chapter 9 resource masters - commack schools 9... · 2018-04-24 · chapter 9 test, form 3 .....63...

Chapter 9 Resource Masters

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CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.

MHID ISBNStudy Guide and Intervention Workbook 0-07-660292-3 978-0-07-660292-6Homework Practice Workbook 0-07-660291-5 978-0-07-660291-9

Spanish VersionHomework Practice Workbook 0-07-660294-X 978-0-07-660294-0

Answers For Workbooks The answers for Chapter 9 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.

Spanish Assessment Masters (MHID: 0-07-660289-3, ISBN: 978-0-07-660289-6) These masters contain a Spanish version of Chapter 9 Test Form 2A and Form 2C.

connected.mcgraw-hill.com

Copyright © by The McGraw-Hill Companies, Inc.

All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 1, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.

Send all inquiries to:McGraw-Hill Education8787 Orion PlaceColumbus, OH 43240

ISBN: 978-0-07-660283-4MHID: 0-07-660283-4

Printed in the United States of America.

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iii

Teacher’s Guide to Using the Chapter 9 Resource Masters ........................................iv

Chapter Resources Chapter 9 Student-Built Glossary ...................... 1Chapter 9 Anticipation Guide (English) ............. 3Chapter 9 Anticipation Guide (Spanish) ............ 4

Lesson 9-1Graphing Quadratic FunctionsStudy Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice ............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10

Lesson 9-2Solving Quadratic Equations by GraphingStudy Guide and Intervention ...........................11Skills Practice .................................................. 13Practice ........................................................... 14Word Problem Practice .................................... 15Enrichment ...................................................... 16

Lesson 9-3Transformations of Quadratic FunctionsStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ........................................................... 20Word Problem Practice .................................... 21Enrichment ...................................................... 22

Lesson 9-4Solving Quadratic Equations by Completing the SquareStudy Guide and Intervention .......................... 23Skills Practice .................................................. 25Practice ........................................................... 26Word Problem Practice .................................... 27Enrichment ...................................................... 28

Lesson 9-5Solving Quadratic Equations by Using the Quadratic FormulaStudy Guide and Intervention .......................... 29Skills Practice .................................................. 31Practice ........................................................... 32Word Problem Practice .................................... 33Enrichment ...................................................... 34

Lesson 9-6Analyzing Functions with Successive Differences and RatiosStudy Guide and Intervention .......................... 35Skills Practice .................................................. 37Practice ........................................................... 38Word Problem Practice .................................... 39Enrichment ...................................................... 40

Lesson 9-7Special FunctionsStudy Guide and Intervention .......................... 41Skills Practice .................................................. 43Practice ........................................................... 44Word Problem Practice .................................... 45Enrichment ...................................................... 46

AssessmentStudent Recording Sheet ............................... 47Rubric for Scoring Extended Response .......... 48Chapter 9 Quizzes 1 and 2 ............................. 49Chapter 9 Quizzes 3 and 4 ............................. 50Chapter 9 Mid-Chapter Test ............................ 51Chapter 9 Vocabulary Test ............................... 52Chapter 9 Test, Form 1 .................................... 53Chapter 9 Test, Form 2A ................................. 55Chapter 9 Test, Form 2B ................................. 57Chapter 9 Test, Form 2C ................................. 59Chapter 9 Test, Form 2D ................................. 61Chapter 9 Test, Form 3 .................................... 63Chapter 9 Extended-Response Test ................ 65Standardized Test Practice .............................. 66Unit 3 Test ........................................................ 69

Answers ........................................... A1–A39

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iv

Teacher’s Guide to Using the Chapter 9 Resource Masters

The Chapter 9 Resource Masters includes the core materials needed for Chapter 9. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.

Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 9-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson ResourcesStudy Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer a historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI-Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

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v

Assessment OptionsThe assessment masters in the Chapter 9 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.

Extended Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers

• The answers for the Anticipation Guide and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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Chapter 9 1 Glencoe Algebra 1

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 9. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

9 Student-Built Glossary

Vocabulary TermFound

on PageDefi nition/Description/Example

absolute value function

axis of symmetry(SIH·muh·tree)

common ratio

completing the square

compound interest

discriminant

double root

geometric sequence

greatest integer function

maximum

(continued on the next page)

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9 Student-Built Glossary (continued)

Vocabulary TermFound

on PageDefi nition/Description/Example

minimum

nonlinear function

parabola(puh·RA·buh·luh)

piecewise-defined function

piecewise-linear function

Quadratic Formula (kwah·DRA·tihk)

quadratic function

step function

transformation

vertex


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Before you begin Chapter 9

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

STEP 1A, D, o NS

StatementSTEP 2A o D

1. The graph of a quadratic function is a parabola. 2. The graph of y = 4x2 – 2x + 7 will be a parabola opening

downward since the coefficient of x2 is positive. 3. A quadratic function’s axis of symmetry is either the x-axis or

the y-axis.

4. The graph of a quadratic function opening upward has no maximum value.

5. The x-intercepts of the graph of a quadratic function are the solutions to the related quadratic equation.

6. All quadratic equations have two real solutions.

7. Any quadratic expression can be written as a perfect square by a method called completing the square.

8. The quadratic formula can only be used to solve quadratic equations that cannot be solved by factoring or graphing.

9. The graph of a step function is a series of disjointed line segments.

10. It is not possible to identify data as linear based on patterns of behavior of their y-values.

After you complete Chapter 9

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

9 Anticipation GuideQuadratic Functions and Equations

Step 1

Step 2


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Antes de comenzar el Capítulo 9

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).

PASO 1A, D, or NS

EnunciadoPASO 2A or D

1. La gráfica de una función cuadrática es una parábola. 2. La gráfica de y = 4x2 – 2x + 7 será una parábola que se abre

hacia abajo, puesto que el coeficiente de x2 es positivo. 3. El eje de simetría de una función cuadrática es el eje x o el eje y.

4. La gráfica de una función cuadrática que se abre hacia arriba no tiene un valor máximo.

5. Las intersecciones x de la gráfica de una función cuadrática son las soluciones de la ecuación cuadrática relacionada.

6. Todas las ecuaciones cuadráticas tienen dos soluciones reales. 7. Cualquier expresión cuadrática puede escribirse como un

cuadrado perfecto mediante el método denominado completar el cuadrado.

8. La fórmula cuadrática sólo puede usarse para resolver ecuaciones cuadráticas que no pueden resolverse mediante factorización o gráficas.

9. La gráfica de una función de paso es una serie de segmentos de líneas inconexas.

10. No es posible identificar los datos como lineal basado en patrones de comportamiento de sus valores de y.

Después de completar el Capítulo 9

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Ejercicios preparatoriosFunciones y Ecuaciones Cuadráticas

Paso 1

Paso 2

9

Capítulo 9 4 Álgebra 1 de Glencoe

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Characteristics of Quadratic Functions

Quadratic

Function

a function described by an equation of the form f(x) = ax2 + bx + c, where a ≠ 0

Example:

y = 2x2 + 3x + 8

The parent graph of the family of quadratic fuctions is y = x2. Graphs of quadratic functions have a general shape called a parabola. A parabola opens upward and has a minimum point when the value of a is positive, and a parabola opens downward and has a maximum point when the value of a is negative.

a. Use a table of values to graph y = x2 - 4x + 1.

x y

-1 6

0 1

1 -2

2 -3

3 -2

4 1

x

y

O

Graph the ordered pairs in the table and connect them with a smooth curve.

b. What are the domain and range of this function?The domain is all real numbers. The range is all real numbers greater than or equal to -3, which is the minimum.

a. Use a table of values to graph y = -x2 - 6x - 7.

x y

-6 -7

-5 -2

-4 1

-3 2

-2 1

-1 -2

0 -7

x

y

O

Graph the ordered pairs in the table and connect them with a smooth curve.

b. What are the domain and range of this function?The domain is all real numbers. The range is all real numbers less than or equal to 2, which is the maximum.

ExercisesUse a table of values to graph each function. Determine the domain and range.

1. y = x2 + 2 2. y = -x2 - 4 3. y = x2 - 3x + 2

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9-1 Study Guide and Intervention Graphing Quadratic Functions

Example 1 Example 2


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Symmetry and Vertices Parabolas have a geometric property called symmetry. That is, if the figure is folded in half, each half will match the other half exactly. The vertical line containing the fold line is called the axis of symmetry. The axis of symmetry contains the minimum or maximum point of the parabola, the vertex.

a. Write the equation of the axis of symmetry.In y = 2x2 + 4x + 1, a = 2 and b = 4. Substitute these values into the equation of the axis of symmetry.

x = -

b −

2a

x = - 4 −

2(2) = -1

The axis of symmetry is x = -1.

b. Find the coordinates of the vertex.Since the equation of the axis of symmetry is x = -1 and the vertex lies on the axis, the x-coordinate of the vertex is -1.y = 2x2 + 4x + 1 Original equation

y = 2(-1)2 + 4(-1) + 1 Substitute.

y = 2(1) - 4 + 1 Simplify.

y = -1The vertex is at (-1, -1).

c. Identify the vertex as a maximum or a minimum.Since the coefficient of the x2-term is positive, the parabola opens upward, and the vertex is a minimum point.

d. Graph the function.

x

y

O(-1, -1)

x = -1

ExercisesConsider each equation. Determine whether the function has maximum or minimum value. State the maximum or minimum value and the domain and range of the function. Find the equation of the axis of symmetry. Graph the function.

9-1 Study Guide and Intervention (continued)

Graphing Quadratic Functions

Example

Axis ofSymmetry

For the parabola y = ax2 + bx + c, where a ≠ 0,

the line x = - b −

2a is the axis of symmetry.

Example: The axis of symmetry of

y = x2 + 2x + 5 is the line x = -1.

Consider the graph of y = 2x2 + 4x + 1.

1. y = x2 + 3 2. y = -x2 - 4x - 4 3. y = x2 + 2x + 3

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Skills PracticeGraphing Quadratic Functions

Use a table of values to graph each function. State the domain and the range.

1. y = x2 - 4 2. y = -x2 + 3 3. y = x2 - 2x - 6

x

y

O

x

y

O

x

y

O

Find the vertex, the equation of the axis of symmetry, and the y-intercept of the graph of each function.

4. y = 2x2 - 8x + 6 5. y = x2 + 4x + 6 6. y = -3x2 - 12x + 3

Consider each equation.

a. Determine whether the function has a maximum or a minimum value.

b. State the maximum or minimum value.

c. What are the domain and range of the function?

7. y = 2x2 8. y = x2 - 2x - 5 9. y = -x2 + 4x - 1

Graph each function.

10. f(x) = -x2 - 2x + 2 11. f(x) = 2x2 + 4x - 2 12. f(x) = -2x2 - 4x + 6

xO

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Practice Graphing Quadratic Functions

Use a table of values to graph each function. Determine the domain and range.

1. y = -x2 + 2 2. y = x2 - 6x + 3 3. y = -2x2 - 8x - 5

x

y

O

x

y

O

Find the vertex, the equation of the axis of symmetry, and the y-intercept of the graph of each function.

4. y = x2 - 9 5. y = -2x2 + 8x - 5 6. y = 4x2 - 4x + 1

Consider each equation. Determine whether the function has a maximum or a minimum value. State the maximum or minimum value. What are the domain and range of the function?

7. y = 5x2 - 2x + 2 8. y = -x2 + 5x - 10 9. y = 3 −

2 x2 + 4x - 9


10. f(x) = -x2 + 1 11. f(x) = -2x2 + 8x - 3 12. f(x) = 2x2 + 8x + 1

xO

f (x)

xO

f (x)

xO

f (x)

13. BASEBALL The equation h = -0.005x2 + x + 3 describes the path of a baseball hit into the outfield, where h is the height and x is the horizontal distance the ball travels.

a. What is the equation of the axis of symmetry?

b. What is the maximum height reached by the baseball?

c. An outfielder catches the ball three feet above the ground. How far has the ball traveled horizontally when the outfielder catches it?

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1. OLYMPICS Olympics were held in 1896 and have been held every four years except 1916, 1940, and 1944. The winning height y in men’s pole vault at any number Olympiad x can be approximated by the equation y = 0.37x2 + 4.3x + 126. Complete the table to estimate the pole vault heights in each of the Olympic Games. Round your answers to the nearest tenth.

Source: National Security Agency

2. PHYSICS Mrs. Capwell’s physics class investigates what happens when a ball is given an initial push, rolls up, and then back down an inclined plane. The class finds that y = -x2 + 6x accurately predicts the ball’s position y after rolling x seconds. On the graph of the equation, what would be the y value when x = 4?

3. ARCHITECTURE A hotel’s main entrance is in the shape of a parabolic arch. The equation y = -x2 + 10x models the arch height y for any distance x from one side of the arch. Use a graph to determine its maximum height.

4. SOFTBALL Olympic softball gold medalist Michele Smith pitches a curveball with a speed of 64 feet per second. If she throws the ball straight upward at this speed, the ball’s height h in feet after t seconds is given by h = -16t2 + 64t. Find the coordinates of the vertex of the graph of the ball’s height and interpret its meaning.

5. GEOMETRY Teddy is building the rectangular deck shown below.

x + 6

x - 2

a. Write an equation representing the area of the deck y.

b. What is the equation of the axis of symmetry?

c. Graph the equation and label its vertex.

yxO

-4-6-8

-10-12-14-16

-2 1-3-4-5

9-1 Word Problem PracticeGraphing Quadratic Functions

YearOlympiad

(x)

Height

(y inches)

1896 1

1900 2

1924 7

1936 10

1964 15

2008 26


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A cubic function is a polynomial written in the form of f(x) = ax3 + bx2 + cx + n, where a ≠ 0. Cubic functions do not have absolute minimum and maximum values like quadratic functions do, but they can have a local minimum and a local maximum point.

Parent Function: f(x) = x3

Domain: {all real numbers}

Range: {all real numbers }

x

f (x)

Use a table of values to graph y = x3 + 3x2 - 1. Then use the graph to estimate the locations of the local minimum and local maximum points.

x –3 –2 –1 0 1

y –1 2 1 –1 2

Graph the ordered pairs, and connect them to create a smooth curve. The end behavior of the “S” shaped curve shows that as x increases,y increases, and as x decreases, y decreases.The local minimum is located at (0, –1). The local maximum is located at (–2, 2).

ExercisesUse a table of values to graph each equation. Then use the graph to estimate the locations of the local minimum and local maximum points.

1. y = 0.5x3 + x2 - 1 2. y = -2x3 - 3x2 - 1 3. y = x3 + 3x2 + x - 4

x

y

x

y

x

y

9-1 EnrichmentGraphing Cubic Functions

Example

x

y

(-2, 2)

(0, -1)


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Solve by Graphing

Quadratic Equation an equation of the form ax2 + bx + c = 0, where a ≠ 0

The solutions of a quadratic equation are called the roots of the equation. The roots of a quadratic equation can be found by graphing the related quadratic function f(x) = ax2 + bx + c and finding the x-intercepts or zeros of the function.

Solve x2 + 4x + 3 = 0 by graphing.

Graph the related function f(x) = x2 + 4x + 3. The equation of the axis of symmetry is

x = - 4 −

2(1) or -2. The vertex is at (-2, -1).

Graph the vertex and several other points on either side of the axis of symmetry.

xO

f(x)

To solve x2 + 4x + 3 = 0, you need to know where f(x) = 0. This occurs at the x-intercepts, -3 and -1.The solutions are -3 and -1.

Solve x2 - 6x + 9 = 0 by graphing.

Graph the related function f(x) = x2 - 6x + 9. The equation of the axis of symmetry is x = 6 −

2(1) or 3. The vertex is at (3, 0). Graph

the vertex and several other points on either side of the axis of symmetry.

x

f(x)

O

To solve x2 - 6x + 9 = 0, you need to know where f(x) = 0. The vertex of the parabola is the x-intercept. Thus, the only solution is 3.

ExercisesSolve each equation by graphing.

1. x2 + 7x + 12 = 0 2. x2 - x - 12 = 0 3. x2 - 4x + 5 = 0

xO

f (x)

xO

4

-4

-8

-12

8-4-8

f (x)

4

xO

f (x)

9-2 Study Guide and InterventionSolving Quadratic Equations by Graphing

Example 1 Example 2


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9-2

Estimate Solutions The roots of a quadratic equation may not be integers. If exact roots cannot be found, they can be estimated by finding the consecutive integers between which the roots lie.

Solve x2 + 6x + 6 = 0 by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.

Graph the related function f(x) = x2 + 6x + 6.

x f(x)

-5 1

-4 -2

-3 -3

-2 -2

-1 1

Notice that the value of the function changes

x

f(x)

O

from negative to positive between the x-values of -5 and -4 and between -2 and -1.

The x-intercepts of the graph are between -5 and -4 and between -2 and -1. So one root is between -5 and -4, and the other root is between -2 and -1.

ExercisesSolve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

1. x2 + 7x + 9 = 0 2. x2 - x - 4 = 0 3. x2 - 4x + 6 = 0

xO

f (x)

xO

f(x)

xO

f (x)

4. x2 - 4x - 1 = 0 5. 4x2 - 12x + 3 = 0 6. x2 - 2x - 4 = 0

xO

f (x)

xO

f (x)

xO

f (x)

Study Guide and Intervention (continued)

Solving Quadratic Equations by Graphing

Example


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Solve each equation by graphing.

1. x2 - 2x + 3 = 0 2. c2 + 6c + 8 = 0

xO

f (x)

cO

f (c)

3. a2 - 2a = -1 4. n2 - 7n = -10

aO

f (a)

nO

f (n)

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

5. p2 + 4p + 2 = 0 6. x2 + x - 3 = 0

pO

f (p)

xO

f (x)

7. d2 + 6d = -3 8. h2 + 1 = 4h

dO

f (d )

hO

f (h)

9-2 Skills PracticeSolving Quadratic Equations by Graphing


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Solve each equation by graphing.

1. x2 - 5x + 6 = 0 2. w2 + 6w + 9 = 0 3. b2 - 3b + 4 = 0

xO

f (x)

wO

f (w)

bO

f (b)

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

4. p2 + 4p = 3 5. 2m2 + 5 = 10m 6. 2v2 + 8v = -7

pO

f (p) mO

f (m)

vO

f (v)

7. NUMBER THEORY Two numbers have a sum of 2and a product of -8. The quadratic equation -n2 + 2n + 8 = 0 can be used to determine the two numbers.

a. Graph the related function f(n) = -n2 + 2n + 8 and determine its x-intercepts.

b. What are the two numbers?

8. DESIGN A footbridge is suspended from a parabolic

support. The function h(x) = -

1

−

25 x2 + 9 represents

the height in feet of the support above the walkway, where x = 0 represents the midpoint of the bridge.

a. Graph the function and determine its x-intercepts.

b. What is the length of the walkway between the two supports?

xO 6 12

12

6

-6

-12

-12 -6

h (x)

9-2 PracticeSolving Quadratic Equations by Graphing

nO

f (n)


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1. FARMING In order for Mr. Moore to decide how much fertilizer to apply to his corn crop this year, he reviews records from previous years. His crop yield y depends on the amount of fertilizer he applies to his fields x according to the equation y = -x2 + 4x + 12. Graph the function, and find the point at which Mr. Moore gets the highest yield possible.

2. LIGHT Ayzha and Jeremy hold a flashlight so that the light falls on a piece of graph paper in the shape of a parabola. Ayzha and Jeremy sketch the shape of the parabola and find that the equation y = x2 - 3x - 10 matches the shape of the light beam. Determine the zeros of the function.

3. FRAMING A rectangular photograph is 7 inches long and 6 inches wide. The photograph is framed using a material that is x inches wide. If the area of the frame and photograph combined is 156 square inches, what is the width of the framing material?

Photograph

Frame

6 in.

7 in.

x

x

4. WRAPPING PAPER Can a rectangular piece of wrapping paper with an area of 81 square inches have a perimeter of 60 inches? (Hint: Let length = 30 – w.) Explain.

5. ENGINEERING The shape of a satellite dish is often parabolic because of the reflective qualities of parabolas. Suppose a particular satellite dish is modeled by the following equation. 0.5x2 = 2 + y

a. Approximate the zeros of this function by graphing.

y

xO

34

21

-2-3-4

4321-2-3-4

b. On the coordinate plane above, translate the parabola so that there is only one zero. Label this curve A.

c. Translate the parabola so that there are no zeros. Label this curve B.

9-2 Word Problem Practice Solving Quadratic Equations by Graphing

y

xO

1416

12108642

4 5321


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Parabolas Through Three Given PointsIf you know two points on a straight line, you can find the equation of the line. To find the equation of a parabola, you need three points on the curve.

Here is how to approximate an equation of the parabola through the points (0, -2), (3, 0), and (5, 2).

Use the general equation y = ax2 + bx + c. By substituting the given values for x and y, you get three equations.(0, -2): -2 = c(3, 0): 0 = 9a + 3b + c(5, 2): 2 = 25a + 5b + c

First, substitute -2 for c in the second and third equations. Then solve those two equations as you would any system of two equations. Multiply the second equation by 5 and the third equation by -3.0 = 9a + 3b - 2 Multiply by 5. 0 = 45a + 15b - 10

2 = 25a + 5b - 2 Multiply by -3. -6 = -75a - 15b + 6

-6 = -30a - 4 a = 1 −

15

To find b, substitute 1 −

15 for a in either the second or third equation.

0 = 9 (

1 −

15 )

+ 3b - 2

b = 7 −

15

The equation of a parabola through the three points is

y = 1 −

15 x2

+

7 −

15 x - 2.

Find the equation of a parabola through each set of three points.

1. (1, 5), (0, 6), (2, 3) 2. (-5, 0), (0, 0), (8, 100)

3. (4, -4), (0, 1), (3, -2) 4. (1, 3), (6, 0), (0, 0)

5. (2, 2), (5, -3), (0, -1) 6. (0, 4), (4, 0), (-4, 4)

9-2 Enrichment


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9-3

Translations A translation is a change in the position of a figure either up, down, left, right, or diagonal. Adding or subtracting constants in the equations of functions translates the graphs of the functions.

The graph of g(x) = x2 + k translates the graph of f(x) = x2 vertically.

If k > 0, the graph of f(x) = x2 is translated k units up.

If k < 0, the graph of f(x) = x2 is translated |k| units down.

The graph of g(x) = (x - h) 2 is the graph of f(x) = x 2 translated horizontally.

If h > 0, the graph of f(x) = x 2 is translated h units to the right.

If h < 0, the graph of f(x) = x 2 is translated |h| units to the left.

Study Guide and InterventionTransformations of Quadratic Functions

Describe how the graph of each function is related to the graph of f(x) = x2.

Example

a. g(x) = x2 + 4The value of k is 4, and 4 > 0. Therefore, the graph of g(x) = x2 + 4 is a translation of the graph of f(x) = x2 up 4 units.

g(x)

f(x)

x

b. g(x) = (x + 3) 2 The value of h is –3, and –3 < 0. Thus, the graph of g(x) = (x + 3) 2 is a translation of the graph of f(x) = x2 to the left 3 units.

y

xO

f(x)g(x)

ExercisesDescribe how the graph of each function is related to the graph of f(x) = x2.

1. g(x) = x2 + 1 2. g(x) = (x – 6) 2 3. g(x) = (x + 1) 2

4. g(x) = 20 + x2 5. g(x) = (–2 + x) 2 6. g(x) = - 1 −

2 + x2

7. g(x) = x2 + 8 −

9 8. g(x) = x2 – 0.3 9. g(x) = (x + 4) 2


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Dilations and Reflections A dilation is a transformation that makes the graph narrower or wider than the parent graph. A reflection flips a figure over the x- or y-axis.

The graph of f(x) = ax2 stretches or compresses the graph of f(x) = x2.

If |a| > 1, the graph of f(x) = x2 is stretched vertically.

If 0 < |a| < 1, the graph of f(x) = x2 is compressed vertically.

The graph of the function –f(x) fl ips the graph of f(x) = x2 across the x-axis.

The graph of the function f(–x) fl ips the graph of f(x) = x2 across the y-axis.



Transformations of Quadratic Functions

x

0 < a < 1

a = 1a > 1

x

y

O

f(x) = -x2

f(x) = x2

Example

a. g(x) = 2x2

The function can be written as f(x) = ax2 where a = 2. Because |a| > 1, the graph of y = 2x2 is the graph of y = x2 that is stretched vertically.

b. g(x) = - 1 −

2 x2 - 3

The negative sign causes a reflection across the x-axis. Then a dilation occurs in which a = 1 −

2 and

a translation in which k = –3. So the graph of g(x) = - 1 −

2 x2 - 3

is reflected across the x-axis, dilated wider than the graph of f(x) = x2, and translated down 3 units.

ExercisesDescribe how the graph of each function is related to the graph of f(x) = x2.

1. g(x) = -5x2 2. g(x) = - (x + 1) 2 3. g(x) = - 1 −

4 x2 - 1

x

y

O

f(x) = x2

f(x) = 2x2 x

f(x)

g(x)

y

O


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1. g(x) = x2 + 2 2. g(x) = (x - 1) 2 3. g(x) = x2 - 8

4. g(x) = 7x2 5. g(x) = 1 −

5 x2 6. g(x) = -6x2

7. g(x) = -x2 + 3 8. g(x) = 5 - 1 −

2 x2 9. g(x) = 4 (x - 1) 2

Match each equation to its graph.

10. y = 2x2 - 2

11. y = 1 −

2 x2 - 2

12. y = - 1 −

2 x2 + 2

13. y = -2x2 + 2

9-3 Skills PracticeTransformations of Quadratic Functions

x

y

x

y

x

y

x

yB.

A.

D.

C.


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1. g(x) = (10 + x)2 2. g(x) = - 2 −

5 + x2 3. g(x) = 9 - x2

4. g(x) = 2x2 + 2 5. g(x) = - 3 −

4 x2 - 1 −

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

Match each equation to its graph.

A.

x

y B.

x

y C.

x

y

7. y = -3x2 - 1 8. y = 1 −

3 x2 - 1 9. y = 3x2 + 1

9-3 PracticeTransformations of Quadratic Functions

List the functions in order from the most vertically stretched to the least vertically stretched graph.

10. f(x) = 3x2, g(x) = 1 −

2 x2, h(x) = -2x2 11. f(x) = 1 −

2 x2, g(x) = -

1 −

6 x2, h(x) = 4x2

12. PARACHUTING Two parachutists jump at the same time from two different planes as part of an aerial show. The height h1 of the first parachutist in feet after t seconds is modeled by the function h1 = -16t2 + 5000. The height h2 of the second parachutist in feet after t seconds is modeled by the function h2 = -16t2 + 4000.

a. What is the parent function of the two functions given?

b. Describe the transformations needed to obtain the graph of h1 from the parent function.

c. Which parachutist will reach the ground first?


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1. SPRINGS The potential energy stored in

a spring is given by U s = 1 −

2 kx2 where k is

a constant known as the spring constant,

and x is the distance the spring is stretched or compressed from its initial position. How is the graph of the function for a spring where k = 2 newtons/meter related to the graph of the function for a spring where k = 10 newtons/meter?

2. BLUEPRINTS The bottom left corner of a rectangular park is located at (–3, 5) on a construction blueprint. The park is 8 units long and has an area of 48 units2

on the blueprint. Suppose the park is translated on the blueprint so that the top right corner is now located at the origin. Describe the translation of the graph.

3. PHYSICS A ball is dropped from a height of 20 feet. The function h = -16t2

+ 20 models the height of the ball in feet after t seconds. Graph the function and compare this graph to the graph of its parent function.

Heig

ht (f

t)

3

0

6

9

12

15

18

21

24

Time (s)0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

9-3 Word Problem PracticeTransformations of Quadratic Functions

4. ACCELERATION The distance d in feet a car accelerating at 6 ft/s2 travels after t seconds is modeled by the function d = 3t2. Suppose that at the same time the first car begins accelerating, a second car begins accelerating at 4 ft/s2 exactly 100 feet down the road from the first car. The distance traveled by second car is modeled by the function d = 2t2

+ 100.

a. Graph and label each function on the same coordinate plane.

Dist

ance

(ft)

100

0

200

300

400

500

600

700

800

Time (s)2 4 6 8 10 12 14 16

b. Explain how each graph is related to the graph of d = t2.

c. After how many seconds will the first car pass the second car?


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A polynomial function is a continuous function that can be described by a polynomial equation in one variable.

Polynomial Function

If n is a nonnegative integer, a0, a1, a2, … , an - 1, an are real numbers, and an ≠ 0, then

f(x) = anxn + an - 1x

n - 1 + … + a2x2 + a1x + a0

is a polynomial function of degree n.

Notice that a quadratic function is a polynomial function of degree 2.

Create a table of values and a graph for y = x3 + 2x2 - x. Then describe its end behavior.

Create a table of values, and graph the ordered pairs. Connect the points with a smooth curve. Find and plot additional points to better approximate the curve’s shape.

From the table and the graph we see that as x decreases, y decreases and as x increases, y increases.

Exercises

Create a table of values and a graph for each function. Then describe its end behavior.

1. f(x) = x3 + 3x2 - 1 2. f(x) = -2x5 + 4x3 3. f(x) = 2x4 - 4x3 + 3x

9-3 EnrichmentGraphing Polynomial Functions

Example

x x3 + 2x2 – x y

-4 (-4)3 + 2(-4)2 - (-4) -28

-3 (-3)3 + 2(-3)2 - (-3) -6

-2 (-2)3 + 2(-2)2 - (-2) 2

-1 (-1)3 + 2(-1)2 - (-1) 2

0 (0)3 + 2(0)2 - 0 0

1 13 + 2(1)2 - 1 2

2 23 + 2(2)2 - 2 14

3 33 + 2(3)2 - 3 42

As x decreases, y decreases

As x increases, y increases.

x f(x)

-4

-3

-2

-1

0

1

x f(x)

-2

-1

-0.5

0

0.5

1

2

x f(x)

-2

-1

-0.5

0

0.5

1

2

y

xO

As you moveleft, the graphgoes down.

As you moveright, the graphgoes up.

(−1, 2)(1, 2)

(−2, 2)

(0, 0)

xO

8

4

−2−4 2 4

−4

−8

f(x)

xO

8

4

−2−4 2 4

−4

−8

f (x)

xO

8

4

−2−4 2 4

−4

−8

f(x)


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Complete the Square Perfect square trinomials can be solved quickly by taking the square root of both sides of the equation. A quadratic equation that is not in perfect square form can be made into a perfect square by a method called completing the square.

