitution. (Lesson 2-3)

20. f (x ) = 9 x 5 + 4 x 4 - 3 x 3 + 18 x 2 - 16x + 8; c = 2

21. f (x ) = 6 x 6 - 3 x 5 + 8 x 4 + 12 x 2 - 6x + 4; c = -3

22. f (x ) = -2 x 6 + 8 x 5 - 12 x 4 + 9 x 3 - 8 x 2 + 6x - 3; c = -2

Use the Factor Theorem to determine if the binomials given are factors

of f (x ). Use the binomials that are factors to write a factored form of

f (x ). (Lesson 2-3)

23. f (x ) = x 3 + 2 x 2 - 25x - 50; (x + 5)

24. f (x ) = x 4 - 6 x 3 + 7 x 2 + 6x - 8; (x - 1), (x - 2)

25. MULTIPLE CHOICE Find the remainder when f (x ) = x 3 - 4x + 5 is divided by x + 3. (Lesson 2-3)

F -10 H 20

G 8 J 26

1–4. See

Chapter 2

Answer

Appendix.

See margin.y = 1.37 x 2.3

149.26 m 2

32.4

1, 3

-2

13

10. 2 real zeros and 1 turning

point; -2 and 13


points; -1, 0, and 1

_ 3


points; -1 and 1

D

See margin.

See margin.

a–b. See margin.

5 x 2 - 2x + 6 - 7

_ x-1

x 3 + x 2 + 2x - 5 + 8

_ x - 2

x 2 - 5x + 2 - 4

_ 2x - 1

376

5881

-695


F

Mid-Chapter QuizMid-Chapter QuizLessons 2-1 through 2-3Lessons 2-1 through 2-3

118 | Chapter 2 | Mid-Chapter Quiz

0118_0118_AM_S_C02_MCQ_664291.indd 118 4/30/12 6:15 PMIntervention PlannerIntervention PlannerIInntteerrvveennttiioonnI t ti PPllaannnneerrPl

1 On Level OL 2 Strategic Intervention ALapproaching grade level

If students miss about 25% of the exercises or less, If students miss about 50% of the exercises,

Then choose a resource: Then choose a resource:

SE Lessons 2-1, 2-2, and 2-3Study Guide and Intervention

TE Chapter Project

connectED.mcgraw-hill.comcococco Self-Check Quiz connectED.mcgraw-hill.comcococco Extra Examples, Homework Help


Lessons 2-1 to 2-3

Formative AssessmentUse the Mid-Chapter Quiz to assess students’ progress in the first half of the chapter.

For problems answered incorrectly, have students review the lessons indicated in parentheses.

Additional Answers

5a.

[0, 10] scl: 1 by [0, 100] scl: 10

14. The degree is 4 and the leading coefficient is -7. Because the degree is even and the leading coefficient is negative, lim x→∞

f (x ) = -∞ and lim x→-∞

= -∞.

15. The degree is 5 and the leading coefficient is -5. Because the degree is odd and the leading coefficient is negative, lim x→-∞


f (x ) = -∞.

16a. Sample answer: f (x ) = -2.707 x 3 + 41.392 x 2 - 141.452x + 238.176

16b. Sample answer: -177.273 kWh; This answer does not make sense because it is not possible to consume negative kWh in a month.

23. yes; f (x ) = (x + 5)(x - 5)(x + 2)

24. yes, yes; f (x ) = (x - 1)(x - 2) · (x - 4)(x + 1)

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Lesson 2-4





Chapter

Resource

Masters

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Graphing Calculator Activity

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Enrichment■ Graphing Calculator Activity

■ Practice■ Word Problem Practice■ Enrichment■ Graphing Calculator Activity

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Graphing Calculator Activity






Before Lesson 2-4 Learn that a polynomial function of degree n can have at most n real zeros.

Lesson 2-4 Find real zeros of polynomial functions. Find complex zeros of polynomial functions.

After Lesson 2-4 Solve polynomial inequalities.




Ask:■ Why do you solve for P (x ) = 1500

and not P (x ) = 1,500,000? x stands for the number of thousands of dollars spent. Since 1,500,000 = 1500 × 1000, solve for P (x ) = 1500.

■ What equation will need to be solved when P (x ) = 1500? -0.0007 x 2 + 2.45x = 1500


Zeros of Polynomial FunctionsZeros of Polynomial Functions

1 Real Zeros Recall that a polynomial function of degree n can have at most n real zeros. These real zeros are either rational or irrational.

Rational Zeros Irrational Zeros

f (x ) = 3x 2 + 7x - 6 or f (x ) = (x + 3)(3x - 2)

There are two rational zeros, -3 or 2 _ 3 .

g(x ) = x 2 - 5 or g (x) = (x + √ � 5 )(x - √ � 5 )

There are two irrational zeros, ± √ � 5 .

The Rational Zero Theorem describes how the leading coefficient and constant term of a polynomial function with integer coefficients can be used to determine a list of all possible rational zeros.

If f is a polynomial function of the form f (x) = a n x n + a n - 1 x n - 1 + … + a 2 x 2 + a 1 x + a 0 , with degree n ≥ 1, integer

coefficients, and a 0 ≠ 0, then every rational zero of f has the form p _ q , where

• p and q have no common factors other than ±1,

• p is an integer factor of the constant term a 0 , and

• q is an integer factor of the leading coefficient a n .

Corollary If the leading coefficient a n is 1, then any rational zeros of f are integer factors of the constant term a 0 .

Key Concept Rational Zero Theorem

Once you know all of the possible rational zeros of a polynomial function, you can then use direct or synthetic substitution to determine which, if any, are actual zeros of the polynomial.

Example 1 Leading Coefficient Equal to 1

List all possible rational zeros of each function. Then determine which, if any, are zeros.

a. f (x) = x 3 + 2x + 1

Step 1 Identify possible rational zeros.

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 1. Therefore, the possible rational zeros of f are 1 and -1.

Step 2 Use direct substitution to test each possible zero.

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

f (-1) = (-1 ) 3 + 2(-1) + 1 or -2

Because f ( 1) ≠ 0 and f (-1) ≠ 0, you can conclude thatf has no rational zeros. From the graph of f you cansee that f has one real zero. Applying the RationalZeros Theorem shows that this zero is irrational.

New VocabularyRational Zero Theorem

lower bound

upper bound

Descartes’ Rule of Signs

Fundamental Theorem of

Algebra

Linear Factorization

Theorem

Conjugate Root Theorem

complex conjugates

irreducible over the reals

Why?A company estimates that the profit P in thousands of dollars from a certain model of video game controller is given by P(x) = -0.0007 x 2 + 2.45x, where x is the number of thousands of dollars spent marketing the controller. To find the number of advertising dollars the company should spend to make a profit of $1,500,000, you can use techniques presented in this lesson to solve the polynomial equation P (x ) = 1500.

ThenYou learned that a polynomial function of degree n can have at most n real zeros. (Lesson 2-1)

Now

1Find real zeros of polynomial functions.

2Find complex zeros of polynomial functions.

[-5, 5] scl: 1 by [-4, 6] scl: 1


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120 | Lesson 2-4 | Zeros of Polynomial Functions

■ How can this equation be solved? Set it equal to 0: -0.0007 x 2 + 2.45x - 1500 = 0. Then factor or use the Quadratic Formula.

■ What is the greatest number of real zeros this function can have? How do you know? 2; The degree is 2.

1 Real ZerosExamples 1–3 show how to find all rational zeros of a polynomial function. Example 4 shows how to use upper and lower bounds to help find real zeros of a polynomial. Example 5 uses Descartes’ Rule of Signs to help find real zeros of a polynomial.


Additional Examples


a. f (x ) = x 3 - 3 x 2 - 2x + 4 ±1, ±2, ±4; 1

b. f (x ) = x 3 - 2x - 1 ±1; -1

List all possible rational zeros of f (x ) = 2 x 3 - 5 x 2 - 28x + 15. Then determine which, if any, are

zeros. ±1, ±3, ±5, ±15, ± 1 _ 2 ,

±

3 _ 2

, ± 5 _ 2 , ±

15 _ 2 ; 1 _

2 , -3, 5


1

2

GuidedPractice

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b. g(x) = x 4 + 4 x 3 - 12x - 9

Step 1 Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term -9. Therefore, the possible rational zeros of g are ±1, ±3, and ±9.

Step 2 Begin by testing 1 and -1 using synthetic substitution.

1 1 4 0 -12 -9

1 5 5 -7

1 5 5 -7 -16

-1 1 4 0 -12 -9

-1 -3 3 9

1 3 -3 -9 0

Because g(-1) = 0, you can conclude that -1 is a zero of g. Testing -3 on the depressed polynomial shows that -3 is another rational zero.

-3 1 3 -3 -9

-3 0 9

1 0 -3 0

Thus, g(x) = (x + 1)(x + 3)( x 2 - 3). Because the factor ( x 2 - 3) yields no rational zeros, we can conclude that g has only two rational zeros, -1 and -3.

CHECK The graph of g(x) = x 4 + 4 x 3 - 12x - 9 in Figure 2.4.1 has x-intercepts at -1 and -3, and close to (2, 0) and (-2, 0). By the Rational Zeros Theorem, we know that these last two zeros must be irrational. In fact, the factor ( x 2 - 3) yields two irrational zeros, √ � 3 and - √ � 3 . �



1A. f (x) = x 3 + 5 x 2 - 4x - 2 1B. h(x) = x 4 + 3 x 3 - 7 x 2 + 9x - 30

When the leading coefficient of a polynomial function is not 1, the list of possible rational zeros can increase significantly.

Example 2 Leading Coefficient not Equal to 1

List all possible rational zeros of h(x) = 3 x 3 - 7 x 2 - 22x + 8. Then determine which, if any, are zeros.

Step 1 The leading coefficient is 3 and the constant term is 8.

Possible rational zeros: Factors of 8 _ Factors of 3

= ±1, ±2, ±4, ±8

__ ±1, ±3

or ±1, ±2, ±4, ±8, ±

1 _ 3 , ±

2 _ 3 , ±

4 _ 3 , ±

8 _ 3

Step 2 By synthetic substitution, you can determine that -2 is a rational zero.

-2 3 -7 -22 8

-6 26 -8

3 -13 4 0

By the division algorithm, h(x) = (x + 2)(3 x 2 - 13x + 4). Once 3 x 2 - 13x + 4 is factored, the polynomial becomes h(x) = (x + 2)(3x - 1)(x - 4), and you can conclude that the

rational zeros of h are -2, 1 _ 3 , and 4. Check this result by graphing.



2A. g(x) = 2 x 3 - 4 x 2 + 18x - 36 2B. f (x) = 3 x 4 - 18 x 3 + 2x - 21

[-5, 5] scl: 1 by [-20, 10] scl: 3

Figure 2.4.1

1A. ±1, ±2; 1

1B. ±1, ±2, ±3, ±5,

±6, ±10, ±15,

±30; 2, -5

2A. ±1, ±2, ±3, ±4,

±6, ±9, ±12,

±18, ±36, ± 1

_ 2 ,

± 3

_ 2 , ±

9

_ 2 ; 2

2B. ±1, ±3, ±7,

±21, ± 1

_ 3 , ±

7

_ 3 ;

no rational zeros


0119_0129_AM_S_C02_L04_664291.indd 120 4/30/12 6:17 PM



Additional Example

WATER LEVEL The water level in a bucket sitting on a patio can be modeled by f (x ) = x 3 + 4 x 2 -

2x + 7, where f (x ) is the height of the water in millimeters and xis the time in days. On what day(s) will the water reach a height of 10 millimeters? day 1

3

Teach with TechBlog Have students write an entry on a class blog to summarize how to find the possible rational zeros of a polynomial function. Make sure that students use the concept of

p _ q in their explanations.

BUSINESS After the first half-hour, the number of video games that were sold by a company on their release date can be modeled by g(x) = 2 x 3 + 4 x 2 - 2x, where g(x) is the number of games sold in hundreds and x is the number of hours after the release. How long did it take to sell 400 games?

Because g(x) represents the number of games sold in hundreds, you need to solve g(x) = 4 to determine how long it will take to sell 400 games.

g(x) = 4 Write the equation.

2 x 3 + 4 x 2 - 2x = 4 Substitute 2 x 3 + 4 x 2 - 2x for g (x ).

2 x 3 + 4 x 2 - 2x - 4 = 0 Subtract 4 from each side.

Apply the Rational Zeros Theorem to this new polynomial function, f (x) = 2 x 3 + 4 x 2 - 2x - 4.

Step 1 Possible rational zeros: Factors of 4 _

Factors of 2 =

±1, ±2, ±4 _

±1, ±2

= ±1, ±2, ±4, ± 1 _ 2

Step 2 By synthetic substitution, you can determine that 1 is a rational zero.

1 2 4 -2 -4

2 6 4

2 6 4 0

Because 1 is a zero of f, x = 1 is a solution of f (x) = 0. The depressed polynomial 2 x 2 + 6x + 4 can be written as 2(x + 2)(x + 1). The zeros of this polynomial are -2 and -1. Because time cannot be negative, the solution is x = 1. So, it took 1 hour to sell 400 games.


3. VOLLEYBALL A volleyball that is returned after a serve with an initial speed of 40 feet per second at a height of 4 feet is given by f (t) = 4 + 40t - 16 t 2 , where f (t) is the height the ball reaches in feet and t is time in seconds. At what time(s) will the ball reach a height of 20 feet?

0.5 seconds and 2 seconds

Real-World Example 3 Solve a Polynomial Equation

One way to narrow the search for real zeros is to determine an interval within which all real zeros of a function are located. A real number a is a lower bound for the real zeros of f if f (x) ≠ 0 for x < a. Similarly, b is an upper bound for the real zeros of f if f (x) ≠ 0 for x > b.

a b x

= ( )

The real zeros of f are in the interval [a, b].

You can test whether a given interval contains all real zeros of a function by using the following upper and lower bound tests.

Let f be a polynomial function of degree n ≥ 1, real coefficients, and a positive leading coefficient. Suppose f (x ) is divided by x - c using synthetic division.

• If c ≤ 0 and every number in the last line of the division is alternately nonnegative and nonpositive, then c is a lower bound for the real zeros of f.

• If c ≥ 0 and every number in the last line of the division is nonnegative, then c is an upper bound for the real zeros of f.

Key Concept Upper and Lower Bound Tests

Real-World LinkA recent study showed that almost a third of frequent video game players are between 6 and 17 years old.

Source: NPD Group Inc

Reading MathNonnegative and Nonpositive

Remember that a nonnegative value is one that is either positive or zero, and a nonpositive value is one that is either negative or zero.


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GuidedPractice

GuidedPractice

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Focus on Mathematical ContentZeros An upper bound for the zeros of a polynomial function is a number for which no real zero greater than that number exists for that function. A lower bound for the zeros of a polynomial function is a number for which no real zero less than that number exists. The upper and lower bounds chosen are not unique.

Testing the chosen bounds is done by synthetically dividing the polynomial by the linear expressions x - c, where c is each number chosen. If every number in the last line of the division is alternately nonnegative and nonpositive, the divisor is a lower bound. If every number in the last line of the division is nonnegative, the divisor is an upper bound. Finding upper and lower bounds can help eliminate values from the list of possible zeros found when using the Rational Zero Theorem.

Additional Example

Determine an interval in which all real zeros of f (x ) = x 4 - 4 x 3 - 11 x 2 - 4x - 12 must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. Upper and lower bounds may vary. Sample answer: [ -3, 7 ]; With synthetic division, the values alternate signs when testing -3, and are all nonnegative when testing 7. So, -3 is a lower bound and 7 is an upper bound. The zeros are -2 and 6.

4

GuidedPractice

0119_0129_AM_S_C02_L04_664291.indd 123 5/31/12 6:40 PM

To make use of the upper and lower bound tests, follow these steps.

Step 1 Graph the function to determine an interval in which the zeros lie.

Step 2 Using synthetic substitution, confirm that the upper and lower bounds of your interval are in fact upper and lower bounds of the function by applying the upper and lower bound tests.

Step 3 Use the Rational Zero Theorem to help find all the real zeros.

Example 4 Use the Upper and Lower Bound Tests

Determine an interval in which all real zeros of h(x) = 2 x 4 - 11 x 3 + 2 x 2 - 44x - 24 must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros.

Step 1 Graph h(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [-1, 7].

Step 2 Test a lower bound of c = -1 and an upper bound of c = 7.

-1 2 -11 2 -44 -24

-2 13 -15 59

2 -13 15 -59 35 Values alternate signs in the

last line, so -1 is a lower bound.

7 2 -11 2 -44 -24

14 21 161 819

2 3 23 117 795 Values are all nonnegative in

last line, so 7 is an upper bound.

Step 3 Use the Rational Zero Theorem.

Possible rational zeros: Factors of 24 __ Factors of 2

= ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

___ ±1, ±2

= ±1, ±2, ±4, ±6, ±8, ±12, ±24, ± 1 _ 2 , ± 3 _

2

Because the real zeros are in the interval [-1, 7], you can narrow this list to just ±1, ± 1 _ 2 ,

± 3 _ 2 , 2, 4, or 6. From the graph, it appears that only 6 and - 1 _

2 are reasonable.

Begin by testing 6. Now test - 1 _ 2 in the depressed polynomial.

6 2 -11 2 -44 -24

12 6 48 24

2 1 8 4 0

- 1 _ 2 2 1 8 4

-1 0 -4

2 0 8 0

By the division algorithm, h(x) = 2(x - 6) (x + 1 _ 2 ) ( x 2 + 4). Notice that the factor ( x 2 + 4)

has no real zeros associated with it because x 2 + 4 = 0 has no real solutions. So, f has two

real solutions that are both rational, 6 and - 1 _ 2 . The graph of h(x) = 2 x 4 - 11 x 3 + 2 x 2 -

44x - 24 supports this conclusion.


Determine an interval in which all real zeros of the given function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros.

4A. g (x) = 6 x 4 + 70 x 3 - 21 x 2 + 35x - 12 4B. f (x) = 10 x 5 - 50 x 4 - 3 x 3 + 22 x 2 - 41x + 304A. Sample answer:

[-13, 1]; -12, 1

_ 3

Sample answer: [-2, 6]; 5, 0.71, -1.16

Study Tip Upper and Lower Bounds

Upper and lower bounds of a function are not necessarily unique.


0119_0129_AM_S_C02_L04_664291.indd 122 4/30/12 6:17 PM



Additional Example

Describe the possible real zeros of f (x ) = x 4 - 3 x 3 - 5 x 2 + 2 x + 7. 2 or 0 positive real zeros, 2 or 0 negative real zeros

5

2 Complex ZerosExample 6 shows how to write a polynomial function in standard form given its real and complex zeros. Example 7 shows how to find real and complex zeros of a polynomial and to write the polynomial as a product of linear and irreducible quadratic factors. Example 8 shows how to find the rest of the real zeros of a polynomial when one is given.

Another way to narrow the search for real zeros is to use Descartes’ Rule of Signs. This rule gives us information about the number of positive and negative real zeros of a polynomial function by looking at a polynomial’s variations in sign.

If f (x ) = a n x n + a n - 1 x n - 1 + … + a 1 x + a 0 is a polynomial function with real coefficients, then

• the number of positive real zeros of f is equal to the number of variations in sign of f (x ) or less than that number by some even number and

• the number of negative real zeros of f is the same as the number of variations in sign of f (-x ) or less than that number by some even number.

Key Concept Descartes’ Rule of Signs

Example 5 Use Descartes’ Rule of Signs

Describe the possible real zeros of g(x) = -3 x 3 + 2 x 2 - x - 1.

Examine the variations in sign for g(x) and for g(-x).

g(x) = -3 x 3 + 2 x 2 - x - 1 g(-x) = -3 (-x) 3 + 2 (-x) 2 - (-x) - 1

= 3 x 3 + 2x 2 + x - 1

The original function g(x) has two variations in sign, while g(-x) has one variation in sign. By Descartes’ Rule of Signs, you know that g(x) has either 2 or 0 positive real zeros and 1 negative real zero.

From the graph of g(x) shown, you can see that the function has

[-5, 5] scl: 1 by [-6, 4] scl: 1

one negative real zero close to x = -0.5 and no positive real zeros.


Describe the possible real zeros of each function.

5A. h(x) = 6 x 5 + 8 x 2 - 10x - 15 5B. f (x) = -11 x 4 + 20 x 3 + 3 x 2 - x + 18

- to +

+ to -

+ to -

5A. 1 positive real zero, 2 or 0 negative real zeros

5B. 3 or 1 positive real zeros, 1 negative real zero

When using Descartes’ Rule of Signs, the number of real zeros indicated includes any repeated zeros. Therefore, a zero with multiplicity m should be counted as m zeros.

2 Complex Zeros Just as quadratic functions can have real or imaginary zeros, polynomials of higher degree can also have zeros in the complex number system. This fact, combined with

the Fundamental Theorem of Algebra, allows us to improve our statement about the number of zeros for any nth-degree polynomial.

A polynomial function of degree n, where n > 0, has at least one zero (real or imaginary) in the complex number system.

Corollary A polynomial function of degree n has exactly n zeros, including repeated zeros, in the complex number system.

Key Concept Fundamental Theorem of Algebra

Reading MathVariation in Sign A variation in sign occurs in a polynomial written in standard form when consecutive coefficients have opposite signs.

Math History LinkRené Descartes

(1596–1650)

A French mathematician, scientist, and philosopher, Descartes wrote many philosophical works such as Discourse on Method and mathematical works such as Geometry.


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Differentiated Instruction AL OL

Logical Learners Ask students to compare the products for (x - 3)(x - 2)(x - 1) and (x + 3)(x + 2)(x + 1). Encourage students to notice that one of these polynomials has only positive roots while the other has only negative roots. Then pose the following question: What features of the polynomials themselves give us a clue to this fact? The second product has all positive coefficients, while the first has alternating signs.


Additional Example

Write a polynomial function of least degree with real coefficients in standard form that has -1, 2, and 2 - i as zeros. Sample answer: f (x ) = x 4 - 5 x 3 +

7 x 2 + 3x - 10

6

Tips for New Teachers- i 2 Remind students that i 2 = -1, and therefore, - i 2 = -(-1) or 1.

Focus on Mathematical ContentExistence Theorems The Fundamental Theorem of Algebra and the Linear Factorization Theorems are called existence theorems. They tell you that the zeros or linear factors of a polynomial exist, but not how to find them.

GuidedPractice

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By extending the Factor Theorem to include both real and imaginary zeros and applying the Fundamental Theorem of Algebra, we obtain the Linear Factorization Theorem.

If f (x ) is a polynomial function of degree n > 0, then f has exactly n linear factors and

f (x ) = a n (x - c 1 )(x - c 2 ) … (x - c n )

where a n is some nonzero real number and c 1 , c 2 , …, c n are the complex zeros (including repeated zeros) of f.

Key Concept Linear Factorization Theorem

According to the Conjugate Root Theorem, when a polynomial equation in one variable with real coefficients has a root of the form a + bi, where b ≠ 0, then its complex conjugate, a - bi, is also a root. You can use this theorem to write a polynomial function given its complex zeros.

Example 6 Find a Polynomial Function Given Its Zeros

Write a polynomial function of least degree with real coefficients in standard form that has -2, 4, and 3 - i as zeros.

Because 3 - i is a zero and the polynomial is to have real coefficients, you know that 3 + i must also be a zero. Using the Linear Factorization Theorem and the zeros -2, 4, 3 - i, and 3 + i, you can write f (x) as follows.

f (x) = a[x - (-2)](x - 4)[x - (3 - i)][x - (3 + i)]

While a can be any nonzero real number, it is simplest to let a = 1. Then write the function in standard form.

f (x) = (1)(x + 2)(x - 4)[x - (3 - i)][x - (3 + i)] Let a = 1.

= ( x 2 - 2x - 8)( x 2 - 6x + 10) Multiply.

= x 4 - 8 x 3 + 14 x 2 + 28x - 80 Multiply.

Therefore, a function of least degree that has -2, 4, 3 - i, and 3 + i as zeros is f (x) = x 4 - 8 x 3 + 14 x 2 + 28x - 80 or any nonzero multiple of f (x).


Write a polynomial function of least degree with real coefficients in standard form with the given zeros.

6A. -3, 1 (multiplicity: 2), 4i 6B. 2 √ � 3 , -2 √ � 3 , 1 + i

Sample answers given.

6A. f (x ) = x 5 + x 4 + 11 x 3 + 19 x 2 - 80x + 48 6B. f (x ) = x 4 - 2 x 3 - 10 x 2 + 24x - 24

In Example 6, you wrote a function with real and complex zeros. A function has complex zeros when its factored form contains a quadratic factor which is irreducible over the reals. A quadratic expression is irreducible over the reals when it has real coefficients but no real zeros associated with it. This example illustrates the following theorem.

Every polynomial function of degree n > 0 with real coefficients can be written as the product of linear factors and irreducible quadratic factors, each with real coefficients.

Key Concept Factoring Polynomial Functions Over the Reals

As indicated by the Linear Factorization Theorem, when factoring a polynomial function over the complex number system, we can write the function as the product of only linear factors.

Study Tip Infinite Polynomials Because a can be any nonzero real number, there are an infinite number of polynomial functions that can be written for a given set of zeros.

Study Tip Prime Polynomials Note the difference between expressions which are irreducible over the reals and expressions which are prime. The expression x 2 - 8 is prime because it cannot be factored into expressions with integral coefficients. However, x 2 - 8 is not irreducible over the reals because there are real zeros associated with it,

√ � 8 and -

√ � 8 .


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Additional Example

Consider k (x ) = x 5 + x 4 -

13 x 3 - 23 x 2 - 14x - 24.

a. Write k (x ) as the product of linear and irreducible quadratic factors. k (x ) = (x - 4)(x + 2) ·

(x + 3)( x 2 + 1)

b. Write k (x ) as the product of linear factors. k (x ) = (x - 4)(x + 2) ·

(x + 3)(x + i )(x - i )

c. List all the zeros of k (x ). 4, -2, -3, i, -i

7

Example 7 Factor and Find the Zeros of a Polynomial Function

Consider k(x) = x 5 - 18 x 3 + 30 x 2 - 19x + 30.

a. Write k(x) as the product of linear and irreducible quadratic factors.

The possible rational zeros are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30. The original polynomial has 4 sign variations.

k(-x) = (-x) 5 - 18 (-x) 3 + 30 (-x) 2 - 19(-x) + 30

= - x 5 + 18x 3 + 30x 2 + 19x + 30

k(-x) has 1 sign variation, so k(x) has 4, 2, or 0 positive real zeros and 1 negative real zero.

The graph shown suggests -5 as one real zero of k(x). Use synthetic substitution to test this possibility.

-5 1 0 -18 30 -19 30

-5 25 -35 25 -30

1 -5 7 -5 6 0[-8, 8] scl: 1 by [-100, 800] scl: 50

Because k(x) has only 1 negative real zero, you do not need to test any other possible negative rational zeros. Zooming in on the positive real zeros in the graph suggests 2 and 3 as other rational zeros. Test these possibilities successively in the depressed quartic and then cubic polynomials.

2 1 -5 7 -5 6

2 -6 2 -6

1 -3 1 -3 0

Begin by

testing 2.

3 1 -3 1 -3

3 0 3

1 0 1 0

Now test 3 on the

depressed polynomial.

The remaining quadratic factor ( x 2 + 1) yields no real zeros and is therefore irreducible over the reals. So, k(x) written as a product of linear and irreducible quadratic factors is k(x) = (x + 5)(x - 2)(x - 3)( x 2 + 1).

b. Write k(x) as the product of linear factors.

You can factor x 2 + 1 by writing the expression first as a difference of squares x 2 - ( √ �� -1 ) 2 or x 2 - i 2 . Then factor this difference of squares as (x - i)(x + i). So, k(x) written as a product of linear factors is as follows.

k(x) = (x + 5)(x - 2)(x - 3)(x - i)(x + i)

c. List all the zeros of k(x).

Because the function has degree 5, by the corollary to the Fundamental Theorem of Algebra k(x) has exactly five zeros, including any that may be repeated. The linear factorization gives us these five zeros: -5, 2, 3, i, and -i.


Write each function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all of its zeros.

7A. f (x) = x 4 + x 3 - 26x 2 + 4x - 120 7B. f (x) = x 5 - 2 x 4 - 2 x 3 - 6 x 2 - 99x + 108

[-8, 8] scl: 1 by [-20, 20] scl: 4

7A. a. f (x ) = ( x 2 + 4) ·

(x - 5)(x + 6)

b. f (x ) = (x - 2i ) ·

(x + 2i )(x - 5) ·

(x + 6)

c. ±2i, 5, -6

7B. a. f (x ) = ( x 2 + 9) ·

(x - 1)(x - 4) ·

(x + 3)

b. f (x ) = (x - 3i) ·

(x + 3i )(x - 1) ·

(x - 4)(x + 3)

c. ±3i, 1, 4, -3

Study Tip Using Multiplicity Sometimes a rational zero will be a repeated zero of a function. Use the graph of the function to determine whether a rational zero should be tested using synthetic substitution in succession.

Study Tip Quadratic Formula You could also use the Quadratic Formula to find the zeros of x 2 + 1 in order to factor the expression.

x = -0 ± √

�� 0 2 - 4(1)(1) __

2(1)

= ± √ �� -4

_ 2

= ± 2i _

2 or ± i

So, i and -i are zeros and (x - i ) and (x + i ) are factors.


