Math Binder 4
Algebra 2 By Michael Buckley
Three Watson Irvine, CA 92618-2767 Web site: www.sdlback.com Development a...
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Math Binder 4
Algebra 2 By Michael Buckley
Three Watson Irvine, CA 92618-2767 Web site: www.sdlback.com Development and Production: Frishco Ltd. and Pearl Production
ISBN 1-59905-024-2 Copyright © 2006 by Saddleback Educational Publishing. All rights reserved. No part of this book may be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without the written permission of the publisher, with the exception below. Pages labeled with the statement Saddleback Educational Publishing ©2006 are intended for reproduction. Saddleback Educational Publishing grants to individual purchasers of this book the right to make sufficient copies of reproducible pages for use by all students of a single teacher. This permission is limited to a single teacher, and does not apply to entire schools or school systems. Printed in the United States of America 10 09 08 07 06 05 9 8 7 6 5 4 3 2 1
Table of Contents Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Compound Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Graphing Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Introduction to Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Matrix Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Matrix Subtraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Writing the Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Solving a Matrix Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Direct Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Slope-Intercept Form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Point-Slope Form I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Point-Slope Form II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Linear Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Writing an Exponential Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Solving an Exponential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Properties of Rational Exponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Doubling Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 The Number e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Properties of Logarithms: Product Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Properties of Logarithms: Quotient Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Properties of Logarithms: Power Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Properties of Logarithms: Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Solving Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Solving a Natural Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Graphing Quadratic Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Properties of a Graph of a Quadratic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Writing a Quadratic Function from Its Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Quadratic Functions in Intercept Form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Solving Quadratic Equations Using Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Solving a Quadratic Equation by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Solving a Quadratic Equation by Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Using the Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Methods for Solving Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Writing an Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Foci of an Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Margin of Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Writing a System of Equations as a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Using Matrices to Solve a System of Two Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Simplifying Radical Expressions by Removing Perfect Squares . . . . . . . . . . . . . . . . . . . . . . . . . 54 Simplifying Radical Expressions with Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Adding Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Subtracting Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Multiplying Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Dividing Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Rationalizing the Denominator of a Radical Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Square Root of a Negative Real Number: Imaginary Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 61 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Adding Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Subtracting Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Multiplying Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Dividing Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 Absolute Value and Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Finding a Complex Solution to a Simple Quadratic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Finding a Complex Solution to a Quadratic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Solving Cubic Equations: Finding x-Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Factoring Cubic Equations: Sum of Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Factoring Cubic Equations: Difference of Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Writing Cubic Equations in Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Evaluating Polynomials Using Synthetic Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Dividing Polynomials Using Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Algebra 2
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End Behavior of a Polynomial Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Inverse Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Simplifying Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Multiplying Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Dividing Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Rational Functions—Finding Vertical Asymptotes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 End Behavior of Rational Functions: m < n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 End Behavior of Rational Functions: m = n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 End Behavior of Rational Functions: m > n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 End Behavior of Rational Functions: Using All Three Conditions . . . . . . . . . . . . . . . . . . . . . . . 86 Recursive Formulas: Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Recursive Formulas: Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Finding the Sum of a Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Finding the Sum of a Finite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Finding the Sum of an Infinite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Function Operations—Adding Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Function Operations—Subtracting Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Function Operations—Multiplying Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Function Operations—Dividing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Inverse of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Finding Trigonometric Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
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Properties of Real Numbers Real numbers include all rational and irrational numbers. You will use the properties of real numbers listed below throughout your study of algebra. Complete the table by filling in the last column. Summary of Properties of Real Numbers Property Using Symbols
Example
Summary
Commutative Addition: a + b = b + a
–2 + 3 = 3 + (–2)
Multiplication: ab = ba
–2(3) = 3(–2)
Associative Addition: (a + b) + c = a + (b + c)
(–2 + 3) + 4 = –2 + (3 + 4)
Multiplication: (ab) c = a (bc)
(–2 × 3) × 4 = –2 × (3 × 4)
Distributive a(b + c) = ab + ac
–2(3 + 4) = (–2 × 3) + (–2 × 4)
Example Identify the property being used. (5x)(8y) = (5x × 8)y Step 1 Describe the change from the left side
to the right side of the equal sign. Step 2 Match the description of the change
Numbers being multiplied have been regrouped. associative property
with one of the properties.
Practice Identify the property being used.
1. (2a × 7)b = (7 × 2a)b Describe the change from the left side to the right side of the equal sign. Match the description of the change with one of the properties. 2. 5x + (2 − 4x) = 5x + (−4x + 2) 3. (3 × 6x)y = (3 × 6)(xy) Complete the right side of each equation by using the identified property.
4. Associative: (3x + 2y) + 5 = 5. Distributive: x(2 + 5) = Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Absolute Value The absolute value of a number is the distance between the origin of a number line and the point representing that number. Look at the number line below. Both 7 and −7 are 7 units from the origin. 7 units –7
7 units 7
0
The notation for absolute value is | a | and is read as “the absolute value of a.” Rules for the Absolute Value of a Number 1. If a is a positive number, then | a | = a (e.g. | 4 | = 4). 2. If a is zero, then | a | = 0 (e.g. | 0 | = 0). 3. If a is a negative number, then | −a | = a (e.g. | −4 | = 4).
Example Solve the equation. |x | = 15 Step 1 Which positive number is 15 units
15
from the origin? Step 2 Which negative number is 15 units
−15
from the origin? Step 3 Check that both solutions are true.
| 15| = 15 | −15| = 15
true true
Practice Solve.
1. −| x | = –10 Which positive number is 10 units from the origin? Which negative number is 10 units from the origin? Check that both solutions are true.