Completing the Square

To complete the square for any quadratic equation of the form x2 + bx:

Step 1 Find one-half of b, the coefficient of x.

Step 2 Square the result in Step 1.

Step 3 Add the result of Step 2 to x2 + bx.

x2 + bx +

(

b −

2 )

2

= (

x +

b −

2 )

2

Find the value of c that makes x2 + 2x + c a perfect square trinomial.

Step 1 Find 1 −

2 of 2. 2 −

2 = 1

Step 2 Square the result of Step 1. 12 = 1

Step 3 Add the result of Step 2 to x2 + 2x. x2 + 2x + 1

Thus, c = 1. Notice that x2 + 2x + 1 equals (x + 1)2.

ExercisesFind the value of c that makes each trinomial a perfect square.

1. x2 + 10x + c 2. x2 + 14x + c

3. x2 - 4x + c 4. x2

- 8x + c

5. x2 + 5x + c 6. x2 + 9x + c

7. x2 - 3x + c 8. x2

- 15x + c

9. x2 + 28x + c 10. x2 + 22x + c

Study Guide and InterventionSolving Quadratic Equations by Completing the Square

Example

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Solving Quadratic Equations by Completing the Square

Solve by Completing the Square Since few quadratic expressions are perfect square trinomials, the method of completing the square can be used to solve some quadratic equations. Use the following steps to complete the square for a quadratic expression of the form ax2 + bx.

Solve x2 + 6x + 3 = 10 by completing the square.

x2 + 6x + 3 = 10 Original equation

x2 + 6x + 3 - 3 = 10 - 3 Subtract 3 from each side.

x2 + 6x = 7 Simplify.

x2 + 6x + 9 = 7 + 9 Since (

6 −

2 )

2

= 9, add 9 to each side.

(x + 3)2 = 16 Factor x2 + 6x + 9.

x + 3 = ±4 Take the square root of each side.

x = -3 ± 4 Simplify.

x = -3 + 4 or x = -3 - 4 = 1 = -7The solution set is {-7, 1}.

ExercisesSolve each equation by completing the square. Round to the nearest tenth if necessary.

1. x2 - 4x + 3 = 0 2. x2 + 10x = -9 3. x2 - 8x - 9 = 0

4. x2 - 6x = 16 5. x2 - 4x - 5 = 0 6. x2 - 12x = 9

7. x2 + 8x = 20 8. x2 = 2x + 1 9. x2 + 20x + 11 = -8

10. x2 - 1 = 5x 11. x2 = 22x + 23 12. x2 - 8x = -7

13. x2 + 10x = 24 14. x2 - 18x = 19 15. x2 + 16x = -16

16. 4x2 = 24 + 4x 17. 2x2 + 4x + 2 = 8 18. 4x2 = 40x + 44

Step 1 Find b −

2 .

Step 2 Find (

b −

2 )

2

.

Step 3 Add (

b −

2 )

2

to ax2 + bx.

9-4

Example


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Skills PracticeSolving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 + 6x + c 2. x2 + 4x + c

3. x2 - 14x + c 4. x2 - 2x + c

5. x2 - 18x + c 6. x2 + 20x + c

7. x2 + 5x + c 8. x2 - 70x + c

9. x2 - 11x + c 10. x2 + 9x + c

Solve each equation by completing the square. Round to the nearest tenth if necessary.

11. x2 + 4x - 12 = 0 12. x2 - 8x + 15 = 0

13. x2 + 6x = 7 14. x2 - 2x = 15

15. x2 - 14x + 30 = 6 16. x2 + 12x + 21 = 10

17. x2 - 4x + 1 = 0 18. x2 - 6x + 4 = 0

19. x2 - 8x + 10 = 0 20. x2 - 2x = 5

21. 2x2 + 20x = -2 22. 0.5x2 + 8x = -7

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PracticeSolving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 - 24x + c 2. x2 + 28x + c 3. x2 + 40x + c

4. x2 + 3x + c 5. x2 - 9x + c 6. x2 - x + c


7. x2 - 14x + 24 = 0 8. x2 + 12x = 13 9. x2 - 30x + 56 = -25

10. x2 + 8x + 9 = 0 11. x2 - 10x + 6 = -7 12. x2 + 18x + 50 = 9

13. 3x2 + 15x - 3 = 0 14. 4x2 - 72 = 24x 15. 0.9x2 + 5.4x - 4 = 0

16. 0.4x2 + 0.8x = 0.2 17. 1 −

2 x2 - x - 10 = 0 18. 1 −

4 x2 + x - 2 = 0

19. NUMBER THEORY The product of two consecutive even integers is 728. Find the integers.

20. BUSINESS Jaime owns a business making decorative boxes to store jewelry, mementos, and other valuables. The function y = x2 + 50x + 1800 models the profit y that Jaime has made in month x for the first two years of his business.

a. Write an equation representing the month in which Jaime’s profit is $2400.

b. Use completing the square to find out in which month Jaime’s profit is $2400.

21. PHYSICS From a height of 256 feet above a lake on a cliff, Mikaela throws a rock out over the lake. The height H of the rock t seconds after Mikaela throws it is represented by the equation H = -16t2 + 32t + 256. To the nearest tenth of a second, how long does it take the rock to reach the lake below? (Hint: Replace H with 0.)

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9-4 Word Problem Practice Solving Quadratic Equations by Completing the Square

1. INTERIOR DESIGN Modular carpeting is installed in small pieces rather than as a large roll so that only a few pieces need to be replaced if a small area is damaged. Suppose the room shown in the diagram below is being fitted with modular carpeting. Complete the square to determine the number of 1 foot by 1 foot squares of carpeting needed to finish the room. Fill in the missing terms in the corresponding equation below.

x

xxxxx

x x x xx 2

x2+ 10x + _____ = (x + _____)2

2. FALLING OBJECTS Keisha throws a rock down an old well. The distance d in feet the rock falls after t seconds can be represented by d = 16t2 + 64t. If the water in the well is 80 feet below ground, how many seconds will it take for the rock to hit the water?

3. MARS On Mars, the gravity acting on an object is less than that on Earth. On Earth, a golf ball hit with an initial upward velocity of 26 meters per second will hit the ground in about 5.4 seconds. The height h of an object on Mars that leaves the ground with an initial velocity of 26 meters per second is given by the equation h = -1.9t2

+ 26t. How much longer will it take for the golf ball hit on Mars to reach the ground? Round your answer to the nearest tenth.

4. FROGS A frog sitting on a stump 3 feet high hops off and lands on the ground. During its leap, its height h in feet is given by h = -0.5d2

+ 2d + 3, where d is the distance from the base of the stump. How far is the frog from the base of the stump when it landed on the ground?

5. GARDENING Peg is planning a rectangular vegetable garden using 200 feet of fencing material. She only needs to fence three sides of the garden since one side borders an existing fence.

x

a. Let x = the width of the rectangle. Write an equation to represent the area A of the garden if Peg uses all the fencing material.

b. For what widths would the area of Peg’s garden equal 4800 square feet if she uses all the fencing material?


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Completing the square is a useful tool for factoring quadratic expressions. You can utilize a similar technique to factor simple quartic polynomials of the form x4 + c.

Factor the quartic polynomial x4 + 64.

Step 1 Find the value of the middle term needed to complete the square.

This value is (2 √

��

64 ) (x2) , or 16x2.

Step 2 Rewrite the original polynomial in factorable form.

(

x4 + 16x2 +

(

16 −

2 )

2 )

- 16x2

Step 3 Factor the polynomials. (x2 + 8) 2 - (4x)

2

Step 4 Rewrite using the difference of two squares. (x2 + 8 + 4x)(x2 + 8 - 4x)

The factored form of x2 + 64 is (x2 + 4x + 8) (x2 - 4x + 8) . This could then be factored further, if needed, to find the solutions to a quartic equation.

ExercisesFactor each quartic polynomial.

1. x4 + 4 2. x4 + 324

3. x4 + 2500 4. x4 + 9604

5. x4 + 1024 6. x4 + 5184

7. x4 + 484 8. x4 + 9

9. x4 + 144 10. x8 + 16,384

11. Factor x4 + c to come up with a general rule for factoring quartic polynomials.

9-4 EnrichmentFactoring Quartic Polynomials

Example


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Study Guide and InterventionSolving Quadratic Equations by Using the Quadratic Formula

Quadratic Formula To solve the standard form of the quadratic equation, ax 2 + bx + c = 0, use the Quadratic Formula.

Quadratic Formula The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by x = -b ± √

��

b2 - 4ac −

2a .

Solve x2 + 2x = 3 by using the Quadratic Formula.

Rewrite the equation in standard form. x2 + 2x = 3 Original equation

x2 + 2x - 3 = 3 - 3 Subtract 3 from each side.

x2 + 2x - 3 = 0 Simplify.

Now let a = 1, b = 2, and c = -3 in the Quadratic Formula.

x =

-b ±

√

��

b2 - 4ac −

2a

= -2 ± √

��

(2)2 - 4(1)(-3) −−

2(1)

= -2 ±

√

��

16 −

2

x = -2 + 4 −

2 or x = -2 - 4 −

2

= 1 = -3The solution set is {-3, 1}.

Solve x2 - 6x - 2 = 0 by using the Quadratic Formula. Round to the nearest tenth if necessary.

For this equation a = 1, b = -6, and c = -2.

x = -b ± √

��

b 2 - 4ac −

2a

= 6 ± √

��

(-6) 2 - 4(1)(-2) −−

2(1)

= 6 + √

��

44 −

2

x = 6 + √

��

44 −

2 or x = 6 - √

��

44 −

2

x ≈ 6.3 ≈ -0.3The solution set is {-0.3, 6.3}.

Exercises

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 - 3x + 2 = 0 2. x2 - 8x = -16

3. 16x2 - 8x = -1 4. x2 + 5x = 6

5. 3x2 + 2x = 8 6. 8x2 - 8x - 5 = 0

7. -4x2 + 19x = 21 8. 2x2 + 6x = 5

9. 48x2 + 22x - 15 = 0 10. 8x2 - 4x = 24

11. 2x2 + 5x = 8 12. 8x2 + 9x - 4 = 0

13. 2x2 + 9x + 4 = 0 14. 8x2 + 17x + 2 = 0

9-5

Example 1 Example 2


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Solving Quadratic Equations by Using the Quadratic Formula

The Discriminant In the Quadratic Formula, x = -b ± √

��

b 2 - 4ac −

2a , the expression

under the radical sign, b2 - 4ac, is called the discriminant. The discriminant can be used to determine the number of real solutions for a quadratic equation.

Case 1: b2 - 4ac < 0 no real solutions

Case 2: b2 - 4ac = 0 one real solution

Case 3: b2 - 4ac > 0 two real solutions

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

a. 12x2 + 5x = 4Write the equation in standard form.

12x2 + 5x = 4 Original equation

12x2 + 5x - 4 = 4 - 4 Subtract 4 from each side.

12x2 + 5x - 4 = 0 Simplify.

Now find the discriminant.b2 - 4ac = (5)2 - 4(12)(-4) = 217

Since the discriminant is positive, the equation has two real solutions.

b. 2x2 + 3x = -4

2x2 + 3x = -4 Original equation

2x2 + 3x + 4 = -4 + 4 Add 4 to each side.

2x2 + 3x + 4 = 0 Simplify.

Find the discriminant.

b2 - 4ac = (3)2 - 4(2)(4) = -23

Since the discriminant is negative, the equation has no real solutions.

ExercisesState the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

1. 3x2 + 2x - 3 = 0 2. 3x2 - 7x - 8 = 0 3. 2x2 - 10x - 9 = 0

4. 4x2 = x + 4 5. 3x2 - 13x = 10 6. 6x2 - 10x + 10 = 0

7. 2x2 - 20 = -x 8. 6x2 = -11x - 40 9. 9 - 18x + 9x2 = 0

10. 12x2 + 9 = -6x 11. 9x2 = 81 12. 16x2 + 16x + 4 = 0

13. 8x2 + 9x = 2 14. 4x2 - 4x + 4 = 3 15. 3x2 - 18x = - 14

9-5

Example


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Skills PracticeSolving Quadratic Equations by Using the Quadratic Formula


1. x2 - 49 = 0 2. x2 - x - 20 = 0

3. x2 - 5x - 36 = 0 4. x2 + 11x + 30 = 0

5. x2 - 7x = -3 6. x2 + 4x = -1

7. x2 - 9x + 22 = 0 8. x2 + 6x + 3 = 0

9. 2x2 + 5x - 7 = 0 10. 2x2 - 3x = -1

11. 2x2 + 5x + 4 = 0 12. 2x2 + 7x = 9

13. 3x2 + 2x - 3 = 0 14. 3x2 - 7x - 6 = 0


15. x2 + 4x + 3 = 0 16. x2 + 2x + 1 = 0

17. x2 - 4x + 10 = 0 18. x2 - 6x + 7 = 0

19. x2 - 2x - 7 = 0 20. x2 - 10x + 25 = 0

21. 2x2 + 5x - 8 = 0 22. 2x2 + 6x + 12 = 0

23. 2x2 - 4x + 10 = 0 24. 3x2 + 7x + 3 = 0

9-5


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9-5 Practice Solving Quadratic Equations by Using the Quadratic Formula


1. x2 + 2x - 3 = 0 2. x2 + 8x + 7 = 0 3. x2 - 4x + 6 = 0

4. x2 - 6x + 7 = 0 5. 2x2 + 9x - 5 = 0 6. 2x2 + 12x + 10 = 0

7. 2x2 - 9x = -12 8. 2x2 - 5x = 12 9. 3x2 + x = 4

10. 3x2 - 1 = -8x 11. 4x2 + 7x = 15 12. 1.6x2 + 2x + 2.5 = 0

13. 4.5x2 + 4x - 1.5 = 0 14. 1 −

2 x2 + 2x + 3 −

2 = 0 15. 3x2 - 3 −

4 x = 1 −

2


16. x2 + 8x + 16 = 0 17. x2 + 3x + 12 = 0 18. 2x2 + 12x = -7

19. 2x2 + 15x = -30 20. 4x2 + 9 = 12x 21. 3x2 - 2x = 3.5

22. 2.5x2 + 3x - 0.5 = 0 23. 3 −

4 x2 - 3x = -4 24. 1 −

4 x2 = -x - 1

25. CONSTRUCTION A roofer tosses a piece of roofing tile from a roof onto the ground 30 feet below. He tosses the tile with an initial downward velocity of 10 feet per second.

a. Write an equation to find how long it takes the tile to hit the ground. Use the model for vertical motion, H = -16t2 + vt + h, where H is the height of an object after t seconds, v is the initial velocity, and h is the initial height. (Hint: Since the object is thrown down, the initial velocity is negative.)

b. How long does it take the tile to hit the ground?

26. PHYSICS Lupe tosses a ball up to Quyen, waiting at a third-story window, with an initial velocity of 30 feet per second. She releases the ball from a height of 6 feet. The equation h = -16t2 + 30t + 6 represents the height h of the ball after t seconds. If the ball must reach a height of 25 feet for Quyen to catch it, does the ball reach Quyen? Explain. (Hint: Substitute 25 for h and use the discriminant.)


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Word Problem PracticeSolving Quadratic Equations by Using the Quadratic Formula

1. BUSINESS Tanya runs a catering business. Based on her records, her weekly profit can be approximated by the function f(x) = x2 + 2x - 37, where x is the number of meals she caters. If f(x) is negative, it means that the business has lost money. What is the least number of meals that Tanya needs to cater in order to have a profit?

2. AERONAUTICS At liftoff, the space shuttle Discovery has a constant acceleration of 16.4 feet per second squared and an initial velocity of 1341 feet per second due to the rotation of Earth. The distance Discovery has traveled t seconds after liftoff is given by the equation d(t) = 1341t + 8.2t2. How long after liftoff has Discovery traveled 40,000 feet? Round your answer to the nearest tenth.

3. ARCHITECTURE The Golden Ratio appears in the design of the Greek Parthenon because the width and height of the façade are related by the

equation W + H −

W = W −

H . If the height of a

model of the Parthenon is 16 inches, what is its width? Round your answer to the nearest tenth.

4. CRAFTS Madelyn cut a 60-inch pipe cleaner into two unequal pieces, and then she used each piece to make a square. The sum of the areas of the squares was 117 square inches. Let x be the length of one piece. Write and solve an equation to represent the situation and find the lengths of the two original pieces.

5. SITE DESIGN The town of Smallport plans to build a new water treatment plant on a rectangular piece of land 75 yards wide and 200 yards long. The buildings and facilities need to cover an area of 10,000 square yards. The town’s zoning board wants the site designer to allow as much room as possible between each edge of the site and the buildings and facilities. Let x represent the width of the border.

Buildings and Facilities

Border

75 yd

200 yd

x

x

xx

a. Use an equation similar to A = � × w to represent the situation.

b. Write the equation in standard quadratic form.

c. What should be the width of the border? Round your answer to the nearest tenth.

9-5


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Enrichment

Golden RectanglesA golden rectangle has the property that its sides satisfy the following proportion.

a + b − a = a −

b

Two quadratic equations can be written from the proportion. These are sometimes called golden quadratic equations.

1. In the proportion, let a = 1. Use 2. Solve the equation in Exercise 1 for b. cross-multiplication to write aquadratic equation.

3. In the proportion, let b = 1. Write 4. Solve the equation in Exercise 3 for a. a quadratic equation in a.

5. Explain why 1 −

2 ( √

�

5 + 1) and 1 −

2 ( √

�

5 - 1) are called golden ratios.

Another property of golden rectangles is that a square drawn inside a golden rectangle creates another, smaller golden rectangle.

In the design at the right, opposite vertices of each square have been connected with quarters of circles.

For example, the arc from point B to point C is created by putting the point of a compass at point A. The radius of the arc is the length BA.

6. On a separate sheet of paper, draw a larger version of the design. Start with a golden rectangle with a long side of 10 inches.

b

a

9-5

C

B A


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Study Guide and InterventionAnalyzing Functions with Successive Differences and Ratios

9-6

Identify Functions Linear functions, quadraticfunctions, and exponential functions can all be used to model data. The general forms of the equations are listed at the right.

You can also identify data as linear, quadratic, or exponential based on patterns of behavior of their y-values.

Graph the set of ordered pairs {(–3, 2), (–2, –1), (–1, –2), (0, –1), (1, 2)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

The ordered pairs appear to represent a quadratic function.

Look for a pattern in the table to determine which model best describes the data.

Start by comparing the first differences.

The first differences are not all equal. The table does not represent a linear function.Find the second differences and compare.

The table does not represent a quadratic function. Find the ratios of the y-values.

The ratios are equal. Therefore, the table can be modeled by an exponential function.

Example 1

x

y

Example 2

x –2 –1 0 1 2

y 4 2 1 0.5 0.25

-2 -1 -0.5 -0.25

+ 1 + 0.5 + 0.25

4 2 1 0.5 0.25

× 0.5 × 0.5 × 0.5 × 0.5

4 2 1 0. 5 0.25

-2 -1 -0.5 -0.25

Exercises

Graph each set of ordered pairs. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

1. (0, –1), (1, 1), (2, 3), (3, 5) 2. (–3, –1), (–2, –4), (–1, –5), (0, –4), (1, –1)

x

y

x

y

Look for a pattern in each table to determine which model best describes the data.

3. x –2 –1 0 1 2

y 6 5 4 3 2

4. x –2 –1 0 1 2

y 6.25 2.5 1 0.4 0.16

Linear Function y = mx + b

Quadratic Function y = ax2 + bx + c

Exponential Function y = abx


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Analyzing Functions with Successive Differences and Ratios

9-6

Write Equations Once you find the model that best describes the data, you can write an equation for the function.

Basic Forms

Linear Function y = mx + b

Quadratic Function y = ax2

Exponential Function y = abx

Determine which model best describes the data. Then write an equation for the function that models the data.

x 0 1 2 3 4

y 3 6 12 24 48

Step 1 Determine whether the data is modeled by a linear, quadratic, or exponential function.

First differences: 3 6 12 24 48

+ 3 + 6 + 12 + 24

Second differences: 3 6 12 24

+ 3 + 6 + 12

y-value ratios: 3 6 12 24 48

× 2 × 2 × 2 × 2

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential function.

Step 2 Write an equation for the function that models the data. The equation has the form y = abx. The y-value ratio is 2, so this is the value of the base.

y = abx Equation for exponential function

3 = a(2)0 x = 0, y = 3, and b = 2

3 = a Simplify.

An equation that models the data is y = 3 ․ 2 x . To check the results, you can verify that the other ordered pairs satisfy the function.

ExercisesLook for a pattern in each table of values to determine which model best describes the data. Then write an equation for the function that models the data.

1. x –2 –1 0 1 2

y 12 3 0 3 12

2. x –1 0 1 2 3

y –2 1 4 7 10

3. x –1 0 1 2 3

y 0.75 3 12 48 192

Example


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Skills PracticeAnalyzing Functions with Successive Differences and Ratios

9-6


1. (2, 3), (1, 1), (0, –1), (–1, –3), (–3, –5) 2. (–1, 0.5), (0, 1), (1, 2), (2, 4)

O x

y

x

y

3. (–2, 4), (–1, 1), (0, 0), (1, 1), (2, 4) 4. (–3, 5), (–2, 2), (–1, 1), (0, 2), (1, 5)

x

y

x

y

Look for a pattern in each table of values to determine which model best describes the data. Then write an equation for the function that models the data.

5. x –3 –2 –1 0 1 2

y 32 16 8 4 2 1

6. x –1 0 1 2 3

y 7 3 –1 –5 –9

7. x –3 –2 –1 0 1

y –27 –12 –3 0 –3

8. x 0 1 2 3 4

y 0.5 1.5 4.5 13.5 40.5

9. x –2 –1 0 1 2

y –8 –4 0 4 8


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9-6 Practice Analyzing Functions with Successive Differences and Ratios


1. (4, 0.5), (3, 1.5), (2, 2.5), (1, 3.5), (0, 4.5) 2. (

–1, 1 −

9 )

, (

0, 1 −

3 )

, (1, 1), (2, 3)

x

y

O

x

y

3. (–4, 4), (–2, 1), (0, 0), (2, 1), (4, 4) 4. (–4, 2), (–2, 1), (0, 0), (2, –1), (4, –2)

x

y

x

y

Look for a pattern in each table of values to determine which model best describes the data. Then write an equation for the function that models the data.

5. x –3 –1 1 3 5

y –5 –2 1 4 7

6. x –2 –1 0 1 2

y 0.02 0.2 2 20 200

7. x –1 0 1 2 3

y 6 0 6 24 54

8. x –2 –1 0 1 2

y 18 9 0 –9 –18

9. INSECTS The local zoo keeps track of the number of dragonflies breeding in their insect exhibit each day.

Day 1 2 3 4 5

Dragonfl ies 9 18 36 72 144

a. Determine which function best models the data.

b. Write an equation for the function that models the data.

c. Use your equation to determine the number of dragonflies that will be breeding after 9 days.


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Word Problem PracticeAnalyzing Functions with Successive Differences and Ratios

9-6

1. WEATHER The San Mateo weather station records the amount of rainfall since the beginning of a thunderstorm. Data for a storm is recorded as a series of ordered pairs shown below, where the x value is the time in minutes since the start of the storm, and the y value is the amount of rain in inches that has fallen since the start of the storm.

(2, 0.3), (4, 0.6), (6, 0.9), (8, 1.2), (10, 1.5) Graph the ordered pairs. Determine

whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

Tota

l Rai

nfal

l (in

.)

0.2

0

0.4

0.6

0.8

1.0

1.2

1.4

1.6

Time (min)2 4 6 8 10 12

2. INVESTING The value of a certain parcel of land has been increasing in value ever since it was purchased. The table shows the value of the land parcel over time.

Year Since

Purchasing0 1 2 3 4

Land Value

(thousands $)$1.05 $2.10 $4.20 $8.40 $16.80

Look for a pattern in the table of values to determine which model best describes the data. Then write an equation for the function that models the data.

3. BOATS The value of a boat typically depreciates over time. The table shows the value of a boat over a period of time.

Years 0 1 2 3 4

Boat

Value ($)8250 6930 5821.20 4889.81 4107.44

Write an equation for the function that models the data. Then use the equation to determine how much the boat is worth after 9 years.

4. NUCLEAR WASTE Radioactive material slowly decays over time. The amount of time needed for an amount of radioactive material to decay to half its initial quantity is known as its half-life. Consider a 20-gram sample of a radioactive isotope.

Half-Lives

Elapsed0 1 2 3 4

Amount of Isotope

Remaining (grams)20 10 5 2.5 1.25

a. Is radioactive decay a linear decay, quadratic decay, or an exponential decay?

b. Write an equation to determine how many grams y of a radioactive isotope will be remaining after x half-lives.

c. How many grams of the isotope will remain after 11 half-lives?

d. Plutonium-238 is one of the most dangerous waste products of nuclear power plants. If the half-life of plutonium-238 is 87.7 years, how long would it take for a 20-gram sample of plutonium-238 to decay to 0.078 gram?


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Enrichment 9-6

Sierpinski Triangle Sierpinski Triangle is an example of a fractal that changes exponentially. Start with an equilateral triangle and find the midpoints of each side. Then connect the midpoints to form a smaller triangle. Remove this smaller triangle from the larger one.

Repeat the process to create the next triangle in the sequence. Find the midpoints of the sides of the three remaining triangles and connect them to form smaller triangles to be removed.

1. Find the next triangle in the sequence. How much has been cut out? What is the area of the fourth figure in the sequence?

2. Make a conjecture as to what you need to multiply the previous amount cut by to find the new amount cut.

3. Fill in the chart to represent the amount cut and the area remaining for each triangle in the sequence.

Figure 1 2 3 4 5 6

Amount Cut 0 1 −

4 7 −

16

Area Remaining 1 3 −

4 9 −

16

4. Write an equation to represent the area that is left in the nth triangle in the sequence.

5. If this process is continued, make a conjecture as to the remaining area.

figure 2 figure 3figure 1Cut out: 1

4

Area = 1 - or14

34

Cut out: 14

316

Area = 1 - or716

916

716+

or


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9-7 Study Guide and InterventionSpecial Functions

Step Functions The graph of a step function is a series of disjointed line segments. Because each part of a step function is linear, this type of function is called a piecewise-linear function.One example of a step function is the greatest integer function, written as f(x) = �x�, where f(x) is the greatest integer not greater than x.

Graph f(x) = �x + 3�.

Make a table of values using integer and noninteger values. On the graph, dots represent included points, and circles represent points that are excluded.

x x + 3 �x + 3�

–5 –2 –2

–3.5 –0.5 –1

–2 1 1

–0.5 2.5 2

1 4 4

2.5 5.5 5

Because the dots and circles overlap, the domain is all real numbers. The range is all integers.

ExercisesGraph each function. State the domain and range.

1. f(x) = �x + 1� 2. f(x) = –�x� 3. f(x) = �x – 1�

4. f (x) = �x� + 4 5. f (x) = �x� – 3 6. f (x) = �2x�

x

f (x)

xO

f (x)

xO

f (x)

xO

f (x)

Example

xO

f (x)

xO

f (x)

xO

f (x)


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Special Functions

Absolute Value Functions Another type of piecewise-linear function is the absolute value function. Recall that the absolute value of a number is always nonnegative. So in the absolute value function, written as f (x) = |x|, all of the values of the range are nonnegative.The absolute value function is called a piecewise-defined function because it can be written using two or more expressions.

Graph f (x) = |x + 2|. State the domain and range.

f (x) cannot be negative, so the minimum point is f (x) = 0. f (x) = |x + 2| Original function

0 = x + 2 Replace f(x) with 0.

-2 = x Subtract 2 from each side.

Make a table. Include values for x > -2 and x < -2.

The domain is all real numbers. The range is all real numbers greater than or equal to 0.

Graph

f(x) = {

x + 1 if x > 1 3x if x ≤ 1

. State the domain

and range.

Graph the first expression. When x > 1,f (x) = x + 1. Since x ≠ 1, place an open circle at (1, 2).Next, graph the second expression. When x ≤ 1, f(x) = 3x. Since x = 1, place a closed circle at (1, 3).

The domain and range are both all real numbers.

x

f (x)

ExercisesGraph each function. State the domain and range.

1. f(x) = |x - 1| 2. f(x) = |-x + 2| 3. f(x) = {

- x + 4 if x ≤ 1 x - 2 if x > 1

x

f (x)x f(x)

-5 3

-4 2

-3 1

-2 0

-1 1

0 2

1 3

2 4

Example 1 Example 2

x

f (x)

x

f (x)

x

f (x) y

x


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9-7 Skills PracticeSpecial Functions

Graph each function. State the domain and range.

1. f (x) = �x – 2� 2. f (x) = 3�x� 3. f (x) = �2x�

4. f (x) = |x| – 3 5. f(x) = |2x| 6. f(x) = |2x + 5|

7. f(x) = {

2x if x ≤ 1 –x + 3 if x > 2

8. f(x) = {

x + 4 if x ≤ 1 if x > 1

0.25x + 1

9. f(x) = {

x + 2 if x < 0 if x ≥ 0

-0.5x + 1

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)


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9-7 Practice Special Functions

Graph each function. State the domain and range.

1. f(x) = -2�x + 1� 2. f(x) = �x + 3� - 2 3. f(x) = -| 1 −

2 x| + 1

x

f (x)

x

f (x)

x

f (x)

x

f (x)

4. f(x) = |2x + 4| - 3 5. f(x) = {

2 if x > - 1 x + 4 if x ≤ - 1

6. f(x) = {

–2x + 3 if x > 0 1 −

2 x - 1 if x ≤ 0

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

x

f (x)

Determine the domain and range of each function.

7. y

x

8. y

x

9.

xx

y

10. CELL PHONES Jacob’s cell phone service costs $5 each month plus $0.35 for each minute he uses. Every fraction of a minute is rounded up to the next minute.

a. Draw a graph to represent the cost of using the cell phone.

b. What is Jacob’s monthly bill if he uses 124.8 minutes?

Mon

thly

Bill

($)

5

0

5.50

6

6.50

7

7.50

Minutes Used1 2 3 4 5 6


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9-7 Word Problem PracticeSpecial Functions

1. BABYSITTING Ariel charges $4 per hour as a babysitter. She rounds every fraction of an hour up to the next half-hour. Draw a graph to represent Ariel’s total earnings y after x hours.

2. SHIPPING A package delivery service determines rates for express shipping by the weight of a package, with every fraction of a pound rounded up to the next pound. The table shows the cost of express shipping for packages between 1 and 9 pounds. Write a piecewise-linear function representing the cost to ship a package between 1 and 9 pounds. State the domain and range.