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Additional Example

Find all complex zeros of p (x ) = x 4 - 6 x 3 + 35 x 2 - 50x - 58 given that 2 + 5i is a zero of p. Then write the linear factorization of p (x ). 2 + 5i, 2 - 5i, 1 + √ � 3 , 1 - √ � 3 ; p (x ) = [x - (2 + 5i )][ x - (2 - 5i ) ][ x - (1 + √ � 3 )] ·

[x - (1 - √ � 3 )]

8

Additional Answers

1. ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180; 6, 5, 1, -6

2. ±1, ±2, ±3, ±6, ±

1 _ 2 , ±

3 _ 2 , ±

1 _ 4 ,

±

3 _ 4 ; 6, 1 _

2 , -

1 _ 2

3. ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30; -5, 6, -1, 1

4. ±1, ±2, ±4, ±5, ±10, ±20, ±

1 _ 2 ,

±

5 _ 2 , ±

1 _ 4 , ± 5 _

4 ; 2 (multiplicity: 2),

5, - 1 _ 4

5. ±1, ±2, ±3, ±4, ±5, ±6, ±10,

±12, ±15, ±20, ±30, ±60, ± 1 _ 2 ,

±

3 _ 2 , ±

5 _ 2 , ±

15 _ 2 , ±

1 _ 3 , ±

2 _ 3 , ±

4 _ 3 , ±

5 _ 3 ,

±

10 _ 3 , ±

20 _ 3 , ±

1 _ 6 , ±

5 _ 6 ; -

5 _ 3 , -

1 _ 2

6. ±1, ±2, ±4, ±8, ±16, ±32, ±64,

±

1 _ 2 , ±

1 _ 3 , ±

2 _ 3 , ±

4 _ 3 , ±

8 _ 3 , ±

16 _ 3 , ±

32 _ 3 ,

±

64 _ 3 , ±

1 _ 6 , ±

1 _ 9 , ±

2 _ 9 , ±

4 _ 9 , ±

8 _ 9 , ±

16 _ 9 ,

±

32 _ 9 , ±

64 _ 9 , ±

1 _ 18

; -

4 _ 3 , 2 _

3

7. ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±16, ±18, ±24, ±27, ±36, ±48, ±54, ±72, ±108, ±144, ±216, ±432; 4 (multiplicity: 2), 3

8. ±1, ±3, ±5, ±9, ±15, ±45, ±

1 _ 2 ,

±

3 _ 2 , ±

5 _ 2 , ±

9 _ 2 , ±

15 _ 2 , ±

45 _ 2 , ±

1 _ 4 ,

±

3 _ 4 , ±

5 _ 4 , ±

9 _ 4 , ±

15 _ 4 , ±

45 _ 4 , ±

1 _ 8 ,

±

3 _ 8 , ±

5 _ 8 , ±

9 _ 8 , ±

15 _ 8 , ±

45 _ 8 ; 1 _

4 , -

5 _ 2

18. Sample answer: [ -3, 10 ]; 6, -2,

1, 4

19. Sample answer: [ -6, 5 ]; 3, 2, -4, -

1 _ 2

20. Sample answer: [ -6, 4 ]; 2, -1, 1, -4

21. Sample answer: [ -4, 7 ]; -2, 5, 1, 3 _ 2

22. Sample answer: [ -2, 9 ]; 5, 2

(multiplicity: 2), -

1 _ 2

23. Sample answer: [ -2, 7 ]; 3 (multiplicity: 2),

-1, 3 _ 2

24. Sample answer: [ -3, 5 ]; 1 (multiplicity: 2), -2 (multiplicity: 2), 3

25. Sample answer: [ -3, 5 ]; 1 _ 2 (multiplicity: 2),

-2, 3 (multiplicity: 2)

26. 1 positive zero, 2 or 0 negative zeros

27. 3 or 1 positive zeros, 1 negative zero

28. 2 or 0 positive zeros, 2 or 0 negative zeros



31. 4, 2 or 0 positive zeros, 1 negative zero

0119_0129_AM_S_C02_L04_664291.indd 127 4/30/12 6:16 PM1 1

You can use synthetic substitution with complex numbers in the same way you use it with real numbers. Doing so can help you factor a polynomial in order to find all of its zeros.

Example 8 Find the Zeros of a Polynomial When One is Known

Find all complex zeros of p(x) = x 4 - 6 x 3 + 20 x 2 - 22x - 13 given that 2 - 3i is a zero of p. Then write the linear factorization of p(x).

Use synthetic substitution to verify that 2 - 3i is a zero of p(x).

2 - 3i 1 -6 20 -22 -13

2 - 3i -17 + 6i

1 -4 - 3i

(2 - 3i )(-4 - 3i ) = -8 + 6i + 9 i 2

= -8 + 6i + 9(-1)

= -17 + 6i

2 - 3i 1 -6 20 -22 -13

2 - 3i -17 + 6i 24 + 3i

1 -4 - 3i 3 + 6i 0

(2 - 3i )(3 + 6i ) = 6 + 3i - 18 i 2

= 6 + 3i - 18(-1)

= 24 + 3i

2 - 3i 1 -6 20 -22 -13

2 - 3i -17 + 6i 24 + 3i 13

1 -4 - 3i 3 + 6i 2 + 3i 0

(2 - 3i )(2 + 3i ) = 4 - 9 i 2

= 4 - 9(-1)

= 4 + 9 or 13

Because 2 - 3i is a zero of p, you know that 2 + 3i is also a zero of p. Divide the depressed polynomial by 2 + 3i.

2 + 3i 1 -4 - 3i 3 + 6i 2 + 3i

2 + 3i -4 - 6i -2 - 3i

1 -2 -1 0

Using these two zeros and the depressed polynomial from this last division, you can write p(x) = [x - (2 - 3i)][x - (2 + 3i)]( x 2 - 2x - 1).

Because p(x) is a quartic polynomial, you know that it has exactly 4 zeros. Having found 2, you know that 2 more remain. Find the zeros of x 2 - 2x - 1 by using the Quadratic Formula.

x = -b ± √ ��

b 2 - 4ac __

2a Quadratic Formula

= -(-2) ± √

�� (- 2) 2 - 4(1)(-1) ___

2(1) a = 1, b = -2, and c = -1

= 2 ± √ � 8 _

2 Simplify.

= 1 ± √ � 2 Simplify.

Therefore, the four zeros of p(x) are 2 - 3i, 2 + 3i, 1 + √ � 2 , and

[-4, 6] scl: 1 by [-40, 40] scl: 8

1 - √ � 2 . The linear factorization of p(x) is [x - (2 - 3i)] ·[x - (2 + 3i)][x - (1 + √ � 2 )][x - (1 - √ � 2 )].

Using the graph of p(x), you can verify that the function has two real zeros at 1 + √ � 2 or about 2.41 and 1 - √ � 2 or about -0.41.

GuidedGuided Practice Practice

For each function, use the given zero to find all the complex zeros of the function. Then write the linear factorization of the function.

8A. g(x) = x 4 - 10 x 3 + 35 x 2 - 46x + 10; 2 + √ � 3

8B. h(x) = x 4 - 8 x 3 + 26 x 2 - 8x - 95; 1 - √ � 6

Watch Out!Complex Numbers Recall from Lesson 0-2 that all real numbers are also complex numbers.

8A. 2 + √

�

3 , 2 - √

�

3 , 3

+ i, 3 - i ; (x - 2 - √

�

3 ) ·

(x - 2 + √

�

3 ) ·

(x - 3 - i ) ·(x - 3 + i )

8B. 1 - √

�

6 , 1 + √

�

6 ,

3 + i √

�

10 ,

3 - i √

�

10 ;

(x - 1 + √

�

6 ) ·

(x - 1 - √

�

6 ) ·

(x - 3 - i √

�

10 ) ·

(x - 3 + i √

�

10 )

Study TipDividing Out Common Factors

Before applying any of the methods in this lesson, remember to factor out any common monomial factors. For example, g (x ) = -2 x 4 + 6 x 3 - 4 x 2 - 8x should first be factored as g (x ) = -2x ( x 3 - 3 x 2 + 2x + 4), which implies that 0 is a zero of g.


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Watch Out!

Common Error For Exercise 9, students will find they have many possible rational zeros to test. Remind them that they can eliminate many of the zeros to test by noting that � must be a positive number greater than 4. Remind students to subtract 45 from the given function, set it equal to 0, and then solve for the variable using only the possible zeros that make sense for dimensions of the container.

Additional Answers


33. Sample answer: f (x ) = x 4 + 4 x 3 - 19 x 2 - 106x - 120

34. Sample answer: f (x ) = x 4 - 6 x 3 - 14 x 2 + 154x - 255

35. Sample answer: f (x ) = x 4 - 19 x 3 + 113 x 2 - 163x - 296


37. Sample answer: f (x ) = x 4 - 5 x 3 - 21 x 2 + 119x - 130

38. Sample answer: f (x ) = x 4 + 9 x 2 - 112

39. Sample answer: f (x ) = x 4 - 6 x 3 + 19 x 2 + 36x - 150



55a. V (ℓ ) = 1 _ 3 � 3 - 3 � 2

55b. 6300 = 1 _ 3 � 3 - 3 � 2

55c. base: 30 in. by 30 in.; height: 21 in.



AL Approaching Level 1–56, 73, 74, 76, 77, 82–97

1–55 odd, 95–97 2–56 even, 73, 74, 76, 77, 82–97

OL On Level 1–71 odd, 72–74, 76, 77, 82–97

1–56, 95–97 57–74, 76, 77, 82–97



Exercises

List all possible rational zeros of each function. Then determine which, if any, are zeros. (Examples 1 and 2)

1. g(x) = x 4 - 6 x 3 - 31 x 2 + 216x - 180

2. f (x) = 4 x 3 - 24 x 2 - x + 6

3. g(x) = x 4 - x 3 - 31 x 2 + x + 30

4. g(x) = -4 x 4 + 35 x 3 - 87 x 2 + 56x + 20

5. h(x) = 6 x 4 + 13 x 3 - 67 x 2 - 156x - 60

6. f (x) = 18 x 4 + 12 x 3 + 56 x 2 + 48x - 64

7. h(x) = x 5 - 11 x 4 + 49 x 3 - 147 x 2 + 360x - 432

8. g(x) = 8 x 5 + 18 x 4 - 5 x 3 - 72 x 2 - 162x + 45

9. MANUFACTURING The specifications for the dimensions of a new cardboard container are shown. If the volume of the container is modeled by V(h) = 2 h 3 -9 h 2 + 4h and it will hold 45 cubic inches of merchandise, what are the container’s dimensions? (Example 3)

h - 42h -

h

Solve each equation. (Example 3)

10. x 4 + 2 x 3 - 7 x 2 - 20x - 12 = 0

11. x 4 + 9 x 3 + 23 x 2 + 3x - 36 = 0

12. x 4 - 2 x 3 - 7 x 2 + 8x + 12 = 0

13. x 4 - 3 x 3 - 20 x 2 + 84x - 80 = 0

14. x 4 + 34x = 6 x 3 + 21 x 2 - 48

15. 6 x 4 + 41 x 3 + 42 x 2 - 96x + 6 = -26

16. -12 x 4 + 77 x 3 = 136 x 2 - 33x - 18

17. SALES The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 2 x 3 - 2 x 2 + 4x, where x is the number of days after the first day of the month. How many days will it take the store to make $16,000? (Example 3)

Determine an interval in which all real zeros of each function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. (Example 4)

18. f (x) = x 4 - 9 x 3 + 12 x 2 + 44x - 48

19. f (x) = 2 x 4 - x 3 - 29 x 2 + 34x + 24

20. g(x) = 2 x 4 + 4 x 3 - 18 x 2 - 4x + 16

21. g(x) = 6 x 4 - 33 x 3 - 6 x 2 + 123x - 90

22. f (x) = 2 x 4 - 17 x 3 + 39 x 2 - 16x - 20

23. f (x) = 2 x 4 - 13 x 3 + 21 x 2 + 9x - 27

24. h(x) = x 5 - x 4 - 9 x 3 + 5 x 2 + 16x - 12

25. h(x) = 4 x 5 - 20 x 4 + 5 x 3 + 80 x 2 - 75x + 18

Describe the possible real zeros of each function. (Example 5)

26. f (x) = -2 x 3 - 3 x 2 + 4x + 7

27. f (x) = 10 x 4 - 3 x 3 + 8 x 2 - 4x - 8

28. f (x) = -3 x 4 - 5 x 3 + 4 x 2 + 2x - 6

29. f (x) = 12 x 4 + 6 x 3 + 3 x 2 - 2x + 12

30. g(x) = 4 x 5 + 3 x 4 + 9 x 3 - 8 x 2 + 16x - 24

31. h(x) = -4 x 5 + x 4 - 8 x 3 - 24 x 2 + 64x - 124

Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. (Example 6)

32. 3, -4, 6, -1 33. -2, -4, -3, 5

34. -5, 3, 4 + i 35. -1, 8, 6 - i

36. 2 √ � 5 , -2 √ � 5 , -3, 7 37. -5, 2, 4 - √ � 3 , 4 + √ � 3

38. √ � 7 , - √ � 7 , 4i 39. √ � 6 , - √ � 6 , 3 - 4i

40. 2 + √ � 3 , 2 - √ � 3 , 4 + 5i 41. 6 - √ � 5 , 6 +

√ � 5 , 8 - 3i

Write each function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all of its zeros. (Example 7)

42. g(x) = x 4 - 3 x 3 - 12 x 2 + 20x + 48

43 g(x) = x 4 - 3 x 3 - 12 x 2 + 8

44. h(x) = x 4 + 2 x 3 - 15 x 2 + 18x - 216

45. f (x) = 4 x 4 - 35 x 3 + 140 x 2 - 295x + 156

46. f (x) = 4 x 4 - 15 x 3 + 43 x 2 + 577x + 615

47. h(x) = x 4 - 2 x 3 - 17 x 2 + 4x + 30

48. g(x) = x 4 + 31 x 2 - 180

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. (Example 8)

49. h(x) = 2 x 5 + x 4 - 7 x 3 + 21 x 2 - 225x + 108; 3i

50. h(x) = 3 x 5 - 5 x 4 - 13 x 3 - 65 x 2 - 2200x + 1500; -5i

51. g(x) = x 5 - 2 x 4 - 13 x 3 + 28 x 2 + 46x - 60; 3 - i

52. g(x) = 4 x 5 - 57 x 4 + 287 x 3 - 547 x 2 + 83x + 510; 4 + i

53. f (x) = x 5 - 3 x 4 - 4 x 3 + 12 x 2 - 32x + 96; -2i

54. g(x) = x 4 - 10 x 3 + 35 x 2 - 46x + 10; 3 + i

55. ARCHITECTURE An architect is constructing a scale model of a building that is in the shape of a pyramid.

a. If the height of the scale model is 9 inches less than its length and its base is a square, write a polynomial function that describes the volume of the model in terms of its length.

b. If the volume of the model is 6300 cubic inches, write an equation describing the situation.

c. What are the dimensions of the scale model?

1–8. See margin.

1 in., 5 in., 9 in.

-1, 3, -2 (multiplicity: 2)

-4, 1, -3 (multiplicity: 2)

3, -2, 2, -14, -5, 2 (multiplicity: 2)

-3, 8, 2, -115. 1

_ 2 ,

2

_ 3 , -4

(multiplicity: 2)

16. -

1

_ 4 ,

2

_ 3 , 3

(multiplicity: 2)

2 days




42–48. See Chapter

2 Answer Appendix.


B

a–c. See margin.


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Guided Practice

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Additional Answers

57. Sample answer: f (x ) = x 3 - 6

58. Sample answer: f (x ) = x 3 - 5

59. Sample answer: f (x ) = x 3 + 2

60. Sample answer: f (x ) = x 3 + 7

68. (x - 3 √ � 3 )( x 2 + x 3

√ � 3 + 3 √ � 9 )

69. (x + 2 3 √ � 2 )( x 2 - 2x 3

√ � 2 + 4 3 √ � 4 )

70. (2x + 3 √ � 9 )(4 x 2 - 2x 3

√ � 9 + 3 √ � 81 )

71. (3 x 2 + 3 √ � 4 )(9 x 4 - 3 x 2

3 √ � 4 + 2

3 √ � 2 )



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Study Guide and InterventionZeros of Polynomial Functions

Example

2-4

Finding Real Zeros of Polynomial Functions The Rational Zero Theorem describes how the leading coefficient and constant term of a polynomial function with integer coefficients determine a list of all possible rational zeros. Descartes’ Rule of Signs states that the number of positive real zeros is equal to the number of sign changes in f(x) or less than that by an even number. The number of negative real zeros is the same as the number of sign changes in f(-x) or less than that by an even number.

a. List all possible rational zeros of h(x) = x3 - 5x2 − 17x − 6. Then determine which, if any, are zeros.

The leading coefficient is 1 and the constant term is −6.

Possible rational zeros: Factors of -6 −

Factors of 1 =

±1, ±2, ±3, ±6 −

±1 or ±1, ±2, ±3, ±6

By synthetic substitution, you can determine that x = −2 is a rational zero. -2 1 -5 -17 -6 -2 14 6 1 -7 -3 0 The depressed polynomial is x2 - 7x - 3. You can use the quadratic

formula to find the two irrational roots.

b. Describe the possible real zeros of g(x) = 4x4 − 13x3 - 21x2 + 38x − 8. Examine the variations in sign for g(x) and for g(−x).

g(x) = 4x4 - 13x3 − 21x2 + 38x - 8

g(−x) = 4(−x)

4 − 13(−x)

3 − 21(−x)

2 + 38(−x) - 8 = 4x4 + 13x3 − 21x2 - 38x − 8

By Descartes’ Rule of Signs, g(x) has either 3 or 1 positive real zeros and 1 negative real zero. The graph shows it has 1 negative real zero and 3 positive real zeros.

Exercises

1. List all possible rational zeros of h(x) = x3 - 5x2 - 4x + 20. Then determine which, if any, are zeros. ±1, ±2, ±4, ±5, ±10, ±20; -2, 2, 5

2. Describe the possible real zeros of f(x) = x5 - 10x4 + 7x3 - x2 + 4x + 20.4 or 2 or 0 positive real zeros; 1 negative real zero

+ to -

- to +

+ to -

+ to -

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1. f(x) = x3 - x2 - 8x + 12 ±1, ±2, ±3, ±4, ±6, ±12; -3, 2

2. h(x) = 2x3 - 3x2 - 2x + 3 ±1, ±3, ±

1 −

2 , ±

3 −

2 ; ±1, 3 −

2

3. g(x) = 36x4 - 13x2

+ 1 ±

1 −

36 , ±

1 −

18 , ±

1 −

12 , ±

1 −

9 , ±

1 −

6 , ±

1 −

4 , ±

1 −

3 ; ±

1 −

2 , ±1, ±

1 −

3 , ±

1 −

2

4. g(x) = x3 + 3x2 - 6x - 8 ±1, ±2, ±4, ±8; -4, -1, 2


5. 2x4 + 9x3 - 87x2 - 49x + 45 = 0 6. x3 - 5x2 - 17x = -21

5, -9, -1, 1 −

2 -3, 1, 7

7. Determine an interval in which all real zeros of f(x) = x4 + x3 - 7x2 - x + 6 must lie. Explain your reasoning using the upper and lower bound tests. Then find all real zeros.

Sample answer: [-3, 3]; -3, -1, 1, 2

8. Describe the possible real zeros of f(x) = x4 + 2x3 - 13x2 - 14x + 24.

2 or 0 positive real zeros; 2 or 0 negative real zeros

Write a polynomial function of least degree with real coefficients in standard form that has the given zeros.

9. −5, 1 + √ �� 11 i −

2 , 1 - √ �� 11 i −

2 10. 1, −1, √ � 3 i, − √ � 3 i

f(x) = x3 + 4x2 - 2x + 15 f(x) = x4 + 2x2 - 3

Use the given zero to find all real zeros of each function. Then write the linear factorization of p(x).

11. p(x) = x4 − 5x3 + 8x2 - 20x + 16; 2i

1, 4, 2i, -2i; p(x) = (x - 4)(x - 1)(x + 2i)(x − 2i)

12. p(x) = x4 + 4x3 - 2x2 - 4x + 16; 1 + i

−2, −4, 1 + i, 1 − i; p(x) = (x + 4)(x + 2)[x − (1 + i)][x − (1 − i)]

13. DRIVING An automobile moving at 12 meters per second on level ground begins to decelerate at a rate of 1.6 meters per second squared. The formula for the distance an

object has traveled is d(t) = v0t + 1 −

2 at2, where v0 is the initial velocity and a is the

acceleration. For what value(s) of t does d(t) = 40 meters? 5 seconds and 10 seconds

PracticeZeros of Polynomial Functions

2-4

005_038_PCCRMC02_893803.indd 23 10/1/09 5:46:15 PM


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Word Problem PracticeZeros of Polynomial Functions

1. FLOODS The number of feet a stream is above or below its flood level in a town can be modeled by f(x) = 4x3

− 22x2+ 30x, where x is the

number of days since a storm hit the area.

a. Use Descartes’ Rule of Signs to describe the possible real zeros of the function. 2 or 0 positive real zeros, 0 negative real zeros

b. Determine an interval in which all real zeros of the function must lie. Then find the real zero(s).Sample answer: [0, 3]; 0, 2.5, 3

2. FLOWERS A group of students found experimentally that a population of wildflowers, after the seed is introduced into the area, can be approximated by p(t) = t4 + 3t3 - 4t2 + 18t - 60, where t is the number of years after introduction.

a. List all possible rational zeros of the function.±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60

b. According to Descartes’ Rule of Signs, how many possible positive real zeros are there? 3 or 1

c. Write the linear factorization of the function.

p(t) = (t - 2)(t + 5)(t - i √

�

6 ) ·

(t + i √

�

6 )

3. CREDIT For a period of two weeks, a customer’s credit card balance can be modeled by f(x) = x3 - x2 - 10x, where x is the day number. When was the balance equal to $8?day 4

4. PROFIT The total profit for a small business during its first 10 years, in thousands, can be approximated by f(x) = x3

- 37x - 84, where x is the number of years.

a. Use Descartes’ Rule of Signs to describe the possible real zeros of the function. 1 positive real zero, 2 or 0 negative real zeros

b. Determine an interval in which all real zeros of the function must lie. Then find the real zero(s). Sample answer: [-4, 7]; -4, -3, 7

5. FESTIVAL The table shows the number of people who attended a festival from June 9 through June 15.

Day of the Month

Number of People

9 91

10 220

11 393

12 616

13 895

14 1236

15 1645

a. Use a graphing calculator to find a cubic function that models the data.

f(x) = x3 - 8x2 + 10x - 80

b. Use Descartes’ Rule of Signs to describe the possible real zeros of the function. 3 or 1 positive real zeros, 0 negative real zeros

c. Write the linear factorization of the function.

f(x) = (x - 8)(x - √

��

10 i)(x + √

��

10 i)

2-4

005_038_PCCRMC02_893803.indd 24 10/23/09 5:56:48 PM

Enrichment

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The Secret Cubic EquationYou might have supposed that there existed simple formulas for solving higher-degree equations. After all, there is a simple formula for solving quadratic equations. Might there not be formulas for cubics, quartics, and so on?

There are formulas for some higher-degree equations, but they are certainlynot “simple” formulas.

Here is a method for solving a reduced cubic of the formx3

+ ax + b = 0 published by Jerome Cardan in 1545. Cardan was given the formula by another mathematician, Tartaglia. Tartaglia made Cardan promise to keep the formula secret, but Cardan published it anyway. He did, however, give Tartaglia the credit for inventing the formula!

Let R = (1−

2b)

2+ a

3−

27

ThThenThen xx == [[[[[[- 111−222

bbb− ++ √√√�√√RRR √√√√ ]]]]]]� 11−

3++ [[[[[[- 111−

222bbb−− - √√√�√√RRR √√√√ ]]]]]]�

11−

3

U C dUse CardUse Card ’ tan’s metan s meth d t fhod to fhod to fi d thind theind the lreal rooreal root ft of eact of each bih cubich cubic tiequationequation.. R dRound anRound an tswers toswers to th dthree d three d i lecimal pecimal pl Tlaces Tlaces. Th k then skethen sket hch a grach a gra h f thph of thph of thee correspocorrespo di fnding funding fu tinction onction o thn the grn the grid iid proviid provid ddedded.

EnrichmentEnrichmentEnrichmentEnrichmentE i h2 42-42 42 4

0119_0129_AM_S_C02_L04_664291.indd 129 4/30/12 6:15 PM

56. CONSTRUCTION The height of a tunnel that is under construction is 1 foot more than half its width and its length is 32 feet more than 324 times its width. If the volume of the tunnel is 62,231,040 cubic feet and it is a rectangular prism, find the length, width, and height.

Write a polynomial function of least degree with integer coefficients that has the given number as a zero.

57. 3 √ � 6 58. 3

√ � 5

59. -

3 √ � 2 60. -

3 √ � 7

Use each graph to write g as the product of linear factors. Then list all of its zeros.

61. g(x) = 3 x 4 - 15 x 3 + 87 x 2 - 375x + 300

[-2, 8] scl: 1 by [-300, 200] scl: 50

62. g(x) = 2 x 5 + 2 x 4 + 28 x 3 + 32 x 2 - 64x

[-4, 4] scl: 1 by [-40, 80] scl: 12

Determine all rational zeros of the function.

63. h(x) = 6 x 3 - 6 x 2 + 12

64. f ( y) = 1 _ 4 y 4 + 1 _

2 y 3 - y 2 + 2y - 8

65. w(z) = z 4 - 10 z 3 + 30 z 2 - 10z + 29

66. b(a) = a 5 - 5 _ 6 a 4 + 2 _

3 a 3 - 2 _

3 a 2 - 1 _

3 a + 1 _

6

67 ENGINEERING A steel beam is supported by two pilings 200 feet apart. If a weight is placed x feet from the piling on the left, a vertical deflection represented by d = 0.0000008x 2 (200 - x) occurs. How far is the weight from the piling if the vertical deflection is 0.8 feet?

x

200 ft

d

Write each polynomial as the product of linear and irreducible quadratic factors.

68. x 3 - 3 69. x 3 + 16

70. 8 x 3 + 9 71. 27 x 6 + 4

72. MULTIPLE REPRESENTATIONS In this problem, you will explore even- and odd-degree polynomial functions.

a. ANALYTICAL Identify the degree and number of zeros of each polynomial function.

i. f (x) = x 3 - x 2 + 9x - 9

ii. g(x) = 2 x 5 + x 4 - 32x - 16

iii. h(x) = 5 x 3 + 2 x 2 - 13x + 6

iv. f (x) = x 4 + 25 x 2 + 144

v. h(x) = 3 x 6 + 5 x 5 + 46 x 4 + 80 x 3 - 32 x 2

vi. g(x) = 4 x 4 - 11 x 3 + 10 x 2 - 11x + 6

b. NUMERICAL Find the zeros of each function.

c. VERBAL Does an odd-degree function have to have a minimum number of real zeros? Explain.


73. ERROR ANALYSIS Angie and Julius are using the Rational Zeros Theorem to find all the possible rational zeros of f (x) = 7 x 2 + 2 x 3 - 5x - 3. Angie thinks the possible zeros

are ± 1 _ 7 , ±

3 _ 7 , ±1, ±3. Julius thinks they are ±

1 _ 2 , ± 3 _

2 , ±1,

±3. Is either of them correct? Explain your reasoning.

74. REASONING Explain why g(x) = x 9 - x 8 + x 5 + x 3 - x 2 + 2 must have a root between x = -1 and x = 0.

75. CHALLENGE Use f (x) = x 2 + x - 6, f (x) = x 3 + 8 x 2 + 19x + 12, and f (x) = x 4 - 2 x 3 - 21 x 2 + 22x + 40 to make a conjecture about the relationship between the graphs and zeros of f (x) and the graphs and zeros of each of the following.

a. -f (x) b. f (-x)

76. OPEN ENDED Write a function of 4 th degree with an imaginary zero and an irrational zero.

77. REASONING Determine whether the statement is true or false. If false, provide a counterexample. A third-degree polynomial with real coefficients has at least one nonreal zero.

CHALLENGE Find the zeros of each function if h(x) has zeros at x 1 , x 2 , and x 3 .

78. c(x) = 7h(x) 79. k(x) = h(3x)

80. g(x) = h(x - 2) 81. f (x) = h(-x)

82. REASONING If x - c is a factor of f (x) = a 1 x 5 - a 2 x 4 + …, what value must c be greater than or equal to in order to be an upper bound for the zeros of f (x)? Assume a ≠ 0. Explain your reasoning.

83. WRITING IN MATH Explain why a polynomial with real coefficients and one imaginary zero must have at least two imaginary zeros.

� = 23,360 ft, w = 72 ft, h = 37 ft


g (x) =

3(x - 4) ·

(x - 1) ·

(x - 5i ) ·

(x + 5i );

4, 1, ±5i

g (x ) = 2x ·

(x - 1) ·

(x + 2) ·

(x - 4i ) ·

(x + 4i );

0, 1, -2,

± 4i

-1

-4, 2

no rational zeros

-

1

_ 2 ,

1

_ 3 , 1

100 ft or about 161.8 ftC


3, 3 zeros

5, 5 zeros

3, 3 zeros

4, 4 zeros

6, 6 zeros

4, 4 zeros

b–c. See Chapter 2 Answer Appendix.

See margin.

See margin.


Sample answer: f (x ) = x 4 - 2 x 2 - 3


x 1 , x 2 , x 3 x 1

_ 3 ,

x 2

_ 3 ,

x 3

_ 3

x 1 + 2, x 2 + 2, x 3 + 2 - x 1 , - x 2 , - x 3




0119_0129_AM_S_C02_L04_664291.indd 128 5/25/12 6:13 PM



Watch Out!

Error Analysis For Exercise 73, remind students that the first step in finding all the possible zeros is to write the polynomial function in standard form. Then identify the factors of the leading coefficient and the factors of the constant term.

4 AssessYesterday’s News Ask students to write how the synthetic division teachings in Lesson 2-3 helped them test the possible rational zeros of a polynomial function.

Additional Answers

73. Julius; sample answer: Angie did not divide by the factors of the leading coefficient.

74. Sample answer: An input of -1 produces a negative output, whereas an input of 0 results in a positive output.

88. The degree is 7, and the leading coefficient is -4. Because the degree is odd and the leading coefficient is negative, lim x→-∞


f (x ) = -∞.

89. The degree is 6, and the leading coefficient is 4. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

90. The degree is 5, and the leading coefficient is 5. Because the degree is odd and the leading coefficient is positive, lim x→-∞

f (x ) = -∞ and lim x→∞

f (x ) = ∞.

91. It appears that f (x ) has a relative minimum of -3 at x = 0 and a relative maximum of 3 at x = -2.

92. It appears that f (x ) has a relative maximum of 8 at x = 1 and a relative minimum of -16 at x = 3 and x = -1.

93. It appears that f (x ) has a relative maximum of 0 at x = 0 and x = -4 and a relative minimum of -80 at x = -2 and -150 at x = 2.