−| −|
| = −10 |=
2. | x | = 3
5. | −x | = 13
3. | x | = 0
6. −| 5 | = x
4. | −7 | = x
7. −| −6 | = x
Algebra 2
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Absolute Value Equations The absolute value of a number is the distance between the origin of a number line and the point representing that number. To solve an absolute value equation you need to account for the value inside the absolute value symbol being positive, and the value inside the absolute value symbol being negative. Rules for Solving an Absolute Value Equation 1. Account for the value of the expression inside the absolute value symbol being positive. 2. Account for the value of the expression inside the absolute value symbol being negative.
Example Solve. |x + 2| = 10 Step 1 Account for the value of the expression What value of x will result in an amount of
inside the absolute value symbol being 10? 8 positive. | x + 2 | = 10 | 8 + 2 | = 10 True | 10 | = 10 Step 2 Account for the value of the expression What value of x will result in an answer of inside the absolute value symbol being −10? −12 negative. | x + 2 | = 10 | −12 + 2 | = 10 | −10 | = 10 True The solution is −12 and 8.
Practice Solve.
1. |2x − 1| = 9 Account for the value of the expression inside the absolute value symbol being positive.
Account for the value of the expression inside the absolute value symbol being negative.
2. | x − 4 | = 8 3. | x + 3 | = 15
What value of x will result in an answer of 9?
| 2x − 1| = 9 − 1| = 9 |2 | |= 9 What value of x will result in a value of −9?
| 2x − 1| = 9 −1 | = 9 |2 | |= 9 4. | 3x | = 21 5. | 3x + 3 | = 30
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Compound Inequalities A compound inequality is made of two inequalities connected by “and” or “or.” A compound inequality with “and”: −4 < x < 10 A compound inequality with “or”: x < −2 or x > 5 Rules for Solving a Compound Inequality
Compound inequality with “and” (x is in the middle of the expression). 1. Write the original inequality as two inequalities. 2. Solve for the left side of the inequality. 3. Solve for the right side of the inequality. Compound inequality with “or.” 1. Write the original inequality. 2. Solve the left side of the inequality. 3. Solve the right side of the inequality.
Example Solve. −4 2x < 10 (Compound inequality with “and.”) Step 1 Write the original inequality as two
inequalities.
−4 2x < 10 −4 2x and 2x < 10
Step 2 Solve for the left side of the inequality.
−4 2x → −2 x
Step 3 Solve for the right side of the
2x < 10 → x < 5 −2 x < 5
inequality.
Practice Solve.
1. 3x < 6 or 2x + 2 > 10 (Compound inequality with “or.”) Write the original inequality.
3x < 6 or 2x + 2 > 10
Solve the left side of the inequality.
3x < 6 →
Solve the right side of the inequality.
2x + 2 > 10 → The solution is
.
2. 6x + 2 < −10 or 4x > 16 3. −3x + 3 > 12 or 4x > 4
Algebra 2
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Absolute Value Inequalities When solving an absolute inequality you can apply what you know from solving absolute value equations. As with absolute value equations, you look for values that are positive and negative that make the inequality true. Rules for Solving Absolute Value Inequalities 1. Write the absolute value inequality as an inequality with the original number
to the right of the inequality symbol. 2. Write the absolute value inequality as an inequality with the inequality symbol reversed and the inverse of the number to the right of the symbol. 3. Solve each inequality. Write the solutions as a compound inequality.
Example Solve. |x + 2| > 10 Step 1 Write the absolute value inequality as
| x + 2 | > 10
an inequality with the original number x + 2 > 10 to the right of the inequality symbol. Step 2 Write the absolute value inequality as
an inequality with the inequality symbol reversed and the inverse of the number to the right of the symbol. Step 3 Solve each inequality. Write the
solutions as a compound inequality.
| x + 2 | > 10 x + 2 < −10
x + 2 > 10 → x > 8 x + 2 < −10 → x < −12 x < −12 or x > 8
Practice Solve.
1. | 2x + 4| < 12 Write the absolute value inequality as an inequality with the original number to the right of the inequality symbol.
| 2x + 4| < 12
Write the absolute value inequality as an inequality with the inequality symbol reversed and the inverse of the number to the right of the symbol.
| 2x + 4| < 12
→x<
Solve each inequality. Write the solutions as a compound inequality.
→x> The solution is
2. | x − 3| 3
4. | 5x − 15| > 5
3. | x + 5| 10
5. | x + 1| + 3 > 5
.
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Graphing Absolute Value Inequalities The absolute value of a number is the distance between the origin of a number line and the point representing that number. When graphing an absolute value inequality, you first have to solve the inequality, treating it as a compound inequality. Then graph each solution. Rules for Graphing an Absolute Value Inequality 1. Rewrite the inequality as two inequalities. 2. Solve each inequality. 3. Graph each solution. If the solution is connected by “or,” then the
graph is away from the two points. If the solution is connected by “and,” then the graph is between the two points.
Example Solve, then graph the solution. |x + 6| < 11 Step 1 Rewrite the inequality as two
x + 6 < 11 and x + 6 > −11
inequalities. Step 2 Solve each inequality.
x < 5 and x > −17
Step 3 Graph each solution. If the solution
is connected by “or,” then the graph is away from the two points. If the solution is connected by “and,” then the graph is between the two points.
–17
0
5
Practice Solve, then graph the solution.
1. | x + 5| 3
Rewrite the inequality as two inequalities.
x + 5 3 or
Solve each inequality.
x
or x
Graph each solution. If the solution is connected by “or,” then the graph is away from the two points. If the solution is connected by “and,” then the graph is between the two points. 2. | x + 4 | > 5 3. | x + 7 | 10
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Introduction to Matrices A matrix is a rectangular array of numbers written within brackets. element X12 A matrix is identified by a capital letter. A matrix is classified by its dimensions—the number of columns and rows it contains. 29,300 2,900 Matrix X to the right has 3 rows and 2 columns. It is a X= 23,200 2,100 3 rows 3 × 2 matrix. 15,400 1,200 A matrix element is a number in the matrix. 2 columns Each matrix element is identified by its location within the matrix. Rules for Reading a Matrix 1. The dimensions of a matrix are given in terms of rows and columns. 2. A matrix element is identified by (1) using the letter of the matrix, and (2) using
a subscript to identify the position of the element by row and column.