Weight

(pounds)

Rate

(dollars)

1 17.40

2 19.30

3 22.40

4 25.50

5 28.60

6 31.70

7 34.80

8 37.90

9 41.00

3. DISEASE PREVENTION Body Mass Index (BMI) is used by doctors to determine weight categories that may lead to health problems. According to the Centers for Disease Control and Prevention, 21.7 is a healthy BMI for adults. If the BMI differs from the desired 21.7 by more than x, there may be health risk involved. Write an equation that can be used to find the highest and lowest BMI for a healthy adult. Then solve if x = 3.2.

4. FUNDRAISING Students are selling boxes of cookies at a fund-raiser. The boxes of cookies can only be ordered by the case, with 12 boxes per case. Draw a graph to represent the number of cases needed y when x boxes of cookies are sold.

5. WAGES Kelly earns $8 per hour the first 8 hours she works in a day and $11.50 per hour each hour thereafter.

a. Organize the information into a table. Include a column for hours worked x, and a column for daily earnings f (x).

b. Write the piecewise equation describing Kelly’s daily earnings f (x) for x hours.

c. Draw a graph to represent Kelly’s daily earnings.

Daily

Ear

ning

s ($

)

24

0

48

72

96

120

144

Hours Worked2 4 6 8 10 12 x

f x

Arie

l’s E

arni

ngs

($)

10

15

50

20

25

30

Hours Babysitting1 2 3

y

x

Case

s N

eede

d

2

3

10

4

5

6

Boxes Sold24 36 6012 48 72

y

x


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9-7 EnrichmentParametric Equations

A parametric equation is a pair of functions x = f (t) and y = g (t) that describe both the x- and y-coordinates for the graph as a whole. Parametric functions allow the drawing of many complex curves and figures.

Graph the parametric function given by x = t - 2 andy = t + 1 for -2 ≤ t ≤ 2.

Step 1 Create a table of values for –2 ≤ t ≤ 2 by evaluating x and y for each value of t.

t –2 –1 0 1 2

x –4 –3 –2 –1 0

y –1 0 1 2 3

Step 2 Plot the points.

Step 3 Draw a line to connect the points between –2 ≤ t ≤ 2.

Exercises

Graph each pair of parametric equations over the given range of t values.

1. x = t + 2 2. x = 2t 3. x = t - 3

y = t + 2 y = 1 −

2 t - 1 y = 2t - 1

–2 ≤ t ≤ 2 –2 ≤ t ≤ 2 –2 ≤ t ≤ 2

y

x

y

x

y

x

4. x = 4t 5. x = t2 6. x = t2

y = 2t - 2 y = t + 1 y = t2 - t - 1

–1 ≤ t ≤ 1 –2 ≤ t ≤ 2 –2 ≤ t ≤ 2

y

x

y

x

y

x

y

x

Example


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9 Student Recording SheetUse this recording sheet with pages 614–615 of the Student Edition.

7.

Read each question. Then fill in the correct answer.

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.

Record your answers for Question 11 on the back of this paper.

Multiple Choice

Extended Response

Short Response/Gridded Response

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

7. (grid in)

8a.

8b.

8c.

8d.

9a.

9b.

10a.

10b.

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9 Rubric for Scoring Extended Response

General Scoring Guidelines• If a student gives only a correct numerical answer to a problem but does not show how

he or she arrived at the answer, the student will be awarded only 1 credit. All extended-response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.

• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.

Exercise 11 Rubric

Score Specifi c Criteria

4 A correct solution that is supported by well-developed, accurate explanations. In part a, the student correctly factors the polynomial as (x - 2)(x - 5). In part b, x is found to be 2 and 5. Each step in solving the equations is shown. In part c, students correctly note that the graph of the equation crosses the x-axis at x = 2 and x = 5, the solutions to the equation (where y = 0).

3 A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0 An incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given.


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9

9 Chapter 9 Quiz 2(Lessons 9-4 and 9-5)

Chapter 9 Quiz 1 (Lessons 9-1 through 9-3)

4.

5.

1. Find the value of c that makes p2 + 9p + c a perfect square.


2. w2 + 8w - 10 = 10

3. 3x2 - 2x = 27

4. MULTIPLE CHOICE Solve x2 + x - 7 = 0 by using the Quadratic Formula.

A 6, 7 C 1 ±

√

��

29 −

2

B -1 ±

√

��

-27 −

2 D -1 ±

√

��

29 −

2

5. State the value of the discriminant for 2x2 - 3x + 8 = 0. Then determine the number of real solutions of the equation.

1. State the domain and range of y = 2x2- 8x + 4.

2. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of y = -x2

- 4x + 5. Identify the vertex as a maximum or a minimum.

Solve each equation by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.

3. x2+ x - 2 = 0 4. 3x2

- 5x + 1 = 0

5. MULTIPLE CHOICE Which is an equation for the function shown in the graph?

A y = -

1 −

2 x2 - 4 C y =

1 −

2 x2 - 4

B y = 2x2 - 4 D y = 2x2 - 2

x

y

1.

2.

3.

4.

5.

x

y

x

y

SCORE

SCORE

1.

2.

3.

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9 Chapter 9 Quiz 3(Lesson 9-6)

1. Graph the set of ordered pairs: {(-3, 3), (-2, 0), (-1, -1), (0, 0), (1, 3)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

2. Determine which model best describes the data. Then write an equation for the function that models the data.

x -1 0 1 2 3

y -27 9 -3 1 -

1 −

3

3. MULTIPLE CHOICE Which equation best models the data graphed?A y = 0.8xB y = 100 - 2x C y = 100 (0.97) x

D y = 0.16x2 - 2.03x + 98.5

Chapter 9 Quiz 4 (Lesson 9-7)

1.

2.

3.

x

y

SCORE

SCORE

9

1. If f(x) = 2 �x - 1�, find f (1.5).

2. Graph f(x) = |2x + 1|.

3. Determine the domain and range of f(x) = |2x + 1|.

4. MULTIPLE CHOICE Mr. Aronsohn wants to rent a car on vacation. The rate the car rental company charges is $19 per day or any fraction thereof. How much would it cost for Mr. Aronsohn to rent a car for 6.4 days?

A $114.00 C $124.00 B $121.60 D $133.00

5. Graph g (x) =

{

2x - 3 if x ≤ 2 x - 1 if x > 2

1.

2.

3.

4.

5.

x

x

O

y

x

Daylight in Ocean Water

Perc

ent o

f Lig

ht

20

30

10

0

40

5060

70

8090

100

Depth Below Surface (m)20 3010 40 50 60 70 80 90 100


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9

Write the letter for the correct answer in the blank at the right of each question.

1. Which equation corresponds to the graph shown? A y = x2 - 1 C y = x2 + 1

B y = -(x - 1)2 D y = -(x + 1)2

2. Find the coordinates of the vertex of the graph of y = x2 - 8x + 10. Identify the vertex as a maximum or a minimum.

F (4, -6); minimum H (4, 6); maximum

G (-4, 58); maximum J (-4, 26); minimum

3. Solve x2 - 24x + 144 = 36 by taking the square root of each side.

A -6, 18 B 6, 18 C 6, 12 D -6, 6

4. Which equation can be used to solve 5b2 + 30b - 10 = 0 by completing the square?

F (b + 6)2 = 38 G (b + 6)2

= 46 H (b + 3)2 = 11 J (b + 3)2

= 19

5. Which step is not performed in the process of solving r2 + 8r + 5 = 0 by completing the square?

A Subtract 5 from each side. C Add 16 to each side.

B Factor r2 + 8r. D Take the square root of each side.

Part II

Solve each equation by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.

6. x2 - 7x - 8 = 0

7. x2 + 1 = 5x

For Questions 8 and 9, round to the nearest tenth if necessary.

8. Solve x2 + 4x = 20 by completing the square.

9. Solve -2x2 + 18 = 7x by using the Quadratic Formula.

10. The base of a rectangle is 4 more than the height. The area of the rectangle is 15 square inches. What are the dimensions of the rectangle?

Part I

1.

2.

3.

4.

5.

Chapter 9 Mid-Chapter Test (Lessons 9-1 through 9-5)

x

y

6.

7.

8.

9.

10.

SCORE


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9 Chapter 9 Vocabulary Test

absolute value function

axis of symmetry

completing the square

discriminant

double root


maximum

minimum

nonlinear function

parabola

piecewise-defi ned function

piecewise-linear function

Quadratic Formula

quadratic function

step function

transformation

translation

vertex

Choose a term from the vocabulary list above to complete each sentence.

1. is a change in the position of a figure either up, down, or diagonal.

2. Symmetry is a geometric property of a(n) .

3. The vertical line containing the fold line when a

figure is folded in half is called the .

4. The function f(x) = �x + 1� is a special type of step function called a(n) .

5. The is the result of solving the standard form of the quadratic equation for x.

6. The of a parabola is a minimum or maximum point.

7. The graph of a(n) is a parabola.

8. When the vertex of a parabola lies on the x-axis, the related quadratic equation has a(n) .

Define each term in your own words.

9. axis of symmetry

10. discriminant

? 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

?

?

?

?

?

?

?

SCORE


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9 Chapter 9 Test, Form 1

y

xO

y

xO

y

xO

1.

2.

3.

4.

5.

6.

7.

8.

SCORE


1. Consider the equation y = x2 + 3x - 4. Determine whether the function has a maximum or minimum value. State the maximum or minimum value. What are the domain and range of the function?

A min.; (0, 0)

D: {all real numbers}

R: {all real numbers}

C max.; (-1.5, -6.25)

D:

{x| x ≤ -1.5

}

R: {y| y ≥ -6.25

}

B max.; (0, 0)


R: {y|y ≤ 0

}

D min.; (-1.5, -6.25)


R: {y| y ≥ -6.25

}

2. What is the equation of the axis of symmetry of the graph of y = x2 + 6x - 7?

F x = 6 G x = -3 H x = 3 J x = -6

3. Find the coordinates of the vertex of the graph of y = 4 - x2. Identify the vertex as a maximum or a minimum.

A (2, 0); maximum C (0, 4); maximum B (0, 4); minimum D (2, 0); minimum

4. What are the roots of the quadratic equation whose related function is graphed at the right?

F -1, 3 H -3, 1 G -1, 1 J 1, 3

5. One root of the quadratic equation whose related function is graphed lies between which two consecutive integers?

A 1 and 2 C 0 and -1 B 2 and 3 D 0 and 1

6. Which equation corresponds to the graph shown?

F y = x2 + 1 H y = x2 - 1 G y = -x2 - 1 J y = x2

7. Describe how the graph of the function g(x) = -3x2 - 2 is

related to the graph of the function f(x) = -3x2.A translation of f(x) = -3x2 reflected over the x-axis and down 2 units

B translation of f(x) = -3x2 down 2 units C translation of f(x) = -3x2 reflected over the x-axis and up 2 units D translation of f(x) = -3x2 up 2 units

8. Find the value of c that makes x2 - 5x + c a perfect square trinomial.

F -12.25 G -6.25 H 6.25 J 10


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9 Chapter 9 Test, Form 1 (continued)

9. Which value of c makes y2 + 8y + c a perfect square trinomial? A 4 B 16 C 64 D 8

10. Which equation is equivalent to x2 + 2x - 3 = 0? F (x + 1)2 = 2 G (x - 1)2 = 4 H (x - 1)2 = 2 J (x + 1)2 = 4

11. Solve the equation 2x2 + 3x - 5 = 0 by using the Quadratic Formula. A -2 1 −

2 , 1 B -5, 1 C -1, 2 1 −

2 D -1, 5

12. State the value of the discriminant for y = x2 - 8x + 10. F 4.9 G 24 H 104 J 10.2

13. Determine the number of real solutions of n2 - 5n - 6 = 0. A 1 real solution C infinitely many real solutions B 2 real solutions D no real solutions

14. TREE HOUSE Bob tosses his basketball onto the ground from his tree house. He tosses the basketball with an initial downward velocity of 8 feet per second. The equation h = -16t2 - 8t + 20 represents the height h of the basketball after t seconds. How long does the basketball take to hit the ground?

F 0.9 s H 1.0 s G 9 s J 20 s

15. State the value of the discriminant of 5x2 + 9x = 3 A 5 B 12 C 21 D 141

16. Look for a pattern in the table of values x 0 1 2 3

y 0 2 8 18to determine which model best describes the data.

F linear G quadratic H exponential J none of these

17. Which function best models the data in Question 16? A y = 2x B 2x + 1 C y = 2x2 D y = 2x

18. What is the domain of f (x) = -2x + 1 if x ≥ 0

x + 3 if x < 0{ ?

F {all real num.} G {x | x ≥ 3} H {x | x < 2} J {

x | x ≤ 1 −

2 }

19. If f(x) = 2�x�, find f (

-

1 −

4 )

.

A -2 B - 1 −

2 C 0 D 1 −

2

20. What is the range of y = ⎪3x + 1⎥ ? F {all real num.} G {y | y ≥ 0} H {y | y ≥ 1} J

{

y | y ≥ 1 −

3 }

Bonus If b2 - 4ac = 0, determine the number of real solutions of the equation ax2 + bx + c = 0. B:

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.


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9


1. Consider the equation y = x2 + 5x - 6. Determine whether the function has a maximum or minimum value. State the maximum of minimum value. What are the domain and range of the function?

A min.; (0, 0)



C min.; (-2.5, -12.25)


R: {y| y ≥ -12.25}

B max.; (0, 0)


R: {y| y ≤ 0}

D max.; (2.5, -12.25)

D: x| x ≤ 2.5


2. Which equation corresponds to the graph shown?

F y = x2 + 7x - 12 H y = x2 + 5x + 12 G y = x2 - x - 12 J y = x2 + 12x - 1

3. Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of y = 2x2 - 12x + 6.

A x = -3; (-3, 60) C x = -3; (-3, 78) B x = 3; (3, -12) D x = 3; (3, 6)

4. Find the coordinates of the vertex of the graph of y = -2x2 - 8. Identify the vertex as a maximum or a minimum.

F (-2, -16); minimum H (2, -16); maximum G (-2, 8); maximum J (0, -8); maximum

5. What are the root(s) of the quadratic equation whose related function is graphed at the right?

A 2 C 0, 4 B 3 D -4, 0

6. One root of the quadratic equation whose related function is graphed lies between which two consecutive integers?

F -3 and -2 H -2 and -1 G 2 and 3 J 1 and 2

7. How is the graph of g(x) = x2 - 3 related to the graph of f(x) = x2?

A translated down 3 units C translated right 3 units B translated up 3 units D translated left 3 units

8. Find the value of c that makes x2 + 10x + c a perfect square trinomial.

F -25 G -5 H 10 J 25

1.

2.

3.

4.

5.

6.

7.

8.

Chapter 9 Test, Form 2A

y

xO-12 -6

6

-6

-12

6 12

y

xO

y

xO

SCORE


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9

9. What value of c makes 3x2 + 24x + c contain a perfect square trinomial? A 144 B 16 C 64 D 48

10. Which step is not performed in the process of solving n2 - 12n - 10 = 0 by completing the square?

F Add 10 to each side. H Factor n2 - 12n - 10 = 0. G Add 36 to each side. J Take the square root of each side.

11. Which equation is equivalent to 2x2 - 24x - 14 = 0? A (x - 6)2 = 50 B (x - 3)2 = 13 C (x - 3)2 = 20 D (x - 6)2 = 43

12. State the value of the discriminant of 3x2 + 8x = 2. F 3 G 40 H 88 J 100


13. 4x2 + 11x - 3 = 0 A -2.4, -0.3 B - 1 −

4 , 3 C 0.3, 2.4 D -3, 1 −

4

14. y2 + 8y = 2 F -8.2, 0.2 G 8.2, -0.2 H 0.3, 7.7 J -7.7, -0.3

15. Determine the number of real solutions of 7x2 - 18x + 12 = 0. A 2 B infinitely many C none D 1

16. Look for a pattern in the table of values to x 0 1 2 3

y 1 7 49 343determine which model best describes the data.

F linear H exponential G quadratic J none of these

17. Which function best models the data in Question 16? A y = 7x B y = 7 x 2 C y = 7 x D y = 7 x + 1

18. If f (x) = �x + 2�, find f (1.5). F 0.5 G 3 H 3.5 J 4

19. Which is not true about the graph of f (x) = ⎪3x + 2⎥ ? A The range includes all real numbers. B It includes the point (-3, 7). C The domain includes all real numbers. D The graph is “V-shaped.”

20. Which point is located on the graph of f (x) = 1 −

2 x + 1 if x > 1

1 −

3 x + 2 if x ≤ 1{ ?

F (-3, 1) G (0, 1) H (2, 0) J (3, 3)

Bonus What is the equation of the axis of symmetry of a parabolaif its x-intercepts are -3 and 5?

Chapter 9 Test, Form 2A (continued)

B:

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.


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9 SCORE Chapter 9 Test, Form 2B


1. Consider the equation y = -x2 - 7x + 12. Determine whether the function has a maximum or a minimum value. State the maximum or minimum value. What are the domain and range of the function?

2. Which equation corresponds to the graph shown? F y = x2 - 3x - 10 H y = x2 - 10x + 6 G y = x2 + 7x + 10 J y = x2 - 11x - 10

3. Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of y = -x2 - 10x + 17.

A x = -5; (-5, -8) C x = 5; (5, 92)

B x = -5; (-5, 42) D x = 5; (5, 32)

4. Find the coordinates of the vertex of the graph of y = 3 x 2 - 6. Identify the vertex as a maximum or a minimum.

F (0, -6); maximum H (0, -6); minimum G (-6, 0); minimum J (6, 0); minimum

5. What are the root(s) of the quadratic equation whose related function is graphed at the right? A 4, 0 C 2, 3

B -2, 3 D -4, 0

6. One root of the quadratic equation whose related function is graphed lies between which consecutive integers?

F -4 and -3 H -5 and -4

G 0 and 1 J -3 and -2

7. Describe how the graph of the function g(x) = -5x2 - 2 is

related to the graph of the function f(x) = 5x2 + 1. A translation of f(x) = 5x2 + 1 reflected over the x-axis and up 3 units B translation of f(x) = 5x2 + 1 down 2 units C translation of f(x) = 5x2 + 1 reflected over the x-axis and down 3 units D translation of f(x) = 5x2 + 1 up 2 units

1.

2.

3.

4.

5.

6.

7.

y

xO-12

-12

-6 6 12

-6

6

y

xO

y

xO

A min.; (-7, 0)

D: {x|x ≤ 12

}


B

max.; (-3.5, 24.25)


R: {y|y ≤ 24.25

}

C min.; (3.5, -24.25)


R: {y|y ≥ -24.25

}

D max.; (12, 0)

D:

{x|x ≤ 12

}



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9

8. How is the graph of f(x) = x2 translated to create the graph of g(x) = x2 + 4?

F down 4 units H right 4 units G up 4 units J left 4 units

9. What value of c makes 4 x 2 + 24x + c a perfect square trinomial? A 1 B 36 C 144 D 9

10. Which is not performed when solving r2 + 12r - 6 = 0 by completing the square? F Add 6 to each side. H Add 36 to each side. G Factor r2 + 12r - 6. J Take the square root of each side.

11. Which equation is equivalent to 3x2 + 24x + 15 = 0? A (x + 4)2

= 11 B (x + 4)2 = 1 C (x + 2)2

= -1 D (x + 2)2 = -11

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary. 12. 3x2 - 11x - 4 = 0 F -4, 1 −

3 G 0.4, 3.3 H - 1 −

3 , 4 J -3.3, -0.4

13. d2 + 4 = 6d

A -5.2, -0.8 B -6.6, 0.6 C -0.6, 6.6 D 0.8, 5.2

14. Determine the number of real solutions of 6x2 + 19x + 14 = 0. F 2 G infinitely many H 1 J none

15. Determine the number of real solutions of 2x2 + 2x + 3 = 0. A 2 B infinitely many C 1 D none

16. Look for the pattern in the table of values to determine which model best describes the data.

F exponential G quadratic H linear J none of these

17. Which function best models the data in Question 16 ? A y = 5x + 1 B y = 5x C y = 5x2 D y = 5x

18. If f (x) = �x - 2�, find f (4.5). F 2 G 2.5 H 3 J 6.5

19. Which is not true about the graph of f (x) = |2x - 1|? A Domain: all real numbers. C It includes the point (-2, 5). B Range: all real numbers. D The graph is “V-shaped.”

20. Which point is not located on the graph of f (x) =

2x + 3 if x ≤ -1

4 + x if x > -1{ ?

F (-1.5, 0) G (-1, 3) H (0, 4) J (4, 8)

Bonus Find all values of c that make x2 + cx + 64 a perfect

square trinomial.

Chapter 9 Test, Form 2B (continued)

x 0 1 2 3

y 1 5 25 125

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:


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9 SCORE Chapter 9 Test, Form 2C

Use a table of values to graph each function. 1. y = -x2 + 3x + 10

2. y = 2x2 - 3x

3. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of y = 2x2 - 8x + 3. Identify the vertex as a maximum or a minimum.

4. Solve x2 - 6x + 4 = 0 by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.

5. NUMBER THEORY Two numbers have a sum of 1 and a product of -6. Use the quadratic equation -n2 + n + 6 = 0 to determine the two numbers.

Solve each equation by factoring. 6. 2x2 + x = 10 7. r2 - 11r + 28 = 0

Describe how the graph of each function is related to the graph of f(x) = x2. 8. g(x) = -2 + x2 9. h(x) = -x2 + 6

1.

2. y

xO

3.

4.

5.

6.

7.

8.

9.

xO

f (x)

y

xO 6 12

6

-6

12

-12 -6


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9 SCORE


10. 2x2 + 3x = 20 11. p2 - 12p + 9 = 0

Solve each equation by using the Quadratic Formula.Round to the nearest tenth if necessary.

12. 15n2 - 3 = 4n 13. r2 + 16r + 21 = 0


14. 7m2 + 8m = 3 15. 4p2 = 4p - 1

16. The length of a rectangle is 5 inches more than the width. The area is 33 square inches. Find the length and width. Round to nearest tenth if necessary.

Determine an equation that models the data. 17. x 1 2 3 4

y 4 16 36 64

18. x 0 1 2 3 4

y 2 6 18 54 162

19. Determine the domain and range for the graph shown.

20. Graph g (x) = x + 2 if x > 1

2x - 1 if x ≤ 1{ .

Bonus Without graphing, determine the x-intercepts of the graph of f(x) = 7x2 + 9x + 1.

y

x

Chapter 9 Test, Form 2C (continued)9

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

x


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9 SCORE

Use a table of values to graph each function. 1. y = x2 - 7x + 12

2. y = 4x2 - 8x

3. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of y = -2x2 + 4x - 5. Identify the vertex as a maximum or a minimum.

4. Solve x2 - 2x - 1 = 0 by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.

5. NUMBER THEORY Two numbers have a sum of 4 and a product of -5. Use the quadratic equation -n2 + 4n + 5 = 0 to determine the two numbers.

Solve each equation by factoring. 6. x2 + 6x = 27 7. k2 - 13k + 36 = 0

Describe how the graph of each function is related to the graph of f(x) = x2. 8. g(x) = 5 + x2 9. g(x) = -x2 - 3

Chapter 9 Test, Form 2D

x O

f (x)

1. y

xO 4 8

4

28 24

2. y

xO

3.

4.

5.

6.

7.

8.

9.


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9


10. 2x2 + 9x = 18 11. t2

+ 15t + 11 = 0


12. 12v2 - 6 = -v 13. d2 - 14d - 22 = 0

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation. 14. 3b2 + 10 = -8b 15. 9a2 = 6a - 1

16. The width of a rectangle is 3 inches less than the length. The area is 50 square inches. Find the length and width. Round to nearest tenth if necessary.

Determine which model best describes the data. Then write an equation that models the data.

17. x 1 2 3 4

y 2 5 8 11

18. x 1 2 3 4

y 14 28 56 112

19. Determine the domain and range for the graph shown.

20. Graph g (x) = 2x + 1 if x ≤ 2

x - 2 if x > 2{ .

Bonus Find the value of a so that the equation ax2 + 8x + 32 = 0 has 1 real root.

Chapter 9 Test, Form 2D (continued)

y

x

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

y

x


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9 SCORE

Use a table of values to graph each function. 1. y = -2x2 - 3x + 3 2. y = 3x2 - 5x - 2

3. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of y - 1 = 5x2 - 25x + 3. Identify the vertex as a maximum or a minimum.

4. Two points on a parabola are (-3, 5) and (11, 5). What is the equation of the axis of symmetry?

5. NUMBER THEORY Two numbers have a sum of -6 and a product of 9. Use the quadratic equation n2 - 6n + 9 = 0 to determine the two numbers.

Solve each equation by factoring. 6. 12x2 - 5x = 2 7. 18y2 + 39y - 15 = 0

Describe how the graph of each function is related to the graph of f(x) = x2.8. g(x) = 4 - 2x2 9. g(x) = -

1 −

6 x2

+

5 −

6

Solve each equation by graphing. If integral solutions cannot be found, estimate the solutions by stating the consecutive integers between which the solutions lie.

10. 1 −

2 x 2 + x − 4 = 0

11. x2 + 7 = 6x

12. 4x2 + 6x + 5 = 0

Chapter 9 Test, Form 3

1. y

xO

2. y

xO

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.


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9

13. Solve 21u2 = u + 10 by using the Quadratic Formula. Round to the nearest tenth if necessary.

14. Solve -2x2 = -(5x + 4) by using the Quadratic Formula.Round to nearest tenth if necessary.

15. PHYSICAL SCIENCE A projectile is shot vertically from ground level. Its height h, in feet, after t seconds is givenby h = 88t - 16t2. Will the projectile have a height of 125 feet 0, 1, or 2 times after being shot?

16. Find the values for c so that t2 + ct + 81 = 0 has one

real root.

Determine which model best describes the data. Then write an equation that models the data.

17. x 0 1 2 3 4

y 0 -4 -16 -36 -64


18. f (x) = �2x + 1�

19. g (x) = ⎪-x - 1⎪ + 1

20. h (x) = 2x + 1 if x ≤ 2

x - 2 if x > 2{

Bonus Solve the equation ax2 + 7x - 4 = 0 for x by using

the Quadratic Formula.

Chapter 9 Test, Form 3 (continued)

13.

14.

15.

16.

17.

18.

19.

20.

B:

x−1

−2

1

2

−1

1 2−2

x

x


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9 SCORE Chapter 9 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. Write a quadratic function whose graph opens upward. Write an exponential function. Graph both functions on the same coordinate plane. Compare and contrast the domains and ranges of the two functions and any symmetry of the two graphs.

2. a. Write a quadratic equation that has no real roots, and find its discriminant. Explain how the discriminant shows that a quadratic equation has no real roots.

b. Write a quadratic equation that has one real root, and find its discriminant. Explain why the Quadratic Formula yields only one solution when the discriminant of a quadratic equation is equal to zero.

c. Write a quadratic equation that has two real roots. Determine whether completing the square, graphing, or factoring would be the better method to use to solve your quadratic equation.

3. Curt’s Appliance Service charges $125 for the first half hour of each home visit plus $40 for each additional half-hour block of work.

a. Write and graph a step-function to represent the total charges for every hour h of work.

b. How much is charged for 76 minutes of work?

c. Laurie was charged $365. How long did it take the work to get completed?


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9 SCORE Standardized Test Practice (Chapters 1–9)

1. Armando collects baseball cards. He currently has 48 cards, and buys 5 new cards a week. How many cards will Armando have in 10 weeks? (Lesson 3-5)

A 102 B 98 C 94 D 100

2. Wesley carves whistles out of wood and sells them at a gift shop. The equation C = 2n + 15 models his weekly cost C of making n whistles. The equation R = 5n models his weekly revenue R from selling n whistles. How many whistles must Wesley make and sell for his cost and revenue to be equal? (Lesson 6-2)

F 5 G 4 H 25 J 10

3. Solve 2(5x - 4) ≥ 7(x - 2). (Lesson 5-3)

A x ≥ -2 B x ≤ 2 C x ≥ 2 −

3 D x ≤ 2

4. Find 3mt(2m2 - 3mt + t2). (Lesson 8-2)

F 5m3t - 6m2t + 3mt3 H 6m3t - 9m2t2 + 3mt3

G 6m2t - 3mt + 3mt2 J 2m2 + t2

5. Which binomial is a factor of 6ut - 9u + 8yt - 12y? (Lesson 8-5)

A 2t + 3 B 3u + 4 C -2t + 3 D 3u + 4y

6. Factor 16 - n4. (Lesson 8-8)

F (4 - n2)(4 + n2) H (2 - n)(2 + n)(n2 + 4)

G (n + 2)(n - 2)(4 + n2) J (2 - n)(2 + n)(2 - n)(2 + n)

7. Solve 2x2 + 72 = 24x. (Lesson 8-9)

A {2, 6} B {-6, 2} C {6} D {-6}

8. Find the coordinates of the vertex of the graph of y = 3x2 + 2x.

(Lesson 9-1)

F (

1 −

3 , 1

)

G (

-

1 −

3 , -

1 −

3 )

H (

1 −

3 , -

1 −

3 )

J (

-

1 −

3 , 1

)

9. Which equation is equivalent to 2r2 - 28r + 38 = 0? (Lesson 9-4)

A 2(r - 7)2 = 49 C (r - 14)2 = 177 B (r - 14)2 = 60 D (r - 7)2 = 30

10. What number should be inserted into the table for the data to display exponential behavior? (Lesson 9-6)

F 152 G 186 H 248 J 216

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

x 2 3 4 5

y 8 24 72


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Standardized Test Practice (continued)

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

17. A B C D

11. A watch is on sale for 30% off the original price. If the original price of the watch is $14, what is the discounted price? (Lesson 2-7)

A $9.80 B $4.20 C $13.70 D $9.33

12. Solve the proportion n −

500 = 2 −

40 . (Lesson 2-6)

F 125 G 25 H 16 J 80

13. Determine the x-intercept and y-intercept of 4x - 2y = 10. (Lesson 3-1)

A 0, 0 B 2.5, 0 C 0, -5 D 2.5, -5

14. Write a direct variation equation that relates x and y if y = 10when x = 12. (Lesson 3-4)

F y = 6 −

5 x G y = 5 −

6 x H y = 12x J 10y = 12x

15. Write the slope-intercept form of an equation for the line that passes through (4, 0) and is parallel to the graph of 3y - 6x = 4. (Lesson 4-4)

A 2y = -x C y = 2x + 4B y = 2x - 8 D y = -2x + 8

16. Which ordered pair is not a solution of 4x - 8y ≥ 24? (Lesson 5-6)

F (-5, -4) G (2, -2) H (7, -1) J (-8, -8)

17. Which binomial is a factor of 6x2 + x - 12? (Lesson 8-7)

A (3x - 4) B (2x + 6) C (3x + 4) D (2x - 3)

18. If 3x + 2y = 9 and 3x + 3y = 6, what is the value of x? (Lesson 6-3)

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

19. State the value of the discriminant for 4x2

+ 17x + 18 = 0. (Lesson 9-5)

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column box

and then shading in the appropriate circle that corresponds to that entry.