Extension Ask students if it is possible for a polynomial function to have a real zero greater than its greatest possible rational root. If so, ask students to give an example. Yes; sample answer: The function f (x ) = 2 x 2 - 5x + 1 has 1 as its greatest possible rational zero, while its actual positive

zero is 5 + √ � 17

_ 4 or about 2.28.

DDiifffferentiiateddIInsttructtiion

Spiral Review

Divide using synthetic division. (Lesson 2-3)

84. ( x 3 - 9 x 2 + 27x - 28) ÷ (x - 3) 85. ( x 4 + x 3 - 1) ÷ (x - 2)

86. (3 x 4 - 2 x 3 + 5 x 2 - 4x - 2) ÷ (x + 1) 87. (2 x 3 - 2x - 3) ÷ (x - 1)

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test. (Lesson 2-2)

88. f (x) = -4 x 7 + 3 x 4 + 6 89. f (x) = 4 x 6 + 2 x 5 + 7 x 2 90. g(x) = 3 x 4 + 5 x 5 - 11

Estimate to the nearest 0.5 unit and classify the extrema for the graph of each function. Support the answers numerically. (Lesson 1-4)

91. 92. 93. y

x

−4

−8

−4−8

8

4

4 8

( ) = ( - 1)( + 1)( + 3)

y

x

−10

−20

−2−4

10

20

2 4

( ) = ( - 2)( + 2)( - 4)

y

x

−40

−80

−120

−4−8 4 8

= 2( - 3)( + 4) 2( )

x 2 - 6x + 9 - 1

_ x - 3

x 3 + 3 x 2 + 6x + 12 + 23

_ x - 2

3 x 3 - 5 x 2 + 10x - 14 + 12

_ x + 1

2 x 2 + 2x



94. SAT/ACT A circle is inscribed in a square and intersects the square at points A, B, C, and D. If AC = 12, what is the total area of the shaded regions?

A 18 D 24π

B 36 E 72

C 18π

95. REVIEW f (x) = x 2 - 4x + 3 has a relative minimum located at which of the following x-values?

F -2 H 3

G 2 J 4

96. Find all of the zeros of p(x) = x 3 + 2 x 2 - 3x + 20.

A -4, 1 + 2i, 1 - 2i C -1, 1, 4 + i, 4 - i

B 1, 4 + i, 4 - i D 4, 1 + i, 1 - i

97. REVIEW Which expression is equivalent to ( t 2 + 3t - 9)(5 - t) -1 ?

F t + 8 - 31 _ 5 - t

G -t - 8

H -t - 8 + 31 _ 5 - t

J -t - 8 - 31 _ 5 - t

B

D

A CE

G

A

H



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Lesson 2-5

130 | Lesson 2-5 | Rational Functions


Before Lesson 2-5 Identify points of discontinuity and end behavior of graphs of functions using limits.

Lesson 2-5 Analyze and graph rational functions. Solve rational equations.

After Lesson 2-5 Write partial fraction decompositions of rational expressions.




Ask:■ What are some ways salt can be

removed from water? Sample answers: reverse osmosis, distillation, electrodialysis, and vacuum freezing




Chapter

Resource

Masters


■ Study Guide and Intervention ■ Practice■ Word Problem Practice■ Enrichment







Seawater supply

Pretreatmentsystem

Reverseosmosisprocess

Porous layer

Semi-permeable

membrane

Treatedwater

Freshwater storage

Salt water

Meshspacer

Concentrated sea-water disposal

1.

2.

3.

5.Post-treatment4.

130

Why?Water desalination, or removing the salt from sea water, is currently in use in areas of the world with limited water availability and on many ships and submarines. It is also being considered as an alternative for providing water in the future. The cost for various extents of desalination can be modeled using rational functions.

Now

1Analyze and graph rational functions.

2Solve rational equations.

ThenYou identified points of discontinuity and end behavior of graphs of functions using limits. (Lesson 1-3)

New Vocabularyrational function

asymptote

vertical asymptote

horizontal asymptote

oblique asymptote

holes

1Rational Functions A rational function f (x) is the quotient of two polynomial functions a(x) and b(x), where b is nonzero.

f (x) = a(x)

_ b(x)

, b (x) ≠ 0

The domain of a rational function is the set of all real numbers excluding those values for which b (x) = 0 or the zeros of b (x).

One of the simplest rational functions is the reciprocal function y

x

( ) = 1_f (x) = 1 _ x . The graph of the reciprocal function, like many rational

functions, has branches that approach specific x- and y-values. The lines representing these values are called asymptotes.

The reciprocal function is undefined when x = 0, so its domain is

(-∞, 0) or (0, ∞). The behavior of f (x) = 1 _ x to the left ( 0 - ) and right

( 0 + ) of x = 0 can be described using limits.

lim x→ 0 - f (x) = -∞ lim

x→ 0 + f (x) = ∞

From Lesson 1-3, you should recognize 0 as a point of infinite discontinuity in the domain of f. The line x = 0 in Figure 2.5.1 is called a vertical asymptote of the graph of f. The end behavior of f can be also be described using limits.

lim x→-∞ f (x) = 0 lim x→+∞

f (x) = 0

The line y = 0 in Figure 2.5.2 is called a horizontal asymptote of the graph of f.

y

x

( ) = 1_

verticalasymptote: = 0

Figure 2.5.1 Figure 2.5.2

y

x

( ) = 1_

horizontalasymptote: = 0

These definitions of vertical and horizontal asymptotes can be generalized.

Rational FunctionsRational Functions

130 | Lesson 2-5

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Focus on Mathematical ContentAsymptotes Asymptotes can be horizontal, vertical, or slanted lines. Asymptotes can be determined by observing the limits, discontinuity, and end behavior of the rational function.

■ The cost for desalination of water lies mostly in the energy used. As the percent of salt removed increases, what do you think happens to the cost? Sample answer: The cost increases as the percent of salt removed increases.

■ If it costs $1000 to remove each percent of the salt, then the ratio of the cost to remove x % of the salt to the percent of salt remaining would

be 1000x _ 100 - x

. What type of values

could x have? Explain. 0 ≤ x < 100; Denominators cannot equal 0, so x would have to be less than 100. A negative percent is not reasonable, so x would have to be a rational number greater than 0.

1 Rational FunctionsExamples 1–4 show how to analyze and graph the rational function

f (x ) = a (x )

_ b (x )

by analyzing the real zeros

of b(x ) to find the domain and by using limits or by comparing the degree n of a (x ) to the degree m of b (x ) to find the equations of the vertical, horizontal, and oblique asymptotes. Example 5 shows how there can be removable discontinuities in a rational function.



Reading Math

Limit Notation The expression

lim x→ c -

f (x ) is read as the limit of f

of x as x approaches c from the

left and the expression lim x→ c +

f (x ) is read as the limit of f of x

as x approaches c from the right.

You can use your knowledge of limits, discontinuity, and end behavior to determine the vertical and horizontal asymptotes, if any, of a rational function.

Words The line x = c is a vertical asymptote of the graph of f if lim x→ c -

f (x) = ±∞ or lim x→ c +

f (x ) = ±∞.

Example

y

x

−4

−4

8

4

4

=

verticalasymptote: = -2

( ) 5_( + 2)2

Words The line y = c is a horizontal asymptote of the graph of f if lim x→-∞

f (x ) = c or lim x→∞ f (x ) = c.

Example

y

x

−4

−4

8

4

4

=( ) 6 2_( 2 + 2)

horizontalasymptote: = 6

Key Concept Vertical and Horizontal Asymptotes

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

a. f (x) = x + 4 _

x - 3

Step 1 Find the domain.

The function is undefined at the real zero of the denominator b (x) = x - 3. The real zero of b (x) is 3. Therefore, the domain of f is all real numbers except x = 3.

Step 2 Find the asymptotes, if any.

Check for vertical asymptotes.

Determine whether x = 3 is a point of infinite discontinuity. Find the limit as x approaches 3 from the left and the right.

x 2.9 2.99 2.999 3 3.001 3.01 3.1

f (x ) -69 -699 -6999 undefined 7001 701 71

Because lim x→ 3 - f (x) = -∞ and lim

x→ 3 + f (x) = ∞, you know that x = 3 is a vertical

asymptote of f.

Check for horizontal asymptotes.

Use a table to examine the end behavior of f (x).

x -10,000 -1000 -100 0 100 1000 10,000

f (x ) 0.9993 0.9930 0.9320 -1.33 1.0722 1.0070 1.0007

The table suggests that lim x→-∞ f (x) = lim x→∞

f (x) = 1. Therefore, you know that y = 1 is a

horizontal asymptote of f.

CHECK The graph of f (x) = x + 4 _ x - 3

shown supports each of y

−4

−8

−4−8

8

4

4 8 x

these findings. �

Example 1 Find Vertical and Horizontal Asymptotes

131

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Additional Example


a. f (x ) = x _ x - 1

D = {x | x ≠ 1,

x ∈ �}; vertical asymptote at x = 1; horizontal asymptote at y = 1

b. f (x ) = 4 x 2 + 3 _ 2 x 2 + 1

D = {x | x ∈ �}; no vertical asymptotes; horizontal asymptote at y = 2


1

Teach with TechStudent Response System Show students the equation of a rational function. Ask them if the function has a vertical asymptote. Have students respond A for yes and B for no.

Tips for New TeachersRational Functions The relationships described in the Key Concept box assume that the rational function f cannot be reduced to a constant function.


Extension Use after Example 1 Ask students to determine the horizontal asymptotes of

f (x ) = x _ x - 4

, f (x ) = x 2 _ x 2 - 4

, and f (x ) = x 3 _ x 3 - 4

. Then have students make a conjecture about the

asymptotes for f (x ) = x n _ x n - 4

for any positive integer n. Each function has y = 1 as a horizontal asymptote.


b. g (x) = 8 x 2 + 5 _

4 x 2 + 1

Step 1 The zeros of the denominator b (x) = 4 x 2 + 1 are imaginary, so the domain of g is all real numbers.

Step 2 Because the domain of g is all real numbers, the function has no vertical asymptotes.

Using division, you can determine that

g (x) = 8 x 2 + 5 _ 4 x 2 + 1

= 2 + 3 _ 4 x 2 + 1

.

As the value of |x| increases, 4 x 2 + 1 becomes an increasing large positive number

and 3 _ 4 x 2 + 1

decreases, approaching 0. Therefore,

lim x→-∞

g (x) = lim x→∞

g (x) = 2 + 0 or 2.

CHECK You can use a table of values to support this y

x

reasoning. The graph of g (x) = 8 x 2 + 5 _ 4x 2 + 1

shown

also supports each of these findings. �



1A. m(x) = 15x + 3 _ x + 5

1B. h (x) = x 2 - x - 6 _ x + 4

1A. D = {x | x ≠ -5, x ∈ �}; vertical asymptote at x = -5; horizontal asymptote at y = 15

The analysis in Example 1 suggests a connection between the end behavior of a function and its horizontal asymptote. This relationship, along with other features of the graphs of rational functions, is summarized below.

If f is the rational function given by

f (x ) = a (x )

_ b (x )

= a n x n + a n - 1 x n - 1 + … + a 1 x + a 0

___ b m x m + b m - 1 x m - 1 + … + b 1 x + b 0

,

where b (x ) ≠ 0 and a (x ) and b (x ) have no common factors other than ±1, then the graph of f has the following characteristics.

Vertical Asymptotes Vertical asymptotes may occur at the real zeros of b (x ).

Horizontal Asymptote The graph has either one or no horizontal asymptotes as determined by comparing the degree n of a (x ) to the degree m of b (x ).

• If n < m, the horizontal asymptote is y = 0.

• If n = m, the horizontal asymptote is y = a n

_ b m

.

• If n > m, there is no horizontal asymptote.

Intercepts The x-intercepts, if any, occur at the real zeros of a (x ). The y-intercept, if it exists, is the value of fwhen x = 0.

Key Concept Graphs of Rational Functions

Study TipPoles A vertical asymptote in the graph of a rational function is also called a pole of the function.

1B. D = {x | x ≠ -4, x ∈ �}; vertical asymptote

at x = -4; no

horizontal asymptotes


0130_0140_AM_S_C02_L05_664291.indd 132 4/30/12 6:19 PM



2B. vertical asymptotes at x = -2 and x = 1; horizontal asymptote at y = 0; x-intercept: 0; y-intercept: 0; D = {x | x ≠ -2, 1, x ∈ �}

O

−4

−8

−4−8

8

4

4 8 x

y = 0

x = 1

h(x) = x__x 2 + x - 2

x = -2

y

Additional Example

For each function, determine any vertical and horizontal asymptotes and intercepts. Then graph the function and state its domain.

a. k(x ) = 7 _ x + 5

vertical

asymptote at x = -5; horizontal asymptote at y = 0; y-intercept: 1.4 D = {x | x ≠ -5, x ∈ �}

−4

−8

−4−8−12

4

8

4 x

y

b. f (x ) = x + 1 _ x 2 - 4

vertical

asymptotes at x = 2 and x = -2; horizontal asymptote at y = 0; x-intercept: -1; y-intercept: -0.25; D = {x | x ≠ 2, -2, x ∈ �}

x

f(x) =x + 1_x 2 4

y

2

Additional Answers

(Guided Practice)

2A. vertical asymptote at x = -3 and x = 1; horizontal asymptote at

y = 0; y-intercept: -

2 _ 3 ; D =

{x | x ≠ -3, 1, x ∈ �}

−4

−8

−4−8

8

4

4 8 x

y = 0

x = -3

h(x) = 2__x 2 + 2x - 3

y

x = 1


Study TipHyperbola The graphs of the reciprocal functions f (x ) = 1 _

x and

g (x ) = 6 _ x + 3

are called

hyperbolas. You will learn moreabout hyperbolas in Chapter 7.

To graph a rational function, simplify f, if possible, and then follow these steps.


Step 2 Find and sketch the asymptotes, if any.

Step 3 Find and plot the x-intercepts and y-intercept, if any.

Step 4 Find and plot at least one point in the test intervals determined by any x-intercepts and vertical asymptotes.

Example 2 Graph Rational Functions: n < m and n > mFor each function, determine any vertical and horizontal asymptotes and intercepts. Then graph the function, and state its domain.

a. g(x) = 6 _ x + 3

Step 1 The function is undefined at b(x) = 0, so the domain is {x | x ≠ -3, x ∈ �}.

Step 2 There is a vertical asymptote at x = -3.

The degree of the polynomial in the numerator is 0, and the degree of the polynomial in the denominator is 1. Because 0 < 1, the graph of g has a horizontal asymptote at y = 0.

Step 3 The polynomial in the numerator has no real zeros, so g has no x-intercepts. Because g(0) = 2, the y-intercept is 2.

Step 4 Graph the asymptotes and intercepts. Then choose x-values that fall in the test intervals determined by the vertical asymptote to find additional points to plot on the graph. Use smooth curves to complete the graph.

Interval x (x, g (x ))

(-∞, -3)

-8 (-8, -1.2)

-6 (-6, -2)

-4 (-4, -6)

(-3, ∞)-2 (-2, 6)

2 (2, 1.2)

y

xy = 0

x = -3

=( ) 6_+ 3

b. k(x) = x 2 - 7x + 10 __

x - 3

Factoring the numerator yields k(x) = (x - 2)(x - 5)

__ x - 3

. Notice that the numerator and

denominator have no common factors, so the expression is in simplest form.

Step 1 The function is undefined at b(x) = 0, so the domain is {x | x ≠ 3, x ∈ �}.

Step 2 There is a vertical asymptote at x = 3.

Compare the degrees of the numerator and denominator. Because 2 > 1, there is no horizontal asymptote.

Step 3 The numerator has zeros at x = 2 and x = 5, so the x-intercepts are 2 and 5. k(0) = -

10 _ 3 , so the y-intercept is at about -3.3.

Step 4 Graph the asymptotes and intercepts. y

x

−4

−8

−4−8

8

4

4 8

x = 3

( ) =2 - 7 + 10__

- 3

Then find and plot points in the test intervals determined by the intercepts and vertical asymptotes: (-∞, 0), (0, 3), (3, ∞). Use smooth curves to complete the graph.


2A. h(x) = 2 _ x 2 + 2x - 3

2B. n(x) = x _ x 2 + x - 2

2A–B. See margin.

T

S

S

S

S

Study Tip Test Intervals A rational function can change sign at its zeros and its undefined values, so when these x-values are ordered, they divide the domain of the function into test intervals that can help you determine if the graph lies above or below the x-axis.

133

0130_0140_AM_S_C02_L05_664291.indd 133 4/30/12 6:19 PM



Additional Example

Determine any vertical and horizontal asymptotes and

intercepts for f (x ) = x 2 - x - 12 _ 2 x 2 - 8

.

Then graph the function, and state its domain. vertical asymptotes at x = -2 and x = 2; horizontal asymptote at y = 0.5; x-intercepts: 4, -3; y-intercept: 1.5

x

f(x) = x2 - x - 12__2x 2 - 8

y

D = {x | x ≠ -2, 2, x ∈ �}

3

Tips for New TeachersMissing Branches Depending on the window settings students set on their graphing calculators, students may miss some of the branches of the graph of the rational function they are graphing. Explain the importance of finding as much information as possible about the graph of the function (asymptotes, intercepts, domain) before turning to a calculator for the graph.

Additional Answers

(Guided Practice)

3A. vertical asymptote at x = -2 horizontal asymptote at y = 1; x-intercept: 6; y-intercept: -3; D = {x | x ≠ -2, x ∈ �}

y

−4

−8−16

4

8 x

h(x) = x - 6_x + 2

y = 1

x = -2

3B. vertical asymptotes at x = -1 and x = 1;

horizontal asymptote at y = 1 _ 5 ;

x-intercepts: 2, -2; y-intercept: 4 _ 5 ;

D = {x | x ≠ -1, 1, x ∈ �}

y

x

y = 15

h(x) = x2 - 4

5x2_

- 5

x = 1x = -1

Study Tip Nonlinear Asymptotes

Horizontal, vertical, and oblique asymptotes are all linear. A rational function can also have a nonlinear asymptote. For example,

the graph of f (x ) = x 3 _ x - 1

has a

quadratic asymptote.

y

x

f (x) = x3_

x 1

y = x 2 + x + 1

−4

8

12

4

4

In Example 3, the degree of the numerator is equal to the degree of the denominator.

Example 3 Graph a Rational Function: n = m

Determine any vertical and horizontal asymptotes and intercepts for f (x) = 3 x 2 - 3 _ x 2 - 9

. Then graph the function, and state its domain.

Factoring both numerator and denominator yields f (x) = 3(x - 1)(x + 1)

__ (x - 3)(x + 3)

with no common factors.

Step 1 The function is undefined at b(x) = 0, so the domain is {x | x ≠ -3, 3, x ∈ �}.

Step 2 There are vertical asymptotes at x = 3 and x = -3.

There is a horizontal asymptote at y = 3 _ 1 or y = 3, the ratio of the leading coefficients of

the numerator and denominator, because the degrees of the polynomials are equal.

Step 3 The x-intercepts are 1 and -1, the zeros of the numerator. The y-intercept is

1 _ 3 because f (0) = 1 _

3 .

Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals (-∞, -3), (-3, -1), (-1, 1), (1, 3), and (3, ∞).

y

x

y = 3

x = −3

x = 3

=( ) 3 2 - 3_2 - 9


For each function, determine any vertical and horizontal asymptotes and intercepts. Then graph the function and state its domain.

3A. h(x) = x - 6 _ x + 2

3B. h(x) = x 2 - 4 _ 5 x 2 - 5

3A–B. See margin.

When the degree of the numerator is exactly one more than the degree of the denominator, the graph has a slant or oblique asymptote.

If f is the rational function given by

f (x ) = a (x )

_ b (x )

= a n x n + a n - 1 x n - 1 + … + a 1 x + a 0

___ b m x m + b m - 1 x m - 1 + … + b 1 x + b 0

,

where b (x ) has a degree greater than 0 and a (x ) and b (x ) have no common factors other than 1, then the graph of f has an oblique asymptote if n = m + 1. The function for the oblique asymptote is the quotient polynomial q (x ) resulting from the division of a (x ) by b (x ).

f (x ) = a (x )

_ b (x )

= q (x ) + r (x )

_ b (x )

function for oblique asymptote

Example

y

x

=

ObliqueAsymptote: = - 1

( )2_

+ 1

Key Concept Oblique Asymptotes


0130_0140_AM_S_C02_L05_664291.indd 134 4/30/12 6:19 PM



Additional Example

Determine any asymptotes and

intercepts for f (x ) = x 2 + x - 8 _

x + 3 .

Then graph the function, and state its domain. vertical asymptote at x = -3; slant asymptote at y = x - 2;

x-intercepts: -1 + √ � 33 _

2 ,

-1 - √ � 33 _

2 ; y-intercept: -

8 _ 3

x

y = x - 2

f(x) = x2 + x - 8__

x + 3

−4

−8

−4−8

8

4

4 8

y

D = {x | x ≠ -3, x ∈ �}

4

Additional Answers

(Guided Practice)

4A. vertical asymptote at x = -4; oblique asymptote at y = x - 1; x-intercepts: about 0.8, about 3.8;

y-intercept: - 3 _ 4 ;

D = {x | x ≠ -4, x ∈ �}

y

−4

−8

−4−8−12

4

4 x

y = x - 1x = -4

h(x) = x2 + 3x - 3

x + 4__

4B. vertical asymptote at x = 3 _ 2 ; oblique

asymptote at y = 1 _ 2 x - 5 _

4 ; x-intercepts:

about 0.27, about 3.73; y-intercept: - 1 _ 3 ;

D = ⎧

⎨

⎩

x x ≠ 3 _ 2 , x ∈ �

⎫

⎬

⎭

−8

−16

−8−16

16

8

8 16x

x = 1.5

p (x) = x2

-__ 4x + 12x - 3

y = 1_2

x - 5_4

y3

Study Tip End-Behavior Asymptote

In Example 4, the graph of f approaches the slant asymptote y = 2x - 2 as x → ±∞. Between the vertical asymptotes x = -4 and x = 3, however, the graph crosses the line y = 2x - 2. For this reason, a slant or horizontal asymptote is sometimes referred to as an end-behavior asymptote.


Study Tip Removable and Nonremovable

Discontinuities If the function is not continuous at x = a, but could be made continuous at that point by simplifying, then the function has a removable discontinuity at x = a. Otherwise, it has a nonremovable discontinuity x = a.

Example 4 Graph a Rational Function: n = m + 1

Determine any asymptotes and intercepts for f (x) = 2 x 3 _ x 2 + x - 12


Factoring the denominator yields f (x) = 2 x 3 __ (x + 4)(x - 3)

.


Step 2 There are vertical asymptotes at x = -4 and x = 3.

The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote.

Because the degree of the numerator is exactly one more than the degree of the denominator, f has a slant asymptote. Using polynomial division, you can write the following.

f (x) = 2 x 3 _ x 2 + x - 12

= 2x - 2 + 26x - 24 _ x 2 + x - 12

Therefore, the equation of the slant asymptote is y = 2x - 2.

Step 3 The x- and y-intercepts are 0 because 0 is the zero of the numerator and f (0) = 0.

Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals (-∞, -4), (-4, 0), (0, 3), and (3, ∞).

y

x

−20

−40

−8−16

40

20

8 16

= 3

= 2 - 2

=( ) 2 3__2 + - 12

= -4


For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

4A. h(x) = x 2 + 3x - 3 _ x + 4

4B. p(x) = x 2 - 4x + 1 _ 2x - 3

4A–B. See margin.

When the numerator and denominator of a rational function have common factors, the graph of the function has removable discontinuities called holes, at the zeros of the common factors. Be sure to indicate these points of discontinuity when you graph the function.

f (x) = (x - a)(x - a) (x - b)

__ (x - a)(x - a) (x - c)

Divide out the common factor in

the numerator and denominator.

The zero of x - a is a.

y

xa

hole atx = a

f (x)

135

0130_0140_AM_S_C02_L05_664291.indd 135 4/30/12 6:19 PM



Additional Example

Determine any vertical and horizontal asymptotes, holes, and

intercepts for h (x ) = x 2 - 9 _ x 2 - x - 6

.

Then graph the function and state its domain. vertical asymptote at x = -2; horizontal asymptote at y = 1; x-intercept:

-3; y-intercept: 3 _ 2 ; hole: (3, 6 _

5 )

x

h(x) = x2 - 9__

x 2 - x - 6

y

D = {x | x ≠ -2, 3, x ∈ �}

5

Tips for New TeachersHoles in Graphs When the numerator and denominator of a rational function have a common factor (x - c ), the point (c, f (c )) needs to be omitted from the graph. It is indicated by a circle or hole. If x = c is a vertical asymptote, there is no hole in the graph.

2 Rational EquationsExamples 6–8 show how to solve rational equations involving fractions. All terms are multiplied by the LCD and then the variable is isolated. Solutions should be checked in the original equation to identify any extraneous solutions.

Additional Example

Solve x -

4 _ x - 6

= 0. 3 ± √ � 13 6


Verbal/Linguistic Learners Divide students into groups of three or four. Write a rational equation on the board and have each group solve it, writing the steps they use to find the solution. Then have groups compare and contrast the processes they used.


Study Tip Hole For Example 5, x + 2 was divided out of the original expression. Substitute -2 into the new expression.

h (-2) = (-2) - 2

_ (-2) -4

= -4 _ -6

or 2 _ 3

There is a hole at (-2, 2 _ 3 ) .

Study TipCheck for Reasonableness

You can also check the result in Example 6 by using a graphing calculator to graph y = x + 6 _

x - 8 .

Use the CALC menu to locate the zeros. Because the zeros of the graph appear to be at about x = 7.16 and x = 0.84, the solution is reasonable.

[-20, 20] scl: 2 by [-20, 20] scl: 2

Example 5 Graph a Rational Function with Common Factors

Determine any vertical and horizontal asymptotes, holes, and intercepts for h(x) = x 2 - 4 _ x 2 - 2x - 8


Factoring both the numerator and denominator yields h(x) = (x - 2)(x + 2)

__ (x - 4)(x + 2)

or x - 2 _ x - 4

, x ≠ -2.


Step 2 There is a vertical asymptote at x = 4, the real zero of the simplified denominator.

There is a horizontal asymptote at y = 1 _ 1 or 1, the ratio of the leading coefficients of the

numerator and denominator, because the degrees of the polynomials are equal.

Step 3 The x-intercept is 2, the zero of the simplified numerator. The y-intercept is

1 _ 2 because h(0) = 1 _

2 .

Step 4 Graph the asymptotes and intercepts. y

x= 4

= 1

=( )2 - 4__

2 - 2 - 8Then find and plot points in the test intervals (-∞, 2), (2, 4), and (4, ∞).

There is a hole at (-2, 2 _ 3 ) because

the original function is undefined when x = -2.


For each function, determine any vertical and horizontal asymptotes, holes, and intercepts. Then graph the function and state its domain.

5A. g(x) = x 2 + 10x + 24 __ x 2 + x - 12

5B. c(x) = x 2 - 2x - 3 _ x 2 - 4x - 5

(x + 2)(x + 2)

See margin. See margin.

2 Rational Equations Rational equations involving fractions can be solved by multiplying each term in the equation by the least common denominator (LCD) of all the terms of the

equation.

Example 6 Solve a Rational Equation

Solve x + 6 _ x - 8

= 0.

x + 6 _ x - 8

= 0 Original equation

x(x - 8) + 6 _ x - 8

(x - 8) = 0(x - 8) Multiply by the LCD, x - 8.

x 2 - 8x + 6 = 0 Distributive Property

x = 8 ± √ �� (-8) 2 - 4(1)(6)

__ 2(1)

Quadratic Formula

x = 8 ± 2 √ � 10 _

2 or 4 ± √ � 10 Simplify.



6A. 20 _ x + 3

- 4 = 0 6B. 9x _ x - 2

= 6 2 -4


0130_0140_AM_S_C02_L05_664291.indd 136 4/30/12 6:19 PM



Additional Examples

Solve x +

x_x - 1

=

3x - 2_x - 1

. 2

WATER CURRENT The rate of the water current in a river is 4 miles per hour. In 2 hours, a boat travels 6 miles with the current to one end of the river and 6 miles back. If r is the rate of the boat in still water, r + 4 is its rate with the current, r - 4 is its rate against the current, and 6_

r - 4+

6_r + 4

= 2, find r. 8

7

8

Additional Answers

(Guided Practice)

5A. vertical asymptote at x = 3; horizontal asymptote at y = 1; x-intercept: -6; y-intercept: -2;

hole: (-4, -

2 _ 7 )

D = {x | x ≠ -4, 3, x ∈ �}

y

−8

−16

−4

16

8

4 8 x

x = 3

y = 1

g (x) = x2 + 10x + 24x 2__

+ x - 12

5B. vertical asymptote at x = 5; horizontal asymptote at y = 1;

x-intercept: 3; y-intercept: 3 _ 5 ;

hole: (-1, 2 _ 3 ) ; D = {x | x ≠ -1, 5,

x ∈ �}

y

−4

−8

−4

8

4

4 8 12x

x = 5

y = 1

c (x) = x2 - 2x - 3x2__

- 4x - 5


Study Tip Intersection You can use the intersection feature of your graphing calculator to solve a rational equation by graphing each side of the equation and finding all of the intersections of the two graphs.

Solving a rational equation can produce extraneous solutions. Always check your answers in the original equation.

Example 7 Solve a Rational Equation with Extraneous Solutions

Solve 4 _

x 2 - 6x + 8 =

3x _ x - 2

+ 2 _

x - 4 .

The LCD of the expressions is (x - 2)(x - 4), which are the factors of x 2 - 6x + 8.

4 _ x 2 - 6x + 8

= 3x _ x - 2

+ 2 _ x - 4

Original equation

(x - 2)(x - 4) 4 _ x 2 - 6x + 8

= (x - 2)(x - 4) ( 3x _ x - 2

+ 2 _ x - 4

) Multiply by the LCD.

4 = 3x(x - 4) + 2(x - 2) Distributive Property

4 = 3 x 2 - 10x - 4 Distributive Property

0 = 3 x 2 - 10x - 8 Subtract 4 from each side.

0 = (3x + 2)(x - 4) Factor.

x = - 2 _ 3 or x = 4 Solve.