Example State the dimensions of the matrix. Identify element A23. A = Step 1 The dimensions of a matrix are given
4
5
6
–1
0
2
The matrix has 2 rows and 3 columns; it is a 2 × 3 matrix.
in terms of rows and columns. Step 2 A matrix element is identified by
A23 is the element in row 2, column 3. A23 = 2
(1) using the letter of the matrix, and (2) using a subscript to identify the position of the element by the row and column.
Practice State the dimensions of the matrix. Identify the specified element.
B=
1. Identify element B22.
3
9
1
6
0
7
9
7
The dimensions of a matrix are given in terms of rows and columns.
The matrix has
rows and
columns; it is a
matrix.
A matrix element is identified by (1) using the letter of the matrix, and (2) using a subscript to identify the position of the element by the row and column.
B22 is the element in row
2. Identify element Z21. Z =
10
0
–2
1
3. Identify the location of −10. Z =
, column
. B22 =
0
–1
–4
5
3
5
–10
7
6
–3
–1
0
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Matrix Addition When adding matrices, you add the corresponding elements in each matrix. corresponding elements
–2 4
0 2
+
3
–1
–4
7
Rule for Matrix Addition
Add corresponding elements in each matrix to form one large matrix.
Example Add.
–4
2
–10
7
+
5
–9
9
–3
Add corresponding elements in each matrix to form one large matrix.
4
2
−10
7
+
5
−9
9
−3
4 + 5 2 + (–9)
=
(–10) + 9 7 + (–3)
=
9
–7
–1
4
Practice Add.
1. –5
8 –3
3
11
–1
5
–7
Add corresponding elements in each matrix to form one large matrix.
–5
8
3
–3
+
11
–1
5
–7
=
= 2.
3.
4.
2
–9
–4
3
–1
5
–4
7
–9
3
2 17
+
+
4
16
9
11
1
2
+ 7
2
–4
20
5 –12
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Matrix Subtraction When subtracting matrices, you subtract the corresponding elements in each matrix. corresponding elements
–2
0
3
–1
4
2
–4
7
Rule for Matrix Subtraction
Subtract corresponding elements in each matrix to form one large matrix.
Example
–2
5
0
Subtract.
—
–2
–4
6
8
5
Subtract corresponding elements in each matrix to form one large matrix.
–2
5
–
0 −2
–4
6
8
5
=
=
–2 – (–4)
5–6
0–8
–2 – 5
2
–1
–8
–7
Practice Subtract.
1.
3
3
–4
–1
–
6
–2
8
–2
Subtract corresponding elements in each matrix to form one large matrix.
3
3
–4
–1
–
6
–2
8
–2
= =
2.
–3
5
0
–4
9
–12
15
4
7
–5
6
12
3.
4.
–8
–4
–
–
–5
9
10
3
– –9
0
–2
–9
2
4
–3
5
–1
–3
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Scalar Multiplication A matrix is a rectangular arrangement of numbers in rows and columns. You can think of a matrix as a way to organize data, similar to the way data is displayed in a table. A scalar is a real number factor by which all the elements of a matrix are multiplied. Rule for Scalar Multiplication
Create an expanded matrix by multiplying each element by the scalar.
Example Solve. 2
–6
4
7
–3
Create an expanded matrix by multiplying each element by the scalar.
2
–6 7
4 –3
=
=
–6 × 2
4×2
7×2
–3 × 2
–12
8
14
–6
Practice Solve.
1. 5
11
–9
–4
–5
6
3
Create an expanded matrix by multiplying each element by the scalar.
5
11
–9
–4
–5
6
3
11 × 5
= –5 × 5
= 2
16
9
–2
–11
6
2. –3
3. 4
4. –6
5
–12
8
–2
–8
–4
1
0
2
–9
55 –25
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Matrix Multiplication When multiplying matrices, you multiply the elements of a row in the first matrix by the corresponding elements in a column of the second matrix. You then add the products. 5 3
6
5
5
3
8
×
2
=
2
3×5
+
6×2
+
5×2
5×5
+
3×2
+
8×2
Rules for Matrix Multiplication 1. Circle each row of the first matrix; circle each column of the second matrix. 2. Multiply the elements of a row in the first matrix by the elements of each
column in the second matrix. Add the products in each row. The dimensions of the resulting matrix will be the number of rows in the first matrix by the number of columns in the second matrix.
Example Multiply.
3
5
2
–3
×
2 9
Step 1 Identify the elements to be multiplied. Step 2 Multiply the elements of a row in the
first matrix by the elements of each column in the second matrix.
3
5
2
–3
3
5
2
–3
2
×
9 2
×
=
9
Add the products in each row.
=
3×2 + 5×9 2 × 2 + –3 × 9 6 + 45 4 + –27
51
=
−23
Practice Multiply.
1.
2
–4
3
7
×
1
9
3
2
Identify the elements to be multiplied.
Multiply the elements of a row in the first matrix by the elements of each column in the second matrix.
2
–4
3
7
2
–4
3
7
×
×
1
9
3
2
1
9
3
2
Add the products in each row.
=
=
7
2.