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9 Standardized Test Practice (continued)(Chapters 1–9)

20. Write an equation of the line that passes through (2, 4)and (1, -2) in slope-intercept form. (Lesson 4-2)

21. Solve 4y - 3(7y - 2) ≤ -14 - 13y. (Lesson 5-3)

22. Solve the system of inequalities by graphing. (Lesson 5-6)

x + y ≤ 4y ≤ 2x - 4

23. Simplify (4xy + 3x2y-5y2) - (3y2 - 5xy+ 7x2y). (Lesson 8-1)

24. Find (3a2 + 2)(3a2

- 2). (Lesson 8-4)

25. Factor x2 + 12x + 35. (Lesson 8-6)

26. Factor 2m2 + 11m + 15. (Lesson 8-7)

27. Solve 36 - 1 −

4 y 2 = 0 by factoring. (Lesson 8-8)

28. Use the formula h = - 16t2 + 250t to model the height h in

feet of a model rocket t seconds after it is launched. Determine when the rocket will reach a height of 900 feet. (Lesson 9-5)

29. The population of North Carolina has been increasing at an annual rate of 1.7%. If the population of North Carolina was 7,650,789 in 1999, predict its population in 2015. (Lesson 7-6)

30. Consider y = - x2 - 2x + 2. (Lesson 9-1)

a. Find the equation for the axis of symmetry.

b. Find the coordinates of the vertex and determine if it is a maximum or minimum.

c. Graph the equation.

Part 3: Short Response

Instructions: Write your answers in the space provided.

20.

21.

22. y

xO

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

30c.y

xO


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9 SCORE Unit 3 Test (Chapters 7–9)

1. Express the area of the triangle as a monomial.

2. Simplify (3 b -4 g 3 ) -2

−

(4b ) 2

3. Solve 12 3x - 1 = 144.

4. Solve (3.4 × 105)(7.5 × 10-9). Write your answer in both standard and scientific notation.

5. The population of Las Vegas, Nevada has been increasing at an annual rate of 7.0%. If the population of Las Vegas was 478,434 in 1999, predict its population in 2015.

6. A new motor home costs $89,000. It is expected to depreciate 9% each year. Find the value of the motor home in 4 years.

7. Find the total amount of an investment if $1200 is invested at an interest rate of 3.5% compounded quarterly for 7 years.

8. Write an equation for the nth term of the geometric sequence -5, 15, -45, 135, … .

9. Write a recursive formula for 14, 9, 4, −1, ….

10. Find (4a - 2b + c) - (5c + a - 4b).

11. Solve x(x - 3) = x(x + 5) - 16.

Find each product.

12. (p - 3)(p + 5)

13. (a + 2)(a2 - 3a - 7)

14. (5t + r)2

15. (3k - 1)(3k + 1)

8xy2

5x3y4

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.


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ht © G

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ivision o

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NAME DATE PERIOD

PDF Pass


9

Factor each polynomial.

16. 4p2r - 16pr + 8pr2

17. m2 - 11m + 18

18. 3a2 - 13a + 12

19. 27r2 - 12t2

20. The area of a square is 4p2 - 4p + 1 square inches. What

is the length of the side of the square?

Solve each equation.

21. 22y2 = -11y

22. h2 - 13h + 36 = 0

23. 5a2 - 33a = 14

24. m2 + 81 = -18m

25. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of y = x2

+ 8x + 12. Then graph y = x2

+ 8x + 12.

26. Find the value of c that makes x2 - 20x + c a perfect square trinomial.

27. State the value of the discriminant for 3x2 + 2x + 1 = 0.

28. Solve 6 = 3t2 + 2t by using the Quadratic Formula. Round

to the nearest tenth if necessary.

29. Look for a pattern in the table of values to determine which model best describes the data. Then write an equation of the function that models the data.

x 0 1 2 3

y 6 12 24 48

30. Graph y = |2x + 1|.

Unit 3 Test (continued)

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

y

xO

y

xO


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass

Chapter 9 A1 Glencoe Algebra 1

Chapter Resources

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

3 G

len

co

e A

lge

bra

1

Bef

ore

you

beg

in C

ha

pte

r 9

•

Rea

d ea

ch s

tate

men

t.

•

Dec

ide

wh

eth

er y

ou A

gree

(A

) or

Dis

agre

e (D

) w

ith

th

e st

atem

ent.

•

Wri

te A

or

D i

n t

he

firs

t co

lum

n O

R i

f yo

u a

re n

ot s

ure

wh

eth

er y

ou a

gree

or

disa

gree

, wri

te N

S (

Not

Su

re).

ST

EP

1A

, D, o

NS

Sta

tem

ent

ST

EP

2A

o D

1.

Th

e gr

aph

of

a qu

adra

tic

fun

ctio

n i

s a

para

bola

. 2

. T

he

grap

h o

f y

= 4x

2 – 2

x +

7 w

ill

be a

par

abol

a op

enin

g do

wn

war

d si

nce

th

e co

effi

cien

t of

x2

is p

osit

ive.

3.

A q

uad

rati

c fu

nct

ion

’s a

xis

of s

ymm

etry

is

eith

er t

he

x-ax

is o

r th

e y-

axis

.

4.

Th

e gr

aph

of

a qu

adra

tic

fun

ctio

n o

pen

ing

upw

ard

has

no

max

imu

m v

alu

e.

5.

Th

e x-

inte

rcep

ts o

f th

e gr

aph

of

a qu

adra

tic

fun

ctio

n a

re t

he

solu

tion

s to

th

e re

late

d qu

adra

tic

equ

atio

n.

6.

All

qu

adra

tic

equ

atio

ns

hav

e tw

o re

al s

olu

tion

s.

7.

An

y qu

adra

tic

expr

essi

on c

an b

e w

ritt

en a

s a

perf

ect

squ

are

by

a m

eth

od c

alle

d co

mpl

etin

g th

e sq

uar

e. 8

. T

he

quad

rati

c fo

rmu

la c

an o

nly

be

use

d to

sol

ve q

uad

rati

c eq

uat

ion

s th

at c

ann

ot b

e so

lved

by

fact

orin

g or

gra

phin

g.

9.

Th

e gr

aph

of

a st

ep f

un

ctio

n i

s a

seri

es o

f di

sjoi

nte

d li

ne

segm

ents

.10

. It

is n

ot p

ossi

ble

to id

enti

fy d

ata

as li

near

bas

ed o

n pa

tter

ns o

f be

havi

or o

f th

eir

y-va

lues

.

Aft

er y

ou c

omp

lete

Ch

ap

ter

9

•

Rer

ead

each

sta

tem

ent

and

com

plet

e th

e la

st c

olu

mn

by

ente

rin

g an

A o

r a

D.

•

Did

an

y of

you

r op

inio

ns

abou

t th

e st

atem

ents

ch

ange

fro

m t

he

firs

t co

lum

n?

•

For

th

ose

stat

emen

ts t

hat

you

mar

k w

ith

a D

, use

a p

iece

of

pape

r to

wri

te a

n

exam

ple

of w

hy

you

dis

agre

e.

9A

ntic

ipat

ion

Gui

deQ

uad

rati

c F

un

cti

on

s a

nd

Eq

uati

on

s

Step

1

Step

2

A D D A A D A D A A

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7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-1

Cha

pte

r 9

5 G

len

co

e A

lge

bra

1

Ch

arac

teri

stic

s o

f Q

uad

rati

c Fu

nct

ion

s

Qu

adra

tic

Fun

ctio

na

fu

nctio

n d

escri

be

d b

y a

n e

qu

atio

n o

f th

e fo

rm f

(x)

= a

x2 +

bx

+ c

, w

he

re a

≠ 0

Exa

mp

le:

y =

2x2

+ 3

x +

8

Th

e pa

ren

t gr

aph

of

the

fam

ily

of q

uad

rati

c fu

ctio

ns

is y

= x

2 . G

raph

s of

qu

adra

tic

fun

ctio

ns

hav

e a

gen

eral

sh

ape

call

ed a

par

abol

a. A

par

abol

a op

ens

upw

ard

and

has

a m

inim

um

p

oin

t w

hen

th

e va

lue

of a

is

posi

tive

, an

d a

para

bola

ope

ns

dow

nw

ard

and

has

a m

axim

um

p

oin

t w

hen

th

e va

lue

of a

is

neg

ativ

e.

a. U

se a

tab

le o

f va

lues

to

grap

h

y =

x2

- 4

x +

1.

xy

-1

6

01

1-

2

2-

3

3-

2

41

x

y

O

Gra

ph t

he

orde

red

pair

s in

th

e ta

ble

and

con

nec

t th

em w

ith

a s

moo

th c

urv

e.

b.

Wh

at a

re t

he

dom

ain

an

d r

ange

of

this

fu

nct

ion

?T

he

dom

ain

is

all

real

nu

mbe

rs. T

he

ran

ge i

s al

l re

al n

um

bers

gre

ater

th

an o

r eq

ual

to

-3,

wh

ich

is

the

min

imu

m.

a. U

se a

tab

le o

f va

lues

to

grap

h

y =

-x2

- 6

x -

7.

xy

-6

-7

-5

-2

-4

1

-3

2

-2

1

-1

-2

0-

7

x

y

O

Gra

ph t

he

orde

red

pair

s in

th

e ta

ble

and

con

nec

t th

em w

ith

a s

moo

th c

urv

e.

b.

Wh

at a

re t

he

dom

ain

an

d r

ange

of

this

fu

nct

ion

?T

he

dom

ain

is

all

real

nu

mbe

rs. T

he

ran

ge i

s al

l re

al n

um

bers

les

s th

an o

r eq

ual

to

2, w

hic

h i

s th

e m

axim

um

.

Exer

cise

sU

se a

tab

le o

f va

lues

to

grap

h e

ach

fu

nct

ion

. Det

erm

ine

the

dom

ain

an

d r

ange

.

1. y

= x

2 +

2

2. y

= -

x2 -

4

3. y

= x

2 -

3x

+ 2

x

y

O

x

yO

x

y

O

D

: {al

l rea

l nu

mb

ers}

D

: {

all

rea

l nu

mb

ers}

D:

{a

ll r

eal n

um

ber

s}

R:

{y |

y ≥

2}

R:

{y |

y ≤

-4}

R:

{y |

y ≥

- 1 −

4 }

9-1

Stud

y G

uide

and

Inte

rven

tion

G

rap

hin

g Q

uad

rati

c F

un

cti

on

s

Exam

ple

1Ex

amp

le 2

001-

010_

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10:5

7 A

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Answers (Anticipation Guide and Lesson 9-1)

A01_A14_ALG1_A_CRM_C09_AN_660283.indd A1A01_A14_ALG1_A_CRM_C09_AN_660283.indd A1 12/23/10 2:56 PM12/23/10 2:56 PM

Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

6 G

len

co

e A

lge

bra

1

Sym

met

ry a

nd

Ver

tice

s P

arab

olas

hav

e a

geom

etri

c pr

oper

ty c

alle

d sy

mm

etry

. Th

at

is, i

f th

e fi

gure

is

fold

ed i

n h

alf,

each

hal

f w

ill

mat

ch t

he

oth

er h

alf

exac

tly.

Th

e ve

rtic

al l

ine

con

tain

ing

the

fold

lin

e is

cal

led

the

axis

of

sym

met

ry. T

he

axis

of

sym

met

ry c

onta

ins

the

min

imu

m o

r m

axim

um

poi

nt

of t

he

para

bola

, th

e ve

rtex

.

a. W

rite

th

e eq

uat

ion

of

the

axis

of

sym

met

ry.

In y

= 2

x2 +

4x

+ 1

, a =

2 a

nd

b =

4.

Su

bsti

tute

th

ese

valu

es i

nto

th

e eq

uat

ion

of

th

e ax

is o

f sy

mm

etry

.

x =

-

b −

2a

x =

-

4 −

2(

2) =

-1

Th

e ax

is o

f sy

mm

etry

is

x =

-1.

b.

Fin

d t

he

coor

din

ates

of

the

vert

ex.

Sin

ce t

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

is

x =

-1

and

the

vert

ex l

ies

on t

he

axis

, th

e x-

coor

din

ate

of t

he

vert

ex

is -

1.y

= 2

x2 +

4x

+ 1

O

rigin

al equation

y =

2(-

1)2

+ 4

(-1)

+ 1

S

ubstitu

te.

y =

2(1

) -

4 +

1

Sim

plif

y.

y =

-1

Th

e ve

rtex

is

at (

-1,

-1)

.

c. I

den

tify

th

e ve

rtex

as

a m

axim

um

or

a m

inim

um

.S

ince

th

e co

effi

cien

t of

th

e x2 -

term

is

posi

tive

, th

e pa

rabo

la o

pen

s u

pwar

d, a

nd

the

vert

ex i

s a

min

imu

m p

oin

t.

d. G

rap

h t

he

fun

ctio

n.

x

y O( -

1, -

1)

x =

-1

Exer

cise

sC

onsi

der

eac

h e

qu

atio

n. D

eter

min

e w

het

her

th

e fu

nct

ion

has

ma

xim

um

or

min

imu

m v

alu

e. S

tate

th

e m

axim

um

or

min

imu

m v

alu

e an

d t

he

dom

ain

an

d r

ange

of

th

e fu

nct

ion

. Fin

d t

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

. Gra

ph

th

e fu

nct

ion

.

9-1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Gra

ph

ing

Qu

ad

rati

c F

un

cti

on

s

Exam

ple

Axi

s o

fS

ymm

etry

Fo

r th

e p

ara

bo

la y

= a

x2 +

bx

+ c

, w

he

re a

≠ 0

,

the

lin

e x

= -

b

−

2a i

s t

he

axis

of

sym

me

try.

Exa

mp

le:

Th

e a

xis

of

sym

me

try o

f

y =

x2 +

2x +

5 is t

he

lin

e x

= -

1.

Con

sid

er t

he

grap

h o

f y

= 2

x2 +

4x

+ 1

.

1. y

= x

2 +

3

2. y

= -

x2 -

4x

- 4

3.

y =

x2

+ 2

x +

3

x

y

O

x

y

O

x

y

O

min

.; (0

, 3);

D

: {al

l rea

l nu

mb

ers}

, R

: {y |

y ≥

3}; x

= 0

max

.; (-

2, 0

);

D:

{al

l rea

l nu

mb

ers}

, R

: {y |

y ≤

0}; x

= -

2

min

.; (-

1, 2

);

D: {

all r

eal n

um

ber

s},

R: {

y |

y ≥

2};

x =

-1

001-

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NA

ME

DAT

E

P

ER

IOD

Lesson 9-1

Cha

pte

r 9

7 G

len

co

e A

lge

bra

1

Skill

s Pr

acti

ceG

rap

hin

g Q

uad

rati

c F

un

cti

on

s

Use

a t

able

of

valu

es t

o gr

aph

eac

h f

un

ctio

n. S

tate

th

e d

omai

n a

nd

th

e ra

nge

.

1. y

= x

2 - 4

2.

y =

-x2

+ 3

3.

y =

x2

- 2

x -

6

x

y

O

x

y

O

x

y

O

Fin

d t

he

vert

ex, t

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

, an

d t

he

y-in

terc

ept

of t

he

grap

h o

f ea

ch f

un

ctio

n.

4. y

= 2

x2 -

8x

+ 6

5.

y =

x2

+ 4

x +

6

6. y

= -

3x2

- 1

2x +

3

Con

sid

er e

ach

eq

uat

ion

.

a. D

eter

min

e w

het

her

th

e fu

nct

ion

has

a m

axi

mu

m o

r a

min

imu

m v

alu

e.

b. S

tate

th

e m

axim

um

or

min

imu

m v

alu

e.

c. W

hat

are

th

e d

omai

n a

nd

ran

ge o

f th

e fu

nct

ion

?

7. y

= 2

x2 8.

y =

x2

- 2

x -

5

9. y

= -

x2 +

4x

- 1

Gra

ph

eac

h f

un

ctio

n.

10. f

(x)

= -

x2 -

2x

+ 2

11

. f(x

) =

2x2

+ 4

x -

2

12. f

(x)

= -

2x2

- 4

x +

6

xO

f(x)

xO

f(x)

xO

f(x)

9-1

D =

{al

l rea

l nu

mb

ers}

R =

{y

| y ≥

– 4

} D

= {

all r

eal n

um

ber

s}

R =

{y

| y ≤

3}

D =

{al

l rea

l nu

mb

ers}

R =

{y

| y ≥

– 7

}

(2, -

2 ) ;

x =

2;

(0, 6

) (-

2, 2

) ; x

= -

2; (

0, 6

) (-

2, 1

5 ) ;

x =

-2;

(0,

3)

min

imu

m; (

0, 0

) ;

D =

{al

l rea

l nu

mb

ers}

, R

= {

y | y

≥ 0

}

min

imu

m; (

1, -

6 ) ;

D =

{al

l rea

l nu

mb

ers}

, R

= {

y | y

≥ -

6 }

max

imu

m; (

2, 3

) ;

D =

{al

l rea

l nu

mb

ers}

, R

= {

y | y

≤ 3

}

001-

010_

ALG

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_CR

M_C

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/23/

10

9:42

PM

Answers (Lesson 9-1)


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

8 G

len

co

e A

lge

bra

1

Prac

tice

G

rap

hin

g Q

uad

rati

c F

un

cti

on

sU

se a

tab

le o

f va

lues

to

grap

h e

ach

fu

nct

ion

. Det

erm

ine

the

dom

ain

an

d r

ange

.

1. y

= -

x2 +

2

2. y

= x

2 -

6x

+ 3

3.

y =

-2x

2 -

8x

- 5

x

y

O

x

y

O

D

: {al

l rea

l nu

mb

ers}

D

: {al

l rea

l nu

mb

ers}

D:

{al

l rea

l nu

mb

ers}

R:

{y |

y ≤

2}

R:

{y |

y ≥

-6}

R:

{y |

y ≤

3}

Fin

d t

he

vert

ex, t

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

, an

d t

he

y-in

terc

ept

of t

he

grap

h o

f ea

ch f

un

ctio

n.

4. y

= x

2 -

9

5. y

= -

2x2

+ 8

x -

5

6. y

= 4

x2 -

4x

+ 1

(

0, -

9); x

= 0

; (0,

-9)

(

2, 3

); x

= 2

; (0,

-5)

(

0.5,

0);

x =

0.5

; (0,

1)

Con

sid

er e

ach

eq

uat

ion

. Det

erm

ine

wh

eth

er t

he

fun

ctio

n h

as a

ma

xim

um

or

a m

inim

um

val

ue.

Sta

te t

he

max

imu

m o

r m

inim

um

val

ue.

Wh

at a

re t

he

dom

ain

an

d r

ange

of

the

fun

ctio

n?

7. y

= 5

x2 -

2x

+ 2

8.

y =

-x2

+ 5

x -

10

9. y

= 3 −

2 x2

+ 4

x -

9

Gra

ph

eac

h f

un

ctio

n.

10. f

(x)

= -

x2 +

1

11. f

(x)

= -

2x2

+ 8

x -

3

12. f

(x)

= 2

x2 +

8x

+ 1

xO

f(x)

xO

f(x)

xO

f(x)

13. B

ASE

BA

LL T

he e

quat

ion

h =

-0.

005x

2 +

x +

3 d

escr

ibes

the

pat

h of

a b

aseb

all h

it in

to

the

outf

ield

, whe

re h

is t

he h

eigh

t an

d x

is t

he h

oriz

onta

l dis

tanc

e th

e ba

ll t

rave

ls.

a.

Wh

at i

s th

e eq

uat

ion

of

the

axis

of

sym

met

ry?

x =

10

0

b

. W

hat

is

the

max

imu

m h

eigh

t re

ach

ed b

y th

e ba

seba

ll?

53 f

t

c. A

n o

utf

ield

er c

atch

es t

he

ball

th

ree

feet

abo

ve t

he

grou

nd.

How

far

has

th

e ba

ll

trav

eled

hor

izon

tall

y w

hen

th

e ou

tfie

lder

cat

ches

it?

20

0 ft

9-1

x

y O

min

.; (0

.2, 1

.8);

D

: {al

l rea

l nu

mb

ers}

, R

: {y |

y ≥

1.8

}

max

.; (2

.5, -

3.75

);

D:

{al

l rea

l nu

mb

ers}

, R

: {y |

y ≤

-3.

75}

min

.; (-

1 1 −

3 ,

-11

2 −

3 ; )

D:

{al

l rea

l nu

mb

ers}

,

R:

{y |

y ≥

-11

2 −

3 }

001-

010_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

812

/23/

10

9:43

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 9-1

Cha

pte

r 9

9 G

len

co

e A

lge

bra

1

1.O

LYM

PIC

S O

lym

pics

wer

e h

eld

in

1896

an

d h

ave

been

hel

d ev

ery

fou

r ye

ars

exce

pt 1

916,

194

0, a

nd

1944

. Th

e w

inn

ing

hei

ght

y in

men

’s p

ole

vau

lt a

t an

y n

um

ber

Oly

mpi

ad x

can

be

app

roxi

mat

ed b

y th

e eq

uat

ion

y

= 0

.37x

2 + 4

.3x

+ 1

26. C

ompl

ete

the

tabl

e to

est

imat

e th

e po

le v

ault

hei

ghts

in

eac

h o

f th

e O

lym

pic

Gam

es. R

oun

d yo

ur

answ

ers

to t

he

nea

rest

ten

th.

Sour

ce: N

atio

nal S

ecur

ity A

genc

y

2. P

HY

SIC

S M

rs. C

apw

ell’s

ph

ysic

s cl

ass

inve

stig

ates

wh

at h

appe

ns

wh

en a

bal

l is

gi

ven

an

in

itia

l pu

sh, r

olls

up,

an

d th

en

back

dow

n a

n i

ncl

ined

pla

ne.

Th

e cl

ass

fin

ds t

hat

y =

-x2 +

6x

accu

rate

ly

pred

icts

th

e ba

ll’s

pos

itio

n y

aft

er r

olli

ng

x se

con

ds. O

n t

he

grap

h o

f th

e eq

uat

ion

, w

hat

wou

ld b

e th

e y

valu

e w

hen

x =

4?

8

3. A

RC

HIT

ECTU

RE

A h

otel

’s m

ain

en

tran

ce i

s in

th

e sh

ape

of a

par

abol

ic

arch

. Th

e eq

uat

ion

y =

-x2 +

10x

mod

els

the

arch

hei

ght

y fo

r an

y di

stan

ce x

fro

m

one

side

of

the

arch

. Use

a g

raph

to

dete

rmin

e it

s m

axim

um

hei

ght.

25 f

t

4.SO

FTB

ALL

Oly

mpi

c so

ftba

ll g

old

med

alis

t M

ich

ele

Sm

ith

pit

ches

a

curv

ebal

l w

ith

a s

peed

of

64 f

eet

per

seco

nd.

If

she

thro

ws

the

ball

str

aigh

t u

pwar

d at

th

is s

peed

, th

e ba

ll’s

hei

ght

h i

n f

eet

afte

r t

seco

nds

is

give

n b

y h

= -

16t2 +

64t

. Fin

d th

e co

ordi

nat

es o

f th

e ve

rtex

of

the

grap

h o

f th

e ba

ll’s

h

eigh

t an

d in

terp

ret

its

mea

nin

g.

(2, 6

4); A

fter

2 s

eco

nd

s, t

he

bal

l re

ach

es it

s h

igh

est

po

int,

64 f

eet

abov

e th

e g

rou

nd

.

5. G

EOM

ETRY

Ted

dy i

s bu

ildi

ng

the

rect

angu

lar

deck

sh

own

bel

ow.

x +

6

x -

2

a. W

rite

an

equ

atio

n r

epre

sen

tin

g th

e ar

ea o

f th

e de

ck y

.

y =

(x -

2)(

x +

6)

or

y =

x2 +

4x -

12

b. W

hat

is

the

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

? x

= -

2

c. G

raph

th

e eq

uat

ion

an

d la

bel

its

vert

ex.

yx

O

-4

-6

-8

-10

-12

-14

-16

-2

1-

3-

4-

5

( -2,

-16

)

9-1

Wor

d Pr

oble

m P

ract

ice

Gra

ph

ing

Qu

ad

rati

c F

un

cti

on

s

Year

Oly

mp

iad

(x

)H

eig

ht

(y in

ches

)

18

96

1

19

00

2

19

24

7

19

36

10

19

64

15

20

08

26

130.

713

6.1

174.

220

6.0

273.

848

7.9

001-

010_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

912

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

10

Gle

nc

oe

Alg

eb

ra 1

A c

ub

ic f

un

ctio

n i

s a

poly

nom

ial

wri

tten

in

th

e fo

rm o

f f(

x) =

ax3

+ b

x2 +

cx

+ n

, w

her

e a

≠ 0

. Cu

bic

fun

ctio

ns

do n

ot h

ave

abso

lute

min

imu

m a

nd

max

imu

m v

alu

es l

ike

quad

rati

c fu

nct

ion

s do

, bu

t th

ey c

an h

ave

a lo

cal

min

imu

m a

nd

a lo

cal

max

imu

m p

oin

t.

Pa

ren

t F

un

ctio

n: f(

x) =

x3

Do

ma

in: {a

ll re

al nu

mb

ers

}

Ra

ng

e: {a

ll re

al nu

mb

ers

}

x

f(x)

U

se a

tab

le o

f va

lues

to

grap

h y

= x

3 +

3x2

- 1

. Th

en u

se t

he

grap

h

to e

stim

ate

the

loca

tion

s of

th

e lo

cal

min

imu

m a

nd

loc

al m

axim

um

poi

nts

.

x–3

–2

–1

01

y–1

21

–1

2

Gra

ph t

he

orde

red

pair

s, a

nd

con

nec

t th

em t

o cr

eate

a s

moo

th c

urv

e.

Th

e en

d be

hav

ior

of t

he

“S”

shap

ed c

urv

e sh

ows

that

as

x in

crea

ses,

y in

crea

ses,

an

d as

x d

ecre

ases

, y d

ecre

ases

.T

he

loca

l m

inim

um

is

loca

ted

at (

0, –

1). T

he

loca

l m

axim

um

is

loca

ted

at (

–2, 2

).

Exer

cise

sU

se a

tab

le o

f va

lues

to

grap

h e

ach

eq

uat

ion

. Th

en u

se t

he

grap

h t

o es

tim

ate

the

loca

tion

s of

th

e lo

cal

min

imu

m a

nd

loc

al m

axim

um

poi

nts

.

1. y

= 0

.5x3

+ x

2 -

1

2. y

= -

2x3

- 3

x2 -

1

3. y

= x

3 +

3x2

+ x

- 4

x

y

x

y

x

y

l

oca

l max

imu

m:

l

oca

l max

imu

m:

l

oca

l max

imu

m:

(-1.

3, -

0.4)

; (

0, -

1);

(-

1.8,

-1.

9);

loca

l min

imu

m:

l

oca

l min

imu

m:

l

oca

l min

imu

m:

(0, 1

) (

-1,

-2)

(

-0.

2, -

4.1)

9-1

Enri

chm

ent

Gra

ph

ing

Cu

bic

Fu

ncti

on

s

Exam

ple

x

y

(-2,

2)

(0, -

1)

001-

010_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1012

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-2

Cha

pte

r 9

11

Gle

ncoe

Alg

ebra

1

Solv

e b

y G

rap

hin

g

Qu

adra

tic

Eq

uat

ion

an

eq

ua

tio

n o

f th

e fo

rm a

x2 +

bx +

c =

0,

wh

ere

a ≠

0

Th

e so

luti

ons

of a

qu

adra

tic

equ

atio

n a

re c

alle

d th

e ro

ots

of t

he

equ

atio

n. T

he

root

s of

a

quad

rati

c eq

uat

ion

can

be

fou

nd

by g

raph

ing

the

rela

ted

quad

rati

c fu

nct

ion

f(

x) =

ax2

+ b

x +

c a

nd

fin

din

g th

e x-

inte

rcep

ts o

r ze

ros

of t

he

fun

ctio

n.

S

olve

x2

+ 4

x +

3 =

0 b

y gr

aph

ing.

Gra

ph t

he

rela

ted

fun

ctio

n f

(x)

= x

2 +

4x

+ 3

. T

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

is

x =

-

4 −

2(

1) o

r -

2. T

he

vert

ex i

s at

(-

2, -

1).

Gra

ph t

he

vert

ex a

nd

seve

ral

oth

er p

oin

ts o

n

eith

er s

ide

of t

he

axis

of

sym

met

ry.

xO

f(x)

To

solv

e x2

+ 4

x +

3 =

0, y

ou n

eed

to k

now

w

her

e f(

x) =

0. T

his

occ

urs

at

the

x-in

terc

epts

, -3

and

-1.

Th

e so

luti

ons

are

-3

and

-1.

S

olve

x2

- 6

x +

9 =

0 b

y gr

aph

ing.

Gra

ph t

he

rela

ted

fun

ctio

n f

(x)

= x

2 -

6x

+ 9

. T

he

equ

atio

n o

f th

e ax

is o

f sy

mm

etry

is

x =

6

−

2(1)

or

3. T

he

vert

ex i

s at

(3,

0).

Gra

ph

the

vert

ex a

nd

seve

ral

oth

er p

oin

ts o

n e

ith

er

side

of

the

axis

of

sym

met

ry.

x

f(x) O

To

solv

e x2

- 6

x +

9 =

0, y

ou n

eed

to k

now

w

her

e f(

x) =

0. T

he

vert

ex o

f th

e pa

rabo

la i

s th

e x-

inte

rcep

t. T

hu

s, t

he

only

sol

uti

on i

s 3.

Exer

cise

sS

olve

eac

h e

qu

atio

n b

y gr

aph

ing.