Because the original equation is not defined when x = 4, you can eliminate this extraneous solution. So, the only solution is - 2 _

3 .



7A. 2x _ x + 3

+ 3 _ x - 6

= 27 __ x 2 - 3x - 18

7B. - 12 _ x 2 + 6x

= 2 _ x + 6

+ x - 2 _ x -

3

_ 2 no solution

ELECTRICITY The diagram of an electric circuit shows three parallel

321

resistors. If R is the equivalent resistance of the three resistors,

then 1 _ R

= 1 _

R 1 + 1

_ R 2

+ 1 _

R 3 . In this circuit, R 1 is twice the resistance

of R 2 , and R 3 equals 20 ohms. Suppose the equivalent resistance is equal to 10 ohms. Find R 1 and R 2 .

1 _ R

= 1 _ R 1

+ 1 _ R 2

+ 1 _ R 3

Original equation

1 _ 10

= 1 _ 2 R 2

+ 1 _ R 2

+ 1 _ 20

R = 10, R 1 = 2 R 2 , and R 3 = 20

1 _ 20

= 1 _ 2 R 2

+ 1 _ R 2

Subtract 1

_ 20

from each side.

(20 R 2 ) 1 _ 20

= ( 20 R 2 ) ( 1 _ 2 R 2

+ 1 _ R 2

) Multiply each side by the LCD, 20 R 2 .

R 2 = 10 + 20 or 30 Simplify.

R 2 is 30 ohms and R 1 = 2 R 2 or 60 ohms.


8. ELECTRONICS Suppose the current I, in amps, in an electric circuit is given by the formula I = t + 1 _

10 - t , where t is time in seconds. At what time is the current 1 amp?

about 10.1 or about 0.9 seconds

Real-World Example 8 Solve a Rational Equation

Real-World Career Electrician Electricians install and maintain various components of electricity, such as wiring and fuses. They must maintain compliance with national, state, and local codes. Most electricians complete an apprenticeship program that includes both classroom instruction and on-the-job training.

137

A. R

amey

/Pho

toEd

it

0130_0140_AM_S_C02_L05_664291.indd 137 5/31/12 6:45 PM




Then use the table below to customize assignments for your students.

Watch Out!

Common Error Rational equations, such as those in Exercises 35 and 41, have an expression in the denominator of a term that is not written with the variable first. Suggest that students first rewrite these equations with the denominators written in standard form. For example, a _

a + 3 -

3 _ 4 - a

= 2a - 2 _ a 2 - a - 12

can be rewritten as

a _ a + 3

-

3 _ - a + 4

= 2a - 2 _ a 2 - a - 12

before they determine the LCD.

Additional Answers

1. D = {x | x ≠ 2, -2, x ∈ �}; x = 2, x = -2, y = 1

2. D = {x | x ≠ -4, x ∈ �}; x = -4

3. D = {x | x ≠ 4, -3, x ∈ �}; x = 4, x = -3

4. D = {x | x ≠ -3, -5, x ∈ �}; x = -3, x = -5, y = 0

5. D = {x | x ≠ 0, -2, x ∈ �}; x = 0, x = -2, y = 2

6. D = {x | x ≠ 4, x ∈ �}; x = 4

7. D = {x | x ≠ 2, -4, x ∈ �}; x = 2, x = -4, y = 0

8. D = {x | x ≠ 3, -1, x ∈ �}; x = 3, x = -1, y = 1



AL Approaching Level 1–42, 54, 55, 57, 59–76 1–41 odd 2–42 even, 54, 55, 57, 59–76

OL On Level 1–51 odd, 52–55, 57, 59–76

1–42 43–55, 57, 59–76



Exercises

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any. (Example 1)

1. f (x) = x 2 - 2 _ x 2 - 4

2. h(x) = x 3 - 8 _ x + 4

3. f (x) = x(x - 1) (x + 2) 2

__ (x + 3)(x - 4)

4. g(x) = x - 6 __ (x + 3)(x + 5)

5. h(x) = 2 x 2 - 4x + 1 __ x 2 + 2x

6. f (x) = x 2 + 9x + 20 __ x - 4

7. h(x) = (x - 1)(x + 1)

__ (x - 2) 2 (x + 4) 2

8. g(x) = (x - 4)(x + 2)

__ (x + 1)(x - 3)

For each function, determine any asymptotes and intercepts. Then graph the function, and state its domain. (Examples 2–5)

9. f (x) = (x + 2)(x - 3)

__ (x + 4)(x - 5)

10. g(x) = (2x + 3)(x - 6)

__ (x + 2)(x - 1)

11. f (x) = 8 __ (x - 2)(x + 2)

12. f (x) = x + 2 _ x(x - 6)

13. g(x) = (x + 2)(x + 5)

__ (x + 5) 2 (x - 6)

14. h(x) = (x + 6)(x + 4)

__ x(x - 5)(x + 2)

15. h(x) = x 2 (x - 2)(x + 5)

__ x 2 + 4x + 3

16. f (x) = x (x + 6) 2 (x - 4)

__ x 2 - 5x - 24

17. f (x) = x - 8 _ x 2 + 4x + 5

18. g(x) = -4 _ x 2 + 6

19. SALES The business plan for a new car wash projects that profits in thousands of dollars will be modeled by the

function p(z) = 3 z 2 - 3 __ 2 z 2 + 7z + 5

, where z is the week of

operation and z = 0 represents opening. (Example 4)

a. State the domain of the function.

b. Determine any vertical and horizontal asymptotes and intercepts for p(z).

c. Graph the function.

For each function, determine any asymptotes, holes, and intercepts. Then graph the function and state its domain. (Examples 2–5)

20. h(x) = 3x - 4 _ x 3

21. h(x) = 4 x 2 - 2x + 1 __ 3 x 3 + 4

22. f (x) = x 2 + 2x - 15 __ x 2 + 4x + 3

23. g(x) = x + 7 _ x - 4

24. h(x) = x 3 _ x + 3

25. g(x) = x 3 + 3 x 2 + 2x __ x - 4

26. f (x) = x 2 - 4x - 21 __ x 3 + 2 x 2 - 5x - 6

27 g(x) = x 2 - 4 __ x 3 + x 2 - 4x - 4

28. f (x) = (x + 4)(x - 1)

__ (x - 1)(x + 3)

29. g(x) = (2x + 1)(x - 5)

__ (x - 5) (x + 4) 2

30. STATISTICS A number x is said to be the harmonic mean of y

and z if 1 _ x is the average of 1 _ y and 1 _ z . (Example 7)

a. Write an equation for which the solution is the harmonic mean of 30 and 45.

b. Find the harmonic mean of 30 and 45.

31. OPTICS The lens equation is 1 _ f = 1 _

d i + 1 _

d o , where f is the

focal length, d i is the distance from the lens to the image, and d o is the distance from the lens to the object. Suppose the object is 32 centimeters from the lens and the focal length is 8 centimeters. (Example 7)

LensObject

Image

o i

f

a. Write a rational equation to model the situation.

b. Find the distance from the lens to the image.

Solve each equation. (Examples 6–8)

32. y + 6 _ y = 5 33. 8 _ z - z = 4

34. x - 1 _ 2x - 4

+ x + 2 _ 3x

= 1 35. 2 _ y + 2

- y _

2 - y = y 2 + 4

_ y 2 - 4

36. 3 _ x + 2 _

x + 1 = 23 _

x 2 + x 37. 4 _

x - 2 - 2 _

x = 14 _

x 2 - 2x

38. x _ x + 1

- x - 1 _ x = 1 _

20 39. 6 _

x - 3 - 4 _

x + 2 = 12 _

x 2 - x - 6

40. x - 1 _ x - 2

+ 3x + 6 _ 2x + 1

= 3 41. 2 _ a + 3

- 3 _ 4 - a

=

2a - 2 _ a 2 - a - 12

42. WATER The cost per day to remove x percent of the salt

from seawater at a desalination plant is c(x) = 994x _ 100 - x

, where 0 ≤ x < 100.


b. Graph the line y = 8000 and find the intersection with the graph of c(x) to determine what percent of salt can be removed for $8000 per day.

c. According to the model, is it feasible for the plant to remove 100% of the salt? Explain your reasoning.

Write a rational function for each set of characteristics.

43. x-intercepts at x = 0 and x = 4, vertical asymptotes at x = 1 and x = 6, and a horizontal asymptote at y = 0

44. x-intercepts at x = 2 and x = -3, vertical asymptote at x = 4, and point discontinuity at (-5, 0)

1–8. See margin.


a. D = {z | z ≥ 0, z ∈ R}

b. horizontal asymptote:

y = 3

_ 2 ; x-intercept: 1;

y-intercept: -

3

_ 5

c. See Chapter 2

Answer Appendix.


1

_ x = 1

_ 30

+ 1

_ 45

_ 2

36

1

_ 8 =

1

_ d i

+ 1

_ 32

10 2

_ 3 cm

35. no solution

3, 2 ±2 √ � 3 - 2

1, 8

4 5

4, -5 -6

7, 1

-1B

a–c. See Chapter 2 Answer Appendix.



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Additional Answers

43. Sample answer: f (x ) = x (x - 4)

__ (x - 1) (x - 6) 2

44. Sample answer: f (x ) = (x - 2)(x + 3) (x + 5) 2

__ (x - 4)(x + 5)

46. Sample answer: f (x ) = x 2 - 3x - 4 _ x 2 + 5x - 6

47. Sample answer: f (x ) = 2 x 2 - 14x + 20 __

x 2 + x - 12



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Study Guide and InterventionRational Functions

Graphs of Rational Functions Asymptotes are lines that a graph approaches.

For f(x) = a(x)

−

b(x)

, where a(x) and b(x) have no common factors other than one,

and n is the degree of a(x) and m is the degree of b(x):• Vertical asymptotes: May occur at the real zeros of b(x).• Horizontal asymptotes: - If n < m , the asymptote is y = 0. - If n = m, the asymptote is y = c, where c is the ratio of the leading

coefficients of the numerator and denominator. - If n > m, there is no horizontal asymptote.• Oblique asymptotes: If n = m + 1, the equation is the quotient

polynomial q(x) of f(x), or a(x)

−

b(x)

= q(x) + r(x)

−

b(x)

.

• x-intercepts, if any, occur at the real zeros of a(x). The y-intercept, if it exists, is the value of f when x = 0.

Determine any asymptotes and intercepts for

f(x) = 2x - 1 − x + 3

. Then graph the function and state its domain.Vertical asymptotes: The zero of the denominator is -3. The vertical asymptote is x = -3.

Horizontal asymptotes: The degree of the numerator equals the degree of the denominator, so the horizontal asymptote is y = 2.

Intercepts: To find x-intercepts, find the zeros of the numerator by solving 2x - 1 = 0. So, the x-intercept is 1 −

2 . To find the y-intercept, substitute 0 for x.

The y-intercept is -

1 −

3 .

Graph the asymptotes and intercepts. Find and plot points in each interval.D = {x | x ≠ -3, x ∈ �}

Exercise

Determine any asymptotes and intercepts for f(x) = x2 + 1 − x - 2

.

Then graph the function and state its domain.

vertical asymptote: x = 2,

oblique asymptote: y = x + 2, y-int: - 1 − 2

D = {x | x ≠ 2, x ∈ �}

Example

y

x4 8−4−8

4

8

−8

−4

y

x4 8−8

8

16

−16

−8

2-5

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Practice

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PracticeRational Functions


1. f(x) = 4 −

x2 + 1 D = {x | x ∈ �}; no vertical asymptotes: horizontal

asymptote: y = 0

2. f(x) = 2x + 1 −

x + 1 D = {x | x ≠ -1, x ∈ �}; vertical asymptote: x = -1;

horizontal asymptote: y = 2

3. f(x) = x + 3 −

(x + 1)(x - 2)

D = {x | x ≠ -1, 2, x ∈ �}; vertical asymptotes:

x = -1, and x = 2; horizontal asymptote: y = 0

For each function, determine any asymptotes, holes, and intercepts. Then graph the function and state its domain.

4. f(x) = 5x2 - 10x + 1 −

x - 2 5. f(x) = x

2 + x - 6 −

x + 1

x4 8−4−8

−12

y

8

16 y

x


6. 1 −

2n + 6n - 9 −

3n = 2 −

n 7. t - 4 − t = 3

9 −

4 -1, 4

8. 3a −

2a + 1 - 4 −

2a - 1 = 1 9.

2p −

p + 1 + 3 −

p - 1 =

15 - p −

p2 - 1

11 ± √ �� 145

− 4 -3, 2

10. FUNDRAISER A school is sponsoring a fundraiser by selling athletic calendars. The total cost of printing the calendar is $440 plus an additional charge of $2.75 per calendar. How many calendars must be sold to keep the cost of printing under $5 per calendar? 196 calendars

2-5

x = 2; y = 5x;

x-int: 5 + 2 √ � 5

− 5 ;

y-int: - 1 − 2

D = {x | x ≠ 2, x ∈ �}

x = -1; y = x;

x-int: -3, 2;

y-int: -6

D = {x | x ≠ -1, x ∈ �}

005_038_PCCRMC02_893803.indd 29 10/20/09 8:13:09 PM


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Word Problem PracticeRational Functions

1. POLLUTION A cost-benefit model may be used to express the cost of cleaning up environmental pollution as a function of the percent of pollution removed from the environment. A typical model is

C(x) = 8.9x−

100 - x.

a. Determine any asymptotes and intercepts for the function.y = -8.9, x = 100; y-int: 0; x-int: 0

b. State the relevant domain of the function. D = {x | 0 ≤ x < 100, x ∈ �}

c. Graph the function.

x20 40 60 80 100

y

20

40

60

80

100

2. MEDICINE The concentration of a certain medicine in a person’s body after x hours is modeled by f(x) = 35x −

x2 + 10 .

Determine any asymptotes and intercepts for the function.

y = 0; y-int: 0; x-int: 0

3. ADVERTISING The function f(x) = 6x2 + x −

x + 1 gives the approximate

number of new Web site hits a company will receive after x minutes of television advertising. Determine any asymptotes for the function.

x = -1; y = 6x - 5

4. PETS Candice is designing a play center for the cats in a shelter. She is considering a center that is multi-level and cylindrical in shape. She would carpet each level and has 4000 square inches of available carpet. The function f(r) =

4000−

πr2 gives the number of levels

she can make if each level has a radiusof r inches.


[0, 15] scl: 1 by [1, 1000] scl: 50

b. Graph the line y = 8 and find the intersection to determine the radius of each level if she has enough carpet to make at most 8 levels. Round to the nearest tenth. 12.6 in.

5. PACKAGING The surface area of a box with a volume of 625 cubic inches is given by S(x) = 2x2 + 2500 − x , where x is a side length of the square base.

a. Write the function as the quotient of two functions.

S(x) = 2x3 + 2500

− x

b. Determine any asymptotes and intercepts for the function.

x = 0, y = 2x2

c. Find the side length of the base when the surface area is 512.5 square inches.

5.55 in., 12.5 in.

2-5

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EnrichmentE i h t

Write Rational FunctionsSuppose you are given information about the asymptotes and/or intercepts of a rational function. You could work in reverse to write a rational function that meets those requirements. Read the example below.

A rational function has asymptotes y = x - 5 and x = -1. Write a rational function that fits this description.

Because x = -1 is a vertical asymptote, the denominator is undefined for x = -1. That is, the value of the denominator is 0 when x = -1. Therefore,one possible expression for the denominator is x + 1.

Because the oblique asymptote is y = x - 5, the quotient of the numeratorand denominator, without the remainder, must be x - 5. Therefore, you canmultiply your chosen denominator, x + 1, by x - 5 to find a dividend.

(x + 1)(x - 5) = x2- 4x44 - 5

Because the remainder part does not matter, choose any quadratic expression with a quadratic term of x2 and a linear term of -4x44 .

A tiA rationA ration l f tal functal functi th tion thation that t tmeets t meets th ihe givenhe giveng d idescrip descrippti ition istion is ffff((((ffffff xx)))) == xx22

- 44 4xx444 ++ 44 4−−

xx +++ 11 1 .

ExerciExerciExercisessesses

W itWrite aWrite a ti lrationalrational f tifunctio functio th t fn that fn that fit hits eachits each d idescrip descriptitiontion.

2-5

45 TRAVEL When distance and time are held constant, the average rates, in miles per hour, during a round trip can

be modeled by r 2 = 30 r 1

_ r 1 - 30

, where r 1 represents the

average rate during the first leg of the trip and r 2 represents the average rate during the return trip.

a. Find the vertical and horizontal asymptotes of the function, if any. Verify your answer graphically.

b. Copy and complete the table shown.

r 1 45 50 55 60 65 70

r 2

c. Is a domain of r 1 > 30 reasonable for this situation? Explain your reasoning.

Use your knowledge of asymptotes and the provided points to express the function represented by each graph.

46. 47.

y

x

−4

−8

−8−16

4

8

8 16

= 1

= −6

(4, 0)(−5, −6)

= 1

y

x

−4

−8

−4−8

4

8

4 8

= 3

(5, 0)(−1, −3)

= 2

= −4

Use the intersection feature of a graphing calculator to solve each equation.

48. x 4 - 2 x 3 + 1 __ x 3 + 6

= 8 49. 2 x 4 - 5 x 2 + 3 __ x 4 + 3 x 2 - 4

= 1

50. 3 x 3 - 4 x 2 + 8 __ 4 x 4 + 2x - 1

= 2 51. 2 x 5 - 3 x 3 + 5x __ 4 x 3 + 5x - 12

= 6

52. CHEMISTRY When a 60% acetic acid solution is added to 10 liters of a 20% acetic acid solution in a 100-liter tank, the concentration of the total solution changes.

10 L, 20% acetic acid

a L, 60% acetic

V = 100 L

a. Show that the concentration of the solution is f (a) =

3a + 10 _ 5a + 50

, where a is the volume of the 60% solution.

b. Find the relevant domain of f (a) and the vertical or horizontal asymptotes, if any.

c. Explain the significance of any domain restrictions or asymptotes.

d. Disregarding domain restrictions, are there any additional asymptotes of the function? Explain.

53. MULTIPLE REPRESENTATIONS In this problem, you will investigate asymptotes of rational functions.

a. TABULAR Copy and complete the table. Determine the horizontal asymptote of each function algebraically.

FunctionHorizontal

Asymptote

f (x ) = x 2 - 5x + 4 _

x 3 + 2 y = 0

h (x ) = x 3 - 3 x 2 + 4x - 12 __

x 4 - 4 y = 0

g (x ) = x 4 - 1 _ x 5 + 3

y = 0

b. GRAPHICAL Graph each function and its horizontal asymptote from part a.

c. TABULAR Copy and complete the table below. Use the Rational Zero Theorem to help you find the real zeros of the numerator of each function.

FunctionReal Zeros of

Numerator

f (x ) = x 2 - 5x + 4 _

x 3 + 2 1, 4

h (x ) = x 3 - 3 x 2 + 4x - 12 __

x 4 - 4 3

g (x ) = x 4 - 1 _ x 5 + 3

-1, 1

d. VERBAL Make a conjecture about the behavior of the graph of a rational function when the degree of the denominator is greater than the degree of the numerator and the numerator has at least one real zero.


54. REASONING Given f (x) = a x 3 + b x 2 + c __ d x 3 + e x 2 + f

, will f (x) sometimes,

always, or never have a horizontal asymptote at y = 1 if a, b, c, d, e, and f are constants with a ≠ 0 and d ≠ 0. Explain.

55. PREWRITE Design a lesson plan to teach the graphing rational functions topics in this lesson. Make a plan that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion.

56. CHALLENGE Write a rational function that has vertical asymptotes at x = -2 and x = 3 and an oblique asymptote y = 3x.

57. WRITING IN MATH Use words, graphs, tables, and equations to show how to graph a rational function.

58. CHALLENGE Solve for k so that the rational equation has exactly one extraneous solution and one real solution.

2 _ x 2 - 4x + k

= 2x _ x - 1

+ 1 _ x - 3

59. WRITING IN MATH Explain why all of the test intervals must be used in order to get an accurate graph of a rational function.

90 75 66 60 55.7 52.5

a, c. See Chapter 2 Answer Appendix.


≈-1.59, ≈10.05≈-2.65, ≈2.65

≈-0.98, ≈0.90 ≈-3.87, ≈1.20, ≈3.70

C

a–d. See Chapter 2

Answer Appendix.

b, d. See Chapter 2 Answer Appendix.



Sample answer: f (x ) = 3 x 3 - 3 x 2

_ x 2 - x + 6


3



0130_0140_AM_S_C02_L05_664291.indd 139 4/30/12 6:19 PM



4 AssessTicket Out the Door Have students

solve 4 _ x (x - 1)

+ 4 _ x = x _

x - 1 . 4

Additional Answers

69.

70. y

−4

−8

−4−8

8

4 8 x

f (x) = (x - 4) 3

71.

73b. y

x

−40

−80

80

40

4 8 12

( ) = 4 2 - 56 + 96

y

−4−8

8

12

4

x−12

f(x) = (x + 7) 2

y

−8

−4−8

8

4

4 8 x

f (x) = x 4 - 5

Follow-upStudents have explored modeling using power, radical, polynomial, and rational functions.

Ask:■ What are the limitations of mathematical

modeling? Sample answer: Not all real-world situations can be modeled. For those that can be modeled, predictions that are made using the model may not be accurate when based

on data values that are outside the range of data values used to create the model. Therefore, after a model is created, it should be carefully analyzed before it is used to make predictions/decisions.

Spiral Review

List all the possible rational zeros of each function. Then determine which, if any, are zeros. (Lesson 2-4)

60. f (x) = x 3 + 2 x 2 - 5x - 6 61. f (x) = x 3 - 2 x 2 + x + 18 62. f (x) = x 4 - 5 x 3 + 9 x 2 - 7x + 2

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x). (Lesson 2-3)

63. f (x) = x 4 - 2 x 3 - 13 x 2 + 14x + 24; x - 3, x - 2

64. f (x) = 2 x 4 - 5 x 3 - 11 x 2 - 4x; x - 4, 2x - 1

65. f (x) = 6 x 4 + 59 x 3 + 138 x 2 - 45x - 50; 3x - 2, x - 5

66. f (x) = 4 x 4 - 3 x 3 - 12 x 2 + 17x - 6; 4x - 3; x - 1

67. f (x) = 4 x 5 + 15 x 4 + 12 x 3 - 4 x 2 ; x + 2, 4x + 1

68. f (x) = 4 x 5 - 8 x 4 - 5 x 3 + 10 x 2 + x - 2; x + 1, x - 1

Graph each function. (Lesson 2-2)

69. f (x) = (x + 7) 2 70. f (x) = (x - 4) 3 71. f (x) = x 4 - 5

72. RETAIL Sara is shopping at a store that offers $10 cash back for every $50 spent. Let

f (x) = x _ 50

and h(x) = 10x, where x is the amount of money Sara spends. (Lesson 1-6)

a. If Sara spends money at the store, is the cash back bonus represented by f [h(x)] or h [ f (x)]? Explain your reasoning.

b. Determine the cash back bonus if Sara spends $312.68 at the store.

±1, ±2, ±3, ±6; -3, -1, 2 ±1, ±2, ±3, ±6, ±9, ±18; -2 ±1, ±2; 1 (multiplicity: 3), 2

no; yes; f (x) = (x - 2)( x 3 - 13x - 12)

yes; no; f (x ) = (x - 4)(2 x 3 + 3 x 2 + x ) or f (x ) = x (2x + 1)(x + 1)(x - 4)

65. yes; no; f (x ) = (3x - 2)(2 x 3 +

21 x 2 + 60x + 25) or f (x ) =

(3x - 2)(2x + 1)(x + 5 ) 2

66. yes; yes; f (x ) = (4x - 3)(x − 1)( x 2 + x - 2)

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

yes; no; f (x ) = (x + 2)(4 x 4 + 7 x 3 - 2 x 2 ) or f (x ) = x 2 (x + 2 ) 2 (4x - 1)

yes; yes; f (x ) = (x + 1)(x - 1) (4 x 3 - 8 x 2 - x + 2) or

f (x ) = (x - 2)(2x - 1)(2x + 1)(x + 1)(x - 1)69–71. See margin.

a. h [ f (x )]; sample answer: The cash

back bonus is found after calculating

the amount of times $50 was spent.

$60

73. SAT/ACT A company sells ground coffee in two sizes of cylindrical containers. The smaller container holds 10 ounces of coffee. If the larger container has twice the radius of the smaller container and 1.5 times the height, how many ounces of coffee does the larger container hold? (The volume of a cylinder is given by the formula V = π r 2 h.)

A 30 C 60 E 90

B 45 D 75

74. What are the solutions of 1 = 2 _ x 2

+ 2 _ x ?

F x = 1, x = -2 H x = 1 + √ � 3 , x = 1 - √ � 3

G x = -2, x = 1 J x = 1 + √ � 3 _

2 , x = 1 - √ � 3

_ 2

75. REVIEW Alex wanted to determine the average of his 6 test scores. He added the scores correctly to get T but divided by 7 instead of 6. The result was 12 less than his actual average. Which equation could be used to determine the value of T?

A 6T + 12 = 7T C T _ 7 + 12 = T _

6

B T _ 7 = T - 12 _

6 D T _

6 = T - 12 _

7

76. Diana can put a puzzle together in three hours. Ella can put the same puzzle together in five hours. How long will it take them if they work together?

F 1 3 _ 8 hours H 1 3 _

4 hours

G 1 5 _ 8 hours J 1 7 _

8 hours

C

H

C

J



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Lesson 2-6





Chapter

Resource

Masters

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Graphic Calculator Activity

■ Study Guide and Intervention ■ Practice■ Word Problem Practice■ Enrichment■ Graphic Calculator Activity

■ Practice■ Word Problem Practice■ Enrichment■ Graphic Calculator Activity

■ Study Guide and Intervention■ Practice■ Word Problem Practice■ Graphic Calculator Activity






Before Lesson 2-6 Solve polynomial and rational equations.

Lesson 2-6 Solve polynomial inequalities.Solve rational inequalities.

After Lesson 2-6 Solve systems of inequalities in linear optimization.




Ask:■ Solve 2x - 6 = 0. 3

■ Does x = 2 make 2 x - 6 positive or negative? negative

■ Does x = 3 make 2 x - 6 positive or negative? Neither, it makes it equal to 0.

■ Does x = 4 make 2 x - 6 positive or negative? positive


-2-4 2 5

(-)(+) (+) (-)(-)

y =

x

f (x)

f (x)

y

x

New Vocabularypolynomial inequality

sign chart

rational inequality

Why?Many factors are involved when starting a new business, including the amount of the initial investment, maintenance and labor costs, the cost of manufacturing the product being sold, and the actual selling price of the product. Nonlinear inequalities can be used to determine the price at which to sell a product in order to make a specific profit.

Now

1Solve polynomial inequalities.

2Solve rational inequalities.

ThenYou solved polynomial and rational equations. (Lessons 2-3 and 2-4)

1Polynomial Inequalities If f (x) is a polynomial function, then a polynomial inequality has the general form f (x) ≤ 0, f (x) < 0, f (x) ≠ 0, f (x) > 0, or f (x) ≥ 0. The inequality f (x) < 0 is

true when f (x) is negative, while f (x) > 0 is true when f (x) is positive.

In Lesson 1-2, you learned that the x-intercepts of a polynomial function are the real zeros of the function. When ordered, these zeros divide the x-axis into intervals for which the value of f (x) is either entirely positive (above the x-axis) or entirely negative (below the x-axis).

By finding the sign of f (x) for just one x-value in each interval, you can determine on which intervals the function is positive or negative. From the test intervals represented by the sign chart at the right, you know that:

• f (x) < 0 on (-4, -2) ∪ (2, 5) ∪ (5, ∞),

• f (x) ≤ 0 on [-4, -2] ∪ [2, ∞),

• f (x) = 0 at x = -4, -2, 2, 5,

• f (x) > 0 on (-∞, -4) ∪ (-2, 2), and

• f (x) ≥ 0 on (-∞, -4] ∪ [-2, 2] ∪ [5, 5].

Example 1 Solve a Polynomial Inequality

Solve x 2 - 6x - 30 > -3.

Adding 3 to each side, you get x 2 - 6x - 27 > 0. Let f (x) = x 2 - 6x - 27. Factoring yields f (x) = (x + 3)(x - 9), so f (x) has real zeros at -3 and 9. Create a sign chart using these zeros. Then substitute an x-value in each test interval into the factored form of the polynomial to determine if f (x) is positive or negative at that point.

(+)(-)(-)(-) (+)(+)

f (x) = (x + 3)(x - 9)

Test x = -4. Test x = 0. Test x = 10.

x-3 9

Think: (x + 3)and (x - 9) are both negativewhen x = -4.

(-)(+) (+)

x

f (x) = (x + )(x - )

-3 9

Because f (x) is positive on the first and last intervals, the y

x

−20

−4 4 8

f (x) = x 2 - 6x - 30

solution set of x 2 - 6x - 30 > -3 is (-∞, -3) ∪ (9, ∞). The graph of f (x) supports this conclusion, because f(x) is above the x-axis on these same intervals.



1A. x 2 + 5x + 6 < 20 1B. (x - 4) 2 > 4(-7, 2) (-∞, 2) � (6, ∞)

Nonlinear InequalitiesNonlinear Inequalities


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142 | Lesson 2-6 | Nonlinear Inequalities

■ Based on your answers, what conclusion can you make about the solutions of 2x - 6 > 0? x > 3

■ Why do you think 3 is a critical number for the inequality 2x - 6 > 0? Sample answer: When you know what value makes the equation true, you can test intervals on either side of that value to find what makes the inequality true. This inequality is positive for values greater than 3, so x > 3 is the solution.

1 Polynomial InequalitiesExamples 1–3 show how to solve polynomial inequalities using the real zeros of the related polynomial function, their multiplicity, and the function’s end behavior along with a sign chart.


Additional Examples

Solve x 2 - 8x + 16 ≤ 1. [ 3, 5 ]

Solve x 3 - 22 x > 3 x 2 - 24. (-4, 1) ∪ (6, ∞)


a. x 2 + 2 x + 3 < 0 ∅

b. x 2 + 2 x + 3 ≥ 0 (-∞, ∞)

c. x 2 + 12 x + 36 > 0(-∞, -6) ∪ (-6, ∞)

d. x 2 + 12x + 36 ≤ 0 {-6}


1

2

3

Study Tip Polynomial Inequalities You can check the solution of a polynomial inequality by graphing the function and evaluating the truth value of the inequality for each interval of the solution.