6 9
3 2
8 4
×
4 2
3. 2
5
4
6
×
9
4
2
3
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Writing the Inverse of a Matrix A matrix is a rectangular array of numbers written within brackets. A matrix is identified by a capital letter. You will use the inverse of a matrix to help solve systems of equations. Rules for Finding the Inverse of a 2 × 2 Matrix 1. In a 2 × 2 matrix, assign a, b, c, and d to the elements in the matrix as follows: 2. Find the value of ad − bc. 1 3. Plug the values from Steps 1 and 2 into the formula ______
ad − bc
d
–b
–c
a
a
b
c
d
. Simplify.
Example Find the inverse of
3
–1
7
1
.
Step 1 In a 2 × 2 matrix, assign a, b, c, and d
a = 3, b = −1, c = 7, d = 1
to the elements in the matrix. Step 2 Find the value of ad − bc.
ad − bc = (3)(1) − (−1)(7) = 3 + 7 = 10
Step 3 Plug the values from Steps 1 and 2 into
1 ______ ad − bc
the formula. Simplify.
d
–b
–c
a
1 = __ 10
1
1
–7
3
=
0.1
0.1
–0.7
0.3
Practice Find the inverse of each matrix.
1.
2.
3.
–2
–5
1
3
In a 2 × 2 matrix, assign a, b, c, and d to the elements in the matrix.
a=
Find the value of ad − bc.
ad − bc =
Plug the values from Steps 1 and 2 into the formula. Simplify.
1 ______ ad − bc
2
3
1
2
–2
–5
–3
–8
4.
5.
,b=
,c= −
d
–b
–c
a
8
2
7
3
5
–2
–2
1
=
,d= = =
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Solving a Matrix Equation A matrix is a rectangular array of numbers written within brackets. A matrix is identified by a capital letter. You can use the inverse of a matrix to solve matrix equations. Rules for Solving a Matrix Equation 1. A matrix equation has the form AX = B. Find the inverse of the matrix A. 2. Solve for the unknown, matrix X: Multiply matrix B by the inverse of
matrix A. Then simplify.
Example Solve.
–11
25
4
–9
X=
3 –7
Step 1 Find the inverse of matrix A.
d 1 ______ ad − bc –c
–b a
–9 1 _____________ (9 × 11)–(25 × 4) –4
Step 2 Solve for the unknown, matrix X:
Multiply matrix B by the inverse of matrix A. Then simplify.
X=
9
25
3
4
11
–7
–25 –11
9(3) + 25(–7)
X=
4(3) + 11(–7)
=
=
9
25
4
11
–148 –65
Practice 1.
6
–5
–2
2
X=
8 3
Find the inverse of matrix A.
d 1 ______ ad − bc –c
–b a
1 ________________ _______ – ________
Solve for the unknown, matrix X: Multiply matrix B by the inverse of matrix A. Then simplify.
=
X=
+
X=
=
+
2.
–3
–4
8
11
X=
–2 6
3.
–5
–3
3
2
X=
3 4
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Relations and Functions In a relation, you have input values known as the domain of the relation (the x-coordinates). The output values are known as the range of the relation (the y-coordinates). Rules for Identifying a Function 1. Take a set of ordered pairs and place the x-coordinates in the domain box; place
the y-coordinates in the range box. 2. Use arrows to pair domain elements with range elements. 3. If there is more than one arrow from any domain element, then there is more
than one range element for that domain element. The relation is not a function.
Example Determine if the relation is a function. [(−1, −2), (4, 7), (−6, 10), (4, 1)] Step 1 Take a set of ordered pairs and place
The first number in each ordered pair goes into the domain box; the second number in the x-coordinate in the domain box; place the y-coordinate in the range box. each pair goes into the range box.
Step 2 Use arrows to pair domain elements
Domain –6 –1 4
with range elements.
Step 3 If there is more than one arrow from
a domain element, then there is more than one range element for that domain element.
Range –2 1 7 10
Two arrows point from 4; the relation is not a function.
Practice Determine whether each relation is a function.
1. (0, 8), (−2, 3), (2, 8), (−1, 5) Domain
Take a set of ordered pairs and place the x-coordinate in the domain box; place the y-coordinate in the range box.
Range
Use arrows to pair domain elements with range elements. If there is more than one arrow from a domain element, then there is more than one range element for that domain element.
The relation
2. (−3, 3), (2, 2), (−2, −2), (0, 4), (1, −2) The relation 3. (−1, 3), (−2, 1), (−3, 3), (−2, 5) The relation
a function.
a function. a function. Algebra 2
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Types of Functions As you know, a function is defined as a relationship between two variables—an input variable and an output variable. There are many types of functions. A linear function is a function whose graph is a straight line. Another function, a quadratic function, is a function whose graph is U-shaped. By looking carefully at an equation you can determine the type of function. Types of Function 1. Linear
General Form
Description
y = ax + b
2. Quadratic
y = ax2 + bx + c
3. Exponential
y = ax
4. Rational
y = ___ g(x)
Equation with variables y = 2x + 2 but no exponents. Equation with a squared y = x2 + 3x – 2 term. Equation with a variable y = (2)x as an exponent. Equation with a variable y = – _5x + 4 in the denominator.
f(x)
Example
Example a Classify the following function as linear, quadratic, exponential or rational. __ +3=6 4 No Step 1 Does any variable have a 2 as an
exponent? Step 2 Is the variable an exponent?
No
Step 3 Is the variable in the denominator?
No
Step 4 Identify the function.
It is a linear function.
Practice Classify each function as linear, quadratic, exponential, or rational.
1. y = _1x − 1 Does the variable have a 2 as an exponent? Is the variable an exponent? Is the variable in the denominator? Identify the function. 2
It is a
x 2. y = __ 2 + 4x − 3
5 1 _ 5. y = __ 6x × 2
3. y = 4x +2
6. y = 3(0.5)x
1 4. y = ___ 2 +3
7. y = 3x + 22 − 1
2x
function.