1. x

2 +

7x

+ 1

2 =

0

2. x

2 -

x -

12

= 0

3.

x2

- 4

x +

5 =

0

xO

f(x)

xO4

-4

-8

-12

8-4

-8

f(x)

4

xO

f(x)

-

3, -

4 4

, -3

no

rea

l ro

ots

9-2

Stud

y G

uide

and

Inte

rven

tion

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

Exam

ple

1Ex

amp

le 2

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1112

/23/

10

10:5

7 A

M

Answers (Lesson 9-1 and Lesson 9-2)


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

12

Gle

ncoe

Alg

ebra

1

9-2

Esti

mat

e So

luti

on

s T

he

root

s of

a q

uad

rati

c eq

uat

ion

may

not

be

inte

gers

. If

exac

t ro

ots

can

not

be

fou

nd,

th

ey c

an b

e es

tim

ated

by

fin

din

g th

e co

nse

cuti

ve i

nte

gers

bet

wee

n

wh

ich

th

e ro

ots

lie.

S

olve

x2

+ 6

x +

6 =

0 b

y gr

aph

ing.

If

inte

gral

roo

ts c

ann

ot b

e fo

un

d,

esti

mat

e th

e ro

ots

by

stat

ing

the

con

secu

tive

in

tege

rs b

etw

een

wh

ich

th

e ro

ots

lie.

Gra

ph t

he

rela

ted

fun

ctio

n f

(x)

= x

2 +

6x

+ 6

.

xf(

x)

-5

1

-4

-2

-3

-3

-2

-2

-1

1

Not

ice

that

th

e va

lue

of t

he

fun

ctio

n c

han

ges

x

f(x)

O

from

neg

ativ

e to

pos

itiv

e be

twee

n t

he

x-va

lues

of

-5

and

-4

and

betw

een

-2

and

-1.

Th

e x-

inte

rcep

ts o

f th

e gr

aph

are

bet

wee

n -

5 an

d -

4 an

d be

twee

n -

2 an

d -

1.

So

one

root

is

betw

een

-5

and

-4,

an

d th

e ot

her

roo

t is

bet

wee

n -

2 an

d -

1.

Exer

cise

sS

olve

eac

h e

qu

atio

n b

y gr

aph

ing.

If

inte

gral

roo

ts c

ann

ot b

e fo

un

d, e

stim

ate

the

root

s to

th

e n

eare

st t

enth

.

1. x

2 +

7x

+ 9

= 0

2.

x2

- x

- 4

= 0

3.

x2

- 4

x +

6 =

0

xO

f(x)

xO

f(x

)

xO

f(x)

-

6 <

x <

-5,

-

2 <

x <

-1,

n

o r

eal r

oo

ts-

2 <

x <

-1

2 <

x <

3

4. x

2 -

4x

- 1

= 0

5.

4x2

- 1

2x +

3 =

0

6. x

2 -

2x

- 4

= 0

xO

f(x)

xO

f(x)

xO

f(x)

-

1 <

x <

0,

0 <

x <

1,

-2

< x

< -

1,4

< x

< 5

2

< x

< 3

3

< x

< 4

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

Exam

ple

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1212

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-2

Cha

pte

r 9

13

Gle

ncoe

Alg

ebra

1

Sol

ve e

ach

eq

uat

ion

by

grap

hin

g.

1. x

2 -

2x

+ 3

= 0

∅

2. c

2 +

6c

+ 8

= 0

-4,

-2

xO

f(x)

cO

f(c)

3. a

2 -

2a

= -

1 1

4. n

2 -

7n

= -

10 2

, 5

aO

f(a)

nO

f(n)

Sol

ve e

ach

eq

uat

ion

by

grap

hin

g. I

f in

tegr

al r

oots

can

not

be

fou

nd

, es

tim

ate

the

root

s to

th

e n

eare

st t

enth

.

5. p

2 +

4p

+ 2

= 0

6.

x2

+ x

- 3

= 0

pO

f(p)

xO

f(x)

-

3.4,

-0.

6 -

2.3,

1.3

7. d

2 +

6d

= -

3 8.

h2

+ 1

= 4

h

dO

f(d

)

hO

f(h)

-

5.4,

-0.

6 0

.3, 3

.7

9-2

Skill

s Pr

acti

ceS

olv

ing

Qu

ad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1312

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

14

Gle

ncoe

Alg

ebra

1

Sol

ve e

ach

eq

uat

ion

by

grap

hin

g.

1. x

2 -

5x

+ 6

= 0

2, 3

2.

w2

+ 6

w +

9 =

0 -

3 3.

b2

- 3

b +

4 =

0 ∅

xO

f(x)

wO

f(w

)

bOf(

b)

Sol

ve e

ach

eq

uat

ion

by

grap

hin

g. I

f in

tegr

al r

oots

can

not

be

fou

nd

, est

imat

e th

e ro

ots

to t

he

nea

rest

ten

th.

4. p

2 +

4p

= 3

5.

2m

2 +

5 =

10m

6.

2v2

+ 8

v =

-7

pO

f(p)

mO

f(m

)

vO

f(v)

-

4.6,

0.6

0

.6, 4

.4

-2.

7, -

1.3

7. N

UM

BER

TH

EORY

Tw

o n

um

bers

hav

e a

sum

of

2an

d a

prod

uct

of

-8.

Th

e qu

adra

tic

equ

atio

n

-n

2 +

2n

+ 8

= 0

can

be

use

d to

det

erm

ine

the

two

nu

mbe

rs.

a. G

raph

th

e re

late

d fu

nct

ion

f(n

) =

-n

2 +

2n

+ 8

an

d de

term

ine

its

x-in

terc

epts

. -

2, 4

b.

Wh

at a

re t

he

two

nu

mbe

rs?

-2

and

4

8. D

ESIG

N A

foo

tbri

dge

is s

usp

ende

d fr

om a

par

abol

ic

supp

ort.

Th

e fu

nct

ion

h(x

) =

-

1

−

25 x2

+ 9

rep

rese

nts

the

heig

ht in

fee

t of

the

sup

port

abo

ve t

he w

alkw

ay,

wh

ere

x =

0 r

epre

sen

ts t

he

mid

poin

t of

th

e br

idge

.

a. G

raph

th

e fu

nct

ion

an

d de

term

ine

its

x-in

terc

epts

. -

15, 1

5

b.

Wh

at i

s th

e le

ngt

h o

f th

e w

alkw

ay b

etw

een

th

e tw

o su

ppor

ts?

30 f

t

xO

612

12 6

-6

-12

-12

-6

h(x

)

9-2

Prac

tice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

nO

f(n)

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1412

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-2

Cha

pte

r 9

15

Gle

ncoe

Alg

ebra

1

1.FA

RM

ING

In

ord

er f

or M

r. M

oore

to

deci

de h

ow m

uch

fer

tili

zer

to a

pply

to

his

co

rn c

rop

this

yea

r, h

e re

view

s re

cord

s fr

om p

revi

ous

year

s. H

is c

rop

yiel

d y

depe

nds

on

th

e am

oun

t of

fer

tili

zer

he

appl

ies

to h

is f

ield

s x

acco

rdin

g to

th

e eq

uat

ion

y =

-x2 +

4x

+ 1

2. G

raph

th

e fu

nct

ion

, an

d fi

nd

the

poin

t at

wh

ich

M

r. M

oore

get

s th

e h

igh

est

yiel

d po

ssib

le.

2. L

IGH

T A

yzh

a an

d Je

rem

y h

old

a fl

ash

ligh

t so

th

at t

he

ligh

t fa

lls

on a

pi

ece

of g

raph

pap

er i

n t

he

shap

e of

a

para

bola

. Ayz

ha

and

Jere

my

sket

ch t

he

shap

e of

th

e pa

rabo

la a

nd

fin

d th

at t

he

equ

atio

n y

= x

2 - 3

x -

10

mat

ches

th

e sh

ape

of t

he

ligh

t be

am. D

eter

min

e th

e ze

ros

of t

he

fun

ctio

n.

-2

and

5

3. F

RA

MIN

G A

rec

tan

gula

r ph

otog

raph

is

7 in

ches

lon

g an

d 6

inch

es w

ide.

Th

e ph

otog

raph

is

fram

ed u

sin

g a

mat

eria

l th

at i

s x

inch

es w

ide.

If

the

area

of

the

fram

e an

d ph

otog

raph

com

bin

ed i

s 15

6 sq

uar

e in

ches

, wh

at i

s th

e w

idth

of

the

fram

ing

mat

eria

l?

3 in

.

Phot

ogra

ph

Fram

e

6 in

.

7 in

.

x

x

4. W

RA

PPIN

G P

APE

R C

an a

rec

tan

gula

r pi

ece

of w

rapp

ing

pape

r w

ith

an

are

a of

81

squ

are

inch

es h

ave

a pe

rim

eter

of

60 i

nch

es?

(Hin

t: L

et l

engt

h =

30

– w

.)

Exp

lain

. Y

es;

solv

ing

th

e eq

uat

ion

(3

0 -

w)w

= 8

1 g

ives

w =

3 o

r 27

. A

3 in

. by

27 in

. sh

eet

of

pap

er

has

are

a 81

in2

and

per

imet

er

60 in

.

5. E

NG

INEE

RIN

G T

he

shap

e of

a s

atel

lite

di

sh i

s of

ten

par

abol

ic b

ecau

se o

f th

e re

flec

tive

qu

alit

ies

of p

arab

olas

. Su

ppos

e a

part

icu

lar

sate

llit

e di

sh i

s m

odel

ed b

y th

e fo

llow

ing

equ

atio

n.

0.

5x2 =

2 +

y

a. A

ppro

xim

ate

the

zero

s of

th

is f

un

ctio

n

by g

raph

ing.

-

2 an

d 2

y

xO34 2 1

-2

-3

-4

43

21

-2

-3

-4

AB

b. O

n t

he

coor

din

ate

plan

e ab

ove,

tr

ansl

ate

the

para

bola

so

that

th

ere

is

only

on

e ze

ro. L

abel

th

is c

urv

e A

.

See

stu

den

ts’ w

ork

.

c. T

ran

slat

e th

e pa

rabo

la s

o th

at t

her

e ar

e n

o ze

ros.

Lab

el t

his

cu

rve

B.

S

ee s

tud

ents

’ wo

rk.

9-2

Wor

d Pr

oble

m P

ract

ice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y G

rap

hin

g

y

xO1416 12 10 8 6 4 2

45

32

1

( 2,1

6)

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1512

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

16

Gle

ncoe

Alg

ebra

1

Para

bo

las T

hro

ug

h T

hre

e G

iven

Po

ints

If y

ou k

now

tw

o po

ints

on

a s

trai

ght

lin

e, y

ou c

an f

ind

the

equ

atio

n o

f th

e li

ne.

To

fin

d th

e eq

uat

ion

of

a pa

rabo

la, y

ou n

eed

thre

e po

ints

on

th

e cu

rve.

Her

e is

how

to

appr

oxim

ate

an e

quat

ion

of

the

para

bola

th

rou

gh t

he

poin

ts (

0, -

2), (

3, 0

), an

d (5

, 2).

Use

th

e ge

ner

al e

quat

ion

y =

ax2

+ b

x +

c. B

y su

bsti

tuti

ng

the

give

n

valu

es f

or x

an

d y,

you

get

th

ree

equ

atio

ns.

(0, -

2):

-2

= c

(3, 0

):

0 =

9a

+ 3

b +

c(5

, 2):

2 =

25a

+ 5

b +

c

Fir

st, s

ubs

titu

te -

2 fo

r c

in t

he

seco

nd

and

thir

d eq

uat

ion

s.

Th

en s

olve

th

ose

two

equ

atio

ns

as y

ou w

ould

an

y sy

stem

of

two

equ

atio

ns.

M

ult

iply

th

e se

con

d eq

uat

ion

by

5 an

d th

e th

ird

equ

atio

n b

y -

3.0

= 9

a +

3b

- 2

M

ultip

ly b

y 5

. 0

=

45a

+ 1

5b -

10

2 =

25a

+ 5

b -

2

Multip

ly b

y -

3.

-6

= -

75a

- 1

5b +

6

-6

= -

30a

-

4

a

=

1 −

15

To

fin

d b,

su

bsti

tute

1 −

15

for

a i

n e

ith

er t

he

seco

nd

or t

hir

d eq

uat

ion

.

0 =

9 ( 1 −

15

) + 3

b -

2

b =

7 −

15

Th

e eq

uat

ion

of

a pa

rabo

la t

hro

ugh

th

e th

ree

poin

ts i

s

y =

1 −

15

x2 +

7 −

15

x -

2.

Fin

d t

he

equ

atio

n o

f a

par

abol

a th

rou

gh e

ach

set

of

thre

e p

oin

ts.

1. (

1, 5

), (0

, 6),

(2, 3

) 2.

(-

5, 0

), (0

, 0),

(8, 1

00)

y

= -

1 −

2 x

2 -

1 −

2 x

+ 6

y =

25

−

26 x

2 +

125

−

26 x

3. (

4, -

4), (

0, 1

), (3

, -2)

4.

(1,

3),

(6, 0

), (0

, 0)

y

= -

1 −

4 x

2 -

1 −

4 x

+ 1

y

= -

3 −

5 x

2 +

18

−

5 x

5. (

2, 2

), (5

, -3)

, (0,

-1)

6.

(0,

4),

(4, 0

), (-

4, 4

)

y

= -

19

−

30 x

2 +

83

−

30 x

- 1

y

= -

1 −

8 x

2 -

1 −

2 x

+ 4

9-2

Enri

chm

ent

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1612

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-3

Cha

pte

r 9

17

Gle

ncoe

Alg

ebra

1

9-3

Tran

slat

ion

s A

tra

nsl

atio

n i

s a

chan

ge i

n t

he

posi

tion

of

a fi

gure

eit

her

up,

dow

n,

left

, rig

ht,

or

diag

onal

. Add

ing

or s

ubt

ract

ing

con

stan

ts i

n t

he

equ

atio

ns

of f

un

ctio

ns

tran

slat

es t

he

grap

hs

of t

he

fun

ctio

ns.

Th

e g

rap

h o

f g

(x)

= x

2 +

k t

ran

sla

tes t

he

gra

ph

of

f(x)

= x

2 v

ert

ica

lly.

If k

> 0

, th

e g

rap

h o

f f(

x)

= x

2 is t

ran

sla

ted

k u

nits u

p.

If k

< 0

, th

e g

rap

h o

f f(

x)

= x

2 is t

ran

sla

ted

|k

| u

nits d

ow

n.

Th

e g

rap

h o

f g

(x)

= (

x -

h) 2

is t

he

gra

ph

of

f(x)

= x

2 t

ran

sla

ted

ho

rizo

nta

lly.

If h

> 0

, th

e g

rap

h o

f f(

x)

= x

2 is t

ran

sla

ted

h u

nits t

o t

he

rig

ht.

If h

< 0

, th

e g

rap

h o

f f(

x)

= x

2 is t

ran

sla

ted

|h

| u

nits t

o t

he

le

ft.

Stud

y G

uide

and

Inte

rven

tion

Tran

sfo

rmati

on

s o

f Q

uad

rati

c F

un

cti

on

s

D

escr

ibe

how

th

e gr

aph

of

each

fu

nct

ion

is

rela

ted

to

the

grap

h o

f f(

x) =

x2 .

Exam

ple

a. g

(x)

= x

2 +

4T

he

valu

e of

k i

s 4,

an

d 4

> 0

. Th

eref

ore,

th

e gr

aph

of

g(x)

= x

2 +

4 i

s a

tran

slat

ion

of

th

e gr

aph

of

f(x)

= x

2 u

p 4

un

its.

g(x)

f(x)

x

b.

g(x)

= (

x +

3) 2

Th

e va

lue

of h

is

–3,

an

d –3

< 0.

Th

us,

th

e gr

aph

of

g(x)

= (

x +

3) 2

is a

tra

nsl

atio

n o

f th

e gr

aph

of

f(x)

= x

2 to

th

e le

ft 3

un

its.

y

xO

f(x)

g(x)

Exer

cise

sD

escr

ibe

how

th

e gr

aph

of

each

fu

nct

ion

is

rela

ted

to

the

grap

h o

f f(

x) =

x2 .

1. g

(x)

= x

2 +

1

2. g

(x)

= (

x – 6

) 2 3.

g(x

) =

(x

+ 1

) 2

T

ran

slat

ion

of

T

ran

slat

ion

of

f(x)

= x

2

Tran

slat

ion

of

f(x)

= x

2

f(x)

= x

2 u

p 1

un

it.

to

th

e ri

gh

t 6

un

its.

to t

he

left

1 u

nit

.

4. g

(x)

= 2

0 +

x2

5. g

(x)

= (

–2

+ x

) 2 6.

g(x

) =

- 1 −

2 + x

2

T

ran

slat

ion

of

Tran

slat

ion

of

f(x)

= x

2

Tran

slat

ion

of

f(x)

= x

2 u

p 2

0 u

nit

s.

to

th

e ri

gh

t 2

un

its.

f(x)

= x

2 d

ow

n 1

−

2 un

it.

7. g

(x)

= x

2 +

8 −

9

8. g

(x)

= x

2 – 0

.3

9. g

(x)

= (

x +

4) 2

T

ran

slat

ion

of

Tran

slat

ion

of

Tran

slat

ion

of

f(x)

= x

2

f(x)

= x

2 u

p 8

−

9 un

it.

f(

x)

= x

2 d

ow

n 0

.3 u

nit

.

to t

he

left

4 u

nit

s.

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1712

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

18

Gle

ncoe

Alg

ebra

1

Dila

tio

ns

and

Ref

lect

ion

s A

dil

atio

n i

s a

tran

sfor

mat

ion

th

at m

akes

th

e gr

aph

n

arro

wer

or

wid

er t

han

th

e pa

ren

t gr

aph

. A r

efle

ctio

n f

lips

a f

igu

re o

ver

the

x- o

r y-

axis

.

Th

e g

rap

h o

f f(

x)

= a

x2 s

tre

tch

es o

r co

mp

resse

s t

he

gra

ph

of

f(x)

= x

2.

If |

a|

> 1

, th

e g

rap

h o

f f(

x)

= x

2 is s

tre

tch

ed

ve

rtic

ally

.

If 0

< |

a|

< 1

, th

e g

rap

h o

f f(

x)

= x

2 is c

om

pre

sse

d v

ert

ica

lly.

Th

e g

rap

h o

f th

e f

un

ctio

n –

f(x)

fl ip

s t

he

gra

ph

of

f(x)

= x

2 a

cro

ss t

he

x-a

xis

.

Th

e g

rap

h o

f th

e f

un

ctio

n f

(–x)

fl ip

s t

he

gra

ph

of

f(x)

= x

2 a

cro

ss t

he

y-a

xis

.

D

escr

ibe

how

th

e gr

aph

of

each

fu

nct

ion

is

rela

ted

to

the

grap

h o

f f(

x) =

x2 .

9-3

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Tran

sfo

rmati

on

s o

f Q

uad

rati

c F

un

cti

on

s

x

0 <

a <

1

a =

1a

> 1

x

y O

f(x) =

-x2

f(x) =

x2

Exam

ple

a. g

(x)

= 2

x2

Th

e fu

nct

ion

can

be

wri

tten

as

f(x)

= a

x2 w

here

a =

2. B

ecau

se |a

| > 1

, the

gra

ph o

f y

= 2

x2 is

th

e gr

aph

of

y =

x2

that

is

stre

tch

ed v

erti

call

y.

b. g

(x)

= -

1 −

2 x

2 -

3

Th

e n

egat

ive

sign

cau

ses

a re

flec

tion

ac

ross

th

e x-

axis

. Th

en a

dil

atio

n o

ccu

rs

in w

hic

h a

= 1 −

2 a

nd

a tr

ansl

atio

n i

n w

hic

h

k =

–3.

So

the

grap

h

of g

(x)

= -

1 −

2 x2

- 3

is

ref

lect

ed a

cros

s th

e x-

axis

, dil

ated

wid

er

than

th

e gr

aph

of

f(x)

= x

2 , an

d tr

ansl

ated

do

wn

3 u

nit

s.

Exer

cise

sD

escr

ibe

how

th

e gr

aph

of

each

fu

nct

ion

is

rela

ted

to

the

grap

h o

f f(

x) =

x2 .

1. g

(x)

= -

5x2

2. g

(x)

= -

(x +

1) 2

3. g

(x)

= -

1 −

4 x2

- 1

C

om

pre

ssio

n o

f

Tran

slat

ion

of

f(x)

= x

2 D

ilati

on

of

f(x)

= x

2

f(x)

= x

2 n

arro

wer

th

an

to t

he

left

1 u

nit

an

d

wid

er t

han

th

e g

rap

h o

fth

e g

rap

h o

f f(

x)

= x

2

refl

ecte

d o

ver

the

x-a

xis.

f

(x)

= x

2 re

fl ec

ted

ove

rre

fl ec

ted

ove

r th

e

t

he

x-a

xis

tran

slat

edx-a

xis.

do

wn

1 u

nit

.

x

y O

f(x) =

x2

f(x) =

2x2

x

f(x)

g(x)

y O

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1812

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-3

Cha

pte

r 9

19

Gle

ncoe

Alg

ebra

1

Des

crib

e h

ow t

he

grap

h o

f ea

ch f

un

ctio

n i

s re

late

d t

o th

e gr

aph

of

f(x)

= x

2 .

1. g

(x)

= x

2 +

2

2. g

(x)

= (

x -

1) 2

3. g

(x)

= x

2 -

8

T

ran

slat

ion

of

f(x)

= x

2

Tra

nsl

atio

n o

f f(

x)

= x

2 T

ran

slat

ion

of

up

2 u

nit

s t

o t

he

rig

ht

1 u

nit

f

(x)

= x

2 d

own

8 u

nit

s

4. g

(x)

= 7

x2 5.

g(x

) =

1 −

5 x

2 6.

g(x

) =

-6x

2

S

tret

ch o

f f(

x)

= x

2 C

om

pre

ssio

n o

f f(

x)

= x

2

Str

etch

of

n

arro

wer

th

an t

he

gra

ph

w

ider

th

an t

he

gra

ph

of

f(x

) = x

2 nar

row

er th

an

of

f(x)

= x

2

f(x

) =

x2

th

e g

rap

h o

f f(

x)

= x

2

refl

ect

ed o

ver

the

x-a

xis

7. g

(x)

= -

x2 +

3

8. g

(x)

= 5

- 1 −

2 x

2 9.

g(x

) =

4 (x

- 1

) 2

R

efl e

ctio

n o

f f(

x)

= x

2

Com

pres

sion

of

f(x)

= x

2

Str

etch

of

over

th

e x-a

xis

w

ider

tha

n th

e gr

aph

of

f(x

) =

x2

nar

row

ertr

ansl

ated

up

3 u

nit

s f(

x)

= x

2 re

fl ec

ted

ove

r

th

an t

he

gra

ph

of

th

e x-a

xis,

an

d t

ran

slat

ed f

(x)

= x

2 tr

ansl

ated

u

p 5

un

its

to

th

e ri

gh

t 1

un

it

Mat

ch e

ach

eq

uat

ion

to

its

grap

h.

10. y

= 2

x2 -

2

11. y

= 1 −

2 x

2 -

2

12. y

= -

1 −

2 x

2 +

2

13. y

= -

2x2

+ 2

9-3

Skill

s Pr

acti

ceTr

an

sfo

rmati

on

s o

f Q

uad

rati

c F

un

cti

on

s

CB D A

x

y

x

y

x

y

x

yB

.

A.

D.

C.

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

1912

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

20

Gle

ncoe

Alg

ebra

1

Des

crib

e h

ow t

he

grap

h o

f ea

ch f

un

ctio

n i

s re

late

d t

o th

e gr

aph

of

f(x)

= x

2 .

1. g

(x)

= (

10 +

x)2

2. g

(x)

= -

2 −

5 +

x2

3. g

(x)

= 9

- x

2

T

ran

slat

ion

of

f(x)

= x

2 T

ran

slat

ion

of

f(x)

= x

2 R

efl e

ctio

n o

f

to

th

e le

ft 1

0 u

nit

s.

do

wn

2 −

5 u

nit

. f

(x)

= x

2 ac

ross

th

e

x

-axi

s tr

ansl

ated

up

9

un

its.

4. g

(x)

= 2

x2 +

2

5. g

(x)

= -

3 −

4 x

2 -

1 −

2

6. g

(x)

= -

3(x

+ 4

)2

S

tret

ch o

f f(

x)

= x

2 C

om

pre

ssio

n o

f f(

x)

= x

2 S

tret

ch o

fn

arro

wer

th

an t

he

wid

er t

han

th

e g

rap

h o

f

f(x

) =

x2

nar

row

er

gra

ph

of

f(x)

= x

2 f

(x)

= x

2 , re

fl ec

ted

ove

r t

han

th

e g

rap

h o

f tr

ansl

ated

up

2 u

nit

s.

th

e x-a

xis,

tra

nsl

ated

f

(x)

= x

2 , re

fl ec

ted

d

ow

n 1 −

2 u

nit

. o

ver

the

x-a

xis

tra

nsl

ated

to

th

e le

ft

4 u

nit

s.

Mat

ch e

ach

eq

uat

ion

to

its

grap

h.

A.

x

y

B.

x

y

C.

x

y

7. y

= -

3x2

- 1

8.

y =

1 −

3 x2

- 1

9.

y =

3x2

+ 1

9-3

Prac

tice

Tran

sfo

rmati

on

s o

f Q

uad

rati

c F

un

cti

on

s

AC

B

Lis

t th

e fu

nct

ion

s in

ord

er f

rom

th

e m

ost

vert

ical

ly s

tret

ched

to

the

leas

t ve

rtic

ally

str

etch

ed g

rap

h.

10. f

(x)

= 3

x2 , g(

x) =

1 −

2 x

2 , h

(x)

= -

2x2

11. f

(x)

= 1 −

2 x

2 , g(

x) =

- 1 −

6 x

2 , h

(x)

= 4

x2

f

(x),

h(x

), g

(x)

h(x

), f(

x),

g(x

)

12. P

AR

AC

HU

TIN

G T

wo

para

chu

tist

s ju

mp

at t

he

sam

e ti

me

from

tw

o di

ffer

ent

plan

es a

s pa

rt o

f an

aer

ial

show

. Th

e h

eigh

t h

1 of

th

e fi

rst

para

chu

tist

in

fee

t af

ter

t se

con

ds i

s m

odel

ed b

y th

e fu

nct

ion

h

1 =

-1

6t2

+ 5

000.

Th

e h

eigh

t h

2 of

th

e se

con

d pa

rach

uti

st i

n

feet

aft

er t

sec

onds

is

mod

eled

by

the

fun

ctio

n h

2 =

-1

6t2

+ 4

000.

a. W

hat

is

the

pare

nt

fun

ctio

n o

f th

e tw

o fu

nct

ion

s gi

ven

? h

= t

2

b.

Des

crib

e th

e tr

ansf

orm

atio

ns

nee

ded

to o

btai

n t

he

grap

h o

f h

1 fr

om t

he

pare

nt

fun

ctio

n.

Str

etch

of

y =

x2

nar

row

er t

han

th

e g

rap

h o

f f(

x)

= x

2 , re

fl ec

ted

ov

er t

he

x-a

xis,

tra

nsl

ated

up

50

00

un

its.

c. W

hic

h p

arac

hu

tist

wil

l re

ach

th

e gr

oun

d fi

rst?

th

e se

con

d p

arac

hu

tist

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2012

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-3

Cha

pte

r 9

21

Gle

ncoe

Alg

ebra

1

1. S

PRIN

GS

Th

e po

ten

tial

en

ergy

sto

red

in

a sp

rin

g is

giv

en b

y U

s =

1 −

2 k

x2 w

her

e k

is

a co

nst

ant

know

n a

s th

e sp

rin

g co

nst

ant,

and

x is

th

e di

stan

ce t

he

spri

ng

is

stre

tch

ed o

r co

mpr

esse

d fr

om i

ts i

nit

ial

posi

tion

. How

is

the

grap

h o

f th

e fu

nct

ion

fo

r a

spri

ng

wh

ere

k=

2 n

ewto

ns/

met

er

rela

ted

to t

he

grap

h o

f th

e fu

nct

ion

for

a

spri

ng

wh

ere

k=

10

new

ton

s/m

eter

?

Th

e g

rap

h o

f U

s=

1 − 2 (1

0)x

2 is

a

s

tret

ch o

f th

e o

ther

gra

ph

.

2. B

LUEP

RIN

TS T

he

bott

om l

eft

corn

er o

f a

rect

angu

lar

park

is

loca

ted

at (

–3, 5

) on

a

con

stru

ctio

n b

luep

rin

t. T

he

park

is

8 u

nit

s lo

ng

and

has

an

are

a of

48

un

its2

on t

he

blu

epri

nt.

Su

ppos

e th

e pa

rk i

s tr

ansl

ated

on

th

e bl

uep

rin

t so

th

at t

he

top

righ

t co

rner

is

now

loc

ated

at

the

orig

in. D

escr

ibe

the

tran

slat

ion

of

the

grap

h.

T

ran

slat

ed 5

un

its

left

an

d 1

1 u

nit

s d

ow

n

3. P

HY

SIC

S A

bal

l is

dro

pped

fro

m a

hei

ght

of 2

0 fe

et. T

he

fun

ctio

n h

=-

16t2

+ 2

0 m

odel

s th

e h

eigh

t of

th

e ba

ll i

n f

eet

afte

r t

seco

nds

. Gra

ph t

he

fun

ctio

n a

nd

com

pare

th

is g

raph

to

the

grap

h o

f it

s pa

ren

t fu

nct

ion

.

Height (ft)3 0691215182124

Tim

e (s

)0.

20.

40.

60.

81.

01.

21.

41.

6

9-3

Wor

d Pr

oble

m P

ract

ice

Tran

sfo

rmati

on

s o

f Q

uad

rati

c F

un

cti

on

s

4. A

CC

ELER

ATI

ON

Th

e di

stan

ce d

in

fee

t a

car

acce

lera

tin

g at

6 f

t/s2

trav

els

afte

r t

seco

nds

is

mod

eled

by

the

fun

ctio

n

d=

3t2 .

Su

ppos

e th

at a

t th

e sa

me

tim

e th

e fi

rst

car

begi

ns

acce

lera

tin

g, a

sec

ond

car

begi

ns

acce

lera

tin

g at

4 f

t/s2

exac

tly

100

feet

dow

n t

he

road

fro

m t

he

firs

t ca

r. T

he

dist

ance

tra

vele

d by

sec

ond

car

is

mod

eled

by

the

fun

ctio

n d

= 2

t2+

100

.

a. G

raph

an

d la

bel

each

fu

nct

ion

on

th

e sa

me

coor

din

ate

plan

e.