If you know the real zeros of a function, including their multiplicity, and the function’s end behavior, you can create a sign chart without testing values.

Example 2 Solve a Polynomial Inequality Using End Behavior

Solve 3x 3 - 4x 2 - 13x - 6 ≤ 0.

Step 1 Let f (x) = 3x 3 - 4x 2 - 13x - 6. Use the techniques from Lesson 2-4 to determine that f has real zeros with multiplicity 1 at -1, - 2 _

3 , and 3. Set up a sign chart.

x-1 3

-

23

Step 2 Determine the end behavior of f (x). Because the degree of f is odd and its leading coefficient is positive, you know lim x→-∞


f (x) = ∞. This means that the function starts off negative at the left and ends positive at the right.

(-

) (+

)

x-1 3

-

23

Step 3 Because each zero listed is the location of a sign change, you can complete the sign chart.

(+

)(-

) (-

) (+

)

x-1 3

-

23

The solutions of 3x 3 - 4x 2 - 13x - 6 ≤ 0 are x-values such that f (x) is negative or equal

to 0. From the sign chart, you can see that the solution set is (-∞, -1] ∪ ⎡

⎢

⎣

-

2 _ 3 , 3 ⎤

�

⎦

.

CHECK The graph of f (x) = 3x 3 - 4 x 2 - 13x - 6 is on or below

[-4, 6] scl: 1 by [-25, 5] scl: 3

the x-axis on (-∞, -1] ∪ ⎡

⎢

⎣

-

2 _ 3 , 3 ⎤

�

⎦

. �



2A. 2 x 2 - 10x ≤ 2x - 16 2B. 2 x 3 + 7x 2 -12x - 45 ≥ 0[2, 4] {-3} � ⎡

⎢

⎣ 5

_ 2 , ∞

⎞ ⎥

⎠

When a polynomial function does not intersect the x-axis, the related inequalities have unusual solutions.

Example 3 Polynomial Inequalities with Unusual Solution Sets


a. x 2 + 5x + 8 < 0

The related function f (x) = x 2 + 5x + 8 has no real zeros, so there are no sign changes. This function is positive for all real values of x. Therefore, x 2 + 5x + 8 < 0 has no solution. The graph of f (x) supports this conclusion, because the graph is never on or below the x-axis. The solution set is �.

b. x 2 + 5x + 8 ≥ 0

Because the related function f (x) = x 2 + 5x + 8 is positive for all real values of x, the solution set of x 2 + 5x + 8 ≥ 0 is all real numbers or (-∞, ∞).

[-12, 8] scl: 1 by [-5, 10] scl: 1

= x 2+ 5x + 8f (x)


0141_0147_AM_S_C02_L06_664291.indd 142 4/30/12 6:21 PM



2 Rational InequalitiesExamples 4 and 5 show how to solve rational inequalities by first writing the inequality in general form with a single rational expression on the left and 0 on the right. The zeros of the function are used with a sign chart to determine on which intervals expressions in the function are positive, negative, or equal to zero.

Additional Example

Solve 3x + 4 _

x + 2 - 3 ≥ 0.

(-∞, -2)

4

Tips for New TeachersChanging Signs A rational function can change signs at its real zeros or at its points of discontinuity, so the zeros of both the numerator and the denominator are included in a sign chart.

Teach with TechDocument Camera Choose several students to demonstrate and explain to the class how to use a sign chart to test intervals when solving an inequality.

Focus on Mathematical ContentSolving Rational Inequalities To solve a rational inequality, first find the zeros from the numerator and the undefined points from the denominator. Use these zeros and undefined points to divide a number line into intervals represented in a sign chart. Choose one number from each interval and evaluate f (x ) to determine if f (x ) is positive or negative in that interval. If f (x ) < 0, then the inequality is true when f (x ) is negative. If f (x ) > 0, then the inequality is true when f (x ) is positive.


Extension Have students determine the positive values of n for which the cube of n will be greater than 10 times the square of n. (10, ∞)


c. x 2 - 10x + 25 > 0

The related function f (x) = x 2 - 10x + 25 has one real zero, 5, with multiplicity 2, so the value of f (x) does not change signs. This function is positive for all real values of x except x = 5. Therefore, the solution set of x 2 - 10x + 25 > 0 is (-∞, 5) ∪ (5, ∞). The graph of f (x) supports this conclusion.

[-2, 8] scl: 1 by [-2, 8] scl: 1

f (x) = x 2 - 10x + 25

d. x 2 - 10x + 25 ≤ 0

The related function f (x) = x 2 - 10x + 25 has a zero at 5. For all other values of x, the function is positive. Therefore, the solution set of x 2 - 10x + 25 ≤ 0 is {5}.



3A. x 2 + 2x + 5 > 0 3B. x 2 + 2x + 5 ≤ 0

3C. x 2 - 2x - 15 ≤ -16 3D. x 2 - 2x - 15 > -16

(-∞, ∞) �

{1} (-∞, 1) � (1, ∞)

2Rational Inequalities Consider the rational function at the right. Notice the intervals on which f (x) is positive and negative. While

a polynomial function can change signs only at its real zeros, a rational function can change signs at its real zeros or at its points of discontinuity. For this reason, when solving a rational inequality, you must include the zeros of both the numerator and the denominator in your sign chart.

y

x

y = f (x)

You can begin solving a rational inequality by first writing the inequality in general form with a single rational expression on the left and a zero on the right.

Example 4 Solve a Rational Inequality

Solve 4 _ x - 6

+ 2 _ x + 1

> 0.

4 _ x - 6

+ 2 _ x + 1

> 0 Original inequality

4x + 4 + 2x - 12 __ (x - 6)(x + 1)

> 0 Use the LCD, (x - 6)(x + 1), to rewrite each fraction. Then add.

6x - 8 __ (x - 6)(x + 1)

> 0 Simplify.

Let f (x) = 6x - 8 __ (x - 6)(x + 1)

. The zeros and undefined points of the inequality are the zeros of the

numerator, 4 _ 3 , and denominator, 6 and -1. Create a sign chart using these numbers. Then choose

and test x-values in each interval to determine if f (x) is positive or negative.

x-1 64

3

x -

(x - )(x + )f (x) =

Test x = -2 Test x = 0 Test x = 2 Test x = 7

undef. zero undef.(-)(-)(-)

(-)(-)(+)

(+)(-)(+)

(+)(+)(+)

x

-1 643

undef. zero undef.(+)(-) (-) (+)

x -

(x - )(x + )f (x) =

The solution set of the original inequality is the union of those intervals for which f (x) is positive,

(-1, 4 _ 3 ) ∪ (6, ∞). The graph of f (x) = 4 _

x - 6 + 2 _

x + 1 in Figure 2.6.1 supports this conclusion.

ardm

Study Tip Rational Inequalities Remember to include all zeros and undefined points of a rational function when creating a sign chart.

Figure 2.6.1

y

x

−4

−8

−4−8

8

4

4 8


0141_0147_AM_S_C02_L06_664291.indd 143 4/30/12 6:21 PM


Differentiated Instruction AL OL BL

Interpersonal Learners Have students work together in mixed-ability groups of three to factor

the numerator in the rational inequality x 2 - x - 12 _ x + 4

≥ 0 . Ask each member of the group to name

a different critical number. Then ask groups to explain why the solution can be written as (-4, -3 ] ∪ [ 4, ∞). -3, 4, -4; It is interval notation for {x |-4 < x ≤ -3 or 4 ≤ x, x ∈ �}.


Additional Example

CARPENTRY A carpenter is making tables. The tables have rectangular tops with a perimeter of 20 feet and an area of at least 24 square feet. Write and solve an inequality that can be used to determine the possible lengths to which the tables can be made. �(10 - � ) ≥ 24; 4 ft to 6 ft

5

Additional Answers

1. [ -4, 2 ]

2. (-∞, -1) ∪ (6, ∞)

3. (-∞, -

1 _ 3 ⎤

⎦ ∪ [ 8, ∞)

4. (-∞, 5 _ 2 ) ∪ (4, ∞)

5. (-∞, -

1 _ 2 ) ∪ ( 2 _

3 , ∞)

6. (-∞, -2] ∪ ⎡

⎣ 1 _ 2 , 6

⎤

⎦

7. (-3, -

3 _ 4 ) ∪ (0, ∞)

8. (-

2 _ 5 , 3) ∪ (6, ∞)

9. (-∞, ∞)

10. ∅

12. (-∞, ∞)

14. {-3}

11. ∅

13. {4}

15. {-2}

16. (-∞, ∞)

17. -0.0004 x 2 + 80x - 1,000,000 ≥ 2,000,000; [50,000, 150,000] or 50,000 ≤ x ≤ 150,000

18. (-

15 _ 2 , -4)

19. (-∞, 5)

20. (6, 25 _ 2 ⎤

⎦

21. (-∞, -

20 _ 3 ) ∪ (-3, ∞)

22. (-∞, -

2 _ 5 ) ∪ (-

7 _ 27

, ∞)

23. (-∞, 14 _ 13

⎤

⎦

∪ ( 5 _ 3 , ∞)

24. (-∞, -1) ∪ ⎡

⎣

-

3 _ 4 , 3) ∪ [ 4, ∞)

25. (-3, 2 _ 3 ⎤

⎦

∪ (1, ∞)

26. [-5, -4) ∪ (-4, -

3 _ 5 ⎤

⎦




4A. x + 6 _ 4x - 3

≥ 1 4B. x 2 - x - 11 _ x - 2

≤ 3 4C. 1 _ x > 1 _

x + 5

⎛

⎪

⎝

3

_ 4 , 3

⎤

�

⎦

(-∞, -1] (2, 5] (-∞, -5) (0, ∞)

You can use nonlinear inequalities to solve real-world problems.

AMUSEMENT PARKS A group of high school students is renting a bus for $600 to take to an amusement park the day after prom. The tickets to the amusement park are $60 less an extra $0.50 group discount per person in the group. Write and solve an inequality that can be used to determine how many students must go on the trip for the total cost to be less than $40 per student.

Let x represent the number of students.

Ticket cost per student + bus cost per student must be less than $40.

60 - 0.5x + 600 _ x < 40

60 - 0.5x + 600 _ x < 40 Write the inequality.

60 - 0.5x + 600 _ x - 40 < 0 Subtract 40 from each side.

60x - 0.5 x 2 + 600 - 40x __ x < 0 Use the LCD, x, to rewrite each fraction. Then add.

-0.5 x 2 + 20x + 600 __ x < 0 Simplify.

x 2 - 40x - 1200 __ x > 0 Multiply each side by -2. Reverse the inequality sign.

(x + 20)(x - 60)

__ x > 0 Factor.

Let f (x) = (x + 20)(x - 60)

__ x . The zeros of this inequality are -20, 60, and 0. Use these numbers to create and complete a sign chart for this function.

x-20 600

(x + )(x - )x

f (x) =

Test x = -30 Test x = -10 Test x = 10 Test x = 70

zero undef. zero(-)(-)(-)

(+)(-)(-)

(+)(-)(+)

(+)(+)(+)

x

-20 600

zero undef. zero(+)(-) (-) (+)

(x + )(x - )x

f (x) =

So, the solution set of 60 - 0.5x + 600 _ x < 40 is (-20, 0) ∪ (60, ∞).

Because there cannot be a negative number of students, more than 60 students must go to the amusement park for the total cost to be less than $40 per student.


5. LANDSCAPING A landscape architect is designing a fence that will enclose a rectangular garden that has a perimeter of 250 feet. If the area of the garden is to be at least 1000 square feet, write and solve an inequality to find the possible lengths of the fence.

l ( 250 - 2l _ 2 ) ≥1000; about 8.6 feet to 116.4 feet

Real-World Example 5 Solve a Rational Inequality

Real-World LinkThe Kingda Ka roller coaster at Six Flags Great Adventure in New Jersey is the tallest and fastest roller coaster in the world. The ride reaches a maximum height of 456 feet in the air and then plunges vertically into a 270° spiral, while reaching speeds of up to 128 miles per hour.

Source: Six Flags


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Then use the table below to customize assignments for your students.

Watch Out!

Common Error In Exercises 40–43, remind students that when solving inequalities that involve eliminating a radical by raising each side to a power, it is important to test every possible solution interval using the original inequality. While a sign chart can aid in developing a solution set, it may not account for all solutions.

Additional Answers

27. (-∞, 13 _ 6 ) ∪ (4, ∞)

29. 750 + 25x _ x < 120; 8 to 14 people

30. (-∞, -3] ∪ [-2, ∞)

31. (-∞, -5] ∪ [8, ∞)

32. [-4, 4]

33. (-∞, -3] ∪ [3, ∞)

34. (-5, 0] ∪ (5, ∞)

35. (-∞, -6) ∪ (-6, 6) ∪ (6, ∞)

39a. �(56 - � ) ≥ 588 or - � 2 + 56� - 588 ≥ 0

39b. [ 14, 42 ]; The length of the playing field is at least 14 and at most 42 feet.

39c. 0 < �(56 - �) ≤ 588; (0, 14] ∪ [42, 56); Sample answer: The area of the playing field must be greater than 0 square feet but at most 588 square feet, so 0 < � ≤ 14 or 42 ≤ � < 56.

46. (-∞, -3) ∪ (-

1 _ 2 , 2) ∪ (6, ∞)

47. (-5, -

4 _ 3 ) ∪ (0, 4)

48. [ -6, -4 ] ∪ [ 0, ∞)

49. (-∞, -8 ] ∪ ⎡ ⎣

-3, -

1 _ 3 ⎤ ⎦

∪ [ 1, 6 ]



AL Approaching Level 1–35, 63, 64, 66, 68–81 1–35 odd, 80–81 2–34 even, 63, 64, 66, 68–79

OL On Level 1–37 odd, 38, 39, 41–49 odd, 50, 51, 53, 55–57, 59, 61–64, 66, 68–81

1–35, 80–81 36–64, 66, 68–79



Exercises

Solve each inequality. (Examples 1–3)

1. (x + 4)(x - 2) ≤ 0 2. (x - 6)(x + 1) > 0

3. (3x + 1)(x - 8) ≥ 0 4. (x - 4)(-2x + 5) < 0

5. (4 - 6y)(2y + 1) < 0 6. 2 x 3 - 9 x 2 - 20x + 12 ≤ 0

7. -8 x 3 - 30 x 2 - 18x < 0 8. 5 x 3 - 43 x 2 + 72x + 36 > 0

9. x 2 + 6x > -10 10. 2 x 2 ≤ -x - 4

11. 4 x 2 + 8 ≤ 5 - 2x 12. 2 x 2 + 8x ≥ 4x - 8

13. 2 b 2 + 16 ≤ b 2 + 8b 14. c 2 + 12 ≤ 3 - 6c

15. - a 2 ≥ 4 a + 4 16. 3 d 2 + 16 ≥ - d 2 + 16d

17. BUSINESS A new company projects that its first-year revenue will be r (x) = 120x - 0.0004 x 2 and the start-up cost will be c(x) = 40x + 1,000,000, where x is the number of products sold. The net profit p that they will make the first year is equal to p = r - c. Write and solve an inequality to determine how many products the company must sell to make a profit of at least $2,000,000. (Example 1)

Solve each inequality. (Example 4)

18. x - 3 _ x + 4

> 3 19. x + 6 _ x - 5

≤ 1

20. 2x + 1 _ x - 6

≥ 4 21 3x - 2 _ x + 3

< 6

22. 3 - 2x _ 5x + 2

< 5 23. 4x + 1 _ 3x - 5

≥ -3

24. (x + 2)(2x - 3)

__ (x - 3)(x + 1)

≤ 6 25. (4x + 1)(x - 2)

__ (x + 3)(x - 1)

≤ 4

26. 12x + 65 _ (x + 4) 2

≥ 5 27. 2x + 4 _ (x - 3) 2

< 12

28. CHARITY The Key Club at a high school is having a dinner as a fundraiser for charity. A dining hall that can accommodate 80 people will cost $1000 to rent. If each ticket costs $20 in advance or $22 the day of the dinner, and the same number of people bought tickets in advance as bought the day of the dinner, write and solve an inequality to determine the minimum number of people that must attend for the club to make a profit of at least $500. (Example 5)

29. PROM A group of friends decides to share a limo for prom. The cost of rental is $750 plus a $25 fee for each occupant. There is a minimum of two passengers, and the limo can hold up to 14 people. Write and solve an inequality to determine how many people can share the limo for less than $120 per person. (Example 5)

Find the domain of each expression.

30. √ ��

x 2 + 5x + 6 31. √ ��

x 2 - 3x - 40

32. √ ��

16 - x 2 33. √ ��

x 2 - 9

34. √ ��

x _ x 2 - 25

35. 3 √ ��

x _ 36 - x 2

Find the solution set of f (x) - g (x) ≥ 0.

36. 37. y

x

(2, -2)

(-1, -5)

fg

yx

−6

−12

−18

−24

g

f

−2−4 2 4

(3, -3)

(-4, -24)

38. SALES A vendor sells hot dogs at each school sporting event. The cost of each hot dog is $0.38 and the cost of each bun is $0.12. The vendor rents the hot dog cart that he uses for $1000. If he wants his costs to be less than his profits after selling 400 hot dogs, what should the vendor charge for each hot dog?

39. PARKS AND RECREATION A rectangular playing field for a community park is to have a perimeter of 112 feet and an area of at least 588 square feet.

a. Write an inequality that could be used to find the possible lengths to which the field can be constructed.

b. Solve the inequality you wrote in part a and interpret the solution.

c. How does the inequality and solution change if the area of the field is to be no more than 588 square feet? Interpret the solution in the context of the situation.

Solve each inequality. (Hint: Test every possible solution interval that lies within the domain using the original inequality.)

40. √ �� 9y + 19 - √ �� 6y - 5 > 3

41. √ �� 4x + 4 - √ �� x - 4 ≤ 4

42. √ �� 12y + 72 - √ �� 6y - 11 ≥ 7

43. √ �� 25 - 12x - √ �� 16 - 4x < 5

Determine the inequality shown in each graph.

44. 45. y

x

(4, -6)

(2, -12)

−4

−8

−12

−2 2 4 6

y

x

−4

−8

8

4

(0, −8)(−2, −8)

−2−4 2 4


46. 2 y 4 - 9 y 3 - 29 y 2 + 60y + 36 > 0

47. 3 a 4 + 7 a 3 - 56 a 2 - 80a < 0

48. c 5 + 6 c 4 - 12 c 3 - 56 c 2 + 96c ≥ 0

49. 3 x 5 + 13 x 4 - 137 x 3 - 353 x 2 + 330x + 144 ≤ 0

1–16. See margin.

See margin.18–27. See margin.

500 ≤ 20 ( x _ 2 ) + 22

( x _ 2 ) -1000; 72 people

See margin.30–35. See margin.

B

[-4, 3](-∞, -1] � [2, ∞)

more than $3

a–c. See margin.

40. ⎡

⎢

⎣

5

_ 6 , 5

⎞

⎥

⎠

� (9, ∞)

41. ⎡

⎢

⎣

40

_ 9 , 8

⎤

�

⎦

42. ⎡

⎢

⎣

11

_ 6 , 6

⎤

�

⎦

� ⎡

⎢

⎣

46

_ 3 , ∞

⎞

⎥

⎠

43. ⎛

⎪

⎝

-12, 25

_ 12

⎤

�

⎦

x 2 - 3x - 10 ≥ 0 x 2 + 2x - 8 < 0



0141_0147_AM_S_C02_L06_664291.indd 145 4/30/12 6:20 PM



Additional Answers

50a. S (r ) = 2π r 2 + 4000 _ r

50b. 2π r 2 + 4000 _ r < 2400 or

2π r 2 + 4000 _ r - 2400 < 0

50c. (-20.33, 0) ∪ (1.68, 18.65); Because the radius cannot be negative, the only possible lengths for the radius are between 1.68 cm and 18.65 cm.



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Study Guide and InterventionNonlinear Inequalities

Polynomial Inequalities Real zeros divide the x-axis into intervals for which the value of f(x) is positive (above the x-axis) or negative (below the x-axis). A sign chart shows these real zeros and the sign of f(x) in that interval as either positive (+) or negative (-).


a. x 2 - 3x - 15 > 3 x 2 - 3x - 15 > 3 Original inequality

x 2 − 3x − 18 > 0 Subtract 3 from each side of the inequality.

(x + 3) (x − 6) > 0 Factor.

Let f(x) = (x + 3) (x − 6) .f(x) has real zeros at x = − 3 and x = 6.Create a sign chart using these real zeros. Substitute any number from each interval forx in the factored function rule. Write the signof each factor and the sign of the product ofthe factors. Because we are solving for when x 2 − 3x − 18 > 0, we choose the intervals for which the product is positive. The solution is(-∞, -3) or (6, ∞) .

b. x 2 + 2x + 2 < 0 The related function has no real zeros. The graph is always above the x-axis so the values of f(x) are always positive. Because we are solving for when the values are negative, the equation has no real solution.

Exercises


1. x 3 − 3x − 2 < 0 2. x 2 + 2x + 1 > 0

(−∞, −1) or (−1, 2) (−∞, −1) or (−1, ∞)

3. x 3 + 6 x 2 + 5x > 12 4. 3 x 2 + x < 2

(−4, −3) or (1, ∞) (−1, 2 − 3 )

Example

6 (+)

(+) (+)

(+)

(-) (-)

(-)

(+) (-)Test = 10.Test = -4. Test = 0.

f ( ) = ( + 3)( - 6)

-3

[-10, 10] scl: 1 by [-10, 10] scl: 1

2-6

005_038_PCCRMC02_893803.indd 32 10/1/09 5:46:52 PM

Practice

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PracticeNonlinear Inequalities


1. (3 - 2y) (2y + 5) < 0 2. x 2 - 2x + 1 ≥ 0

(−∞, − 5 − 2 ) or (

3 −

2 , ∞) (−∞, ∞)

3. x 3 + 3 x 2 ≤ 4 4. x 2 - 12 > −2 x 2 − 16x

(−∞, 1] (−∞, −6) or ( 2 − 3 , ∞)

5. 3 x 3 − 5 x 2 + 10 ≤ 74x + 34 6. 4 x 2 + 6x ≤ 12 - 7x

(−∞, − 4] or ⎡ ⎢

⎣ − 1 −

3 , 6) [− 4,

3 −

4 ]

7. 2n + 1 −

3n + 1 ≤

n - 1 −

3n + 1 8. 1 +

3y −

1 - y > 2

⎡ ⎢

⎣ −2, − 1 −

3 ) ( 1 −

4 , 1)

9. 2x −

4 - 5x + 1 −

3 > 3 10. 4x - 2 −

x + 1 ≥ - 2

(−∞, − 20 −

7 ) (−∞,−1) or [0, ∞)

11. 4x - 28 −

(x - 5) (x + 1) ≥ 2 12. x + 1 −

(x - 4) (x + 8) ≥ 0

(−1, 5) (−8, −1] or (4, ∞)

13. GREENHOUSE You want to build a greenhouse from at most 500 square feet of material in the shape of a half cylinder. The surface area of the greenhouse is given by the formula S = πrh + π r 2 , where r is the radius and h is the length of the greenhouse. Write an expression for h in terms of r. Then find all of the possible values of r.

h ≤ 500 - π r 2

− πr ; (0, 12.6)

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Word Problem PracticeNonlinear Inequalities

1. COMMUTING Rosea drives her car30 kilometers to the train station,where she boards the train tocomplete her trip. The total trip is120 kilometers. The average speed of the train is 20 kilometers per hour faster than that of the car.

a. Write a rational inequality that can be solved to find the speed she must drive if the total time for the trip is less than 2.5 hours.30−x

+ 90−x + 20

< 2.5

b. What is the minimum speed she must drive?

35 km/h

2. CONSTRUCTION The Kingsley family wants to construct an addition to its house with a minimum area of 240 square feet. The room will be rectangular and the length will be 10 feet longer than the width.

a. Write an inequality that can be used to find the possible widths.

w 2 + 10w - 240 ≥ 0

b. To the nearest quarter foot, what will be the minimum width of the addition?

11 1−4

ft

3. ENROLLMENT The projected revenue at a school can be modeled byf(x) = 2.8 x 2 - 1450x + 147,500, where x is the number of students. If the school cannot operate unless its revenue is over $750,000, how many students must be enrolled?

791 students

4. DOG RUN Hector is building a rectangular dog run with 100 feet of fencing and an area of at least 500 square feet. The dog run will have three sides and use a house wall for the fourth side.

a. Write an inequality that could be used to find the possible lengths of the dog run.

100x - 2 x 2 - 500 ≥ 0

b. Sketch the graph of the related function.

200

300

100

0

400

500

600

700

800

15105 25 3520 30 40 45 50

c. To the nearest tenth, what could be the lengths of the sides perpendicular to the house?

between 5.6 and 44.4 ft

5. CLIMBING The cost for a group of climbers to take a bus to a trailhead is $280. The group permit costs $12, minus $0.25 per member because the climbers are frequent users. How many climbers must be in the group for the cost per person to be $20 or less?

15 climbers

2-6

005_038_PCCRMC02_893803.indd 35 10/1/09 5:47:03 PM

Enrichment

p. 36 OL BL

NAME DATE PERIOD

EnrichmentEnrichmentEnrichmentEnrichmentE i h

Systems of Nonlinear InequalitiesA system of nonlinear inequalities is a set of two or more inequalities inwhich at least one is nonlinear. As with systems of linear inequalities, the solutions lie in the overlapping regions of the graphs.

Graph the solutions of ⎧

⎨ ⎧⎧

⎩

⎨⎨

y < - √��√√x√√ + 3 + 4 y > 1−

2x - 6

.

Graph the boundaries, which are the graphs of the related

equations, y = - √��√√x √√ + 3 + 4 and y = 1−2

x - 6.

The intersection of the half-planes contains the solutions.

Exercises

Graph thGraph thp e solutie solutions of eons of each systach systy emem.

111. yy y >> 11−−222

(( ((−− xx -- 4)4) 4))22++ 33 3 222. yy y ≤≤ ⎪⎪⎪⎪xx33

⎥⎥⎥⎥ ++ 22 2

y y y ≤ -≤ -((((xx -- 6)6) 6))33 yy y ≥≥ --

111−−

44xx

2 62-62 62 6

y

84−4−88

44

88

−88

−4

88

50. PACKAGING A company sells cylindrical oil containers like the one shown.

2 L

a. Use the volume of the container to express its surface area as a function of its radius in centimeters. (Hint: 1 liter = 1000 cubic centimeters)

b. The company wants the surface area of the container to be less than 2400 square centimeters. Write an inequality that could be used to find the possible radii to meet this requirement.

c. Use a graphing calculator to solve the inequality you wrote in part b and interpret the solution.


51. (x + 3) 2 (x - 4) 3 (2x + 1) 2 < 0

52. ( y - 5) 2 ( y + 1) (4y - 3) 4 ≥ 0

53. (a - 3) 3 (a + 2) 3 (a - 6) 2 > 0

54. c 2 (c + 6) 3 (3c - 4) 5 (c - 3) ≤ 0

55. STUDY TIME Jarrick determines that with the information that he currently knows, he can achieve a score of a 75% on his test. Jarrick believes that for every 5 complete minutes he spends studying, he will raise his score by 1%.

a. If Jarrick wants to obtain a score of at least 89.5%, write an inequality that could be used to find the time t that he will have to spend studying.

b. Solve the inequality that you wrote in part a and interpret the solution.

56. GAMES A skee ball machine pays out 3 tickets each time a person plays and then 2 additional tickets for every 80 points the player scores.

a. Write a nonlinear function to model the amount of tickets received for an x-point score.

b. Write an inequality that could be used to find the score a player would need in order to receive at least 11 tickets.

c. Solve the inequality in part b and interpret your solution.

57. The area of a region bounded by a parabola and a horizontal line is A = 2 _

3 bh, where b represents the base of

the region along the horizontal line and h represents the height of the region. Find the area bounded by f and g.

y

x

(5, -7)(-1, -7)−6

6

−12

−18

−4 4 8

y = f (x)

y = g(x)

If k is nonnegative, find the interval for x for which each inequality is true.

58. x 2 + kx + c ≥ c 59. (x + k)(x - k) < 0

60. x 3 - k x 2 - k 2 x + k 3 > 0 61. x 4 - 8 k 2 x 2 + 16 k 4 ≥ 0

62. MULTIPLE REPRESENTATIONS In this problem, you will investigate absolute value nonlinear inequalities.

a. TABULAR Copy and complete the table below.

Function ZerosUndefined

Points

f (x ) = x - 1 _ |x + 2|

g (x ) = |2x - 5| _ x - 3

h (x ) = |x + 4| _

|3x - 1|

b. GRAPHICAL Graph each function in part a.

c. SYMBOLIC Create a sign chart for each inequality. Include zeros and undefined points and evaluate the sign of the numerators and denominators separately.

i. x - 1 _

|x + 2|

< 0

ii. |2x - 5|

_ x - 3

≥ 0

iii. |x + 4|

_ |3x - 1|

> 0

d. NUMERICAL Write the solution for each inequality in part c.


63. ERROR ANALYSIS Ajay and Mae are solving x 2 _ (3 - x) 2

≥ 0.

Ajay thinks that the solution is (-∞, 0] or [0, ∞), and Mae thinks that the solution is (-∞, ∞). Is either of them correct? Explain your reasoning.

64. REASONING If the solution set of a polynomial inequality is (-3, 3), what will be the solution set if the inequality symbol is reversed? Explain your reasoning.

65. CHALLENGE Determine the values for which (a + b) 2 > (c + d) 2 if a < b < c < d.

66. REASONING If 0 < c < d, find the interval on which (x - c)(x - d) ≤ 0 is true. Explain your reasoning.

67 CHALLENGE What is the solution set of (x - a) 2n > 0 if n is a natural number?

68. REASONING What happens to the solution set of (x + a)(x - b) < 0 if the expression is changed to -(x + a)(x - b) < 0, where a and b > 0? Explain your reasoning.