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Direct Variation When the ratio of two variables is constant, they show direct variation. In a direct variation, when one variable increases the other variable increases proportionally. Similarly, when one variable decreases the other variable decreases proportionally. Rules for Direct Variation 1. Examine the data. When one variable varies does the other variable vary in proportion? 2. Does the ratio of the two variables stay constant? 3. Does the graph of the data go through the origin? 4. If the answer is “yes” to all questions, then the data show a direct variation.
Example Tell whether the data show a direct variation.
Time (hours)
1
2
3
4
Distance (miles)
40
80
120
160
Step 1 Examine the data. When one variable
varies does the other variable vary in proportion? Step 2 Does the ratio of the two variables stay constant? Step 3 Does the graph of the data go through the origin? Step 4 If the answer is “yes” to all questions, then the data show a direct variation.
Yes, as hours increase, the distance increases. distance 80 Yes, the ratio of ______ is 40 (e.g. __ 2 = 40). time
Yes, the graph does go through the origin. Each question was answered “yes,” so the data show a direct variation.
Practice Tell whether the data show a direct variation.
1.
Time (hours)
0.5
2.0
3.5
5.0
Distance (miles)
30
40
55
80
Step 1 As the time increases the distance distance Step 2 The ratio of ______ time
Step 3 The graph Step 4 The data
2.
3.
.
constant. go through the origin. show a direct variation.
Drop Height (cm)
10
20
30
40
50
Bounce (cm)
9
18
27
36
45
Age (months)
1
2
3
4
5
Weight (lbs)
15
30
45
60
75 Algebra 2
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Slope-Intercept Form Looking at an equation can tell you certain pieces of information about the graph of that equation. An equation written with y isolated on one side of the equal sign and x on the other side of the equation is in slope-intercept form. An equation in slope-intercept form is written as: y-intercept
y = mx + b slope
The y-intercept is the point on the y-axis through which the line passes.
Example Find the slope and the y-intercept of the line y = 4x − 2. Step 1 Find the coefficient of the x-term. Be
sure to include the negative sign if necessary. This is the slope. Step 2 Find the term without a variable. This
number is the y-coordinate where the line crosses the y-axis. Be sure to include a negative if necessary.
y = mx + b y = 4x − 2 m = slope = 4 y = 4x − 2 b = y-intercept = −2
Practice Find the slope and y-intercept for each equation.
1. y = −3x + 7 Find the coefficient of the x-term. Be sure to include the negative sign if necessary. This is the slope.
y = mx + b y = −3x + 7 m = slope =
Find the term without a variable. This number is the y-coordinate where the line crosses the y-axis. Be sure to include a negative if necessary.
y = −3x + 7 b = y-intercept =
2. y = _13x − 3
4. y = _34 x − 13
3. 2y = 2x + 2
5. 3y = −2x + 9
Use the graphs to the right to write equations in slope-intercept form.
6. Line A 7. Line B
y A x B
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Point-Slope Form I There are instances in which you are given the slope and an ordered pair. For example, you may know that the slope of a line is −2 and the graph of the equation passes through (−2, 1). You can use the point-slope form of a linear equation to write an equation of the line. slope
Point-slope form: y − y1 = m(x − x1) y-coordinate
x-coordinate
Rules for Using the Point-Slope Form 1. Identify the slope (m). 2. From the ordered pair identify the x-coordinate and the y-coordinate. 3. Use the point-slope form to write the equation: y − y1 = m(x − x1).
Example Write the equation of the line that has a slope of 3 and passes through the point (2, 5). Step 1 Identify the slope (m).
The slope is 3.
Step 2 From the ordered pair, identify the
The ordered pair is (2, 5). The x-coordinate is 2; the y-coordinate is 5.
x-coordinate and the y-coordinate. Step 3 Use the point-slope form to write the
equation.
y − y1 = m(x − x1) y − 5 = 3(x − 2)
Practice Write the equation of the line.
1. Slope = 6; point: (−3, −1) Identify the slope (m).
The slope is
.
From the ordered pair, identify the x-coordinate and the y-coordinate.
The ordered pair is (–3, –1) The x-coordinate is ; the y-coordinate is .
Use the point-slope form to write the equation.
y − y1 = m(x − x1)
2. Slope = – _12 ; point: (7, 1) 3. Slope = 2; point: (−3, −3) 4. Slope = _23 ; point: (4, −5) 5. Slope = −3; point: (−1, 3)
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Point-Slope Form ll When you are given the slope of a line and an ordered pair identifying a point on the graph of the line, you can use the point-slope form. You can also use the point-slope form when given two sets of ordered pairs. To use the two ordered pairs you will need to first use the ordered pairs to find the slope. Rules for Using Point-Slope Form Using Two Points
y2 – y1 vertical change 1. Use the formula for slope (slope = _____________ = _____ x – x to find the slope. horizontal change
2
1
2. Use one set of ordered pairs for the x-coordinate and y-coordinate. 3. Use point-slope form to write the equation.
Example Write the equation of the line that passes through (−3, −3) and (1, 5). y2 – y1 Step 1 Use the formula for slope _____ x – x to
8 2 1 ______ _ Slope = _____ x – x = 1 – (–3) = 4 = 2
Step 2 Use one set of ordered pairs for the
Use the ordered pair (1, 5). The x-coordinate is 1; the y-coordinate is 5.
(
find the slope.
2
1
)
x-coordinate and the y-coordinate. Step 3 Use point-slope form to write the
equation.
y –y 2
5 – (–3)
1
y − y1 = m(x − x1) y − 5 = 2(x − 1)
Practice Use the point-slope form to write an equation.
1. (−2, −2), (0, −4) y2 – y1 Use the formula for slope _____ x2 – x1 to find the slope.