Distance (ft)

100 0

200

300

400

500

600

700

800

Tim

e (s

)2

46

810

1214

16

d =

2t2 +

100

d =

3t2

b.

Exp

lain

how

eac

h g

raph

is

rela

ted

to

the

grap

h o

f d

=t2 .

d=

3t2

is a

str

etch

of

d=

t2 ; d

= 2

t2+

10

0 is

a s

tret

ch

of

d=

t2 tr

ansl

ated

up

10

0 u

nit

s (f

eet)

.

c. A

fter

how

man

y se

con

ds w

ill

the

firs

t ca

r pa

ss t

he

seco

nd

car?

10 s

eco

nd

s

h =

-16

t2 +

20

is a

str

etch

o

f h

= t

2 re

fl ec

ted

ove

r th

e x-a

xis

and

tra

nsl

ated

up

20

un

its

(fee

t).

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2112

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

22

Gle

ncoe

Alg

ebra

1

A p

olyn

omia

l fu

nct

ion

is

a co

nti

nu

ous

fun

ctio

n t

hat

can

be

desc

ribe

d by

a p

olyn

omia

l eq

uat

ion

in

on

e va

riab

le.

Po

lyn

om

ial F

un

ctio

n

If n

is

a n

onn

egat

ive

inte

ger,

a 0, a 1,

a 2, …

, a n

- 1, a

n a

re r

eal

nu

mbe

rs,

and

a n ≠

0, t

hen

f(x)

= a

nxn

+ a

n -

1xn

- 1 +

… +

a2x

2 +

a1x

+ a

0

is a

pol

ynom

ial

fun

ctio

n o

f de

gree

n.

Not

ice

that

a q

uad

rati

c fu

nct

ion

is

a po

lyn

omia

l fu

nct

ion

of

degr

ee 2

.

C

reat

e a

tab

le o

f va

lues

an

d a

gra

ph

for

y =

x3

+ 2

x2 -

x. T

hen

d

escr

ibe

its

end

beh

avio

r.

Cre

ate

a ta

ble

of v

alu

es, a

nd

grap

h t

he

orde

red

pair

s. C

onn

ect

the

poin

ts w

ith

a s

moo

th

curv

e. F

ind

and

plot

add

itio

nal

poi

nts

to

bett

er a

ppro

xim

ate

the

curv

e’s

shap

e.

F

rom

th

e ta

ble

and

the

grap

h w

e se

e th

at a

s x

decr

ease

s, y

dec

reas

es

and

as x

in

crea

ses,

y i

ncr

ease

s.

Exer

cise

s

Cre

ate

a ta

ble

of

valu

es a

nd

a g

rap

h f

or e

ach

fu

nct

ion

. Th

en d

escr

ibe

its

end

b

ehav

ior.

1. f

(x)

= x

3 +

3x2

- 1

2.

f(x)

= -

2x5

+ 4

x3 3.

f(x)

= 2

x4 -

4x3

+ 3

x

9-3

Enri

chm

ent

Gra

ph

ing

Po

lyn

om

ial

Fu

ncti

on

s

Exam

ple

xx

3 +

2x

2 – x

y

-4

(-4)

3 +

2(-

4)2

- (

-4)

-28

-3

(-3)

3 +

2(-

3)2

- (

-3)

-

6

-2

(-2)

3 +

2(-

2)2

- (

-2)

2

-1

(-1)

3 +

2(-

1)2

- (

-1)

2

0

(0)3

+ 2

(0)2

- 0

0

1

13 +

2(1

)2 -

1

2

2

23 +

2(2

)2 -

2

14

3

33 +

2(3

)2 -

3

42

As

x de

crea

ses,

y

decr

ease

s

As

x in

crea

ses,

y

incr

ease

s.

xf(

x)

-4

-17

-3

-1

-2

3

-1

1

0-

1

13

xf(

x)

-2

32

-1

-2

-0.5

-0.

4

00

0.5

-0.

4

12

2-

32

xf(

x)

-2

58

-1

3

-0.5

-0.

1

00

0.5

1.1

11

26

y

xO

As y

ou m

ove

left,

the

grap

hgo

es d

own.

As y

ou m

ove

right

, the

gra

phgo

es u

p.( −

1, 2

)( 1

, 2)

( −2,

2)

( 0, 0

)

xO8 4

−2

−4

24

−4

−8

f(x)

xO8 4

−2

−4

24

−4

−8

f(x)

xO8 4

−2

−4

24

−4

−8

f(x)

As

x d

ecre

ases

, y d

ecre

ases

an

d, a

s x in

crea

ses,

y in

crea

ses.

As

x d

ecre

ases

, y in

crea

ses,

as

x in

crea

ses,

y d

ecre

ases

.A

s x d

ecre

ases

, y in

crea

ses,

as

x in

crea

ses,

y in

crea

ses.

011-

022_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2212

/28/

10

7:31

AM

Lesson X-4


NA

ME

DAT

E

P

ER

IOD

Lesson 9-4

Cha

pte

r 9

23

Gle

ncoe

Alg

ebra

1

Co

mp

lete

th

e Sq

uar

e P

erfe

ct s

quar

e tr

inom

ials

can

be

solv

ed q

uic

kly

by t

akin

g th

e sq

uar

e ro

ot o

f bo

th s

ides

of

the

equ

atio

n. A

qu

adra

tic

equ

atio

n t

hat

is

not

in

per

fect

squ

are

form

can

be

mad

e in

to a

per

fect

squ

are

by a

met

hod

cal

led

com

ple

tin

g th

e sq

uar

e.

Co

mp

leti

ng

th

e S

qu

are

To

co

mp

lete

th

e s

qu

are

fo

r a

ny q

ua

dra

tic e

qu

atio

n o

f th

e fo

rm x

2 +

bx:

Ste

p 1

Fin

d o

ne

-ha

lf o

f b

, th

e c

oe

ffic

ien

t o

f x.

Ste

p 2

Sq

ua

re t

he

re

su

lt in

Ste

p 1

.

Ste

p 3

Ad

d t

he

re

su

lt o

f S

tep

2 t

o x

2 +

bx.

x2 +

bx

+ ( b

−

2 ) 2 =

(x +

b

−

2 ) 2

F

ind

th

e va

lue

of c

th

at m

akes

x2

+ 2

x +

c a

per

fect

sq

uar

e tr

inom

ial.

Ste

p 1

F

ind

1 −

2 o

f 2.

2

−

2 =

1

Ste

p 2

S

quar

e th

e re

sult

of

Ste

p 1.

1

2 =

1

Ste

p 3

A

dd t

he

resu

lt o

f S

tep

2 to

x2

+ 2

x.

x2 +

2x

+ 1

Th

us,

c =

1. N

otic

e th

at x

2 +

2x

+ 1

equ

als

(x +

1)2 .

Exer

cise

sF

ind

th

e va

lue

of c

th

at m

akes

eac

h t

rin

omia

l a

per

fect

sq

uar

e.

1. x

2 +

10x

+ c

25

2. x

2 +

14x

+ c

49

3. x

2 -

4x

+ c

4

4. x

2 -

8x

+ c

16

5. x

2 +

5x

+ c

25

−

4

6. x

2 +

9x

+ c

81

−

4

7. x

2 -

3x

+ c

9 −

4

8. x

2 -

15x

+ c

225

−

4

9. x

2 +

28x

+ c

196

10

. x2

+ 2

2x +

c 1

21

Stud

y G

uide

and

Inte

rven

tion

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y C

om

ple

tin

g t

he S

qu

are

Exam

ple

9-4

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2312

/23/

10

10:5

7 A

M


A01_A14_ALG1_A_CRM_C09_AN_660283.indd A10A01_A14_ALG1_A_CRM_C09_AN_660283.indd A10 12/28/10 7:57 AM12/28/10 7:57 AM

An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

24

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y C

om

ple

tin

g t

he S

qu

are

Solv

e b

y C

om

ple

tin

g t

he

Squ

are

Sin

ce f

ew q

uad

rati

c ex

pres

sion

s ar

e pe

rfec

t sq

uar

e tr

inom

ials

, th

e m

eth

od o

f co

mp

leti

ng

the

squ

are

can

be

use

d to

sol

ve s

ome

quad

rati

c eq

uat

ion

s. U

se t

he

foll

owin

g st

eps

to c

ompl

ete

the

squ

are

for

a qu

adra

tic

expr

essi

on o

f th

e fo

rm a

x2 +

bx.

S

olve

x2

+ 6

x +

3 =

10

by

com

ple

tin

g th

e sq

uar

e.

x2

+ 6

x +

3 =

10

Ori

gin

al equation

x2

+ 6

x +

3 -

3 =

10

- 3

S

ubtr

act

3 f

rom

each s

ide.

x2

+ 6

x =

7

Sim

plif

y.

x2

+ 6

x +

9 =

7 +

9

Sin

ce ( 6

−

2 ) 2

= 9

, add 9

to e

ach s

ide.

(x +

3)2

= 1

6 F

acto

r x2

+ 6

x +

9.

x

+ 3

= ±

4 Take

the s

quare

root

of

each s

ide.

x

= -

3 ±

4

Sim

plif

y.

x =

-3

+ 4

or

x

= -

3 -

4

= 1

=

-7

Th

e so

luti

on s

et i

s {-

7, 1

}.

Exer

cise

sS

olve

eac

h e

qu

atio

n b

y co

mp

leti

ng

the

squ

are.

Rou

nd

to

the

nea

rest

ten

th i

f n

eces

sary

.

1. x

2 -

4x

+ 3

= 0

2.

x2

+ 1

0x =

-9

3. x

2 -

8x

- 9

= 0

1

, 3

-1,

-9

-1,

9

4. x

2 -

6x

= 1

6 5.

x2

- 4

x -

5 =

0

6. x

2 -

12x

= 9

-

2, 8

-

1, 5

-

0.7,

12.

7

7. x

2 +

8x

= 2

0 8.

x2

= 2

x +

1

9. x

2 +

20x

+ 1

1 =

-8

-

10, 2

-

0.4,

2.4

-

19, -

1

10. x

2 -

1 =

5x

11. x

2 =

22x

+ 2

3 12

. x2

- 8

x =

-7

-

0.2,

5.2

-

1, 2

3 1

, 7

13. x

2 +

10x

= 2

4 14

. x2

- 1

8x =

19

15. x

2 +

16x

= -

16

-

12, 2

-

1, 1

9 -

14.9

, -1.

1

16. 4

x2 =

24

+ 4

x 17

. 2x2

+ 4

x +

2 =

8

18. 4

x2 =

40x

+ 4

4

-

2, 3

-

3, 1

-

1, 1

1

Ste

p 1

Fin

d b

−

2 .

Ste

p 2

Fin

d ( b

−

2 ) 2

.

Ste

p 3

Add ( b

−

2 ) 2

to a

x2 +

bx.

9-4

Exam

ple

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2412

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-4

Cha

pte

r 9

25

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceS

olv

ing

Qu

ad

rati

c E

qu

ati

on

s b

y C

om

ple

tin

g t

he S

qu

are

Fin

d t

he

valu

e of

c t

hat

mak

es e

ach

tri

nom

ial

a p

erfe

ct s

qu

are.

1. x

2 +

6x

+ c

9

2. x

2 +

4x

+ c

4

3. x

2 -

14x

+ c

49

4. x

2 -

2x

+ c

1

5. x

2 -

18x

+ c

81

6. x

2 +

20x

+ c

10

0

7. x

2 +

5x

+ c

6.2

5 8.

x2

- 7

0x +

c 1

225

9. x

2 -

11x

+ c

30.

25

10. x

2 +

9x

+ c

20.

25

Sol

ve e

ach

eq

uat

ion

by

com

ple

tin

g th

e sq

uar

e. R

oun

d t

o th

e n

eare

st

ten

th i

f n

eces

sary

.

11. x

2 +

4x

- 1

2 =

0 2

, -6

12. x

2 -

8x

+ 1

5 =

0 3

, 5

13. x

2 +

6x

= 7

-7,

1

14. x

2 -

2x

= 1

5 -

3, 5

15. x

2 -

14x

+ 3

0 =

6 2

, 12

16. x

2 +

12x

+ 2

1 =

10

-11

, -1

17. x

2 -

4x

+ 1

= 0

0.3

, 3.7

18

. x2

- 6

x +

4 =

0 0

.8, 5

.2

19. x

2 -

8x

+ 1

0 =

0 1

.6, 6

.4

20. x

2 -

2x

= 5

-1.

4, 3

.4

21. 2

x2 +

20x

= -

2 -

9.9,

-0.

1 22

. 0.5

x2 +

8x

= -

7 -

15.1

, -0.

9

9-4

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2512

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

26

Gle

ncoe

Alg

ebra

1

Prac

tice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y C

om

ple

tin

g t

he S

qu

are

Fin

d t

he

valu

e of

c t

hat

mak

es e

ach

tri

nom

ial

a p

erfe

ct s

qu

are.

1. x

2 -

24x

+ c

144

2.

x2

+ 2

8x +

c 1

96

3. x

2 +

40x

+ c

40

0

4. x

2 +

3x

+ c

9 −

4

5. x

2 -

9x

+ c

81

−

4

6. x

2 -

x +

c

1 −

4

Sol

ve e

ach

eq

uat

ion

by

com

ple

tin

g th

e sq

uar

e. R

oun

d t

o th

e n

eare

st t

enth

if

nec

essa

ry.

7. x

2 -

14x

+ 2

4 =

0

8. x

2 +

12x

= 1

3 9.

x2

- 3

0x +

56

= -

25

2

, 12

-13

, 1

3, 2

7

10. x

2 +

8x

+ 9

= 0

11

. x2

- 1

0x +

6 =

-7

12. x

2 +

18x

+ 5

0 =

9

-

6.6,

-1.

4 1

.5, 8

.5

-15

.3, -

2.7

13. 3

x2 +

15x

- 3

= 0

14

. 4x2

- 7

2 =

24x

15

. 0.9

x2 +

5.4

x -

4 =

0

-

5.2,

0.2

-

2.2,

8.2

-

6 2 −

3 ,

2 −

3

16. 0

.4x2

+ 0

.8x

= 0

.2

17.

1 −

2 x

2 -

x -

10

= 0

18

. 1 −

4 x

2 +

x -

2 =

0

-

2.2,

0.2

-

3.6,

5.6

-

5.5,

1.5

19. N

UM

BER

TH

EORY

Th

e pr

odu

ct o

f tw

o co

nse

cuti

ve e

ven

in

tege

rs i

s 72

8. F

ind

the

inte

gers

.

20. B

USI

NES

S Ja

ime

own

s a

busi

nes

s m

akin

g de

cora

tive

box

es t

o st

ore

jew

elry

, mem

ento

s,

and

oth

er v

alu

able

s. T

he

fun

ctio

n y

= x

2 +

50x

+ 1

800

mod

els

the

prof

it y

th

at J

aim

e h

as m

ade

in m

onth

x f

or t

he

firs

t tw

o ye

ars

of h

is b

usi

nes

s.

a. W

rite

an

equ

atio

n r

epre

sen

tin

g th

e m

onth

in

wh

ich

Jai

me’

s pr

ofit

is

$24

00.

x

2 +

50x

+ 1

800

= 2

400

b. U

se c

ompl

etin

g th

e sq

uar

e to

fin

d ou

t in

wh

ich

mon

th J

aim

e’s

prof

it i

s $24

00.

th

e te

nth

mo

nth

21. P

HY

SIC

S F

rom

a h

eigh

t of

256

fee

t ab

ove

a la

ke o

n a

cli

ff, M

ikae

la t

hro

ws

a ro

ck o

ut

over

th

e la

ke. T

he

hei

ght

H o

f th

e ro

ck t

sec

onds

aft

er M

ikae

la t

hro

ws

it i

s re

pres

ente

d by

th

e eq

uat

ion

H =

-16

t2 +

32t

+ 2

56. T

o th

e n

eare

st t

enth

of

a se

con

d, h

ow l

ong

does

it

tak

e th

e ro

ck t

o re

ach

th

e la

ke b

elow

? (H

int:

Rep

lace

H w

ith

0.)

5.1

s

9-4 26

, 28

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2612

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-4

Cha

pte

r 9

27

Gle

ncoe

Alg

ebra

1

9-4

Wor

d Pr

oble

m P

ract

ice

So

lvin

g Q

uad

rati

c E

qu

ati

on

s b

y C

om

ple

tin

g t

he S

qu

are

1.IN

TER

IOR

DES

IGN

Mod

ula

r ca

rpet

ing

is

inst

alle

d in

sm

all

piec

es r

ath

er t

han

as

a la

rge

roll

so

that

on

ly a

few

pie

ces

nee

d to

be

repl

aced

if

a sm

all

area

is

dam

aged

. S

upp

ose

the

room

sh

own

in

th

e di

agra

m

belo

w i

s be

ing

fitt

ed w

ith

mod

ula

r ca

rpet

ing.

Com

plet

e th

e sq

uar

e to

de

term

ine

the

nu

mbe

r of

1 f

oot

by 1

foo

t sq

uar

es o

f ca

rpet

ing

nee

ded

to f

inis

h t

he

room

. Fil

l in

th

e m

issi

ng

term

s in

th

e co

rres

pon

din

g eq

uat

ion

bel

ow.

25;

5

x

xxxxx

xx

xx

x2

x2+

10x

+ _

____

=(x

+ _

____

)2

2. F

ALL

ING

OB

JEC

TS K

eish

a th

row

s a

rock

dow

n a

n o

ld w

ell.

Th

e di

stan

ce

d i

n f

eet

the

rock

fal

ls a

fter

t s

econ

ds

can

be

repr

esen

ted

by d

= 1

6t2 +

64t

. If

the

wat

er i

n t

he

wel

l is

80

feet

bel

ow

grou

nd,

how

man

y se

con

ds w

ill

it t

ake

for

the

rock

to

hit

th

e w

ater

?

1 se

con

d

3.M

AR

S O

n M

ars,

th

e gr

avit

y ac

tin

g on

an

obj

ect

is l

ess

than

th

at o

n E

arth

. On

E

arth

, a g

olf

ball

hit

wit

h a

n i

nit

ial

upw

ard

velo

city

of

26 m

eter

s pe

r se

con

d w

ill

hit

th

e gr

oun

d in

abo

ut

5.4

seco

nds

. T

he

hei

ght

h o

f an

obj

ect

on M

ars

that

le

aves

th

e gr

oun

d w

ith

an

in

itia

l ve

loci

ty

of 2

6 m

eter

s pe

r se

con

d is

giv

en b

y th

e eq

uat

ion

h=

-1.

9t2

+ 2

6t. H

ow m

uch

lo

nge

r w

ill

it t

ake

for

the

golf

bal

l h

it o

n

Mar

s to

rea

ch t

he

grou

nd?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

ab

ou

t 8.

3 se

con

ds

4.FR

OG

S A

fro

g si

ttin

g on

a s

tum

p 3

feet

h

igh

hop

s of

f an

d la

nds

on

th

e gr

oun

d.

Du

rin

g it

s le

ap, i

ts h

eigh

t h

in

fee

t is

gi

ven

by

h=

-0.

5d2

+ 2

d+

3, w

her

e d

is

the

dist

ance

fro

m t

he

base

of

the

stu

mp.

H

ow f

ar i

s th

e fr

og f

rom

th

e ba

se o

f th

e st

um

p w

hen

it

lan

ded

on t

he

grou

nd?

2 +

√ �

�

10 o

r ab

ou

t 5.

16 f

t

5. G

AR

DEN

ING

Peg

is

plan

nin

g a

rect

angu

lar

vege

tabl

e ga

rden

usi

ng

200

feet

of

fen

cin

g m

ater

ial.

Sh

e on

ly

nee

ds t

o fe

nce

th

ree

side

s of

th

e ga

rden

si

nce

on

e si

de b

orde

rs a

n e

xist

ing

fen

ce.

x

a.L

et x

= t

he

wid

th o

f th

e re

ctan

gle.

Wri

te

an e

quat

ion

to

repr

esen

t th

e ar

ea A

of

the

gard

en i

f P

eg u

ses

all

the

fen

cin

g m

ater

ial.

A =

x(2

00

- 2

x)

b. F

or w

hat

wid

ths

wou

ld t

he

area

of

Peg

’s

gard

en e

qual

480

0 sq

uar

e fe

et i

f sh

e u

ses

all

the

fen

cin

g m

ater

ial?

Th

e ar

ea o

f th

e g

ard

en w

ill b

e 48

00

ft 2

for

a w

idth

of

40 f

t an

d a

w

idth

of

60 f

t.

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2712

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

28

Gle

ncoe

Alg

ebra

1

Com

plet

ing

the

squ

are

is a

use

ful

tool

for

fac

tori

ng

quad

rati

c ex

pres

sion

s. Y

ou c

an u

tili

ze a

si

mil

ar t

ech

niq

ue

to f

acto

r si

mpl

e q

uar

tic

pol

ynom

ials

of

the

form

x4

+ c

.

Fac

tor

the

qu

arti

c p

olyn

omia

l x4

+ 6

4.

Ste

p 1

F

ind

the

valu

e of

th

e m

iddl

e te

rm n

eede

d to

com

plet

e th

e sq

uar

e.

Th

is v

alu

e is

(2

√ �

�

64 ) (

x2 ) , o

r 16

x2 .

Ste

p 2

R

ewri

te t

he

orig

inal

pol

ynom

ial

in f

acto

rabl

e fo

rm.

(x4 +

16x

2 +

( 16

−

2 ) 2 ) - 1

6x2

Ste

p 3

Fa

ctor

th

e po

lyn

omia

ls. (

x2 +

8) 2 -

(4x

)2

Ste

p 4

R

ewri

te u

sin

g th

e di

ffer

ence

of

two

squ

ares

. (x

2 +

8 +

4x)

(x2

+ 8

- 4

x)

Th

e fa

ctor

ed f

orm

of

x2 +

64

is (

x2 +

4x

+ 8

) (x2

- 4

x +

8) .

Th

is c

ould

th

en b

e fa

ctor

ed

furt

her

, if

nee

ded,

to

fin

d th

e so

luti

ons

to a

qu

arti

c eq

uat

ion

.

Exer

cise

sF

acto

r ea

ch q

uar

tic

pol

ynom

ial.

1. x

4 +

4

2. x

4 +

324

(

x2

+ 2

x +

2)(

x2

- 2

x +

2)

(x2 +

6x

+ 1

8)(x

2 - 6

x +

18)

3. x

4 +

250

0 4.

x4

+ 9

604

(

x2 + 1

0x +

50)

(x2 -

10x

+ 5

0)

(x2 +

14x

+ 9

8)(x

2 - 1

4x +

98)

5. x

4 +

102

4

6. x

4 +

518

4

(x2 +

8x

+ 3

2)(x

2 - 8

x +

32)

(

x2 + 1

2x +

72)

(x2 -

12x

+ 7

2)

7. x

4 +

484

8.

x4

+ 9

(

x2 + x

√ �

�

44 +

22)

(x2 -

x √

��

44 +

22)

(

x2 + x

√ �

6 +

3)(x

2 - x

√ �

6 +

3)

9. x

4 +

144

10

. x8

+ 1

6,38

4

(

x2 + 2

x √ �

6 +

12)

(x2 -

2x √

� 6 +

12)

(

x4 + 1

6x2 +

128

)(x4 -

16x

2 +

128

)

11. F

acto

r x4

+ c

to

com

e u

p w

ith

a g

ener

al r

ule

for

fac

tori

ng

quar

tic

poly

nom

ials

.

(

x2 + x

√ �

�

2 √

�

c +

√ �

c )

+ (

x2 + x

√ �

�

2 √

�

c +

√ �

c )

9-4

Enri

chm

ent

Facto

rin

g Q

uart

ic P

oly

no

mia

ls

Exam

ple

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2812

/23/

10

10:1

0 P

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-5

Cha

pte

r 9

29

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

So

lvin

g Q

uad

ratic E

qu

atio

ns

by

Usi

ng

th

e Q

uad

ratic F

orm

ula

Qu

adra

tic

Form

ula

To

solv

e th

e st

anda

rd f

orm

of

the

quad

rati

c eq

uat

ion

, ax

2 +

bx

+ c

= 0

, use

th

e Q

uad

rati

c F

orm

ula

.

Qu

adra

tic

Form

ula

The s

olu

tions o

f ax

2 +

bx

+ c

= 0

, w

here

a ≠

0,

are

giv

en

by x

= -

b ±

√ �

��

�

b2 -

4ac

−

2a

.

S

olve

x2

+ 2

x =

3 b

y u

sin

g th

e Q

uad

rati

c F

orm

ula

.

Rew

rite

th

e eq

uat

ion

in

sta

nda

rd f

orm

.

x2

+ 2

x =

3

Ori

gin

al equation

x2 +

2x

- 3

= 3

- 3

S

ubtr

act

3 f

rom

each s

ide.

x2

+ 2

x -

3 =

0

Sim

plif

y.

Now

let

a =

1, b

= 2

, an

d c

= -

3 in

th

e Q

uad

rati

c F

orm

ula

.

x =

-

b ±

√ �

��

�

b2 -

4ac

−

2a

=

-

2 ±

√ �

��

��

�

(2

)2 -

4(1

)(-

3)

−

−

2(

1)

=

-2

± √

��

16

−

2

x =

-

2 +

4

−

2

or

x =

-

2 -

4

−

2

=

1

= -

3T

he

solu

tion

set

is

{-3,

1}.

S

olve

x2

- 6

x -

2 =

0 b

y u

sin

g th

e Q

uad

rati

c F

orm

ula

. Rou

nd

to

the

nea

rest

ten

th i

f n

eces

sary

.

For

th

is e

quat

ion

a =

1, b

= -

6, a

nd

c =

-2.

x =

-b

± √

��

��

b 2 -

4ac

−

2a

=

6

± √

��

��

��

�

(-

6) 2

- 4

(1)(

-2)

−−

2(1)

=

6

+ √

��

44

−

2

x =

6 +

√ �

�

44

−

2

or

x =

6

- √

��

44

−

2

x ≈

6.3

≈

-0.

3T

he

solu

tion

set

is

{-0.

3, 6

.3}.

Exer

cise

s

Sol

ve e

ach

eq

uat

ion

by

usi

ng

the

Qu

adra

tic

For

mu

la. R

oun

d t

o th

e n

eare

st t

enth

if

nec

essa

ry.

1. x

2 -

3x

+ 2

= 0

1, 2

2.

x2

- 8

x =

-16

4

3. 1

6x2

- 8

x =

-1

1 −

4

4. x

2 +

5x

= 6

-6,

1

5. 3

x2 +

2x

= 8

-2,

4 −

3

6. 8

x2 -

8x

- 5

= 0

-0.

4, 1

.4

7. -

4x2

+ 1

9x =

21

7 −

4 ,

3 8.

2x2

+ 6

x =

5 -

3.7,

0.7

9. 4

8x2

+ 2

2x -

15

= 0

- 5

−

6 ,

3 −

8

10. 8

x2 -

4x

= 2

4 -

3 −

2 ,

2

11. 2

x2 +

5x

= 8

-3.

6, 1

.1

12. 8

x2 +

9x

- 4

= 0

-1.

5, 0

.3

13. 2

x2 +

9x

+ 4

= 0

-4,

- 1 −

2

14. 8

x2 +

17x

+ 2

= 0

-2,

- 1 −

8

9-5

Exam

ple

1Ex

amp

le 2

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

2912

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

30

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

So

lvin

g Q

ua

dra

tic

Eq

ua

tio

ns b

y U

sin

g t

he

Qu

ad

rati

c F

orm

ula

The

Dis

crim

inan

t In

th

e Q

uad

rati

c F

orm

ula

, x =

-b

± √

��

��

b 2 -

4ac

−

2a

, th

e ex

pres

sion

un

der

the

radi

cal

sign

, b2

- 4

ac, i

s ca

lled

th

e d

iscr

imin

ant.

Th

e di

scri

min

ant

can

be

use

d to

det

erm

ine

the

nu

mbe

r of

rea

l so

luti

ons

for

a qu

adra

tic

equ

atio

n.

Ca

se

1:

b2 -

4ac

< 0

no

re

al so

lutio

ns

Ca

se

2:

b2 -

4ac

= 0

on

e r

ea

l so

lutio

n

Ca

se

3:

b2 -

4ac

> 0

two

re

al so

lutio

ns

S

tate

th

e va

lue

of t

he

dis

crim

inan

t fo

r ea

ch e

qu

atio

n. T

hen

d

eter

min

e th

e n

um

ber

of

real

sol

uti

ons

of t

he

equ

atio

n.

a. 1

2x2

+ 5

x =

4W

rite

th

e eq

uat

ion

in

sta

nda

rd f

orm

.

12x2

+ 5

x =

4

Ori

gin

al equation

12x2

+ 5

x -

4 =

4 -

4

Subtr

act

4 f

rom

each s

ide.

12x2

+ 5

x -

4 =

0

Sim

plif

y.

Now

fin

d th

e di

scri

min

ant.

b2 -

4ac

= (

5)2

- 4

(12)

(-4)

=

217

Sin

ce t

he

disc

rim

inan

t is

pos

itiv

e, t

he

equ

atio

n h

as t

wo

real

sol

uti

ons.

b.

2x2

+ 3

x =

-4

2x

2 +

3x

= -

4 O

rigin

al equation

2x2

+ 3

x +

4 =

-4

+ 4

A

dd 4

to e

ach s

ide.

2x2

+ 3

x +

4 =

0

Sim

plif

y.

Fin

d th

e di

scri

min

ant.

b2 -

4ac

= (

3)2

- 4

(2)(

4)

= -

23

Sin

ce t

he

disc

rim

inan

t is

neg

ativ

e, t

he

equ

atio

n h

as n

o re

al s

olu

tion

s.

Exer

cise

sS

tate

th

e va

lue

of t

he

dis

crim

inan

t fo

r ea

ch e

qu

atio

n. T

hen

det

erm

ine

the

nu

mb

er

of r

eal

solu

tion

s of

th

e eq

uat

ion

.

1. 3

x2 +

2x

- 3

= 0

2.

3x2

- 7

x -

8 =

0

3. 2

x2 -

10x

- 9

= 0

4

0, 2

rea

l so

luti

on

s 1

45, 2

rea

l so

luti

on

s 1

72, 2

rea

l so

luti

on

s

4. 4

x2 =

x +

4

5. 3

x2 -

13x

= 1

0 6.