69. WRITING IN MATH Explain why you cannot solve 3x + 1 _ x - 2

< 6 by multiplying each side by x - 2.

a–c. See margin.

C(-∞, -3) � (-3, -

1

_ 2 ) � (-

1

_ 2 , 4)

[-1, ∞)

(-∞, -2) � (3, 6) � (6, ∞)

(-∞, -6] � ⎡

⎢

⎣

4

_ 3 , 3

⎤

�

⎦

� {0}

75 + t

_ 5 ≥ 89.5

75; Sample answer: Since t ≥

72.5, Jarrick will have to spend 75 minutes studying for the test.

f (x) = 2 x

_ 80

+ 3

11 ≤ 2 x

_ 80

+ 3

See margin.

36 units 2


b–d. See

Chapter 2

Answer

Appendix. 1 -2

5

_ 2 3

-4 1

_ 3

See margin.

See margin.

a < 0 and |a + b| > |c + d |

See margin.

(-∞, a) � (a, ∞)

See margin.

See margin.


0141_0147_AM_S_C02_L06_664291.indd 146 4/30/12 6:20 PM



66. [ c, d ]; Sample answer: Because c < d, when x < c, x < d, so (x - c ) and (x - d ) will be negative and (x - c )(x - d ) will be positive. When x > d, both factors will be positive, so (x - c )(x - d ) will be positive. When c ≤ x ≤ d, (x - c ) will be positive or zero, (x - d ) will be negative or zero, and (x - c )(x - d ) will be either negative or zero.

68. Sample answer: The zeros will remain the same but the solution set will consist of all numbers outside of the original solution set.

For example, the solution set for the original inequality is (-a, b ), while the solution set for the new inequality is (-∞, -a ) ∪ (b, ∞).

69. Sample answer: There will be instances when x - 2 is a negative value. When this occurs, we are multiplying an inequality by a negative and the inequality symbol will need to be reversed. The problem occurs here because x - 2 could be positive or negative.

Watch Out!

Error Analysis For Exercise 63, caution students to copy the problem accurately. Some students may write 3 - x 2 instead of (3 - x ) 2 in the denominator. Remind students that the numerator can equal 0 if the inequality is ≥ or ≤.

4 AssessFormative AssessmentName the Math Have students describe the mathematical procedures

they would use to solve x - 3 _ x + 2

> 2.

Sample answer: Rewrite the inequality as a single fraction greater than 0. Then use a sign chart to test values in each interval between critical values.

Additional Answers

56c. [ 320, ∞); Any score at or above 320 points will result in the player receiving at least 11 tickets.

58. (-∞, -k ] ∪ [0, ∞)

59. (-k, k )60. (-k, ∞)

61. (-∞, ∞)

63. Neither; Sample answer: The expression is undefined at 3, but it is positive everywhere that it is defined, so the solution set is (-∞, 3) ∪ (3, ∞).

64. (-∞, -3) ∪ (3, ∞); Sample answer: If the inequality symbol is reversed, the solution set will consist of all real numbers not in the original solution set.

Spiral Review

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any. (Lesson 2-5)

70. f (x) = 2x _ x + 4

71. h(x) = x 2 _ x + 6

72. f (x) = x - 1 __ (2x + 1)(x - 5)

73. GEOMETRY A cone is inscribed in a sphere with a radius of 15 centimeters. If the volume of 15

x

15the cone is 1152π cubic centimeters, find the length represented by x. (Lesson 2-4)

Divide using long division. (Lesson 2-3)

74. ( x 2 - 10x - 24) ÷ (x + 2) 75. (3 a 4 - 6 a 3 - 2 a 2 + a - 6) ÷ (a + 1)

76. ( z 5 - 3 z 2 - 20) ÷ (z - 2) 77. ( x 3 + y 3 ) ÷ (x + y)

78. FINANCE The closing prices in dollars for a share of stock during a one-month period are shown. (Lesson 2-2)

a. Graph the data.

b. Use a graphing calculator to model the data using a polynomial function with a degree of 3.

c. Use the model to estimate the closing price of the stock on day 25.


about 0.37 cm or 9 cm

x - 12 3 a 3 - 9 a 2 + 7a - 6

z 4 + 2 z 3 + 4 z 2 + 5z + 10

x 2 - xy + y 2

a–c. See Chapter 2 Answer Appendix.Day Price(s) Day Price(s)

1 30.15 15 15.64

5 27.91 20 10.38

7 26.10 21 9.56

10 22.37 28 9.95

12 19.61 30 12.25

79. SAT/ACT Two circles, A and B, lie in the same plane. If the center of circle B lies on circle A, then in how many points could circle A and circle B intersect?

I. 0 II. 1 III. 2

A I only C I and III only E I, II, and III

B III only D II and III only

80. A rectangle is 6 centimeters longer than it is wide. Find the possible widths if the area of the rectangle is more than 216 square centimeters.

F w > 12 H w > 18

G w < 12 J w < 18

E F


81. FREE RESPONSE The amount of drinking water reserves in millions of gallons available for a town is modeled by f (t) = 80 + 10t - 4 t 2 . The minimum amount of water needed by the residents is

modeled by g (t) = (2t) 4 _ 3 , where t is the time in years.

a. Identify the types of functions represented by f (t) and g (t).

b. What is the relevant domain and range for f (t) and g (t)? Explain.

c. What is the end behavior of f (t) and g (t)?

d. Sketch f (t) and g (t) for 0 ≤ t ≤ 6 on the same graph.

e. Explain why there must be a value c for [0, 6] such that f (c) = 50.

f. For what value in the relevant domain does f have a zero? What is the significance of the zero in this situation?

g. If this were a true situation and these projections were accurate, when would the residents be expected to need more water than they have in reserves?

a–g. See Chapter 2 Answer Appendix.


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Chapter 2 Study Guide and Review

Formative Assessment

Key Vocabulary The page references after each word denote where that term was first introduced. If students have difficulty answering Questions 1–10, remind them that they can use these page references to refresh their memories about the vocabulary.

Additional Answers

21. lim x→∞

f (x ) = -∞; lim x→-∞

f (x ) =

-∞; The degree is even and the leading coefficient is negative.

22. lim x→∞

f (x ) = -∞; lim x→-∞

f (x ) = ∞;

The degree is odd and the leading coefficient is negative.

23. lim x→∞

f (x ) = ∞ and lim x→-∞

f (x ) =

∞; The degree is even and the leading coefficient is positive.

24. lim x→∞

f (x ) = ∞ and lim x→-∞

f (x ) =

-∞; The degree is odd and the leading coefficient is positive.

25. 3 real zeros and 2 turning points; 0, 3, and 4

26. 5 real zeros and 4 turning points; -10, 0, and 2

27. 4 real zeros and 3 turning points; -3, -1, 1, and 3

28. 4 real zeros and 3 turning points; √ � 5 and - √ � 5

Chapter Summary

Study Guide and ReviewStudy Guide and Review

Key Concepts

Power and Radical Functions (Lesson 2-1)

• A power function is any function of the form f (x ) = a x n , where a and n are nonzero real numbers.

• A monomial function is any function that can be written as f (x ) = a or f (x ) = a x n , where a and n are nonzero constant real numbers.

• A radical function is a function that can be written as f (x ) = n √ �

n p , where n and p are positive integers greater than 1 that have no common factors.

Polynomial Functions (Lesson 2-2)

• A polynomial function is any function of the form f (x ) = a n x n + a n - 1 x n - 1 + … + a 1 x + a 0 , where a n ≠ 0. The degree is n.

• The graph of a polynomial function has at most n distinct real zeros and at most n - 1 turning points.

• The behavior of a polynomial graph at its zero c depends on the multiplicity of the factor (x - c ).

The Remainder and Factor Theorems (Lesson 2-3)

• Synthetic division is a shortcut for dividing a polynomial by a linear factor of the form x - c.

• If a polynomial f is divided by x - c, the remainder is equal to f (c ).

• x - c is a factor of a polynomial f if and only if f (c ) = 0.

Zeros of Polynomial Functions (Lesson 2-4)

• If f (x ) = a n x n + … + a 1 x + a 0 with integer coefficients, then any

rational zero of f (x ) is of the form p _ q , where p and q have no common

factors, p is a factor of a 0 , and q is a factor of a n .

• A polynomial of degree n has n zeros, including repeated zeros, in the complex system. It also has n factors:

f (x ) = a n (x - c 1 ) (x - c 2 ) … (x - c n ).

Rational Functions (Lesson 2-5)

• The graph of f has a vertical asymptote x = c if lim x→ c -

f (x ) = ±∞ or lim x→ c +

f (x ) = ±∞.

• The graph of f has a horizontal asymptote y = c if lim x→∞

f (x ) = c or lim x→-∞

f (x ) = c.

• A rational function f (x ) = a (x )

_ b (x )

may have vertical asymptotes,

horizontal asymptotes, or oblique asymptotes, x-intercepts, and y-intercepts. They can all be determined algebraically.

Nonlinear Inequalities (Lesson 2-6)

• The sign chart for a rational inequality must include zeros and undefined points.

Key Vocabulary

complex conjugates (p. 124)

extraneous solution (p. 91)

horizontal asymptote (p. 131)

irreducible over the reals (p. 124)

leading coefficient (p. 97)

leading-term test (p. 98)

lower bound (p. 121)

multiplicity (p. 102)

oblique asymptote (p. 134)

polynomial function (p. 97)

power function (p. 86)

quartic function (p. 99)

rational function (p. 130)

repeated zero (p. 101)

sign chart (p. 141)

synthetic division (p. 111)

synthetic substitution (p. 113)

turning point (p. 99)

upper bound (p. 121)

vertical asymptote (p. 131)

Vocabulary Check

Identify the word or phrase that best completes each sentence.

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

2. A (polynomial function, power function) is a function of the form f (x ) = a n x n + a n - 1 x n - 1 + … + a 1 x + a 0 , where a 1 , a 2 , …, a n are real numbers and n is a natural number.

3. A function that has multiple factors of (x - c) has (repeated zeros, turning points).

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 and a numerator with degree one greater than its denominator.

10. A (quartic function, power function) is a function of the form f (x ) = a x n , where a and n are nonzero constant real numbers.

leading coefficient

polynomial function

repeated zeros

Synthetic division

Factor Theorem

Rational Zeros

Vertical

numerator

Oblique

power function

148 | Chapter 2 | Study Guide and Review

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Lesson-by-Lesson ReviewIntervention If the given examples are not sufficient to review the topics covered by the questions, remind students that the lesson references tell them where to review that topic in their textbook.

Two-Day Option Have students complete the Lesson-by-Lesson Review on pp. 149–152. Then you can use to customize another review worksheet that practices all the objectives of this chapter or only the objectives on which your students need more help.

Additional Answers

29a. lim x →∞

f (x ) = ∞; lim x →-∞

f (x ) = ∞

29b. 0 (multiplicity 3), 3, -4 (multiplicity 2)

29c. Sample answer: (-1, 36), (0, 0), (1, -50), (2, -288)

29d. y

x

−160

−320

−4−8

320

160

4 8

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

30a. lim x →∞

f (x ) = ∞; lim x →-∞

f (x ) = ∞

30b. 5 (multiplicity 2), 1(multiplicity 2)

30c. Sample answer: (2, 9), (3, 16), (4, 9), (5, 0)

30d. y

x−4−8

16

12

8

4

4 8

(x 5)2(x 1)2f (x) =

Study Guide and Review

0148_0153_AM_S_C02_SGR_664291.indd 148 5/3/12 11:29 AM

Lesson-by-Lesson Review

22222---11111222222 11111 Power and Radical Functions

Graph and analyze each function. Describe the domain, range,

intercepts, end behavior, continuity, and where the function is increasing

or decreasing.

11. f (x ) = 5 x 6

12. f (x ) = -8 x 3

13. f (x ) = x -9

14. f (x ) = 1 _ 3 x -4

15. f (x ) = √ �� 5x - 6 - 11

16. f (x ) = -

3 _ 4

3 √ ��

6 x 2 - 1 + 2


17. 2x = 4 + √ �� 7x - 12

18. √ �� 4x + 5 + 1 = 4x

19. 4 = √ �� 6x + 1 - √ �� 17 - 4x

20. 4 √ ��

x 2 + 31 - 1 = 3

Example 1 Graph and analyze f (x ) = -4 x -5 . Describe the domain, range,


increasing or decreasing.

x f (x )

-3 0.016

-2 0.125

-1 4

0 undefi ned

1 -4

2 -0.125

3 -0.016

y

x

Domain: (-∞, 0) ∪ (0, ∞) Range: (-∞, 0) ∪ (0, ∞)

Intercepts: none



f (x ) = 0


Increasing: (-∞, 0) Increasing: (0, ∞)


4

1

4

15, -15

22222---22222222222 22222 Polynomial Functions


using limits. Explain your reasoning using the leading term test.

21. f (x ) = -4 x 4 + 7 x 3 - 8 x 2 + 12x - 6

22. f (x ) = -3 x 5 + 7 x 4 + 3 x 3 - 11x - 5

23. f (x ) = 2 _ 3 x 2 - 8x -3

24. f (x ) = x 3 (x - 5)(x + 7)


function. Then determine all of the real zeros by factoring.

25. f (x ) = x 3 - 7 x 2 + 12x 26. f (x ) = x 5 + 8 x 4 - 20 x 3

27. f (x ) = x 4 - 10 x 2 + 9 28. f (x ) = x 4 - 25

For each function, (a) apply the leading term test, (b) find the zeros and

state the multiplicity of any repeated zeros, (c) find a few additional

points, and then (d) graph the function.

29. f (x ) = x 3 (x - 3)(x + 4 ) 2 30. f (x ) = (x - 5 ) 2 (x - 1 ) 2

Example 2 Describe the end behavior of the graph of

f (x ) = -2 x 5 + 3 x 3 - 8 x 2 - 6 using limits. Explain your

reasoning using the leading term test.

The degree is 5 and the leading coefficient is -2. Because the degree is odd and the leading coefficient is negative, lim x→-∞


f (x ) = -∞.

Example 3 State the number of possible real zeros and turning points for

f (x ) = x 3 + 6 x 2 + 9x. Then find all the real zeros by factoring.

The degree of f is 3, so f has at most 3 distinct real zeros and at most 3 - 1 or 2 turning points. To find the real zeros, solve the related equation f (x ) = 0 by factoring.

x 3 + 6 x 2 + 9x = x ( x 2 + 6x + 9)

= x (x + 3)(x + 3) or x (x + 3 ) 2

The expression has 3 factors but only 2 distinct real zeros, 0 and -3.


2 5–28. See margin.



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149

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Additional Answers

50. D = {x | x ≠ -4, x ∈ �}; x = -4

51. D = {x | x ≠ 5, -5, x ∈ �}; x = 5, x = -5, y = 1

52. D = {x | x ≠ 5, -3, x ∈ �}; x = 5, x = -3, y = 0

53. D = {x | x ≠ -3, -9, x ∈ �}; x = -3, x = -9, y = 1

54. asymptotes: x = 5, y = 1; x-intercept: 0; y-intercept: 0; D = {x | x ≠ 5, x ∈ �}

y

x

−8

−4−8

8

4

4 8

f(x) = x_x 5

y = 1x = 5

55. asymptotes: x = -4, y = 1; x-intercept: 2; y-intercept: -

1 _ 2 ;

D = {x | x ≠ -4, x ∈ �}

y

x

−4

−8

−4−8

8

4

4 8

x = -4f(x) = x 2_

x + 4

y = 1

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Study Guide and Review Study Guide and Review ContinuedContinued

22222---33333222222 33333 The Remainder and Factor Theorems


31. ( x 3 + 8 x 2 - 5) ÷ (x - 2)

32. (-3 x 3 + 5 x 2 - 22x + 5) ÷ ( x 2 + 4)

33. (2 x 5 + 5 x 4 - 5 x 3 + x 2 - 18x + 10) ÷ (2x - 1)


34. ( x 3 - 8 x 2 + 7x - 15) ÷ (x - 1)

35. ( x 4 - x 3 + 7 x 2 - 9x - 18) ÷ (x - 2)

36. (2 x 4 + 3 x 3 - 10 x 2 + 16x - 6) ÷ (2x - 1)

Use the Factor Theorem to determine if the binomials given are

factors of f (x ). Use the binomials that are factors to write a factored

form of f (x ).

37. f (x ) = x 3 + 3 x 2 - 8x - 24; (x + 3)

38. f (x ) = 2 x 4 - 9 x 3 + 2 x 2 + 9x - 4; (x - 1), (x + 1)

39. f (x ) = x 4 - 2 x 3 - 3 x 2 + 4x + 4; (x + 1), (x - 2)

Example 4 Divide (2 x 3 - 3 x 2 + 5x - 4) ÷ (2x - 1) using synthetic division.

Rewrite the division expression 2 x 3 - 3 x 2 + 5x - 4 __

2x - 1 so that the

denominator is of the form x - c.

2 x 3 - 3 x 2 + 5x - 4 __

2x - 1 = (2 x 3 - 3 x 2 + 5x - 4) ÷ 2 __

(2x - 1) ÷ 2

= x 3 -

3 _ 2 x 2 + 5 _

2 x - 2 __

x - 1 _ 2

Therefore, c = 1 _ 2 . Perform synthetic division.

1 _ 2 1 -

3 _ 2 5 _

2 -2

1 _ 2 -

1 _ 2 1

1 -1 2 -1

(2 x 3 - 3 x 2 + 5x - 4) __ (2x - 1)

= x 2 - x + 2 -

2 _ (2x - 1)

x 2 + 10x + 20 +

35

_ x - 2

-3x + 5 -

10x + 15

_ x 2 + 4

x 4 + 3x 3 - x 2 - 9 +

1

_ 2x - 1

x 2 - 7x -

15

_ x - 1

x 3 + x 2 + 9x + 9

x 3 + 2x 2 - 4x + 6

yes; f (x ) = (x + 3)( x 2 - 8)

38. yes; yes; f (x ) = (x - 1)(x + 1)(x - 4)(2x - 1)

yes; yes; f (x ) = (x - 2 ) 2 (x + 1 ) 2

22222---44444222222 44444 Zeros of Polynomial Functions

List all possible rational zeros of each function.

Then determine which, if any, are zeros.

40. f (x ) = x 3 - x 2 - x + 1

41. f (x ) = x 3 - 14x - 15

42. f (x ) = x 4 + 5 x 2 + 4

43. f (x ) = 3 x 4 - 14 x 3 - 2 x 2 + 31x + 10


44. x 4 - 9 x 3 + 29 x 2 - 39x + 18 = 0

45. 6 x 3 - 23 x 2 + 26x - 8 = 0

46. x 4 - 7 x 3 + 8 x 2 + 28x = 48

47. 2 x 4 - 11 x 3 + 44x = -4 x 2 + 48

Use the given zero to find all complex zeros of each function.

Then write the linear factorization of the function.

48. f (x ) = x 4 + x 3 - 41 x 2 + x - 42; i

49. f (x ) = x 4 - 4 x 3 + 7 x 2 + 16x + 12; -2i

Example 5 Solve x 3 + 2 x 2 - 16x - 32 = 0.

Because the leading coefficient is 1, the possible rational zeros are the factors of -32. So the possible rational zeros are ±1, ±2, ±4, ±8, ±16, and ±32. Using synthetic substitution, you can determine that -2 is a rational zero.

-2 1 2 -16 -32

-2 0 32

1 0 -16 0

Therefore, f (x ) = (x + 2)( x 2 - 16). This polynomial can be written f (x ) = (x + 2)(x - 4)(x + 4). The rational zeros of f are -2, 4, and -4.

± 1; 1 (multiplicity: 2) and -1

± 1, ± 3, ± 5, ± 15 ; -3

± 1, ± 2, ± 4; no rational zeros

±1, ± 2, ± 5, ±10, ±

1

_ 3 , ±

2

_ 3 , ±

5

_ 3 , ±

10

_ 3 ; 2, -

1

_ 3

1, 2, 3 (multiplicity: 2)

2, 1

_ 2 ,

4

_ 3

-2, 2, 3, 4

3

_ 2 , -2, 2, 4

-7, 6i, -i ; f (x ) = (x + 7)(x - 6)(x + i )(x - i )

1, 3, 2i, -2i ; f (x ) = (x - 3)(x - 1)(x - 2i )(x + 2i )


0148_0153_AM_S_C02_SGR_664291.indd 150 5/3/12 11:28 AM





59. asymptotes: x = 0, x = 1, x = 5, y = 0; x-intercepts: 4 and -4; D = {x | x ≠ 0, 1, 5, x ∈ �}

y

x

−8

−16

−4−8

16

8

4 8

f (x) = x 2 16__x 3 6 x 2 + 5x

x = 5

x = 1

x = 0

y = 0

Additional Answers

56. asymptotes: x = -5, x = 6, y = 1; x-intercepts: -3 and 4;

y-intercept: 2 _ 5 ; D = {x | x ≠ -5, 6,

x ∈ �}

y

x

−4

−8

−8−16

8

4

8 16

f (x) = (x + 3)(x 4)__(x + 5)(x 6)

x = -5

x = 6y = 1

57. asymptotes: x = -6, x = 3, y = 1; x-intercepts: 0 and -7; y-intercept: 0; D = {x | x ≠ -6, 3, x ∈ �}

y

x

−4

−8

−4−8

8

4

4 8

f (x) =

x(x + 7) __

(x + 6)(x 3)

x = -6y = 1

x = 3

58. asymptotes: x = -1, x = 1, y = 0; x-intercept: -2; y-intercept: -2; D = {x | x ≠ -1, 1, x ∈ �}

y

x

−4

−8

−4−8

8

4

4 8

f (x) =

x + 2 _ x 2 1

x = 1

y = 0

x = 1

22222---55555222222 55555 Rational Functions

Find the domain of each function and the equations of the vertical or

horizontal asymptotes, if any.

50. f (x ) = x 2 - 1 _ x + 4

51. f (x ) = x 2 _ x 2 - 25

52. f (x ) = x (x - 3)

__ (x - 5 ) 2 (x + 3 ) 2

53. f (x ) = (x - 5)(x - 2)

_ (x + 3)(x + 9)

For each function, determine any asymptotes and intercepts. Then

graph the function, and state its domain.

54. f (x ) = x _ x - 5

55. f (x ) = x - 2 _ x + 4

56. f (x ) = (x + 3)(x - 4)

_ (x + 5)(x - 6)

57. f (x ) = x (x + 7)

_ (x + 6)(x - 3)

58. f (x ) = x + 2 _

x 2 - 1 59. f (x ) = x 2 - 16 __

x 3 - 6 x 2 + 5x


60. 12 _ x + x - 8 = 1

61. 2 _ x + 2

+ 3 _ x = -

x _ x + 2

62. 1 _ d + 4

= 2 _ d 2 + 3d - 4

- 1 _ 1 - d

63. 1 _ n - 2

= 2n + 1 _

n 2 + 2n - 8 + 2 _

n + 4

Example 6 Find the domain of f (x ) =

x + 7

_ x + 1

and any vertical or horizontal

asymptotes.


The function is undefined at the zero of the denominator h(x ) = x + 1, which is -1. The domain of f is all real numbers except x = -1.

Step 2 Find the asymptotes, if any.

Check for vertical asymptotes.

The zero of the denominator is -1, so there is a vertical asymptote at x = -1.

Check for horizontal asymptotes.

The degree of the numerator is equal to the degree of the

denominator. The ratio of the leading coefficient is 1 _ 1 = 1.

Therefore, y = 1 is a horizontal asymptote.



9 ± √ � 33

_

2 ≈ 1.63, 7.37

-3

∅

7

_ 3

22222---66666222222 66666 Nonlinear Inequalities (pp. 141–147)


64. (x + 5)(x - 3) ≤ 0 65. x 2 - 6x - 16 > 0

66. x 3 + 5 x 2 ≤ 0 67. 2 x 2 + 13x + 15 < 0

68. x 2 + 12x + 36 ≤ 0 69. x 2 + 4 < 0

70. x 2 + 4x + 4 > 0 71. x - 5 _ x < 0

72. x + 1 __

(12x + 6)(3x + 4) ≥ 0 73. 5 _

x - 3 + 2 _

x - 4 > 0

Example 7 Solve x 3 + 5 x 2 - 36x ≤ 0.

Factoring the polynomial f (x ) = x 3 + 5 x 2 - 36x yields f (x ) = x (x + 9)(x - 4), so f (x ) has real zeros at 0, -9, and 4.

Create a sign chart using these zeros. Then substitute an x-value from each test interval into the function to determine whether f (x ) is positive or negative at that point.

-9 0 4

(+)(-) (-) (+)

Because f (x ) is negative on the first and third intervals, the solution of x 3 + 5 x 2 - 36x ≤ 0 is (-∞, -9] ∪ [0, 4].

[-5, 3] (-∞, -2) ∪ (8, ∞)

[-∞, -5] ∪ {0} (-5, -

3

_ 2 )

{-6} �

(-∞, -2) ∪ (-2, ∞)(0, 5)

(3, 26

_ 7 ) ∪ (4, ∞)

⎛

⎪

⎝

-

4

_ 3 , -1

⎤

�

⎦

∪ (-

1

_ 2 , ∞)


0148_0153_AM_S_C02_SGR_664291.indd 151 4/30/12 6:23 PM

Study Guide and Review Continued

0148_0153_AM_S_C02_SGR_664291.indd 150 5/3/12 11:28 AM

151



Additional Answers

74b. Planetary Motion T

ime

(Ear

th y

ears

)

0

10

20

30

Distance (astronomical units)4 8

T = R3_2

75a.

[0, 3.5] scl: 0.5 by [0, 100] scl: 10

76a. Big Monster

Ele

vatio

n (ft

)

0

10

20

30

40

50

60

70

80

90

Time (s)5 10 15 20 25

Sample answer: cubic function

76b. Sample answer: f (x ) = 0.032 x 3 - 1.171 x 2 + 6.943x + 76; r 2 = 0.997

78a. f (x ) = -50 x 2 + 500x + 10,000

78c. $10; Sample answer: If she raises the price of her books more than $10, she will make less than $10,000.

80b. p (x ) = 200 _ x + 50,000

82a. 3000 _ x + 5 < 10; More than 600 people need to attend the dance.

82b. More than 800 people need to attend the dance.

Practice Test

0148_0153_AM_S_C02_SGR_664291.indd 153 4/30/12 6:23 PM

Study Guide and Review Study Guide and Review ContinuedContinued

74. PHYSICS Kepler’s Third Law of Planetary Motion implies that the time T it takes for a planet to complete one revolution in its orbit

about the Sun is given by T = R 3 _ 2 , where R is the planet’s mean

distance from the Sun. Time is measured in Earth years, and distance is measured in astronomical units. (Lesson 2-1)

a. State the relevant domain and range of the function.


c. The time for Mars to orbit the Sun is observed to be 1.88 Earth years. Determine Mars’ average distance from the Sun in miles, given that one astronomical unit equals 93 million miles.

75. PUMPKIN LAUNCH Mr. Roberts’ technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given. (Lesson 2-1)

t 0.5 1.0 1.5 2.0 2.5 3.0

v 85 50 30 20 15 12



c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds.

d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

76. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table. (Lesson 2-2)

Time (seconds) 5 10 15 20 25

Elevation (feet) 85 62 22 4 17

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data.

b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient.

c. Use the model to estimate a rider’s elevation at 17 seconds.

d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

77. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass increased following the model f (x ) = 0.021 x 3 - 0.336 x 2 + 1.945x - 0.720, where x is the number of years since 2001 and f (x ) is the number of square feet each year. Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth. (Lesson 2-3)

78. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Due to rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. (Lesson 2-4)

a. Write a function for her total sales after raising the price of her books x dollars.

b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250?

c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

79. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure. Assume that the width of the enclosure w will not be greater than the side of the barn. (Lesson 2-4)

w

�

= 750 m2

80. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. (Lesson 2-5)

a. How many gallons of acid are in the pond?

b. Suppose x gallons of pure water was added to the pond. Write p(x ), the percentage of acid in the pond after x gallons of pure water are added.

c. Find the horizontal asymptote of p (x ).

d. Does the function have any vertical asymptotes? Explain.

81. BUSINESS For selling x cakes, a baker will make b (x ) = x 2 - 5x - 150 hundreds of dollars in revenue. Determine the minimum number of cakes the baker needs to sell in order to make a profit. (Lesson 2-6)

82. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. (Lesson 2-6)

a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would like to keep the cost per person under $10.

b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

D = (0, ∞), R = (0, ∞)See margin.

about 141.66 million mi

75a. See margin.

v (t ) = 43.63 t -1.12

about 35.6 mi/h

about 0.94 second

See

margin.

See margin.Sample answer: about 12.7 ft

Sample answer: about 11.4 seconds

6.13 f t 2

See margin.

$5

See margin.

15 m by 50 m or 25 m by 30 m

200 gal

See margin.

y = 0

No; sample answer: p (x ) is defined for

all numbers in the interval [0, ∞).

16 cakes

a-b. See margin.

Applications and Problem Solving


0148_0153_AM_S_C02_SGR_664291.indd 152 4/30/12 6:23 PM




Chapter 2 Practice Test



1 On Level OL 2 Strategic Intervention ALapproaching grade level

If students miss about 25% of the exercises or less, If students miss about 50% of the exercises,

Then choose a resource: Then choose a resource:

SE Lessons 2-1, 2-2, 2-3, 2-4, 2-5, and 2-6Study Guide and Intervention

TE Chapter Project, p. 84

connectED.mcgraw-hill.comcococco Self-Check Quiz connectED.mcgraw-hill.comcococco Extra Examples, Homework Help

Additional Answers

1. y

x

f (x) = 0.25x-3

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); no intercepts; lim x→-∞

f (x ) = 0

and lim x→∞


discontinuity at x = 0; decreasing: (-∞, 0) and (0, ∞)

2.

y

x

f (x) = 8x4_3

D = (-∞, ∞), R = [ 0, ∞); intercept: 0; lim x→-∞

f (x ) = ∞

and lim x→∞

f (x ) = ∞; continuous for

all real numbers; decreasing: (-∞, 0); increasing: (0, ∞)

9. lim x→∞

f (x ) = ∞; lim x→-∞

f (x ) = ∞;

The degree is even and the leading coefficient is positive.