(
)
Use one set of ordered pairs for the x-coordinate and the y-coordinate.
y –y 2
1
=
Use the ordered pair (−2, −2). The x-coordinate is is
Use point-slope form to write the equation.
−4 − ____
2 1 ________ Slope = _____ x – x = 0 – ____ =
; the y-coordinate
.
y − y1 = m(x − x1)
2. (0, 1), (2, 2) 3. (−6, 4), (3, −5) 4. (2, 6), (0, 0) 5. (−1, −4), (5, 2) 6. (6, 0), (3, −2)
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Linear Parametric Equations Parametric equations are equations that express the x- and the y-coordinates as separate functions. These separate functions are connected through a third variable, called the parameter. Parametric equations are used to find the location of an object over time. the parameter, t
x = g(t)
y = h(t)
Parametric Equations Rules for Solving Linear Parametric Equations 1. Choose several values of t. 2. Plug the values of t into the function for the x-coordinate. Solve for x. 3. Plug the values of t into the function for the y-coordinate. Solve for y.
Example Solve the parametric equations for four values of t. x = −2t; y = t + 2 Step 1 Choose several values of t.
Decide on four values for t. Let t = −2, −1, 0, 2.
Step 2 Plug the values of t into the function
for the x-coordinate. Solve for x. Step 3 Plug the values of t into the function
for the y-coordinate. Solve for y.
t
x = –2t
y=t+2
–2
x = –2(–2) = 4
y = –2 + 2 = 0
–1
x = –2(–1) = 2
y = –1 + 2 = 1
0
x = –2(0) = 0
y=0+2=2
2
x = –2(2) = –4
y=2+2=4
The solution is (4, 0), (2, 1), (0, 2), (−4, 4).
Practice Solve the parametric equations for four values of t: −2, −1, 0, and 2.
1. x = 4t ; y = 2t − 2 Choose several values of t.
Decide on four values for t. Let t = −2,
Plug the values of t into the function for the x-coordinate. Solve for x. Plug the values of t into the function for the y-coordinate. Solve for y.
t –2
. x = 4t
y = 2t – 2
x = 4(–2) =
y = 2(–2) – 2 =
x = 4(
) = –4 y = 2(
) – 2 = –4
x = 4(
)=0
y = 2(
) – 2 = –2
x = 4(
)=8
y = 2(
)–2=2
The solution is
.
2. x = −2 − t ; y = t 3. x = t + 3; y = t + 1
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Exponential Growth and Decay An exponential function is a function that uses an exponent as the independent (x) variable. An exponential function has the general form y = abx, where x is a real number, a 0, b > 0, and b 1. An exponential function is used to model growth and decay. Time Final amount after time
y=
abx
Exponential growth: b >1 Exponential decay: 0 < b < 1
Growth or decay factor
Rules for Exponential Growth or Decay 1. Identify the original amount (a) and the time (x). 2. Determine the growth or decay factor. For growth, b = 1 + r, where r
is the rate of growth. For decay, b = 1 − r, where r is the rate of decay. 3. Plug the values for a, b, and r into the equation.
Example Determine the final amount. Initial amount: $400; growth rate: 7.25%; time: 7 years Step 1 Identify the original amount (a) and
the time (x). Step 2 Determine the growth or decay factor.
a = 400 x=7 r = 7.25% b = 1 + r = 1 + 0.0725 = 1.0725
For growth, b = 1 + r, where r is the rate of growth. For decay, b = 1 − r, where r is the rate of decay. Step 3 Plug the values for a, b, and r into the y = abx = (400)(1.0725)7 = $652.89 equation.
Practice Determine the final amount.
1. Original amount: 500; decay rate: 6.6%; time: 20 years Identify the original amount (a) and the time (x).
a= x= r=
Determine the growth or decay factor. For growth, b = 1 + r, where r is the rate of growth. For decay, b = 1 − r, where r is the rate of decay.
b=1−r
Plug the values for a, b, and r into the equation.
y = abx =
=
=
=
2. Original amount: 6,500; decay rate: 14.5%; time: 3 years 3. Original amount: 250; growth rate: 10.4%; time: 10 years 4. Original amount: 10,000; growth rate: 6.3%; time: 15 years Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Writing an Exponential Function An exponential function is a function that uses an exponent as the independent (x) variable. You can write an exponential function from two points on the function’s graph. Rules for Writing an Exponential Function 1. Use one ordered pair for x and y. Plug the values into the general equation
y = abx. Write an expression for a. 2. Use the second ordered pair for x and y and the expression from Step 1 for a. Plug the values into the general equation y = abx. Solve for b. 3. Take the value for b from Step 2 and the expression from Step 1 for a. Solve for a. 4. Use the solutions to Steps 2 and 3 to write an exponential function.
Example Write an exponential function for a graph that includes (2, 1) and (3, 3). Step 1 Use one ordered pair for x and y. Plug
the values into the general equation y = abx. Write an expression for a. Step 2 Use the second ordered pair for x and
Use the first ordered pair. 1 = ab2 1 1 ÷ b2 = ab2 ÷ b2 → __ =a b2 1 Use the second ordered pair; also a = __ 2. b b3 1 3 __ __ 3= 2b = 2→b=3
y and the expression from Step 1 for a. b b Plug the values into the general equation y = abx. Solve for b. 1 1 Step 3 Take the value for b from Step 2 and the a = __ = __ = _19 b2 32 expression from Step 1 for a. Solve for a. y = _19 (3)x Step 4 Use the solutions to Steps 2 and 3 to write an exponential function.
Practice Write an exponential function.