6x2

- 1

0x +

10

= 0

6

5, 2

rea

l so

luti

on

s 2

89, 2

rea

l so

luti

on

s

7. 2

x2 -

20

= -

x 8.

6x2

= -

11x

- 4

0 9.

9 -

18x

+ 9

x2 =

0

1

61, 2

rea

l so

luti

on

s -

839,

no

rea

l so

luti

on

s 0

, 1 r

eal s

olu

tio

n

10. 1

2x2

+ 9

= -

6x

11. 9

x2 =

81

12. 1

6x2

+ 1

6x +

4 =

0

-

396,

no

rea

l so

luti

on

s 2

916,

2 r

eal s

olu

tio

ns

0, 1

rea

l so

luti

on

13. 8

x2 +

9x

= 2

14

. 4x2

- 4

x +

4 =

3

15. 3

x2 -

18x

= -

14

1

45, 2

rea

l so

luti

on

s 0

, 1 r

eal s

olu

tio

n

156

, 2 r

eal s

olu

tio

ns

9-5

Exam

ple

-14

0, n

o r

eal s

olu

tion

s

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3012

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-5

Cha

pte

r 9

31

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceS

olv

ing

Qu

ad

rati

c E

qu

ati

on

s b

y U

sin

g t

he

Qu

ad

rati

c F

orm

ula

Sol

ve e

ach

eq

uat

ion

by

usi

ng

the

Qu

adra

tic

For

mu

la. R

oun

d t

o th

e n

eare

st t

enth

if

nec

essa

ry.

1. x

2 -

49

= 0

-7,

7

2. x

2 -

x -

20

= 0

-4,

5

3. x

2 -

5x

- 3

6 =

0 -

4, 9

4.

x2

+ 1

1x +

30

= 0

-6,

-5

5. x

2 -

7x

= -

3 0.

5, 6

.5

6. x

2 +

4x

= -

1 -

3.7,

-0.

3

7. x

2 -

9x

+ 2

2 =

0 �

8.

x2

+ 6

x +

3 =

0 -

5.4,

-0.

6

9. 2

x2 +

5x

- 7

= 0

- 3

1 −

2 ,

1 10

. 2x2

- 3

x =

-1

1 −

2 ,

1

11. 2

x2 +

5x

+ 4

= 0

�

12. 2

x2 +

7x

= 9

-4

1 −

2 ,

1

13. 3

x2 +

2x

- 3

= 0

-1.

4, 0

.7

14. 3

x2 -

7x

- 6

= 0

- 2 −

3 ,

3

Sta

te t

he

valu

e of

th

e d

iscr

imin

ant

for

each

eq

uat

ion

. Th

en d

eter

min

e th

e n

um

ber

of

rea

l so

luti

ons

of t

he

equ

atio

n.

15. x

2 +

4x

+ 3

= 0

16

. x2

+ 2

x +

1 =

0

4

; 2

real

so

luti

on

s 0

; 1

real

so

luti

on

17. x

2 -

4x

+ 1

0 =

0

18. x

2 -

6x

+ 7

= 0

-

24;

no

rea

l so

luti

on

s 8

; 2

real

so

luti

on

s

19. x

2 -

2x

- 7

= 0

20

. x2

- 1

0x +

25

= 0

3

2; 2

rea

l so

luti

on

s 0

; 1

real

so

luti

on

21. 2

x2 +

5x

- 8

= 0

22

. 2x2

+ 6

x +

12

= 0

8

9; 2

rea

l so

luti

on

s -

60;

no

rea

l so

luti

on

s

23. 2

x2 -

4x

+ 1

0 =

0

24. 3

x2 +

7x

+ 3

= 0

-

64;

no

rea

l so

luti

on

s 1

3; 2

rea

l so

luti

on

s

9-5

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3112

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

32

Gle

ncoe

Alg

ebra

1

9-5

Prac

tice

S

olv

ing

Qu

ad

rati

c E

qu

ati

on

s b

y U

sin

g t

he

Qu

ad

rati

c F

orm

ula

Sol

ve e

ach

eq

uat

ion

by

usi

ng

the

Qu

adra

tic

For

mu

la. R

oun

d t

o th

e n

eare

st t

enth

if

nec

essa

ry.

1. x

2 +

2x

- 3

= 0

-3,

1

2. x

2 +

8x

+ 7

= 0

-7,

-1

3. x

2 -

4x

+ 6

= 0

�

4. x

2 - 6

x +

7 =

0 1

.6, 4

.4

5. 2

x2 + 9

x -

5 =

0 -

5, 1 −

2

6. 2

x2 + 1

2x +

10

= 0

-5,

-1

7. 2

x2 -

9x

= -

12 �

8.

2x2

- 5

x =

12

-1

1 −

2 ,

4 9.

3x2

+ x

= 4

-1

1 −

3 ,

1

10. 3

x2 -

1 =

-8x

-2.

8, 0

.1

11. 4

x2 +

7x

= 1

5 -

3, 1

1 −

4

12. 1

.6x2

+ 2

x +

2.5

= 0

�

13. 4

.5x2

+ 4

x -

1.5

= 0

14

. 1 −

2 x

2 +

2x

+ 3 −

2 =

0

15. 3

x2 -

3 −

4 x

= 1 −

2

-

1.2,

0.3

-

3, -

1 -

0.3,

0.6

Sta

te t

he

valu

e of

th

e d

iscr

imin

ant

for

each

eq

uat

ion

. Th

en d

eter

min

e th

e n

um

ber

of

rea

l so

luti

ons

of t

he

equ

atio

n.

16. x

2 +

8x

+ 1

6 =

0

17. x

2 +

3x

+ 1

2 =

0

18. 2

x2 +

12x

= -

7

0

; 1

real

so

luti

on

-

39;

no

rea

l so

luti

on

s 8

8; 2

rea

l so

luti

on

s

19. 2

x2 +

15x

= -

30

20. 4

x2 +

9 =

12x

21

. 3x2

- 2

x =

3.5

-

15;

no

rea

l so

luti

on

s 0

; 1

real

so

luti

on

4

6; 2

rea

l so

luti

on

s

22. 2

.5x2

+ 3

x -

0.5

= 0

23

. 3 −

4 x2

- 3

x =

-4

24.

1 −

4 x2

= -

x -

1

1

4; 2

rea

l so

luti

on

s -

3; n

o r

eal s

olu

tio

ns

0;

1 re

al s

olu

tio

n

25. C

ON

STR

UC

TIO

N A

roo

fer

toss

es a

pie

ce o

f ro

ofin

g ti

le f

rom

a r

oof

onto

th

e gr

oun

d 30

fee

t be

low

. He

toss

es t

he

tile

wit

h a

n i

nit

ial

dow

nw

ard

velo

city

of

10 f

eet

per

seco

nd.

a. W

rite

an

equ

atio

n t

o fi

nd

how

lon

g it

tak

es t

he

tile

to

hit

th

e gr

oun

d. U

se t

he

mod

el f

or

vert

ical

mot

ion

, H =

-16

t2 +

vt

+ h

, wh

ere

H i

s th

e h

eigh

t of

an

obj

ect

afte

r t

seco

nds

, v

is t

he

init

ial

velo

city

, an

d h

is

the

init

ial

hei

ght.

(H

int:

Sin

ce t

he

obje

ct i

s th

row

n

dow

n, t

he

init

ial

velo

city

is

neg

ativ

e.)

H =

-16

t2 -

10t

+ 3

0

b. H

ow l

ong

does

it

take

th

e ti

le t

o h

it t

he

grou

nd?

ab

ou

t 1.

1 s

26. P

HY

SIC

S L

upe

tos

ses

a ba

ll u

p to

Qu

yen

, wai

tin

g at

a t

hir

d-st

ory

win

dow

, wit

h a

n

init

ial

velo

city

of

30 f

eet

per

seco

nd.

Sh

e re

leas

es t

he

ball

fro

m a

hei

ght

of 6

fee

t. T

he

equ

atio

n h

= -

16t2

+ 3

0t +

6 r

epre

sen

ts t

he

hei

ght

h o

f th

e ba

ll a

fter

t s

econ

ds. I

f th

e ba

ll m

ust

rea

ch a

hei

ght

of 2

5 fe

et f

or Q

uye

n t

o ca

tch

it,

doe

s th

e ba

ll r

each

Qu

yen

? E

xpla

in. (

Hin

t: S

ubs

titu

te 2

5 fo

r h

an

d u

se t

he

disc

rim

inan

t.)

No

; th

e d

iscr

imin

ant,

-31

6, is

neg

ativ

e, s

o t

her

e is

no

rea

l so

luti

on

.

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3212

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-5

Cha

pte

r 9

33

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

So

lvin

g Q

ua

dra

tic

Eq

ua

tio

ns b

y U

sin

g t

he

Qu

ad

rati

c F

orm

ula

1. B

USI

NES

S T

anya

ru

ns

a ca

teri

ng

busi

nes

s. B

ased

on

her

rec

ords

, her

w

eekl

y pr

ofit

can

be

appr

oxim

ated

by

the

fun

ctio

n f

(x)

= x

2 + 2

x -

37,

wh

ere

x is

th

e n

um

ber

of m

eals

sh

e ca

ters

. If

f(x)

is

neg

ativ

e, i

t m

ean

s th

at t

he

busi

nes

s h

as

lost

mon

ey. W

hat

is

the

leas

t n

um

ber

of

mea

ls t

hat

Tan

ya n

eeds

to

cate

r in

ord

er

to h

ave

a pr

ofit

?6

mea

ls

2. A

ERO

NA

UTI

CS

At

lift

off,

the

spac

e sh

utt

le D

isco

very

has

a c

onst

ant

acce

lera

tion

of

16.4

fee

t pe

r se

con

d sq

uar

ed a

nd

an i

nit

ial

velo

city

of

1341

fee

t pe

r se

con

d du

e to

th

e ro

tati

on

of E

arth

. Th

e di

stan

ce D

isco

very

has

tr

avel

ed t

sec

onds

aft

er l

ifto

ff i

s gi

ven

by

the

equ

atio

n d

(t)

= 1

341t

+ 8

.2t2 .

How

lo

ng

afte

r li

ftof

f h

as D

isco

very

tra

vele

d 40

,000

fee

t? R

oun

d yo

ur

answ

er t

o th

e n

eare

st t

enth

.25

.8 s

eco

nd

s

3. A

RC

HIT

ECTU

RE

T

he

Gol

den

Rat

io

appe

ars

in t

he

desi

gn o

f th

e G

reek

P

arth

enon

bec

ause

th

e w

idth

an

d h

eigh

t of

th

e fa

çade

are

rel

ated

by

the

e

quat

ion

W +

H

−

W

= W

−

H

. If

th

e h

eigh

t of

a

mod

el o

f th

e P

arth

enon

is

16 i

nch

es,

wh

at i

s it

s w

idth

? R

oun

d yo

ur

answ

er t

o th

e n

eare

st t

enth

.

25.9

in.

4. C

RA

FTS

Mad

elyn

cu

t a

60-i

nch

pip

e cl

ean

er i

nto

tw

o u

neq

ual

pie

ces,

an

d th

en

she

use

d ea

ch p

iece

to

mak

e a

squ

are.

T

he

sum

of

the

area

s of

th

e sq

uar

es w

as

117

squ

are

inch

es. L

et x

be t

he

len

gth

of

one

piec

e. W

rite

an

d so

lve

an e

quat

ion

to

repr

esen

t th

e si

tuat

ion

an

d fi

nd

the

len

gth

s of

th

e tw

o or

igin

al p

iece

s.

(

60 -

x

−

4 ) 2 +

( x

−

4 ) 2 = 1

17;

2

4 in

. an

d 3

6 in

.

5. S

ITE

DES

IGN

Th

e to

wn

of

Sm

allp

ort

plan

s to

bu

ild

a n

ew w

ater

tre

atm

ent

plan

t on

a r

ecta

ngu

lar

piec

e of

lan

d 75

yar

ds w

ide

and

200

yard

s lo

ng.

Th

e bu

ildi

ngs

an

d fa

cili

ties

nee

d to

cov

er a

n

area

of

10,0

00 s

quar

e ya

rds.

Th

e to

wn

’s

zon

ing

boar

d w

ants

th

e si

te d

esig

ner

to

allo

w a

s m

uch

roo

m a

s po

ssib

le b

etw

een

ea

ch e

dge

of t

he

site

an

d th

e bu

ildi

ngs

an

d fa

cili

ties

. Let

x r

epre

sen

t th

e w

idth

of

th

e bo

rder

.

Build

ings

and

Fac

ilitie

s

Bord

er

75 y

d

200

yd

x x

xx

a.

Use

an

equ

atio

n s

imil

ar t

o A

= �

× w

to

rep

rese

nt

the

situ

atio

n.

10,0

00

= (2

00

- 2

x)(7

5 -

2x)

b

. W

rite

th

e eq

uat

ion

in

sta

nda

rd

quad

rati

c fo

rm.

4 x 2 -

550

x +

50

00

= 0

c.

Wh

at s

hou

ld b

e th

e w

idth

of

the

bord

er?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

9.8

yd

9-5

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3312

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

34

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Go

lden

Recta

ng

les

A g

old

en r

ecta

ngl

e h

as t

he

prop

erty

th

at i

ts s

ides

sat

isfy

th

e fo

llow

ing

prop

orti

on.

a +

b

−

a =

a −

b

Tw

o qu

adra

tic

equ

atio

ns

can

be

wri

tten

fro

m t

he

prop

orti

on.

Th

ese

are

som

etim

es c

alle

d go

lden

qu

adra

tic

equ

atio

ns.

1. I

n t

he

prop

orti

on, l

et a

= 1

. Use

2.

Sol

ve t

he

equ

atio

n i

n E

xerc

ise

1 fo

r b.

cros

s-m

ult

ipli

cati

on t

o w

rite

aqu

adra

tic

equ

atio

n.

b =

−

1 ±

√ �

5

−

2

b2

+ b

- 1

= 0

3. I

n t

he

prop

orti

on, l

et b

= 1

. Wri

te

4. S

olve

th

e eq

uat

ion

in

Exe

rcis

e 3

for

a.

a qu

adra

tic

equ

atio

n i

n a

.

a2

- a

- 1

= 0

a

= 1

± √

�

5

−

2

5. E

xpla

in w

hy

1 −

2 (

√ �

5 +

1)

and

1 −

2 (

√ �

5 -

1)

are

call

ed g

olde

n r

atio

s.

T

hey

are

th

e ra

tio

s o

f th

e si

des

in a

go

lden

rec

tan

gle

. Th

e fi

rst

is t

he

rati

o

of

the

lon

g s

ide

to t

he

sho

rt s

ide;

th

e se

con

d is

sh

ort

sid

e: lo

ng

sid

e.

An

oth

er p

rope

rty

of g

olde

n r

ecta

ngl

es

is t

hat

a s

quar

e dr

awn

in

side

a

gold

en r

ecta

ngl

e cr

eate

s an

oth

er,

smal

ler

gold

en r

ecta

ngl

e.

In t

he

desi

gn a

t th

e ri

ght,

opp

osit

e ve

rtic

es o

f ea

ch s

quar

e h

ave

been

co

nn

ecte

d w

ith

qu

arte

rs o

f ci

rcle

s.

For

exa

mpl

e, t

he

arc

from

poi

nt

B t

o po

int

C i

s cr

eate

d by

pu

ttin

g th

e po

int

of a

com

pass

at

poin

t A

. Th

e ra

diu

s of

th

e ar

c is

th

e le

ngt

h B

A.

6. O

n a

sep

arat

e sh

eet

of p

aper

, dra

w a

lar

ger

vers

ion

of

the

desi

gn. S

tart

wit

h a

gol

den

re

ctan

gle

wit

h a

lon

g si

de o

f 10

in

ches

.

T

he

sho

rt s

ide

sho

uld

be

abo

ut

6 3 −

16

in

ches

.

b

a

9-5

C

BA

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3412

/23/

10

2:43

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 9-6

Cha

pte

r 9

35

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

An

aly

zin

g F

un

ctio

ns

with

Su

ccess

ive D

iffe

ren

ces

an

d R

atio

s

9-6

Iden

tify

Fu

nct

ion

s L

inea

r fu

nct

ion

s, q

uad

rati

cfu

nct

ion

s, a

nd

expo

nen

tial

fu

nct

ion

s ca

n a

ll b

e u

sed

to m

odel

dat

a. T

he

gen

eral

for

ms

of t

he

equ

atio

ns

are

list

ed a

t th

e ri

ght.

You

can

als

o id

enti

fy d

ata

as l

inea

r, qu

adra

tic,

or

expo

nen

tial

bas

ed o

n p

atte

rns

of b

ehav

ior

of t

hei

r y-

valu

es.

G

rap

h t

he

set

of o

rder

ed

pai

rs {

(–3,

2),

(–2,

–1)

, (–1,

–2)

, (0,

–1)

, (1

, 2)}

. Det

erm

ine

wh

eth

er t

he

ord

ered

p

airs

rep

rese

nt

a li

nea

r fu

nct

ion

, a

qu

ad

rati

c fu

nct

ion

, or

an e

xpon

enti

al

fun

ctio

n.

Th

e or

dere

d pa

irs

appe

ar t

o re

pres

ent

a qu

adra

tic

fun

ctio

n.

L

ook

for

a p

atte

rn i

n t

he

tab

le t

o d

eter

min

e w

hic

h m

odel

bes

t d

escr

ibes

th

e d

ata.

Sta

rt b

y co

mpa

rin

g th

e fi

rst

diff

eren

ces.

Th

e fi

rst

diff

eren

ces

are

not

all

equ

al. T

he

tabl

e do

es n

ot r

epre

sen

t a

lin

ear

fun

ctio

n.

Fin

d th

e se

con

d di

ffer

ence

s an

d co

mpa

re.

Th

e ta

ble

does

not

rep

rese

nt

a qu

adra

tic

fun

ctio

n. F

ind

the

rati

os o

f th

e y-

valu

es.

Th

e ra

tios

are

equ

al. T

her

efor

e, t

he

tabl

e ca

n b

e m

odel

ed b

y an

exp

onen

tial

fu

nct

ion

.

Exam

ple

1

x

y

Exam

ple

2

x–2

–1

01

2

y4

21

0.5

0.2

5

-2

-

1

-0.

5

-0.

25

+

1

+ 0

.5

+

0.2

5

4 2

1

0.

5 0.

25

×

0.5

×

0.5

× 0

.5

×

0.5

4 2

1

0.

5

0.

25

-2

-1

-0.

5

-0.

25

Exer

cise

s

Gra

ph

eac

h s

et o

f or

der

ed p

airs

. Det

erm

ine

wh

eth

er t

he

ord

ered

pai

rs r

epre

sen

t a

lin

ear

fun

ctio

n, a

qu

ad

rati

c fu

nct

ion

, or

an e

xpon

enti

al

fun

ctio

n.

1. (

0, –

1), (

1, 1

), (2

, 3),

(3, 5

)

2. (–

3, –

1), (

–2, –

4), (

–1, –

5), (

0, –

4), (

1, –

1)

x

y

x

y

Loo

k f

or a

pat

tern

in

eac

h t

able

to

det

erm

ine

wh

ich

mod

el b

est

des

crib

es t

he

dat

a.

3.

x–2

–1

01

2

y6

54

32

4.

x

–2

–1

01

2

y6

.25

2.5

10

.40

.16

Lin

ea

r F

un

ctio

n

y

= m

x +

b

Qu

ad

ratic F

un

ctio

n

y

= a

x2 + b

x +

c

Exp

on

en

tia

l F

un

ctio

n

y

= a

bx

linea

rq

uad

rati

c

linea

rex

po

nen

tial

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3512

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

36

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

An

aly

zin

g F

un

cti

on

s w

ith

Su

cc

essiv

e D

iffe

ren

ce

s a

nd

Ra

tio

s

9-6

Wri

te E

qu

atio

ns

On

ce y

ou f

ind

the

mod

el t

hat

bes

t de

scri

bes

the

data

, you

can

wri

te

an e

quat

ion

for

th

e fu

nct

ion

.

Bas

ic F

orm

s

Lin

ea

r F

un

ctio

n

y

= m

x +

b

Qu

ad

ratic F

un

ctio

n

y

= a

x2

Exp

on

en

tia

l F

un

ctio

n

y

= a

bx

D

eter

min

e w

hic

h m

odel

bes

t d

escr

ibes

th

e d

ata.

Th

en w

rite

an

eq

uat

ion

for

th

e fu

nct

ion

th

at m

odel

s th

e d

ata.

x 0

1 2

34

y3

612

24

48

Ste

p 1

D

eter

min

e w

het

her

th

e da

ta i

s m

odel

ed b

y a

lin

ear,

quad

rati

c, o

r ex

pon

enti

al f

un

ctio

n.

F

irst

dif

fere

nce

s:

3 6

12

24

48

+

3

+ 6

+ 1

2

+

24

S

econ

d di

ffer

ence

s:

3

6

12

24

+

3

+ 6

+ 1

2

y-

valu

e ra

tios

: 3

6

12

24

48

×

2

× 2

× 2

× 2

Th

e ra

tios

of

succ

essi

ve y

-val

ues

are

equ

al. T

her

efor

e, t

he

tabl

e of

va

lues

can

be

mod

eled

by

an e

xpon

enti

al f

un

ctio

n.

Ste

p 2

W

rite

an

equ

atio

n f

or t

he

fun

ctio

n t

hat

mod

els

the

data

. Th

e eq

uat

ion

has

th

e fo

rm y

= a

bx . T

he

y-va

lue

rati

o is

2, s

o th

is i

s th

e va

lue

of t

he

base

.

y

= a

bx E

quation for

exponential fu

nction

3

= a

(2)0

x =

0,

y =

3,

and b

= 2

3

= a

S

implif

y.

An

equ

atio

n t

hat

mod

els

the

data

is

y =

3 ․

2 x .

To

chec

k th

e re

sult

s, y

ou c

an v

erif

y th

at t

he

oth

er o

rder

ed p

airs

sat

isfy

th

e fu

nct

ion

.

Exer

cise

sL

ook

for

a p

atte

rn i

n e

ach

tab

le o

f va

lues

to

det

erm

ine

wh

ich

mod

el b

est

des

crib

es t

he

dat

a. T

hen

wri

te a

n e

qu

atio

n f

or t

he

fun

ctio

n t

hat

mod

els

the

dat

a.

1.

x

–2

–1

01

2

y12

30

312

qu

adra

tic;

y=

3x

2

2.

x

–1

01

23

y–2

14

710

lin

ear;

y =

3x

+ 1

3.

x

–1

01

23

y0

.75

312

48

19

2

ex

po

nen

tial

; y =

3 �

4x

Exam

ple

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

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dd

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10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-6

Cha

pte

r 9

37

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceA

na

lyzin

g F

un

cti

on

s w

ith

Su

cc

essiv

e D

iffe

ren

ce

s a

nd

Ra

tio

s

9-6

Gra

ph

eac

h s

et o

f or

der

ed p

airs

. Det

erm

ine

wh

eth

er t

he

ord

ered

pai

rs r

epre

sen

t a

lin

ear

fun

ctio

n, a

qu

ad

rati

c fu

nct

ion

, or

an e

xpon

enti

al

fun

ctio

n.

1. (

2, 3

), (1

, 1),

(0, –

1), (

–1,

–3)

, (–3,

–5)

2

. (–

1, 0

.5),

(0, 1

), (1

, 2),

(2, 4

)

Ox

y

lin

ear

x

y

exp

on

enti

al

3. (

–2,

4),

(–1,

1),

(0, 0

), (1

, 1),

(2, 4

) 4

. (–

3, 5

), (–

2, 2

), (–

1, 1

), (0

, 2),

(1, 5

)

x

y

qu

adra

tic

x

y

q

uad

rati

c

Loo

k f

or a

pat

tern

in

eac

h t

able

of

valu

es t

o d

eter

min

e w

hic

h m

odel

bes

t d

escr

ibes

th

e d

ata.

Th

en w

rite

an

eq

uat

ion

for

th

e fu

nct

ion

th

at m

odel

s th

e d

ata.

5.

x–3

–2

–1

01

2

y 3

216

84

21

ex

po

nen

tial

; y =

4 ·

(0.5

) x

6.

x–1

01

23

y7

3–1

–5

–9

lin

ear;

y =

-4x

+ 3

7.

x–3

–2

–1

01

y–2

7–12

–3

0–3

q

uad

rati

c; y

= -

3x2

8.

x0

12

34

y0

.51.

54

.513

.54

0.5

ex

po

nen

tial

; y =

1 −

2 ·

3 x

9.

x–2

–1

01

2

y–8

–4

04

8

lin

ear;

y =

4x

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3712

/23/

10

10:5

7 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

38

Gle

ncoe

Alg

ebra

1

9-6

Prac

tice

A

na

lyzin

g F

un

cti

on

s w

ith

Su

cc

essiv

e D

iffe

ren

ce

s a

nd

Ra

tio

s

Gra

ph

eac

h s

et o

f or

der

ed p

airs

. Det

erm

ine

wh

eth

er t

he

ord

ered

pai

rs r

epre

sen

t a

lin

ear

fun

ctio

n, a

qu

ad

rati

c fu

nct

ion

, or

an e

xpon

enti

al

fun

ctio

n.

1. (

4, 0

.5),

(3, 1

.5),

(2, 2

.5),

(1, 3

.5),

(0, 4

.5)

2.

(–1,

1 −

9 ) ,

(0,

1 −

3 ) ,

(1, 1

), (2

, 3)

x

y O

lin

ear

x

y

ex

po

nen

tial

3. (

–4,

4),

(–2,

1),

(0, 0

), (2

, 1),

(4, 4

) 4

. (–

4, 2

), (–

2, 1

), (0

, 0),

(2, –

1), (

4, –

2)

x

y

q

uad

rati

c

x

y

li

nea

r

Loo

k f

or a

pat

tern

in

eac

h t

able

of

valu

es t

o d

eter

min

e w

hic

h m

odel

bes

t d

escr

ibes

th

e d

ata.

Th

en w

rite

an

eq

uat

ion

for

th

e fu

nct

ion

th

at m

odel

s th

e d

ata.

5.

x –

3 –

11

35

y–5

–2

1

4 7

lin

ear;

y =

3 −

2 x

- 1 −

2

6.

x–2

–1

01

2

y0

.02

0.2

22

02

00

ex

po

nen

tial

; y =

2� 1

0 x

7.

x–1

01

23

y6

06

24

54

q

uad

rati

c; y

= 6

x2

8.

x–2

–1

01

2

y18

90

–9

–18

lin

ear;

y =

-9x

9. I

NSE

CTS

Th

e lo

cal

zoo

keep

s tr

ack

of t

he

nu

mbe

r of

dra

gon

flie

s br

eedi

ng

in t

hei

r in

sect

ex

hib

it e

ach

day

.

Day

12

34

5

Dra

go

nfl i

es9

18

36

72

14

4

a. D

eter

min

e w

hic

h f

un

ctio

n b

est

mod

els

the

data

. ex

po

nen

tial

b

. W

rite

an

equ

atio

n f

or t

he

fun

ctio

n t

hat

mod

els

the

data

. y =

9 �

2 x

c.

Use

you

r eq

uat

ion

to

dete

rmin

e th

e n

um

ber

of d

rago

nfl

ies

that

wil

l be

bre

edin

g af

ter

9 da

ys.

4608

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go

nfl i

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10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-6

Cha

pte

r 9

39

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

An

aly

zin

g F

un

cti

on

s w

ith

Su

ccessiv

e D

iffe

ren

ces a

nd

Rati

os

9-6

exp

on

enti

al

y =

20

� (0

.5)x

exp

on

enti

al;

y =

1.0

5 �

2n

701.

8 ye

ars

y =

825

0 �

(0.8

4)x;

$171

7.78

linea

r

0.0

098

g

1. W

EATH

ER T

he

San

Mat

eo w

eath

er

stat

ion

rec

ords

th

e am

oun

t of

rai

nfa

ll

sin

ce t

he

begi

nn

ing

of a

th

un

ders

torm

. D

ata

for

a st

orm

is

reco

rded

as

a se

ries

of

ord

ered

pai

rs s

how

n b

elow

, wh

ere

the

x va

lue

is t

he

tim

e in

min

ute

s si

nce

th

e st

art

of t

he

stor

m, a

nd

the

y va

lue

is t

he

amou

nt

of r

ain

in

in

ches

th

at h

as f

alle

n

sin

ce t

he

star

t of

th

e st

orm

.

(2, 0

.3),

(4, 0

.6),

(6, 0

.9),

(8, 1

.2),

(10,

1.5

)

Gra

ph t

he

orde

red

pair

s. D

eter

min

e w

het

her

th

e or

dere

d pa

irs

repr

esen

t a

lin

ear

fun

ctio

n, a

qu

adra

tic

fun

ctio

n, o

r an

exp

onen

tial

fu

nct

ion

.

Total Rainfall (in.)

0.2 0

0.4

0.6

0.8

1.0

1.2

1.4

1.6

Tim

e (m

in)

24

68

1012

2. I

NV

ESTI

NG

Th

e va

lue

of a

cer

tain

par

cel

of l

and

has

bee

n i

ncr

easi

ng

in v

alu

e ev

er

sin

ce i

t w

as p

urc

has

ed. T

he

tabl

e sh

ows

the

valu

e of

th

e la

nd

parc

el o

ver

tim

e.

Year

Sin

ce

Pu

rch

asin

g0

12

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d V

alu

e

(th

ou

san

ds

$)$

1.0

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2.1

0$

4.2

0$

8.4

0$

16

.80

Loo

k fo

r a

patt

ern

in

th

e ta

ble

of v

alu

es

to d

eter

min

e w

hic

h m

odel

bes

t de

scri

bes

the

data

. Th

en w

rite

an

equ

atio

n f

or t

he

fun

ctio

n t

hat

mod

els

the

data

.

3. B

OA

TS T

he

valu

e of

a b

oat

typi

call

y de

prec

iate

s ov

er t

ime.

Th

e ta

ble

show

s th

e va

lue

of a

boa

t ov

er a

per

iod

of t

ime.

Year

s0

12

34

Bo

at

Val

ue

($)

82

50

69

30

58

21.

20

48

89

.81

410

7.4

4

Wri

te a

n e

quat

ion

for

th

e fu

nct

ion

th

at

mod

els

the

data

. Th

en u

se t

he

equ

atio

n

to d

eter

min

e h

ow m

uch

th

e bo

at i

s w

orth

af

ter

9 ye

ars.