10. lim x→∞

f (x ) = -∞; lim x→-∞

f (x ) = ∞;

The degree is odd and the leading coefficient is negative.

14. The degree is 2, and the leading coefficient is -32. Because the degree is even and the leading coefficient is negative, lim x→-∞

f (x ) = -∞ and

lim x→∞

f (x ) = -∞.

I t ti PPll

Practice TestPractice Test

Graph and analyze each function. Describe the domain, range,


increasing or decreasing.

1. f (x ) = 0.25 x -3 2. f (x ) = 8 x 4 _ 3


3. x = √ �� 4 - x - 8 4. √ �� 5x + 4 = √ �� 9 - x + 7

5. -2 + √ �� 3x + 2 = x 6. 56 - 8 √ ��

7 x 2 + 4 = 54

7. x 4 - 5 x 3 - 14 x 2 = 0 8. x 3 - 3 x 2 - 10x = -24


using limits. Explain your reasoning using the leading term test.

9. f (x ) = 5 x 4 - 3 x 3 - 7 x 2 + 11x - 8

10. f (x ) = -3 x 5 - 8 x 4 + 7 x 2 + 5


function. Then find all of the real zeros by factoring.

11. f (x ) = 4 x 3 + 8 x 2 - 60x

12. f (x ) = x 5 - 16x

13. MULTIPLE CHOICE Which function has 3 turning points?

A f (x ) = x 4 - 4 C f (x ) = x 3 + 9 x 2 + 20x

B f (x ) = x 4 - 11 x 3 D f (x ) = x 4 - 5 x 2 + 4

14. BASEBALL The height h in feet of a baseball after being struck by a batter is given by h (t ) = -32 t 2 + 128t + 4, where t is the time in seconds after the ball is hit. Describe the end behavior of the graph of the function using limits. Explain using the leading term test.

For each function, (a) apply the leading term test, (b) find the zeros and

state the multiplicity of any repeated zeros, (c) find a few additional

points, and then (d) graph the function.

15. f (x ) = x (x - 1)(x + 3)

16. f (x ) = x 4 - 9 x 2

Use the Factor Theorem to determine if the binomials given are factors of

f (x ). Use the binomials that are factors to write a factored form of f (x ).

17. f (x ) = x 3 - 3 x 2 - 13x + 15; (x + 3)

18. f (x ) = x 4 - x 3 - 34 x 2 + 4x + 120; (x + 5), (x - 2)

19. WEATHER The table shows the average high temperature in Bay Town each month.

Jan Feb Mar Apr May Jun

62.3º 66.5º 73.3º 79.1º 85.5º 90.7º

Jul Aug Sep Oct Nov Dec

93.6º 93.5º 89.3º 82.0º 72.0º 64.6º

a. Make a scatter plot for the data.

b. Use a graphing calculator to model the data using a polynomial function with a degree of 3. Use x = 1 for January and round each coefficient to the nearest thousandth.

c. Use the model to predict the average high temperature for the following January. Let x = 13.

Write a polynomial function of least degree with real coefficients in

standard form that has the given zeros.

20. -1, 4, - √ � 3 21. 5, -5, 1 - i

22. MULTIPLE CHOICE Which function graphed below must have imaginary zeros?

F Hy

x

−8

−16

−4−8

16

8

4 8

y

x

−8

−16

−4−8

16

8

4 8

G Jy

x

−8

−16

−4−8

16

8

4 8

y

x

−8

−16

−4−8

16

8

4 8


23. f (x ) = ( x 3 - 7 x 2 + 13) ÷ (x - 2)

24. f (x ) = ( x 4 + x 3 - 2 x 2 + 3x + 8) ÷ (x + 3)

Determine any asymptotes and intercepts. Then graph the function and

state its domain.

25. f (x ) = 2x - 6 _ x + 5

26. f (x ) = x 2 + x - 6 _

x - 4


27. x 2 - 5x - 14 < 0 28. x 2 _

x - 6 ≥ 0

1–2. See margin.

-5 9

no solution 6, -6

0, 7, -2 2, 4, -3

9–10. See margin.

3 real zeros and 2 turning points; -5, 0, and 3

5 real zeros and 4 turning points; -2, 0, and 2

D

See margin.

15–16. See Chapter 2

Answer Appendix.

yes; f (x ) = (x + 3)(x - 1)(x - 5)

yes; yes; f (x ) = (x - 2)(x + 5)(x - 6)(x + 2)

a–c. See Chapter 2

Answer Appendix.

f (x ) = x 4 - 3 x 3 - 7 x 2 + 9x + 12

21. f (x ) = x 4 - 2 x 3 - 23 x 2 + 50x - 50

J

x 2 - 5x - 10 -

7

_ x - 2

x 3 - 2 x 2 + 4x - 9 +

35

_ x + 3


(-2, 7) {0} ∪ (6, ∞)


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Study Guide and Review Continued

0148_0153_AM_S_C02_SGR_664291.indd 152 4/30/12 6:23 PM

153


Chapter 2 Connect to AP CalculusChapter 2 Connect to AP Calculus

Connect to AP CalculusConnect to AP Calculus

Area Under a CurveArea Under a CurveObjectiveApproximate the area between a curve and the x-axis.

Integral calculus is a branch of calculus that focuses on the processes y

x

of finding areas, volumes, and lengths. In geometry, you learned how to calculate the perimeters, areas, and volumes of polygons, polyhedrons, and composite figures by using your knowledge of basic shapes, such as triangles, pyramids, and cones. The perimeters, areas, and volumes of irregular shapes, objects that are not a combination of basic shapes, can be found in a similar manner. Calculating the area between the curve and the x-axis, as shown to the right, is an application of integral calculus.

Activity 1 Approximate Area Under a Curve

Approximate the area between the curve f (x) = √ �� - x 2 + 8x and the x-axis using rectangles.

Step 1 Draw 4 rectangles with a width of 2 units between f (x) and the x-axis. The height of the rectangle should be determined when the left endpoint of the rectangle intersects f (x), as shown in the figure. Notice that the first rectangle will have a height of f (0) or 0.

Step 2 Calculate the area of each rectangle.

Step 3 Approximate the area of the region by taking the sum of the areas of the rectangles.

x

y

-2

2

4

6

2 4

2

6 8

f (2)

Analyze the Results

1. What is the approximation for the area?

2. How does the area of a rectangle that lies outside the graph affect the approximation?

3. Calculate the actual area of the semicircle. How does the approximation compare to the actual area?

4. How can rectangles be used to find a more accurate approximation? Explain your reasoning.

21.86 unit s 2

See margin.

See margin.

See margin.

Using relatively large rectangles to estimate the area under a curve may not produce an approximation that is as accurate as 3desired. Significant sections of area under the curve may go unaccounted for. Similarly, if the rectangles extend beyond the curve, substantial amounts of areas that lie above a curve may be included in the approximation.

x

y

2

4

2 4 6 8

UndesiredAreaDesired Area

In addition, regions are also not always bound by a curve intersecting the x-axis. You have studied many functions with graphs that have different end behaviors. These graphs do not necessarily have two x-intercepts that allow for obvious start and finish points. In those cases, we often estimate the area under the curve for an x-axis interval.

y

x

154 | Chapter 2

0154_0155_AM_S_C02_CAP_664291.indd 154 4/30/12 6:24 PM

154 | Chapter 2 | Connect to AP Calculus

6. Sample answer: There is no extra area that lies above the curve that is included in the approximation that can account for the area that lies under the curve that has not beenincluded. Therefore, the approximation should be lower than the actual area.

7. Yes; Sample answer: For this example, using right endpoints for the rectangles would produce rectangles that overlap the curve and include area that lies above the curve. This would produce a higher approximation. However, this is not always the case. If the curve is symmetrical, like the previous example, the approximations will be the same regardless of the endpoint used.

1 FocusObjective Approximate the area between a curve and the x-axis.

Teaching TipFor Activity 1, you may wish to have students use a compass to draw the graph. The center of the semi-circle is at (4, 0). The radius is 4. The bottom left vertices of the four rectangles are (0, 0), (2, 0), (4, 0), and (6, 0). The upper left vertices of the rectangles are (0, f (0)), (2, f (2)), (4, f (4)), and (6, f (6)). The width of each rectangle is 2. The lengths are f (0), f (2), f (4), and f (6).

2 TeachWorking in Cooperative GroupsFor Activity 1, pair students with different abilities. Have students work through Steps 1–3 and then answer Analyze the Results Exercises 1–4.

Ask:■ What is the formula for the area of

a circle? A = π r 2

■ What is the formula for the area of

a semi-circle? A = 1 _ 2 π r 2

Additional Answers2. Sample answer: The extra area

found outside the curve that is included in the sum helps to account for the area found under the curve that was not included.

3. 25.13 units 2 ; Sample answer: The approximation is less than the actual area. As more rectangles are used, the approximation approaches the actual area.

4. Sample answer: Using rectangles of a smaller width should result in a more accurate approximation. Smaller rectangles would better fit the shape of the curve and would help to reduce areas that are not being accounted for.

154_155_AM_T_C02_STP_664293.indd 154154_155_AM_T_C02_STP_664293.indd 154 7/26/12 1:57 PM7/26/12 1:57 PM

Chapter 2 Connect to AP CalculusChapter 2 Connect to AP Calculus

Activity 2 Approximate Area Under a Curve

Approximate the area between the curve f (x) = x 2 + 2 and the x-axis on the interval [1, 5] using rectangles.

Step 1 Draw 4 rectangles with a width of 1 unit between f (x) and

x

y

8

16

24

2 4 6

the x-axis on the interval [1, 5], as shown in the figure. Use the left endpoint of each sub interval to determine the height of each rectangle.

Step 2 Calculate the area of each rectangle.

Step 3 Approximate the area of the region by determining the sum of the areas of the rectangles.

Step 4 Repeat Steps 1–3 using 8 rectangles, each with a width of 0.5 unit, and 16 rectangles, each with a width of 0.25 unit.

Analyze the Results

5. What value for total area are the approximations approaching?

6. Using left endpoints, all of the rectangles completely lie under the curve. How does this affect the approximation for the area of the region?

7. Would the approximations differ if each rectangle’s height was determined by its right endpoint? Is this always true? Explain your reasoning.

8. What would happen to the approximations if we continued to increase the number of rectangles being used? Explain your reasoning.

9. Make a conjecture about the relationship between the area under a curve and the number of rectangles used to find the approximation. Explain your answer.

Sample answer: 48 unit s 2

See margin.

See margin.

See margin.

See margin.

Model and Apply 10. In this problem, you will approximate the area

between the curve f (x) = -x2 + 12x and the x-axis.

a. Approximate the area by using 6 rectangles, 12 rectangles, and 24 rectangles. Determine the height of each rectangle using the left endpoints.

b. What value for total area are the approximations approaching?

c. Does using right endpoints opposed to left endpoints for the rectangles’ heights produce a different approximation? Explain your reasoning.

11. In this problem, you will approximate the area between the curve f (x) = 1_

2x3 - 3 x2 + 3x + 6 and the x-axis on the

interval [1, 5].

a. Approximate the area by first using 4 rectangles and then using 8 rectangles. Determine the height of each rectangle using left endpoints.

b. Does estimating the area by using 4 or 8 rectangles give sufficient approximations? Explain your reasoning.

c. Does using right endpoints opposed to left endpoints for the rectangles’ heights produce a different approximation? Explain your reasoning.

x

y

2

4

6

8

10

2 4 6

10a. 6 rectangles: 280 unit s 2 ; 12 rectangles: 286 unit s 2 ; 24 rectangles: 287.5 unit s 2

Sample answer: 288 unit s 2

No; sample answer: The graph is symmetrical. Using right endpoints would produce the same approximation as using left endpoints.

x

y

8

16

24

32

2 4 6 8 10 12

4 rectangles: 14 unit s 2 ; 8 rectangles: 13.75

Study Tip Endpoints Any point within a subinterval may be used to determine the height of the rectangles used to approximate area. The most commonly used are left endpoints, right endpoints, and midpoints.

11b. Sample answer: Approximating the area using 4 or 8 rectangles probably does not give a good representation of the actual area. The nature of the curve prevents rectangles of widths of 1 and 0.5 units from really fitting well beneath it.

11c. Yes; sample answer: The graph is not symmetrical. Using right endpoints to determine the height of the rectangles would produce different rectangles with different areas.


0154_0155_AM_S_C02_CAP_664291.indd 155 4/30/12 6:24 PM


Additional Answers8. Sample answer: Each approximation that is

found using more rectangles should be a better representation of the actual area. Smaller rectangles would fill the curved region better and would help to insure that most of the area under the curve is included in the approximation.

9. Sample answer: The more rectangles that are used, the better the approximation of the area. Smaller rectangles fit the desired region better than larger rectangles, thus producing more accurate approximations.

Have student pairs work through Activity 2 Steps 1–4 and then answer Analyze the Results Exercises 5–9.

Practice Have students complete Exercises 10 and 11.

3 AssessFormative AssessmentUse part a of Exercise 11 to assess whether students understand how to approximate the area between a curve and the x-axis.

From Concrete to AbstractHave students summarize what they learned about area under a curve. Have them include a description of how the width of the rectangles used affects the approximation.

Extending the ConceptShow students the symbolic form for expressing the area under the curve of f (x ) = x 2 + 2 in the interval from 1 to 5

from Activity 2, ∫ 1 5 ( x 2 + 2) dx. Explain

that the symbol ∫ is actually an

elongated S, as in the word “sum”.Ask students to press 9 to open a window that allows them to evaluate an integral. Once they input the values x 2 + 2, x, 1, and 5 and then press ENTER their windows will display the following.

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Page 85, Get Ready for Chapter 2

8. 9.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = 1_2x

y

x

−4

−8

−4−8

8

4

4 8

f (x) = -2

10. 11.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = x 2+ 3

y

x

−4

−8

−12

−4−8 4 8f (x) = -x 2

+ x - 6

12. 13.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = 2x 2- 5x - 3

y

x

−4

−8

−4−8

8

4

4 8

f (x) = 3x 2- x - 2

14.

200

400

600

800

08 16 24 32 t

f (t)

f (t) = 2t + 0.5t 2

15. {x | x ≤ 6, x ∈ �}; (-∞, 6 ]

16. {x | -2 ≤ x, x ∈ �}

17. {x | -2 < x < 9, x ∈ �}; (-2, 9)

18. {x | 1 < x ≤ 4, x ∈ �}; (1, 4]

19. {x | x < -4 or x > 5, x ∈ �}; (-∞, -4) � (5, ∞)

20. {x | x < -1 or x ≥ 7, x ∈ �}; (-∞, -1) � [ 7, ∞)

Pages 87–89, Lesson 2-1 (Guided Practice)

1A. y

x−2−4

16

24

32

8

2 4

f (x) = 3x 6


f (x ) = ∞

and lim x→∞

f (x ) = ∞;

continuous for all real numbers; decreasing: (-∞, 0); increasing: (0, ∞)

1B. y

x

−8

−16

−2−4

16

8

2 4

f (x) = - x 52_3

D = (-∞, ∞), R = (-∞, ∞); intercept: 0; lim x→-∞

f (x ) = ∞ and

lim x→∞

f (x ) = -∞; continuous

for all real numbers; decreasing: (-∞, ∞)

2A. yx

f (x) = - x -41_2

−2 2 4−4

−1

−2

−3

−4

D = (-∞, 0) � (0, ∞), R = (-∞, 0); no intercepts; lim x→-∞


f (x )

= 0; infinite discontinuity at x = 0; decreasing: (-∞, 0), increasing: (0, ∞)

2B. y

x

−4

−8

−4−8

8

4

4 8

f (x) = 4x -3

D = (-∞, 0) � (0, ∞), R = (-∞, 0) � (0, ∞); no intercepts; lim x→-∞

f (x ) = 0 and

lim x→∞



3A. y

x

4

8

12

16

4 8 12 16

f (x) = 2x3_4

D = [ 0, ∞), R = [ 0, ∞); intercept: 0; lim x→∞

f (x ) = ∞;

continuous on [0, ∞); increasing: (0, ∞)

3B. y

x

−20

−40

−1−2

40

20

1 2

f (x) = 10x 5_3


f (x ) = -∞ and

lim x→∞

f (x ) = ∞; continuous for

all real numbers; increasing: (-∞, ∞)

4A.

[0, 100] scl: 10 by [0, 250] scl: 25

155A | Chapter 2 | Answer Appendix

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Pages 92—95, Lesson 2-1

1.

y

xO−2−4

8

12

16

4

2 4

f (x) = 5x2


f (x ) = ∞

and lim x→∞

f (x ) = ∞;


2. y

x

g (x) = 8x 5


f (x ) = -∞

and lim x→∞



3. y

x

h (x) = -x 3


f (x ) = ∞

and lim x→∞

f (x ) = -∞;


4. yx

−4

−8

−12

−16

−2−4 2 4

f (x) = -4x 4

D = (-∞, ∞), R = (-∞, 0 ]; intercept: 0; lim x→-∞

f (x ) = -∞

and lim x→∞

f (x ) = -∞;

continuous for all real numbers; increasing: (-∞, 0); decreasing: (0, ∞)

5. y

x

g (x) = x91_3


f (x ) = -∞

and lim x→∞

f (x ) = ∞;

continuous for all real numbers; increasing: (-∞, ∞)

6. y

x−2−4

8

12

16

4

2 4

f (x) = x 85_8


f (x ) = ∞

and lim x→∞

f (x ) = ∞;


7.

y

x

f (x) = - x71_2


f (x ) = ∞

and lim x→∞

f (x ) = -∞;


8. yx

g (x) = - x 61_4


f (x ) = -∞

and lim x→∞

f (x ) = -∞;

continuous for all real numbers; increasing: (-∞, 0); decreasing: (0, ∞)

9.

y

x−4−8

8

12

16

4

4 8

f (x) = 2x-4

D = (-∞, 0) � (0, ∞),R = (0, ∞); no intercepts; lim x→-∞

f (x ) = 0

and lim x→∞



10.

y

x

−4

−8

−4−8

8

4

4 8

h(x) = -3x -7


f (x ) = 0

and lim x→∞


discontinuity at x = 0; increasing: (-∞, 0) and (0, ∞)

11.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = -8x-5


f (x ) = 0

and lim x→∞



12.

y

x−4−8

8

12

16

4

4 8

g(x) = 7x-2

D = (-∞, 0) � (0, ∞), R = (0, ∞); no intercepts; lim x→-∞

f (x ) = 0 and

lim x→∞



13.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = x-92_5


f (x ) = 0

and lim x→∞



155B

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14.

y

x−4−8

8

12

16

4

4 8

h(x) = 1_6

x-6


f (x ) = 0 and

lim x→∞



15.

y

x

−4

−8

−4−8

8

4

4 8

h(x) = 3_4

x-3


f (x ) = 0

and lim x→∞



16.

yx

−4

−8

−12

−16

−4−8 4 8

f (x) = - 7_10

x-8

D = (-∞, 0) � (0, ∞), R = (-∞, 0); no intercepts; lim x→-∞

f (x ) = 0 and

lim x→∞


discontinuity at x = 0; decreasing: (-∞, 0); increasing: (0, ∞)

18.

y

x2 4 6 8

4

8

12

16

f (x) = 8x1_4

D = [ 0, ∞), R = [ 0, ∞); intercept: 0; lim x→∞

f (x ) = ∞;

continuous on [ 0, ∞); increasing: (0, ∞)

19.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = -6x 1_5


f (x ) = ∞

and lim x→∞

f (x ) = -∞;


20.

y

x

g(x) 1_5x

1_3


f (x ) = 0

and lim x→∞



21.

y

x4 8 12 16

4

8

12

16

f (x) = 10x-

1_6

D = (0, ∞), R = (0, ∞); no intercepts; lim x→∞

f (x ) = 0;

continuous on (0, ∞); decreasing: (0, ∞)

22.

yx

−4

−8

−12

−16

4 8 12 16

g(x) = -3x 5_8

D = [ 0, ∞), R = (-∞, 0 ]; intercept: 0; lim x→∞

f (x ) = -∞;

continuous on [0, ∞); decreasing: (0, ∞)

23. y

x

h(x) = 3_4x

3_5

D = (-∞, ∞), R = (-∞, ∞); intercept: 0 lim x→-∞

f (x ) = -∞

and lim x→∞



24. yx

f (x) 1_2x

3_4

D = (0, ∞), R = (-∞, 0); no intercepts; lim x→∞

f (x ) = 0;

continuous on (0, ∞); increasing: (0, ∞)

25. y

x

f (x) = x- 2_3

D = (-∞, 0) � (0, ∞), R = (0, ∞); no intercepts;

lim x→-∞


f (x ) = 0;

infinite discontinuity at x = 0; increasing: (-∞, 0); decreasing: (0, ∞)

26. y

x

−4

−8

−2−4

8

4

2 4

h(x) = 7x 5_3


f (x ) = -∞

and lim x→∞

f (x ) = ∞;

continuous for all real numbers;increasing: (-∞, ∞)

27. yx

h(x) = -4x 7_4

D = [ 0, ∞), R = (-∞, 0 ]; intercept: 0; lim x→∞

f (x ) = -∞;

continuous on [ 0, ∞); decreasing: (0, ∞)

155C | Chapter 2 | Answer Appendix

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28. yx

h(x) = -5x- 3_2

D = (0, ∞), R = (-∞, 0); no intercepts; lim x→∞

f (x ) = 0;

continuous on (0, ∞); increasing: (0, ∞)

29.

y

x

h(x) = 2_3

x- 8_5


f (x ) = 0 and

lim x→∞



34.

y

x

f (x) = 3 √ 6 + 3x

D = [ -2, ∞), R = [ 0, ∞); x-intercept: -2, y-intercept: 3

√ � 6 ;

lim x→∞

f (x ) = ∞;

continuous on [ -2, ∞); increasing: (-2, ∞)

35.

y

x

−6

−12

12

6

200 400−200−400

g(x) = -2 5√ 1024 + 8x

D = (-∞, ∞), R = (-∞, ∞); x-intercept: -128, y-intercept: -8; lim x→-∞

f (x ) = ∞ and

lim x→∞

f (x ) = -∞;


36.

yx

−1

−2

−3

−4

−2 2 4

f (x) = - 3_8

6√(16x + 48) - 3

D = [ -3, ∞), R = (-∞, -3 ]; y-intercept: -3.71; lim x→∞

f (x ) = -∞;

continuous on [ -3, ∞); decreasing: (-3, ∞)

37.

y

x

h(x) = 4 + √7x - 12

D = ⎡

⎣

12 _ 7 , ∞), R = [ 4, ∞);

no intercepts; lim x→∞

f (x ) = ∞;

continuous on ⎡

⎣

12 _ 7 , ∞);

increasing: (

12 _ 7 , ∞)

38. yx

−4

−8

−12

−16

−2−4 2 4

g(x) = √(1 - 4x)3 - 16

D = (-∞, 0.25 ], R = [ -16, ∞); x-intercept: -1.34, y-intercept: -15; lim x→-∞

f (x ) = ∞;

continuous on (-∞, 0.25 ]; decreasing: (-∞, 0.25)

39. yx

−20

−40

−60

−80

−4−8 4 8

f (x) = - 3√(25x - 7) 2 - 49

D = (-∞, ∞), R = (-∞, -49.00 ]; y-intercept: -52.66; lim x→-∞

f (x ) = -∞

and lim x→∞

f (x ) = -∞;

continuous for all real numbers; increasing: (-∞, 0.28); decreasing: (0.28, ∞)

40. yx

−4

−8

−12

−16

−8−16 8 16

h(x) = 1_2

3√27 - 2x - 8

D = (-∞, ∞), R = (-∞, ∞); x-intercept: -2034.5, y-intercept: -6.5; lim x→-∞

f (x ) = ∞

and lim x→∞

f (x ) = -∞;


41.

y

x

−4

−8

8

4

6 12 18 24

g (x) = √22 - x - √3x - 3

D = [ 1, 22 ],

R = ⎡ ⎣

-

√ � 63 ,

√ � 21

⎤ ⎦

;

x-intercept: 6.25; continuous on [ 1, 22 ]; decreasing: (1, 22)

42a.

Velocity of Water with a Nozzle

0

10

20

30

40

Velo

city

(ft/

sec)

2 4 6 8 P

V(P)

Pressure (lbs/in2)

V (P) = 12.1 √P

42b. D = [ 0, ∞), R = [ 0, ∞); lim x→∞

f (x ) = ∞; continuous on [ 0, ∞);

increasing: (0, ∞)

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80a. Sample answers given.

y

x

f (x) = x 1_3

y

x

f (x) = x 1_2

y

xf (x) = x

y

xf (x) = x 2

y

xf (x) = x 3

80c. Sample answer: When 0 < n < 1, the average rate of change of the function decreases as x approaches infinity. When n = 1, the average rate of change of the function is constant as x approaches infinity. When n > 1, the average rate of change of the function increases as x approaches infinity.

81. √

��

8 n · 2 7 _ 4 -n

= √

��

( 2 3 ) n · 2 7

_ ( 2 2 ) -n

= √

��

2 3n · 2 7 _

2 -2n

= √ ��

2 (3n + 7) - (-2n)

= √ ��

2 5n + 7

= √ ��

2 4n + 6 · 2 n + 1

= √ ��

2 4n + 6 · √ ��

2 n + 1

= 2 2n + 3 √ ��

2 n + 1

82d. Sample answer: If the exponent is less than 0, the power is greater than 0 and less than 1. If the exponent is greater than 0 and less than 1, the power is greater than 1 and less than the base. If the exponent is greater than 1, the power is greater than the base. Any nonzero number to the zero power is 1. Thus, if the exponent is less than 0, the power is less than 1. A power of a positive number is never negative, so the power is greater than 0. Any nonzero number to the zero power is 1 and to the first power is itself. Thus, if the exponent is greater than 0 and less than 1, the power is between 1 and the base. Any number to the first power is itself. Thus, if the exponent is greater than 1, the power is greater than the base.

85. Sample answer: As n increases, the value of 1 _ n approaches 0.

This means that the value of x 1 _ n will approach 1 when x is

positive and -1 when x is negative. Therefore, for positive values of x, f (x ) will approach 1 + 5 or 6 and will resemble the line y = 6. For negative values of x, f (x ) will approach -1 + 5 or 4 and will resemble the line y = 4.

91.

y

xO

−4

−8

−4−8

8

4

4 8

f (x)

y

xO

−4

−8

−4−8

8

4

4 8

g (x)

y

xO

−4

−8

−4−8

8

4

4 8

h (x)

92.

y

x

−4

−8

−4−8

8

4

4 8

f (x)

y

x

−4

−8

−4−8

8

4

4 8

g (x)

y

x

−4

−8

−4−8

8

4

4 8

h (x)

93.

y

xO

−4

−8

−12

−4−8 4 8

f (x)

y

xO−4−8 4 8

4

8

12

g (x)

y

xO

−4

−8

−12

−4−8 4 8

h (x)

155E | Chapter 2 | Answer Appendix

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Page 96, Explore 2-2

3. Sample answer:

LeadCoefficient

Degree ofPolynomial

End Behavior

negative odd lim x→-∞

f (x ) = ∞

lim x→∞

f (x ) = -∞

negativeeven and nonzero

lim x→-∞

f (x ) = -∞

lim x→∞

f (x ) = -∞

positive odd lim x→-∞

f (x ) = -∞

lim x→∞

f (x ) = ∞

positiveeven and nonzero

lim x→-∞

f (x ) = ∞

lim x→∞

f (x ) = ∞

Page 102, Lesson 2-2 (Guided Practice)

6A. a. The degree is 5, and the leading coefficient is -54, so lim x→-∞


f (x ) = -∞.

b. The distinct real zeros are x = 0, x = 4, and x = 1 _ 3 .

The zero at x = 1 _ 3 has multiplicity 3.

c. Interval x-value f (x ) (x, f (x ))

(∞, 0) -1 f (-1) = 640 (-1, 640)

(0, 1 _ 3 ) 1 _

4 f ( 1 _

4 ) ≈ -0.03 ( 1 _

4 , -0.03)

( 1 _ 3 , 4) 1 f (1) = 48 (1, 48)

(4, -∞) 5 f (5) = -27,440 (5, -27,440)

d. y

x−2−4

2000

3000

1000

2 4

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

6B. a. The degree is 3, and the leading coefficient is -1, so lim x→-∞


f (x ) = -∞.

b. The distinct real zeros are x = -2, x = 0, and x = 4. There is no multiplicity.

c. Interval x-value f (x ) (x, f (x ))

(∞, -2) -4 f (-4) = 64 (-4, 64)

(-2, 0) -1 f (-1) = -5 (-1, -5)

(0, 4) 2 f (2) = 16 (2, 16)

(4, -∞) 10 f (10) = -720 (10, -720)

d.

y

x

−8

−4−8

16

8

4 8

h(x) = -x 3 + 2x2 + 8x

Pages 104–106, Lesson 2-2

11.

Volu

me

(gal

lons

)

10

23456789

Time (minutes)0.2 0.4 0.6 0.8 t

v (t)Draining Water

v (t) = 10(1 t) 2



f (x ) = -∞.



f (x ) = ∞.


f (x ) = -∞ and lim x→∞

f (x ) = ∞.


f (x ) = -∞ and lim x→∞

f (x ) = -∞.



f (x ) = -∞.



f (x ) = -∞.


f (x ) = -∞ and lim x→∞

f (x ) = ∞.


f (x ) = -∞ and lim x→∞

f (x ) = -∞.

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f (x ) = -∞.



f (x ) = ∞.

22a.