1. (3, 1) (4, 2) Use one ordered pair for x and y. Plug the Use the first ordered pair. values into the general equation y = abx. = ab Write an expression for a. ÷b = ab
÷b
=a 2= Use the second ordered pair for x and y and the expression from Step 1 for a. Plug the values into the general equation y = abx. Solve for b. Take the value for b from Step 2 and the a = expression from Step 1 for a. Solve for a. Use the solutions to Steps 2 and 3 to write y = an exponential function. 2. (1, 4) (0, 2)
b4 =
=
→b=
=
3. (2, 2) (5, 16) Algebra 2
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Solving an Exponential Equation An exponential equation is the equation of a function with an exponent as the independent (x) variable. An exponential equation has the general form y = abx. It can be solved by taking the logarithm of each side of the equation. Rules for Solving an Exponential Equation 1 Rewrite the equation by taking the logarithm of each side. 2. Use the power property of logarithms. 3. Isolate x by dividing each side of the equation by the
expression with the x-term. 4. Solve.
Example Solve. 5x = 10 Step 1 Rewrite the equation by taking the
logarithm of each side.
5x = 10 log5x = log10
Step 2 Use the power property of logarithms.
xlog5 = log10
Step 3 Isolate x by dividing each side of the
xlog5 ÷ log5 = log10 ÷ log5
equation by the expression with the x-term. Step 4 Solve.
log10
x = ____ log5 1 x = ______ 0.69897 = 1.43
Practice Solve.
1. 42x = 22 Rewrite the equation by taking the logarithm of each side.
42x = 22 log
= log
Use the power property of logarithms.
log
= log
Isolate x by dividing each side of the equation by the expression with the x-term.
log
÷
= log
÷
log ______
x = ________ 00000000 Solve.
x=
=
2. 5x = 17 3. 93x = 77 x 4. _12 = 10
()
5. 5.8−0.2x = 4 Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Negative Exponents An expression with a negative exponent can be rewritten with a positive exponent as follows: 1 x−n = __ xn
In other words, an expression with a negative exponent is equivalent to its inverse with a positive exponent. Rules for Working with Negative Exponents 1. Identify the base and the exponent. 2. Write an equivalent expression by writing its inverse with a positive exponent. 3. Evaluate the expression by first evaluating the denominator, and then simplifying.
Example Simplify. 5−3 Step 1 Identify the base and the exponent.
exponent
5−3 base
Step 2 Write an equivalent expression by
1 5−3 = __ 3 5
writing its inverse with a positive exponent. Step 3 Evaluate the expression by first
evaluating the denominator, and then simplifying.
1 _______ 1 1 __ = = ___ 125 53 5 × 5 × 5
Practice Simplify the following.
1. 2−5 Identify the base and the exponent.
2−5
Write an equivalent expression by writing its inverse with a positive exponent.
1 2−5 = ___ 2___
Evaluate the expression by first evaluating the denominator, and then simplifying.
1 1 1 ___ _______________ = ___ 2___ =
2. 8−1
4. (−5)−3
6. a−2
3. 3−2
5. (−3)−4
7. 4x−3
Algebra 2
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Rational Exponents A rational exponent is an exponent that is expressed as a fraction. The following are examples of rational exponents: 1 _
3 _
1 _
83 644 162 In a rational exponent, the numerator (top number) of the fraction tells you the power to which the number is raised. The denominator (bottom number) tells you the root of the number to take. a
raise a to the power x
_x y
cube 5
3 _ 2
5
then take the square root
take the yth root of a
Rules for Working with Rational Exponents 1. Raise the number to the power indicated by the top number in the exponent. 2. Take the root of the result from Step 1 that is indicated by the bottom
number of the fraction.
Example
3 __
Simplify. 164
Step 1 Raise the number to the power
Raise 16 to the 3rd power: 163 = 4,096
indicated by the top number in the exponent. Step 2 Take the root of the result from
Step 1 that is indicated by the bottom number of the fraction.
Take the 4th root of 4,096: _____ √4,096 = 8
4
3 _
164 = 8
Practice Simplify.
2 _
1. –325
Raise the number to the power indicated by the top number in the exponent.
Raise −32 to the
Take the root of the result from Step 1 that is indicated by the bottom number of the fraction.
Take the
2 _
2. (–27)3
3 _
3. –(6254)
−32
power.
= root of the answer in Step 1. ________
√00000000 = 2 _
4. (–8)3 3 _
5. 1002
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Properties of Rational Exponents A rational exponent is an exponent that is expressed as a fraction. Remember that an integer can be expressed as a fraction with 1 as the denominator. Summary of Properties of Rational Exponents Property Using Symbols
What do you do?
Example
ax × ay = ax+y
add exponents, keep the base
32 × 35 = 32+5 = 37
(ax)y = a(x × y)
multiply the exponents
(32)5 = 3(2×5) = 310
(ab)x = ax × bx
each base is raised to the same exponent
(3 × 4)3 = 33 × 43
1 a−x = __ ax
write the inverse of the expression with a positive exponent
1 3−3 = __ 3
ax __ = ax–y ay
subtract the exponents
x
(__ab)x = __ab
each base is raised to the same exponent
x
3
35 __ = 35–2 = 33 32 3
3 (__34)3 = __ 4 3
Example Simplify. (4x)3 Step 1 What is the operation?
Two bases raised to the same power.
Step 2 What do you do?
Raise each base to the given power.
Step 3 Simplify.
(4x)3 = 43 × x3 = 43 × x3 = 64x3
Practice Simplify. x2 1. ___ –11 x
What is the operation?
Powers with the same base are . the exponents.
What do you do? Simplify. 2. (x15)2 3.
(x−3)(x4)
4. (−3y3)2
x2 ___ = x–11
5.