4. N

UC

LEA

R W

AST

E R

adio

acti

ve m

ater

ial

slow

ly d

ecay

s ov

er t

ime.

Th

e am

oun

t of

ti

me

nee

ded

for

an a

mou

nt

of r

adio

acti

ve

mat

eria

l to

dec

ay t

o h

alf

its

init

ial

quan

tity

is

know

n a

s it

s h

alf-

life

. C

onsi

der

a 20

-gra

m s

ampl

e of

a

radi

oact

ive

isot

ope.

Hal

f-L

ives

Ela

pse

d0

12

34

Am

ou

nt

of

Iso

top

e

Rem

ain

ing

(g

ram

s)2

010

52

.51.

25

a. I

s ra

dioa

ctiv

e de

cay

a li

nea

r de

cay,

qu

adra

tic

deca

y, o

r an

exp

onen

tial

de

cay?

b.

Wri

te a

n e

quat

ion

to

dete

rmin

e h

ow

man

y gr

ams

y of

a r

adio

acti

ve i

soto

pe

wil

l be

rem

ain

ing

afte

r x

hal

f-li

ves.

c. H

ow m

any

gram

s of

th

e is

otop

e w

ill

rem

ain

aft

er 1

1 h

alf-

live

s?

d.

Plu

ton

ium

-238

is

one

of t

he

mos

t da

nge

rou

s w

aste

pro

duct

s of

nu

clea

r po

wer

pla

nts

. If

the

hal

f-li

fe o

f pl

uto

niu

m-2

38 i

s 87

.7 y

ears

, how

lo

ng

wou

ld i

t ta

ke f

or a

20-

gram

sa

mpl

e of

plu

ton

ium

-238

to

deca

y to

0.0

78 g

ram

?

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

3912

/23/

10

10:5

7 A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

40

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

9-6

Sie

rpin

ski Tr

ian

gle

S

ierp

insk

i Tri

angl

e is

an

exa

mpl

e of

a f

ract

al t

hat

ch

ange

s ex

pon

enti

ally

. Sta

rt w

ith

an

eq

uil

ater

al t

rian

gle

and

fin

d th

e m

idpo

ints

of

each

sid

e. T

hen

con

nec

t th

e m

idpo

ints

to

form

a s

mal

ler

tria

ngl

e. R

emov

e th

is s

mal

ler

tria

ngl

e fr

om t

he

larg

er o

ne.

Rep

eat

the

proc

ess

to c

reat

e th

e n

ext

tria

ngl

e in

th

e se

quen

ce. F

ind

the

mid

poin

ts o

f th

e si

des

of t

he

thre

e re

mai

nin

g tr

ian

gles

an

d co

nn

ect

them

to

form

sm

alle

r tr

ian

gles

to

be r

emov

ed.

1. F

ind

the

nex

t tr

ian

gle

in t

he

sequ

ence

. How

mu

ch h

as b

een

cu

t ou

t? W

hat

is

the

area

of

the

fou

rth

fig

ure

in

th

e se

quen

ce?

S

ee s

tud

ents

’ tri

ang

les;

1 −

4 +

3

−

16 +

9

−

64 o

r 37

−

64

; 1

- 37

−

64

or

27

−

64

2. M

ake

a co

nje

ctu

re a

s to

wh

at y

ou n

eed

to m

ult

iply

th

e pr

evio

us

amou

nt

cut

by t

o fi

nd

the

new

am

oun

t cu

t.

3 −

4

3. F

ill

in t

he

char

t to

rep

rese

nt

the

amou

nt

cut

and

the

area

rem

ain

ing

for

each

tri

angl

e in

th

e se

quen

ce. Fi

gu

re1

23

45

6

Am

ou

nt

Cu

t0

1

−

4

7

−

16

37

−

64

175

−

256

781

−

1024

Are

a R

emai

nin

g1

3

−

4

9

−

16

27

−

64

81

−

256

243

−

1024

4. W

rite

an

equ

atio

n t

o re

pres

ent

the

area

th

at i

s le

ft i

n t

he

nth

tri

angl

e in

th

e se

quen

ce.

A =

( 3 −

4 ) n

5. I

f th

is p

roce

ss i

s co

nti

nu

ed, m

ake

a co

nje

ctu

re a

s to

th

e re

mai

nin

g ar

ea.

T

he

rem

ain

ing

are

a g

ets

very

clo

se t

o 0

.

figur

e 2

figur

e 3

figur

e 1

Cut o

ut:

1 4

Area

= 1

-

or1 4

3 4

Cut o

ut:

1 43 16

Area

= 1

-

or7 16

9 16

7 16+

or

023-

040_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4012

/23/

10

10:5

7 A

M


NA

ME

DAT

E

P

ER

IOD

Lesson 9-7

Cha

pte

r 9

41

Gle

ncoe

Alg

ebra

1

9-7

Stud

y G

uide

and

Inte

rven

tion

Sp

ecia

l Fu

ncti

on

s

Step

Fu

nct

ion

s T

he

grap

h o

f a

step

fu

nct

ion

is

a se

ries

of

disj

oin

ted

lin

e se

gmen

ts.

Bec

ause

eac

h p

art

of a

ste

p fu

nct

ion

is

lin

ear,

this

typ

e of

fu

nct

ion

is

call

ed a

p

iece

wis

e-li

nea

r fu

nct

ion

.O

ne

exam

ple

of a

ste

p fu

nct

ion

is

the

grea

test

in

tege

r fu

nct

ion

, wri

tten

as

f(x)

= �

x�, w

her

e f(

x) i

s th

e gr

eate

st i

nte

ger

not

gre

ater

th

an x

.

G

rap

h f

(x)

= �

x +

3�.

Mak

e a

tabl

e of

val

ues

usi

ng

inte

ger

and

non

inte

ger

valu

es. O

n t

he

grap

h, d

ots

repr

esen

t in

clu

ded

poin

ts, a

nd

circ

les

repr

esen

t po

ints

th

at a

re e

xclu

ded.

xx

+ 3

�x +

3�

–5

–2

–2

–3

.5–

0.5

–1

–2

11

–0

.52

.52

14

4

2.5

5.5

5

Bec

ause

th

e do

ts a

nd

circ

les

over

lap,

th

e do

mai

n i

s al

l re

al n

um

bers

. Th

e ra

nge

is

all

inte

gers

.

Exer

cise

sG

rap

h e

ach

fu

nct

ion

. Sta

te t

he

dom

ain

an

d r

ange

.

1. f

(x)

= �

x +

1�

2. f

(x)

= –

�x�

3. f

(x)

= �

x – 1

�

4. f

(x)

= �

x� +

4

5. f

(x)

= �

x� –

3

6. f

(x)

= �

2x�x

f(x)

xO

f(x)

xO

f(x)

xO

f(x)

Exam

ple

xO

f(x)

xO

f(x)

xO

f(x)

1–6.

D =

{al

l rea

l nu

mb

ers}

; R

= {

all i

nte

ger

s}

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4112

/23/

10

2:36

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

42

Gle

ncoe

Alg

ebra

1

9-7

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Sp

ecia

l Fu

ncti

on

s

Ab

solu

te V

alu

e Fu

nct

ion

s A

not

her

typ

e of

pie

cew

ise-

lin

ear

fun

ctio

n i

s th

e ab

solu

te

valu

e fu

nct

ion

. Rec

all

that

th

e ab

solu

te v

alu

e of

a n

um

ber

is a

lway

s n

onn

egat

ive.

So

in

the

abso

lute

val

ue

fun

ctio

n, w

ritt

en a

s f (

x) =

|x|

, all

of

the

valu

es o

f th

e ra

nge

are

n

onn

egat

ive.

Th

e ab

solu

te v

alu

e fu

nct

ion

is

call

ed a

pie

cew

ise-

def

ined

fu

nct

ion

bec

ause

it

can

be

wri

tten

usi

ng

two

or m

ore

expr

essi

ons.

G

rap

h f

(x)

= |

x +

2|.

S

tate

th

e d

omai

n a

nd

ran

ge.

f (x)

can

not

be

neg

ativ

e, s

o th

e m

inim

um

po

int

is f

(x)

= 0

. f (

x) =

|x

+ 2

| O

rigin

al fu

nction

0

= x

+ 2

R

epla

ce f

(x)

with 0

.

-2

= x

S

ubtr

act

2 f

rom

each s

ide.

Mak

e a

tabl

e. I

ncl

ude

val

ues

for

x >

-2

and

x <

-2.

Th

e do

mai

n i

s al

l re

al n

um

bers

. Th

e ra

nge

is

all

real

nu

mbe

rs g

reat

er t

han

or

equ

al t

o 0.

G

rap

h

f(x)

= {x

+ 1

if

x >

1

3

x i

f x

≤ 1

. S

tate

th

e d

omai

n

and

ran

ge.

Gra

ph t

he

firs

t ex

pres

sion

. Wh

en x

> 1

,f (

x) =

x +

1. S

ince

x ≠

1, p

lace

an

ope

n

circ

le a

t (1

, 2).

Nex

t, g

raph

th

e se

con

d ex

pres

sion

. Wh

en

x ≤

1, f

(x)

= 3

x. S

ince

x =

1, p

lace

a c

lose

d ci

rcle

at

(1, 3

).

Th

e do

mai

n a

nd

ran

ge a

re b

oth

all

rea

l n

um

bers

.

x

f(x)

Exer

cise

sG

rap

h e

ach

fu

nct

ion

. Sta

te t

he

dom

ain

an

d r

ange

.

1. f

(x)

= |

x -

1|

2. f(

x) =

|-

x +

2|

3. f(

x) =

{ - x

+ 4

if

x ≤

1

x -

2

if

x >

1

D

= {

all r

eal n

um

ber

s};

D

= {

all r

eal n

um

ber

s};

D =

{al

l rea

l nu

mb

ers}

;

R =

{ y

| y

> 0

} R

= {

y |

y >

0}

R

= {

y |

y >

-1}

x

f(x)

xf(

x)

-5

3

-4

2

-3

1

-2

0

-1

1

02

13

24

Exam

ple

1Ex

amp

le 2

x

f(x)

y =

|x -

1|

x

f(x)

x

f(x)

y =

|-x

+ 2

|

y

x

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4212

/23/

10

9:46

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 9-7

Cha

pte

r 9

43

Gle

ncoe

Alg

ebra

1

9-7

Skill

s Pr

acti

ceS

pecia

l Fu

ncti

on

s

Gra

ph

eac

h f

un

ctio

n. S

tate

th

e d

omai

n a

nd

ran

ge.

1. f

(x)

= �

x – 2

� 2.

f (

x) =

3�x

� 3.

f (

x) =

�2x

�

4. f

(x)

= |

x| –

3

5.

f(x)

= |

2x|

6.

f(x)

= |

2x +

5|

7. f

(x)

= {2x

if x

≤ 1

–x

+ 3

if

x >

2

8. f

(x)

= {x

+ 4

if

x ≤

1

if

x >

1

0.25

x +

1

9. f

(x)

= {x

+ 2

i

f x

< 0

i

f x

≥ 0

-

0.5x

+ 1

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

D =

{al

l rea

l nu

mb

ers}

; R

= {

all i

nte

ger

s}D

= {

all r

eal n

um

ber

s};

R =

{al

l in

teg

ers

even

ly

div

isib

le b

y 3}

D =

{al

l rea

l nu

mb

ers}

; R

= {

all i

nte

ger

s}

D =

{al

l rea

l nu

mb

ers}

; R

= {

y |

y ≥

-3}

D =

{al

l rea

l nu

mb

ers}

; R

= {

y |

y ≥

0}

D =

{al

l rea

l nu

mb

ers}

; R

= {

y |

y ≥

0}

D =

{x |

x ≤

1, x

> 2

};

R =

{y |

y ≤

2}

D =

{al

l rea

l nu

mb

ers}

; R

= {

all r

eal n

um

ber

s}D

= {

all r

eal n

um

ber

s};

R =

{y |

y <

2}

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4312

/23/

10

9:46

PM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

44

Gle

ncoe

Alg

ebra

1

9-7

Prac

tice

S

pecia

l Fu

ncti

on

s

Gra

ph

eac

h f

un

ctio

n. S

tate

th

e d

omai

n a

nd

ran

ge.

1. f

(x)

= -

2�x

+ 1

� 2.

f(x)

= �

x +

3�

- 2

3.

f(x)

= -

| 1 −

2 x

| +

1

x

f(x)

x

f(x)

x

f(x)

x

f(x)

D

= {

all r

eal n

um

sber

s};

D

= {

all r

eal n

um

sber

s};

D =

{al

l rea

l nu

mb

ers}

;

R =

{al

l mu

ltip

les

of

2}

R =

{al

l in

teg

ers}

R

= {

f(x)

� f

(x) ≤

1}

4. f

(x)

= |

2x +

4|

- 3

5.

f(x)

= {2

i

f x

> -

1

x

+ 4

if

x ≤

- 1

6.

f(x)

= {–2

x +

3

if x

> 0

1 −

2 x

- 1

if

x ≤

0

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

D

= {

all r

eal n

um

ber

s};

D =

{al

l rea

l nu

mb

ers}

;

D =

{al

l rea

l nu

mb

ers}

;

R =

{f(

x) �

f(x

) ≥ -

3}

R =

{f(

x)

� f(

x) ≤

3}

R =

{f(

x)

� f(

x) <

3}

Det

erm

ine

the

dom

ain

an

d r

ange

of

each

fu

nct

ion

.

7.

y

x

8.

y

x

9.

xx

y

D

= {

all r

eal n

um

ber

s};

D =

{al

l rea

l nu

mb

ers}

; D

= {

all r

eal n

um

ber

s};

R

= {

y �

y ≤

3}

R =

{al

l in

teg

ers}

R

= {

y �

y ≥

-2}

10. C

ELL

PHO

NES

Jac

ob’s

cel

l ph

one

serv

ice

cost

s $5

eac

h m

onth

pl

us

$0.3

5 fo

r ea

ch m

inu

te h

e u

ses.

Eve

ry f

ract

ion

of

a m

inu

te

is r

oun

ded

up

to t

he

nex

t m

inu

te.

a.

Dra

w a

gra

ph t

o re

pres

ent

the

cost

of

usi

ng

the

cell

ph

one.

b

. W

hat

is

Jaco

b’s

mon

thly

bil

l if

he

use

s 12

4.8

min

ute

s?

$4

8.75

Monthly Bill ($)

5 0

5.506

6.507

7.50

Min

utes

Use

d1

23

45

6

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4412

/23/

10

9:46

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 9-7

Cha

pte

r 9

45

Gle

ncoe

Alg

ebra

1

9-7

Wor

d Pr

oble

m P

ract

ice

Sp

ecia

l Fu

ncti

on

s

1. B

AB

YSI

TTIN

G A

riel

ch

arge

s $4

per

h

our

as a

bab

ysit

ter.

Sh

e ro

un

ds e

very

fr

acti

on o

f an

hou

r u

p to

th

e n

ext

hal

f-h

our.

Dra

w a

gra

ph t

o re

pres

ent

Ari

el’s

tot

al e

arn

ings

y a

fter

x h

ours

.

2. S

HIP

PIN

G A

pac

kage

del

iver

y se

rvic

e de

term

ines

rat

es f

or e

xpre

ss s

hip

pin

g by

th

e w

eigh

t of

a p

acka

ge, w

ith

eve

ry

frac

tion

of

a po

un

d ro

un

ded

up

to t

he

nex

t po

un

d. T

he

tabl

e sh

ows

the

cost

of

expr

ess

ship

pin

g fo

r pa

ckag

es b

etw

een

1

and

9 po

un

ds. W

rite

a p

iece

wis

e-li

nea

r fu

nct

ion

rep

rese

nti

ng

the

cost

to

ship

a

pack

age

betw

een

1 a

nd

9 po

un

ds. S

tate

th

e do

mai

n a

nd

ran

ge.

Wei

gh

t(p

ou

nd

s)R

ate

(do

llars

)

117.

40

219

.30

32

2.4

0

42

5.5

0

52

8.6

0

63

1.7

0

73

4.8

0

83

7.9

0

94

1.0

0

3. D

ISEA

SE P

REV

ENTI

ON

Bod

y M

ass

Inde

x (B

MI)

is

use

d by

doc

tors

to

dete

rmin

e w

eigh

t ca

tego

ries

th

at m

ay

lead

to

hea

lth

pro

blem

s. A

ccor

din

g to

th

e C

ente

rs f

or D

isea

se C

ontr

ol a

nd

Pre

ven

tion

, 21.

7 is

a h

ealt

hy

BM

I fo

r ad

ult

s. I

f th

e B

MI

diff

ers

from

th

e de

sire

d 21

.7 b

y m

ore

than

x, t

her

e m

ay

be h

ealt

h r

isk

invo

lved

. Wri

te a

n

equ

atio

n t

hat

can

be

use

d to

fin

d th

e h

igh

est

and

low

est

BM

I fo

r a

hea

lth

y ad

ult

. Th

en s

olve

if

x =

3.2

.

4. F

UN

DR

AIS

ING

Stu

den

ts a

re s

elli

ng

boxe

s of

coo

kies

at

a fu

nd-

rais

er. T

he

boxe

s of

coo

kies

can

on

ly b

e or

dere

d by

th

e ca

se, w

ith

12

boxe

s pe

r ca

se. D

raw

a

grap

h t

o re

pres

ent

the

nu

mbe

r of

ca

ses

nee

ded

y w

hen

x b

oxes

of

cook

ies

are

sold

.

5. W

AG

ES K

elly

ear

ns

$8 p

er h

our

the

firs

t 8

hou

rs s

he

wor

ks i

n a

day

an

d $1

1.50

per

hou

r ea

ch h

our

ther

eaft

er.

a. O

rgan

ize

the

info

rmat

ion

in

to a

tab

le.

Incl

ude

a c

olu

mn

for

hou

rs w

orke

d x,

an

d a

colu

mn

for

dai

ly e

arn

ings

f (x

).

x0

48

1216

f(x)

032

6411

015

6

b.

Wri

te t

he

piec

ewis

e eq

uat

ion

de

scri

bin

g K

elly

’s d

aily

ear

nin

gs

f (x)

for

x h

ours

.

c. D

raw

a g

raph

to

repr

esen

t K

elly

’s

dail

y ea

rnin

gs.

Daily Earnings ($)

24 0487296120

144

Hour

s W

orke

d2

46

810

12x

fx

( 8, 6

4)

Ariel’s Earnings ($)

1015 5 0202530

Hour

s Ba

bysi

tting

12

3

y

x

Cases Needed

23 1 0456

Boxe

s So

ld24

3660

1248

72

y

x

y =

⎪x

-21

.7⎥ ;

18.

5

f(x)

= ⎧ ⎨

⎩ 8x if

x ≤

864

+ 1

1.5(

x -

8)

if x

> 8

f(x)

= 3

.1�x

-1�

+

16.2

;

D =

{x |

1 <

x <

9}

R =

{f(

x)

| 17

.40

≤

f(x)

≤ 4

1.0

0}

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4512

/23/

10

10:5

8 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 9

46

Gle

ncoe

Alg

ebra

1

9-7

Enri

chm

ent

Para

metr

ic E

qu

ati

on

sA

par

amet

ric

equ

atio

n i

s a

pair

of

fun

ctio

ns

x =

f (t

) an

d y

= g

(t)

that

des

crib

e bo

th t

he

x- a

nd

y-co

ordi

nat

es f

or t

he

grap

h a

s a

wh

ole.

Par

amet

ric

fun

ctio

ns

allo

w t

he

draw

ing

of

man

y co

mpl

ex c

urv

es a

nd

figu

res.

G

rap

h t

he

par

amet

ric

fun

ctio

n g

iven

by

x =

t -

2 a

nd

y =

t +

1 f

or -

2 ≤

t ≤

2.

Ste

p 1

C

reat

e a

tabl

e of

val

ues

for

–2

≤ t

≤ 2

by

eval

uat

ing

x an

d y

for

each

val

ue

of t

.

t–2

–1

0

12

x–4

–3

–2

–1

0

y–1

0

1

2

3

Ste

p 2

P

lot

the

poin

ts.

Ste

p 3

D

raw

a l

ine

to c

onn

ect

the

poin

ts b

etw

een

–2

≤ t

≤ 2

.

Exer

cise

s

Gra

ph

eac

h p

air

of p

aram

etri

c eq

uat

ion

s ov

er t

he

give

n r

ange

of

t va

lues

.

1. x

= t

+ 2

2.

x =

2t

3. x

= t

- 3

y

= t

+ 2

y

= 1 −

2 t

- 1

y

= 2

t -

1

–

2 ≤

t ≤

2

–2

≤ t

≤ 2

–

2 ≤

t ≤

2

y

x

y

x

y

x

4. x

= 4

t 5.

x

= t

2 6.

x

= t

2

y

= 2

t -

2

y =

t +

1

y =

t2

- t

- 1

–

1 ≤

t ≤

1

–2

≤ t

≤ 2

–2

≤ t

≤ 2

y

x

y

x

y

x

y

x

Exam

ple

041-

046_

ALG

1_A

_CR

M_C

09_C

R_6

6028

3.in

dd

4612

/23/

10

10:5

8 A

M



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4.

5.

1.

2.

3.

4.

5.

2, -10

81

− 4

3.4, -2.7

D

-55, no real solutions

x

y

y = x 2 + x - 2

D = {all real numbers}, R = {y | y ≥ -4}

x = -2; (-2, 9);

maximum

-2, 1

x

y

y = 3x 2 - 5x + 1

0 < x < 1, 1 < x < 2

C

1.

2.

3.

1.

2.

3.

C

quadratic

exponential; y = 9. (

− 1 − 3 )

x

x

y

1.

2.

3.

4.

5.

x

g (x)

x

O

f (x)

f (x) = |2x + 1|

D: {all real numbers};

R: {y�y ≥ 0}

0

D

1.

2.

3.

4.

5. B

H

B

F

C

6.

7.

8.

9.

10.

height is 2.4 in.; base is 6.4 in.

0 < x < 1,

4 < x < 5

-1, 8

- 7 ± √ �� 193

− 4

-2 + 2 √ � 6


Quiz 2 (Lessons 9-4 and 9-5)

Page 49

Quiz 4 (Lesson 9-7)

Page 50

Chapter 9 Assessment Answer KeyQuiz 1 (Lessons 9-1 through 9-3) Quiz 3 (Lesson 9-6) Mid-Chapter TestPage 49 Page 50 Page 51


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vertex

Sample answer: The discriminant is the expression, b2 - 4ac, that is under the radical sign in the quadratic formula.

Sample answer: The

axis of symmetry of a

parabola is the line that

divides a parabola into

two matching halves.

quadratic function

Quadratic

Formula


parabola

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

translation

axis of symmetry

double root

1.

2.

3.

4.

5.

6.

7.

8.

B

D

F

C

H

D

G

H B: 1 real solution

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

H

D

F

A

B

J

G

B

F

A

G

C


Chapter 9 Assessment Answer KeyVocabulary Test Form 1Page 52 Page 53 Page 54


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swer

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PDF 2nd

1.

2.

3.

4.

5.

6.

7.

8. J

A

F

C

J

B

G

C

B:

H

C

G

A

F

H

D

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

A

F

x = 1

D

D

H

1.

2.

3.

4.

5.

6.

7. C

F

D

H

B

F

B

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

F

B

F

D

F

D

H

A

G

B

G

-16, 16B:

C

G


Chapter 9 Assessment Answer KeyForm 2A Form 2BPage 55 Page 56 Page 57 Page 58


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1.

2. y

xOy = 2x 2- 3x

3.

4.

5.

6.

7.

8.

9.

x = 2; (2, -5);

minimum

xO

f (x)

f (x ) = x2

-6x +4

0 < x < 1, 5 < x < 6

y

xO 6 12

6

-6

12

-12 -6

y = -x 2 + 3x + 10

refl ected about the x-axis and shifted

6 units up

shifted 2 units down

4, 7

2, -2.5

-2, 3

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

about -1.2 and -0.1

x

g (x)

D: {all real numbers}; R: {y | y ≥ -3}

2.5, -4

0.6, -0.3

-9.4, -14.6

148; 2 real solutions

0; 1 real solution

3.8; 8.8

quadratic; y = 4x2

exponential; y = 2 (3) x

0.8, 11.2


Chapter 9 Assessment Answer KeyForm 2CPage 59 Page 60


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-1 < x <0, 2 < x < 3

x = 1; (1, -3); maximum

-1, 5

x O

f (x)

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

1. y

xO 4 8

4

28 24y = x 2 - 7x + 12

2. y

xO

y = 4x 2 - 8x

3.

4.

5.

6.

7.

8.

9.

3, -9

4, 9

refl ected across the

x-axis and translated

down 3 units

translated up 5 units

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 1 −

2

exponential, y = 7 · 2 n

linear, y = 3x - 1

length is 8.7 in.; width is 5.7 in.

0; 1 real solution

-56; no real solutions

y

x

g (x)

D: {all real numbers}; R: {y | y ≥ -3}

- 3 −

4 , 2 −

3

-14.2, -0.8

-1.4, 15.4

-6, 3 − 2


Chapter 9 Assessment Answer KeyForm 2DPage 61 Page 62


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PDF 2nd

1. y

xO

y = -2x 2 - 3x + 3

2. y

xO

y = 3x 2 - 5x - 2

3.

4.

5.

6.

7.

8.

compression of f(x) = x2, refl ected over the x-axis, translated up 4 units

9.

dilation of f(x) = x2, refl ected over the x-axis,

translated up 5 − 6 unit

10.

11. -0.3, -9.7

12. no real solutions

x = 4

-3, -3

-

1 −

4 , 2 −

3

1 −

3 , -

5 −

2

x = 5 −

2 ; (

5 −

2 , -

109 −

4 ) ;

minimum

-4, -2

13.

14.

15.

16.

17.

18.

19.

20.

B:

- 2 −

3 , 5 −

7 or -0.7, 0.7

-18 or 18

0 times

-0.6, 3.1

4

-7 ± √ �� 49 + 16a −

2a

x

f (x)

−1

−2

1

2

−1

1 2−2

x

g (x)

x

h(x)


Chapter 9 Assessment Answer KeyForm 3 Page 63 Page 64


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Chapter 9 Assessment Answer KeyPage 65, Extended-Response Test

Scoring Rubric

Score General Description Specifi c Criteria

4 Superior

A correct solution that

is supported by well-

developed, accurate

explanations

• Shows thorough understanding of the concepts of

graphing quadratic and exponential functions, solving quadratic equations, using the discriminant, exponential growth, and exponential decay.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• Graphs are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution,

but may contain minor fl aws

in reasoning or computation

• Shows an understanding of the concepts of graphing quadratic and exponential functions, solving quadratic equations, using the discriminant, exponential growth, and exponential decay.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Graphs are mostly accurate and appropriate.

• Satisfi es all requirements of problems.

2 Nearly Satisfactory

A partially correct

interpretation and/or

solution to the problem

• Shows an understanding of most of the concepts of


• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Graphs are mostly accurate.

• Satisfi es the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no

supporting evidence or

explanation

• Final computation is correct.

• No written explanations or work is shown to substantiate

the fi nal computation.

• Graphs may be accurate but lack detail or explanation.

• Satisfi es minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution

indicating no mathematical

understanding of the

concept or task, or no

solution is given

• Shows little or no understanding of most of the concepts of


• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Graphs are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer may be given.


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Chapter 9 Assessment Answer Key Page 65, Extended-Response Test

Sample Answers

1. Sample answer: f(x) = x2; f(x) = 2x

y

xO

y = x2

y = 2x

The domains of the two functions are the same, all real numbers. The range of f(x) = x2 is y ≥ 0, while the range of f(x) = 2 x is y > 0. The quadratic function has a vertical line of symmetry, x = 0, and is symmetrical about that line. However, the exponential function has no symmetry.

2a. Sample answer: x2 + x + 1 = 0; -3; The student should explain that the discriminant is the expression under the radical sign in the Quadratic Formula. Since there is no real square root of a negative number, there are no real roots of an equation whose discriminant is negative.

2b. Sample answer: x2 + 2x + 1 = 0; 0; The student should explain that zero has only one square root. Thus, when the determinant is zero the Quadratic Formula yields only one value when you simplify the numerator of the formula.

2c. Sample answer: x2 + x - 1 = 0; Since this quadratic equation does not factor and the roots are not integers, completing the square would be the better method to use to solve the equation.

3a. C(h) = 125 + 40�h�

3b. $205

3c. 3 ≤ h < 3.5

In addition to the scoring rubric found on page A29, the following sample answers may be used as guidance in evaluating extended-response assessment items.

Tota

l Cha

rge

($)

0

100

200

300

400

500

Hours Worked1 2 3 4 5

Total Charge


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1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

17. A B C D

18.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

5 19.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

1


Chapter 9 Assessment Answer KeyStandardized Test PracticePage 66 Page 67


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20.

21.

22. y

xO

x + y = 4

y = 2x + 4

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

30c.

(-1, 3); maximum

x = -1

about 10,019,394 people

10 s and about 5.6 s

{-12, 12}

(2m + 5)(m + 3)

(x + 7)(x + 5)

9a 4 - 4

-4x 2y + 9xy - 8y 2

{y ⎮ y ≥ 5}

y

xO

y = -x2 - 2x + 2

y = 6x - 8


Chapter 9 Assessment Answer KeyStandardized Test PracticePage 68


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1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

a1 = 14,

an = a

n − 1 − 5,

n ≥ 2

1

b 6 −

144 g 6

20x4y 6

0.00255, 2.55 × 10-3

about 1,412,416 people

about $61,032

$1531.51

an = -5 · (-3)n-1

3a + 2b - 4c

2

p2 + 2p - 15

9k 2 - 1

25t 2 + 10tr + r2

a 3 - a 2 - 13a - 14

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

2p - 1

3(3r - 2t)(3r + 2t)

(3a - 4)(a - 3)

(m - 2)(m - 9)

4pr(p - 4 + 2r)

x = -4; (-4, -4)

{-9}

{

-

2 −

5 , 7

}

{4, 9}

{

-

1 −

2 , 0

}

y

xO

y = x 2 + 8x + 12

y

xO

{-1.8, 1.1}

-8

100

exponential,

y = 6 · 2x


Chapter 9 Assessment Answer KeyUnit 3 TestPage 69 Page 70


chapter 9 resource masters - commack schools 9... · 2018-04-24 · chapter 9 test, form 3 .....63...

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