[0, 5] scl: 1 by [6000, 14000] scl: 1000

22b. The degree is 4, and leading coefficient is 43.77. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

23. 5 real zeros and 4 turning points; 0, -1, and -2

24. 6 real zeros and 5 turning points; 0, 6, and 2

25. 4 real zeros and 3 turning points; ± √ � 3

26. 4 real zeros and 3 turning points; 0, 8, and -4

27. 6 real zeros and 5 turning points; 2 and 3 √ �� -2

28. 8 real zeros and 7 turning points; no real zeros

29. 6 real zeros and 5 turning points; 0, -2, and 2

30. 5 real zeros and 4 turning points; 0, 5, and -5

31. 4 real zeros and 3 turning points; 0, -

1 _ 2 , and 3 _

2

32. 5 real zeros and 4 turning points; 0, 4 _ 3 , and -5

33a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

33b. 0, -4, 1 (multiplicity: 2)

33c. Sample answer: (-5, 180), (-2, -36), (0.5, 0.5625), (2, 12)

33d.

y

x

−20

−40

−4−8

40

20

4 8 f (x) = x(x + 4)(x - 1 ) 2

34a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

34b. 0 (multiplicity: 2), 4, -2

34c. Sample answer: (-3, 63), (-1, -5), (2, -32), (5, 175)

34d.

y

x

−20

−40

−4−8

40

20

4 8

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

35a. The degree is 4, and the leading coefficient is -1. Becausethe degree is even and the leading coefficient is negative, lim x→-∞

f (x ) = -∞ and lim x→∞

f (x ) = -∞.

35b. 0, -3 (multiplicity: 2), 5

35c. Sample answer: (-4, -36), (-1, -24), (2, 150), (6, -486)

35d.

y

x

−100

−200

−4−8

200

100

4 8

f (x) = x(x + 3 ) 2 (x 5)

36a. The degree is 4, and the leading coefficient is 2. Becausethe degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

36b. 0, -5 (multiplicity: 2), 3

36c. Sample answer: (-6, 108), (-1, 128), (1, -144), (4, 648)

36d.

y

x

−100

−200

−4−8

200

100

4 8

f (x) = 2x(x + 5 ) 2 (x 3)

37a. The degree is 5, and the leading coefficient is -1. Becausethe degree is odd and the leading coefficient is negative, lim x→-∞


f (x ) = -∞.

37b. 0, 3, -2 (multiplicity: 3)

37c. Sample answer: (-3, 18), (-1, -4), (1, 54), (4, -864)

37d.

y

x

−60

−120

−4−8

120

60

4 8

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

155G | Chapter 2 | Answer Appendix

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38a. The degree is 4, and the leading coefficient is -1. Becausethe degree is even and the leading coefficient is negative, lim x→-∞

f (x ) = -∞ and lim x→∞

f (x ) = -∞.

38b. -2 (multiplicity: 2), 4 (multiplicity: 2)

38c. Sample answer: (-3, -49), (-1, -25), (5, -49)

38d.

y

x

−60

−120

−4−8

120

4 8

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

39a. The degree is 3, and the leading coefficient is 3. Because the degree is odd and the leading coefficient is positive, lim x→-∞

f (x ) = -∞ and lim x→∞

f (x ) = ∞.

39b. 0, 4, -3

39c. Sample answer: (-4, -96), (-2, 36), (2, -60), (5, 120)

39d.

y

x

−40

−80

−4−8

80

40

4 8

f (x) = 3 x 3 3 x 2 36x

40a. The degree is 3, and the leading coefficient is -2. Becausethe degree is odd and the leading coefficient is negative, lim x→-∞


f (x ) = -∞.

40b. 0, -3, 1

40c. Sample answer: (-4, 40), (-2, -12), (0.5, 1.75), (2, -20)

40d.

y

x

−10

−20

−2−4

20

10

2 4

f (x) = 2 x 3 4 x 2 + 6x

41a. The degree is 4, and the leading coefficient is 1. Becausethe degree is even and the leading coefficient is positive, lim x→-∞


f (x ) = ∞.

41b. 0 (multiplicity: 2), 4, -5

41c. Sample answer: (-6, 360), (-2, -72), (2, -56), (5, 250)

41d.

y

x

−60

−120

−4−8

120

60

4 8

f (x) = x 4 + x 3 20 x 2

42a. The degree is 5, and the leading coefficient is 1. Becausethe degree is odd and the leading coefficient is positive, lim x→-∞

f (x ) = -∞ and lim x→∞

f (x ) = ∞.

42b. 0 (multiplicity: 3), -5, 2

42c. Sample answer: (-6, -1728), (-3, 270), (1, -6), (3, 216)

42d.

y

x

−160

−320

−4−8

320

160

4 8

f (x) = x 5 + 3 x 4 10 x 3

44. Sample answer: f (x ) = -1.25x + 5

45. Sample answer: f (x ) = 0.09 x 3 - 2.70 x 2 + 24.63x - 65.21

46. Sample answer: f (x ) = 0.88x 4 + 0.89x 3 - 1.71x 2 - 2.99x + 4.89

47. Sample answer: f (x ) = 4.05x 4 - 0.09x 3 + 6.69x 2 -

222.03x + 2697.74

49a. According to the data, the values are increasing as x increases, therefore, lim x→∞

f (x ) = ∞.

49b.

[0, 12] scl: 1 by [400, 1400] scl: 200

Sample answer: f (x ) = 0.146 x 4 - 3.526 x 3 + 32.406 x 2 - 63.374x + 473.255; The line is not a good fit. There are many outlying points.

49c. The leading coefficient is 0.146, so lim x→∞

f (x ) = ∞.Sample answer: The prediction was accurate because the leading coefficient is positive, as x → ∞, f (x ) → ∞.

68b. A(r ) = 2π r 2 + 30 _ r

68c. about 33.7 in 2

[0, 10] scl: 1 by [0, 500] scl: 50

69. f (x ) = x 3 - 8 x 2 - 3x + 90

70. f (x ) = x 3 + 6 x 2 - 24x - 64

71. f (x ) = x 5 - 9 x 4 + 23 x 3 - 3 x 2 - 36x

72. f (x ) = x 5 - 4 x 4 - 19 x 3 + 46 x 2 - 24x

73. f (x ) = 12 x 4 + 83 x 3 + 131 x 2 - 54x - 72

74. f (x ) = 6 x 5 - 23 x 4 - 39 x 3 + 15 x 2 + 25x

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76.

Sample answer: f (x ) = (x - 2)(x - 5) ·(x + 1)( x 2 + 4)

y

x−4−8

−80

−160

80

160

4 8

y = f (x)

77.

Sample answer: f (x ) = -(x + 2) ·

(x + 3)(x - 4) ·

(x - 1)( x 2 + 6)

y

x

−320

−2−4

640

320

2 4

f (x) = (x + 2)(x + 3)(x 4)(x 1)(x 2+ 6)

78.

Sample answer: f (x ) = (x + 1) · (x + 2) 2 ( x 2 + 1)

y

x

y = f (x)

79.

Sample answer: f (x ) = -(x - 1) · (x - 2) (x + 1) 2 ·

( x 2 + 2)

y

x

−8

−16

−4−8

8

4 8

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

80a.

[0, 9] scl: 1 by [0, 10] scl: 1

80b. f (x ) = - 0.009x 3 - 0.230x 2 + 2.305x + 3.796

80c. f (x ) = 0.012x 4 - 0.225x 3 + 0.978 x 2 + 0.152x + 4.312

80d. Sample answer: According to the correlation coefficients for each of the models, r 2 = 0.94 for the cubic function and r 2 = 0.96 for the quartic function. Therefore, the quartic function is a better model.

89a. f (x)

g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x) g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x)

g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x) g(x)

h(x)

[-1, 1] scl: 0.1 by [-1, 1] scl: 0.1

f (x) g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x)

g(x)

h(x)

[-1, 1] scl: 0.1 by [-1, 1] scl: 0.1

89b. Sample answer: Very close to the origin, the function approximates the behavior of the term of lower degree, or h (x ).

89c. Sample answer: As x → ∞ and -∞, the function approximates the behavior of the term of higher degree, or g (x ).

89d. Sample answer: As x → ∞ and -∞, the function approximates the behavior of function a and, very close to the origin, the function approximates the behavior of function b.

90. Colleen; sample answer: The data suggest that there are two turning points. Thus, a cubic function is a better representation than a quadratic function. Also, when the table is entered into a graphing calculator, r 2 = 0.99 for the cubic function and r 2 = 0.75 for the quadratic function. This further supports Colleen’s model.

91. Sample answer: No; a polynomial function f (x ) cannot have both an absolute maximum and absolute minimum because as x → ∞, f (x ) will approach either ∞ or -∞. If f (x ) → ∞ as x → ∞, then an absolute maximum is impossible. If f (x ) → -∞ as x → ∞, then an absolute minimum is impossible.

92. Sample answer: The function f (x ) = 0 has no degree because there are no terms for the polynomial. The function g (x ) = c,c ≠ 0, has degree 0 because c = cx 0 for all x.

93. Rearranging the terms gives f (x ) = x 3 - x 2 - 12x + 5x 2 - 5x - 60. Notice how the first set of three terms has a common factor of x and the second set of three terms has a common factor of 5. After factoring using the Distributive Property, f (x ) = x ( x 2 - x - 12) + 5( x 2 - x - 12). Now notice how the factors inside the parentheses are identical. Using the Commutative Property, f (x ) = (x + 5)( x 2 - x - 12). After factoring the second factor, f (x ) = (x + 5)(x - 4)(x + 3). The function is now completely factored and the zeros of the function are -5, 4, and -3. These were determined by setting each factor equal to zero and solving for x.

94. Sample answer: This is possible because one of the graphs could be more compressed than the other. For example, f (x ) = a (x - 1)(x + 4)(x - 5) and g (x ) = 5a (x - 1)(x + 4) · (x - 5) have the same zeros, degree, and end behavior, assuming a > 0. The graph of g (x ) will be more stretched due to the factor of 5.

155I | Chapter 2 | Answer Appendix

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95. Sample answer: There is one turning point at the absolute maximum, one at the relative minimum, and one at the relative maximum. Therefore, the minimum degree is 3 + 1 or 4.

96. Sample answer: Make a scatter plot of the data. Use the scatter plot to determine which degree polynomial most resembles the data. Find the regression equation for the polynomial and compare the absolute value of its correlation coefficient to 1. Graph this equation on the same screen as the scatter plot to confirm their similarity. If the model does not fit the data or the absolute value of the correlation coefficient is not close to 1, correlation coefficients for other polynomials can be found to see if there is a better fit.

104. The graph of g (x ) is the graph of f (x ) translated 4 units right and 3 units up; g (x ) = (x - 4 ) 2 + 3.

105. The graph of g (x ) is the graph of f (x ) reflected in the x -axis and translated 2 units left and 4 units down; g (x ) = - (x + 2) 2 - 4.

106. The graph of g (x ) is the graph of f (x ) translated 6 units right and 4 units down; g (x ) = (x - 6) 2 - 4.

Page 116, Lesson 2-3

62. Sample answer: I would use my graphing calculator to graph the polynomial and to determine the three integral zeros, a, b, and c. I would then use synthetic division to divide the polynomial by a. I would then divide the resulting depressed polynomial by b, and then the new depressed polynomial by c. The third depressed polynomial will have degree 2. Finally, I would factor the second-degree polynomial to find the two non-integral, rational zeros, d and e. So, the polynomial is either the product (x - a )(x - b )(x - c )(x - d )(x - e ) or this product multiplied by some rational number.

68. Sample answer: Both long division and synthetic division can be used to divide a polynomial by a linear factor. Long division can also be used to divide a polynomial by a nonlinear factor. In synthetic division, only the coefficients are used. In both long division and synthetic division, placeholders are needed if a power of a variable is missing.

Page 118, Mid-Chapter Quiz

1.

y

x

−4

−8

−2−4

8

4

2 4

f (x) = 2x 3


f (x ) = -∞

and lim x→∞



2.

yx

−4

−8

−12

−16

−2−4 2 4

f(x) 2_3x 4


f (x ) = -∞ and lim x→∞

f (x ) = -∞; continuous for all real numbers; increasing: (-∞, 0); decreasing: (0, ∞)

3.

y

x−2−4

8

12

16

4

2 4

f (x) = 3x-8


f (x ) = 0 and

lim x→∞



4.

y

x

f (x) = 4x2_5


f (x ) = ∞

and lim x→∞




42a. g (x ) = (x - 4)(x - 3) (x + 2) 2

42b. g (x ) = (x - 4)(x - 3) (x + 2) 2

42c. 4, 3, -2 (multiplicity: 2)

43a. g (x ) = (x + 2)(x + 1) (x − 3 + √ � 5 ) (x − 3 − √ � 5 )

43b. g (x ) = (x + 2)(x + 1) (x - 3 +

√ � 5 ) (x - 3 -

√ � 5 )

43c. -2, -1, 3 -

√ � 5 , 3 +

√ � 5

44a. h (x ) = ( x 2 + 9)(x - 4)(x + 6)

44b. h (x ) = (x - 3i )(x + 3i )(x - 4)(x + 6)

44c. 3i, -3i, 4, -6

45a. f (x ) = ( x 2 - 4x + 13)(4x - 3)(x - 4)

45b. f (x ) = (4x - 3)(x - 4)(x - 2 + 3i )(x - 2 - 3i )

45c. 3 _ 4 , 4, 2 - 3i, 2 + 3i

46a. f (x ) = (4x + 5)(x + 3)( x 2 - 8x + 41)

46b. f (x ) = (4x + 5)(x + 3)(x - 4 + 5i )(x - 4 - 5i )

46c. -

5 _ 4 , -3, 4 + 5i, 4 - 5i

47a. h (x ) = (x + 3)(x - 5) (x + √ � 2 ) (x − √ � 2 )

47b. h (x ) = (x + 3)(x - 5) (x +

√ � 2 ) (x -

√ � 2 )

47c. -3, 5, √ � 2 , -

√ � 2

48a. g (x ) = (x + √ � 5 ) (x − √ � 5 ) ( x 2 + 36)

48b. g (x ) = (x +

√ � 5 ) (x -

√ � 5 ) (x + 6i )(x - 6i )

48c. √ � 5 , -

√ � 5 , 6i, -6i

49. 3, -4, 1 _ 2 , 3i, -3i ; h (x ) = (x - 3)(x + 4)(2x - 1)(x + 3i )

(x - 3i )

50. -5, 6, 2 _ 3 , 5i, -5i ; h (x ) = (x + 5)(x - 6)(3x - 2)(x + 5i )

(x - 5i )

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51. -3, -2, 1, 3 + i, 3 - i ; g (x ) = (x + 3)(x + 2)(x - 1) · (x - 3 + i )(x - 3 - i )

52. 5, -

3 _ 4 , 2, 4 + i, 4 - i ; g (x ) = (x - 5)(4x + 3)(x - 2) ·

(x - 4 - i )(x - 4 + i )

53. 3, 2 √ � 2 , -2

√ � 2 , 2i , -2i ; f (x ) = (x - 3)(x + 2

√ � 2 ) ·

(x - 2 √ � 2 )(x + 2i )(x - 2i )

54. 2 + √ � 3 , 2 -

√ � 3 , 3 + i , 3 - i ; g (x ) = (x - 2 -

√ � 3 ) ·

(x - 2 + √ � 3 )(x - 3 - i )(x - 3 + i )

72b. i. 3i, -3i, 1

ii. -

1 _ 2 , -2i, 2i, -2, 2

iii. -2, 3 _ 5 , 1

iv. -3i, 3i, -4i, 4i,

v. 4i, -4i, -2, 0 (multiplicity: 2), 1 _ 3

vi. i, -i, 3 _ 4 , 2

72c. An odd-degree polynomial function always has an odd number of zeros, and a polynomial function with real coefficients has imaginary zeros that occur in conjugate pairs. Therefore, an odd function with real coefficients will always have at least one real zero.

75a. The graphs of -f (x ) are the graphs of f (x ) reflected in the x-axis. The zeros are the same.

75b. The graphs of f (-x ) are the graphs of f (x ) reflected in the y-axis. The zeros are opposites.

77. False; sample answer: The third-degree polynomial x 3 + x 2 - 17x + 15 has three real zeros and no nonreal zeros.

82. Sample answer: c must be greater than or equal to a 2

_ a 1 . If c is

less than a 2

_ a 1 , then the second term of the depressed

polynomial will be negative and the upper bound test will fail.

83. Sample answer: If a polynomial has an imaginary zero, then its complex conjugate is also a zero of the polynomial. Therefore, any polynomial that has one imaginary zero has at least two imaginary zeros.


9.

asymptotes: x = -4, x = 5, y = 1; x-intercepts: -2, 3; y-intercept: 3 _

10 ;

D = {x | x ≠ -4, 5, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 1

x = 5x = -4

f (x) = (x + 2)(x - 3)__(x + 4)(x - 5)

10.

asymptotes: x = 1, x = -2, y = 2;

x-intercepts: -

3 _ 2 ,

6; y-intercept: 9; D = {x | x ≠ 1, -2, x ∈ �}

−8

−16

−4−8

16

8

4 8 x

y = 2

x = 1

x = -2

g(x) = (2x + 3)(x - 6)__(x + 2)(x - 1)

y

11.

asymptotes: x = -2, x = 2, y = 0; y-intercept: -2; D = {x | x ≠ -2, 2, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x = 2x = -2

f (x) = 8__(x - 2)(x + 2)

12.

asymptotes: x = 0, x = 6, y = 0; x-intercept: -2; D = {x | x ≠ 0, 6, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 0x = 0

x = 6

f (x) = x + 2_x(x - 6)

13.

asymptotes: x = -5, x = 6, y = 0; x-intercept: -2;

y-intercept: -

1 _ 15

;

D = {x | x ≠ -5, 6, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x = -5 x = 6

g(x) = (x + 2) (x + 5)__(x + 5)2 (x - 6)

14.

asymptotes: x = 0, x = 5, x = -2, y = 0; x-intercepts: -6, -4; D = {x | x ≠ -2, 0, 5, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

x = 0

x = 5x = -2

y = 0

h(x) = (x + 6)(x + 4)__x(x - 5)(x + 2)

15.

asymptotes: x = -3, x = -1; x-intercepts: -5, 2, 0; y-intercept: 0; D = {x | x ≠ -3, -1, x ∈ �}

y

−80

−160

−4−8

160

80

4 8 x

x = -3

x = -1

h(x) = x2(x - 2)(x + 5)

x 2 + 4x + 3__

155K | Chapter 2 | Answer Appendix

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16.

asymptotes: x = -3, x = 8; x-intercepts: -6, 0, 4; y-intercept: 0; D = {x | x ≠ -3, 8, x ∈ �}

y

−800

−8−16

1600

800

8 16x

−1600

x = 8x = -3

f (x) = x(x + 6) 2 (x - 4)__x2 - 5x - 24

17.

asymptote: y = 0; x-intercept: 8;

y-intercept: -

8 _ 5 ;

D = {x | x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 0

f (x) = x - 8__x2 + 4x + 5

18.

asymptote: y = 0;

y-intercept: -

2 _ 3 ;

D = {x | x ∈ �}

y

x

y = 0

g(x) = -4_x 2 + 6

19c.

Car Wash

Prof

it (th

ousa

nds)

0

1

-1

2

Week

p (z)

z10 20 30 40

p(z) = 3z 2 32z 2 + 7z + 5__

20.

asymptotes: x = 0, y = 0; x-intercept: 4 _

3 ;

D = {x | x ≠ 0, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

x = 0

y = 0

h(x) = 3x - 4_x 3

21.

asymptotes x = 3

√ ��

-

4 _ 3 ,

y = 0; y-intercept: 1 _ 4 ; D =

⎧

⎨

⎩

x x ≠ 3

√ ��

-

4 _ 3 , x ∈ �

⎫

⎬

⎭

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x =3√- 4_

3

h(x) =4x2 - 2x + 1

3x 3 + 4__

22.

asymptotes: y = 1, x = -3, x = -1; x-intercepts: 3, -5; y-intercept: -5; D = {x | x ≠ -3, -1, x ∈ �}

y

−16

−32

−2−4

32

16

2 x

x = -1

y = 1 f (x) =x2 + 2x - 15__x2 + 4x + 3

x = -3

23.

asymptotes: x = 4, y = 1; x-intercept: -7;

y-intercept: -

7 _ 4 ;

D = {x | x ≠ 4, x ∈ �}

y

−8

−16

−8−16

16

8

8 16x

y = 1

x = 4g(x) =x + 7_x - 4

24.

asymptote: x = -3; x-intercept: 0; y-intercept: 0; D = {x | x ≠ -3, x ∈ �}

y

x−4−8 4 8 x

−120

120

60

−60

x = -3

h(x) = x 3_x + 3

25.

asymptote: x = 4; x-intercepts: -2, -1, 0; y-intercept: 0; D = {x | x ≠ 4, x ∈ �}

y

−240

−8−16

240

120

8 16xx = 4

g (x) =x 3 + 3x2 + 2x

x - 4__

−120

26. asymptotes: x = -1, x = 2; y = 0; hole: (-3, -1);

x-intercept: 7; y-intercept: 7 _ 2 ; D = {x | x ≠ -3, -1, 2,

x ∈ �}

−2−4 2 4

y

x

x = 2

−2

−4

4

2x = -1

f (x) =x 2 - 4x 21

x 3 + 2x 2 5x 6

y = 0

__

27.

asymptotes: x = -1, y = 0; holes: (-2, -1),

(2, 1 _ 3 ) ; y-intercept: 1;

D = {x | x ≠ -2, -1, 2, x ∈ �}

y

−2

−4

−2−4

4

2

2 4 x

x = -1g (x) = x2 - 4

x3 +x 2 - 4x - 4__

y = 0

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28.

asymptotes: x = -3;

y = 1; hole : (1, 5 _ 4 )

x-intercept: -4;

y-intercept: 4 _ 3 ;

D = {x | x ≠ -3, 1, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 1

x = -3

f (x) = (x + 4)(x - 1)__(x - 1)(x + 3)

29.

asymptotes: x = -4,

y = 0; hole: (5, 11 _ 81

) ;

x-intercept: -

1 _ 2 ;

y-intercept: 1 _ 16

;

D = {x | x ≠ -4, 5, x ∈ �}

y

−4

−8

−4−8

8

4

4 8 x

y = 0x = -4

g (x) = (2x + 1)(x - 5)__(x - 5)(x + 4) 2

42a.

[0, 100] scl: 10 by [0, 10,000] scl: 1000

42b.

[0, 100] scl: 10 by [0, 10,000] scl: 1000

about 88.9%

42c. No; sample answer: The function is not defined when x = 100. This suggests that it is not financially feasible to remove 100% of the salt at the plant.

45a.

There is a vertical asymptote at r 1 = 30 and a horizontal asymptote at r 2 = 30.

[0, 240] scl: 30 by [0, 100] scl: 10

45c. No; sample answer: As r 1 approaches infinity, r 2 approaches 30. This suggests that the average speeds reached during the first leg of the trip have no bounds. Being able to reach an infinite speed is not reasonable. The same holds true about r 2 as r 1 approaches 30 from the right.

52a. Sample answer: The concentration of the total solution is the sum of the amount of acetic acid in the original 10 liters and the amount in the a liters of the 60% solution, divided by the

total amount of solution or 0.60a + 0.20(10)

__ a + 10

. Multiplying both

the numerator and the denominator by 5 gives 3a + 10 _

5(a + 10)

or 3a + 10 _

5a + 50 .

52b relevant domain: real numbers a such that 0 ≤ a ≤ 90; horizontal asymptote: y = 0.6

52c. Sample answer: Because the tank already has 10 liters of solution in it and it will only hold a total of 100 liters, the amount of solution added must be less than or equal to 90 liters. It is also impossible to add negative amounts of solution, so the amount added must be greater than or equal to 0. As you add more of the 60% solution, the concentration of the total solution will get closer to 60%, but because the solution already in the tank has a lower concentration, the concentration of the total solution can never reach 60%. Therefore, there is a horizontal asymptote at y = 0.6.

52d. Yes; sample answer: The function is not defined at a = -10, but because the value is not in the relevant domain, the asymptote does not pertain to the function. If there were no domain restrictions, there would be a vertical asymptote at a = -10.

53b.

[-5, 10] scl: 1 by [-1, 1] scl: 0.1 [-5, 10] scl: 1 by [-1, 1] scl: 0.1

[-5, 10] scl: 1 by [-1, 1] scl: 0.1

53d. Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least one real zero, the graph of the function will have y = 0 as an asymptote and will intersect the asymptote at the real zeros of the numerator.

54. Sometimes; sample answer: When a = d, the function will have a horizontal asymptote at y = 1. When a ≠ d, the function will not have a horizontal asymptote at y = 1.

59. Sample answer: The test intervals are used to determine the location of points on the graph. Because many rational functions are not continuous, one interval may include y-values that are vastly different than the next interval. Therefore, at least one, and preferably more than one, point is needed for every interval in order to sketch a reasonably accurate graph of the function.

155M | Chapter 2 | Answer Appendix

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62b.

[-10, 10] scl: 1 by [-10, 10] scl: 1

x - 1|x + 2|

f (x) =

[-10, 10] scl: 1 by [-10, 10] scl: 1

|2x - 5|x - 3

g(x) =

[-10, 10] scl: 1 by [-10, 10] scl: 1

|x + 4||3x - 1|

h(x) =

62c. i.

x

-

| + |=

NegativeNegative Positive

und. 0(-)|+|

(-)|+|

(+)|+|

-2 1

ii.

x35

2

| - |-

=

NegativeNegative Positive

0 und.|+|(-)

|+|(-)

|+|(+)

iii.

x13

| + || - |

=

PositivePositive Positive

0 und.|+||+|

|+||+|

|+||+|

-4

62d. i. f (x ) < 0 on (-∞, -2) ∪ (-2, 1)

ii. g (x ) ≥ 0 on ⎧

⎨

⎩

2 _ 5 ⎫

⎬

⎭

∪ (3, ∞)

iii. h (x ) > 0 on (-∞, -4) ∪ (-4, 1 _ 3 ) ∪ ( 1 _

3 , ∞)

70. D = {x | x ≠ -4, x ∈ }; x = -4; y = 2

71. D = {x | x ≠ -6, x ∈ }; x = -6

72. D = ⎧

⎨

⎩

x x ≠ -

1 _ 2 , 5, x ∈

⎫

⎬

⎭

; x = -

1 _ 2 , x = 5, y = 0

78a.

[0, 35] scl: 5 by [0, 35] scl: 5

78b. f (x ) = 0.003 x 3 - 0.111 x 2 +

0.019x + 30.259

78c. about $8.42

81a. f (t ) is a quadratic polynomial function. g (t ) is a power or radical function.

81b. Sample answer: Since time and the amount of water cannot be negative, the relevant domains for f (t ) and g (t ) are restricted to nonnegative values for t that result in nonnegative values for f (t ) and g (t ). For f (t ), D = [0, 5.89] and R = [0, 86.25]. For g (t ), D = [0, ∞) and R = [0, ∞).

81c. lim t→5.89 f (t ) = 0; lim t→0

f (t ) = 80; lim t→0 g (t ) = 0

lim t→∞

g (t ) = ∞

81d.

y

x

60

90

30

2 4 6

f (t)

g(t)

81e. Sample answer: f (x ) is a polynomial function, so it is also continuous and the Intermediate Value Theorem applies. Therefore, since f (3) = 74 and f (5) = 30, it follows that there is a number c, such that 3 < c < 5 and f (c ) = 50.

81f. about 5.89; Sample answer: This means that the town will run out of water reserves after about 5.89 years.

81g. about 5.23 years

Page 149, Study Guide and Review

11. y

x

f (x) = 5x 6

D = (-∞, ∞), R = [ 0, ∞); intercept: 0; lim x →-∞

f (x ) = ∞

and lim x →∞



12. y

x

−4

−8

−2−4

8

4

2 4

f (x) = -8x 3

D = (-∞, ∞), R = (-∞, ∞); intercept: 0; lim x →-∞

f (x ) = ∞

and lim x →∞

f (x ) = -∞;


13.

x

−4

−8

−4−8

8

4

4 8

y

f (x) = x-9

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); no intercepts; lim x →-∞

f (x ) = 0

and lim x →∞



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14.

y

x−2−4

8

12

16

4

2 4

f (x) = x -413

D = (-∞, 0) ∪ (0, ∞), R = (0, ∞); no intercepts; lim x →-∞

f (x ) = 0 and

lim x →∞


discontinuity at x = 0;

increasing: (-∞, 0); decreasing: (0, ∞)

15.

yx

−4

−8

−12

−16

4 8 12 16

√5x 6 11f (x) =

D = [ 1.2, ∞), R = [ -11, ∞); x-intercept: 25.4; lim x →∞

f (x ) = ∞; continuous on

[ 1.2, ∞); increasing: (1.2, ∞)

16. y

x

−2

−4

−2−4

4

2

2 4

f(x) 3_4

3√6x 2 1 + 2

D = (-∞, ∞), R = (-∞, 2.75 ]; x-intercepts: about -1.82 and 1.82, y-intercept: 2.75; lim x →-∞

f (x ) = -∞ and

lim x →∞

f (x ) = -∞; continuous for

all real numbers; increasing: (-∞, 0); decreasing: (0, ∞)

Page 153, Practice Test

15a. The degree is 3 and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive, lim x→∞

f (x ) = ∞ and lim x→-∞

f (x ) = -∞.

15b. 0, 1, -3

15c. Sample answer: (-1, 4), (-2, 6), (2, 10), (3, 36)

15d.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = x (x - 1)(x + 3)

16a. The degree is 4 and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→∞

f (x ) = ∞ and lim x→-∞

f (x ) = ∞

16b. -3, 3, 0 (multiplicity 2)

16c. Sample answer: (-1, -8), (1, -8), (2, -20), (4, 112)

16d.

y

x

−8

−16

−24

−2−4

8

2 4

f (x) = x 4 - 9x 2

19a.

Average High Temperature

Tem

pera

ture

(°F)

60

0

65

70

75

80

85

90

95

MonthJ F M A M J J A S O N D

19b. f (x ) = -0.071 x 3 + 0.415 x 2 + 5.909x + 54.646

19c. 44.9°

25.

y

x

−8

−16

−8−16

16

8

8 16

x = −5

y = 2f (x) = 2x - 6_

x + 5

asymptotes: x = -5, y = 2; x-intercept: 3; y-intercept: -

6 _ 5 ;

D = {x | x ≠ -5, x ∈ �}

26.

y

x

−20

−40

−10−20

40

20

10 20

f (x) = x2 + x - 6__

x - 4

x = 4

y = x + 5

asymptotes: x = 4, y = x + 5; x-intercepts: -3 and 2;

y-intercept: 3 _ 2 ; D = {x | x ≠ 4,

x ∈ �}

155O | Chapter 2 | Answer Appendix

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power, polynomial, and rational functions chapter...

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