= x2 3 __ x5
( )
6. (x2y2)−2 7. (x4)(x−6)
Algebra 2
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Doubling Time An exponential function is a function that uses an exponent as the independent (x) variable. An exponential function has the general form y = abx, where x is a real number, a 0, b > 0, and b 1. An exponential function is used to model growth and decay. Finding Doubling Time 1. Identify the original amount (a), and the growth factor (b).
Write an exponential equation in the form y = abx. 2. Find two times the original amount (the original amount doubled). Substitute this value for y. 3. Solve for x as you would when solving an exponential equation. (This is the time it takes for the original amount to double.)
Example Find the doubling time. Original amount: $300; growth rate: 5% Step 1 Identify the original amount (a),
and the growth factor (b). Write an exponential equation in the form y = abx. Step 2 Find two times the original amount.
Substitute this value for y. Step 3 Solve for x as you would when solving
an exponential equation.
a = 300 b = 1 + r = 1 + 0.05 = 1.05 y = 300(1.05)x y = 2(a) = 2(300) = 600 600 = 300(1.05)x 2 = 1.05x log2 = xlog1.05 x = 14.21
Practice Find the doubling time.
1. Original amount: $450; growth rate: 5.5% Identify the original amount (a), and the growth factor (b). Write an exponential equation in the form y = abx.
a = 450
Find two times the original amount. Substitute this value for y.
y = 2(a) = 2(450) =
Solve for x as you would when solving an exponential equation.
b=1+r=1+
=
y = 450
= 450
log2 = x log1.055 x=
2. Original amount: 125; growth rate: 10% 3. Original amount: 1,000; growth rate: 2.25% Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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The Number e The number e is an irrational number approximately equal to 2.71828. A common example of the use of e is the continuous compounding of interest. Continuously compounded interest is modeled by the equation A = Pert, where A = final amount, P = original amount, r = interest rate, and t = time. Rules for Using the Number e 1. Identify the variables in the given formula. Assign values to the variables. 2. Plug the values into the formula. 3. Evaluate the part of the formula with e and the exponents with e. 4. Solve for the unknown variable.
Example Find the amount after 5 years in an account that started with $1,000 and an interest rate of 4.5%. Use the formula A = Pert. Step 1 Identify the variables in the given
A = final amount = unknown formula. Assign values to the variables. P = original amount = $1,000 r = interest rate = 4.5% = 0.045 t = time = 5 years Step 2 Plug the values into the formula. A = 1,000e(0.045)(5) Step 3 Evaluate the part of the formula ert = e(0.045)(5) = 1.25 with e and the exponents with e. Step 4 Solve for the unknown variable. A = $1,000(1.25) = $1,250
Practice Find the final amount.
1. Radioactive fluorine decays according to the formula A = ge−0.1386t, where t is measured in seconds and g is the original amount in grams. If you start with 50 g of fluorine, how much is left after 30 seconds? A = final amount = unknown Identify the variables in the given formula. Assign values to the variables. g= = g
Plug the values into the formula. Evaluate the part of the formula with e and the exponents with e. Solve for the unknown variable.
t= = r = decay rate = 0.1386 A= e−0.1386 = e−0.1386 A=
seconds
=
g
2. Find the final amount in an account where $2,000 is deposited with an interest rate of 5.2%, continuously compounded over 10 years. 3. Polonium decays according to the formula A = ge−0.005t. If you start with 75 g, how much is left after 10 days? Algebra 2
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Logarithmic Functions The logarithmic function x = logb y is equivalent to the exponential function y = bx. Rules for Evaluating Logarithms 1. Set the logarithmic expression equal to x. 2. Convert to exponential form. The value of the base moves to the other side of the
equation and x becomes the exponent. Remove the “log.” 3. Write each side of the equation in terms of a base that is common to both b and y. 4. Set the exponents equal to each other. Solve for x.
Example Evaluate. log416 Step 1 Set the logarithmic expression equal
to x. Step 2 Convert to exponential form. The value of the base moves to the other side of the equation and x becomes the exponent. Remove the “log.” Step 3 Write each side of the equation in terms of a base that is common to both b and y. Step 4 Set the exponents equal to each other. Solve for x.
log416 becomes log416 = x. log416 = x 16 = 4x
Use base 2 as both 16 and 4 are powers of 2. 24 = (22)x = 22x 4 = 2x 2=x
Practice Evaluate.
1. log432 Set the logarithmic expression equal to x. Convert to exponential form. The value of the base moves to the other side of the equation and x becomes the exponent. Remove the “log.” Write each side of the equation in terms of a base that is common to both b and y.
log432 becomes log432 = x x
32 = Use base
as 4 and 32 are powers of
. x
32 = Set the exponents equal to each other. Solve for x.
.
2 5=
=( x
)x =
=x
2. log327
4. log464
3. log16256
5. log648
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Properties of Logarithms: Product Property The inverse of an exponential function is a logarithmic function. You write the inverse of the exponential function f(x) = bx as f −1(x) = logbx. In a logarithmic function, b is the base; b is positive and b 1. The expression logbx is called the base-b logarithm of x. Rules for Using the Product Property of Logarithms
To write logarithms in expanded form: 1. Verify that the logarithm is in the form logb(MN). Identify M and N. 2. Rewrite the logarithm as two logarithms, adding logbM and logbN (logb(MN) = logbM + logbN). To write a logarithm as a single expression: 1. Verify that the expression is in the form logbM + logbN. Identify M and N. 2. Rewrite the expression as one logarithm, multiplying M and N (logbM + logbN = logb(MN)).
Example Write the expression logb (p6r5) in terms of logb p and logbr. Step 1 Verify that the logarithm is in the form The expression logb(p6r5) is in the correct form.
logb(MN). Identify M and N.
Step 2 Rewrite the logarithm as two
logarithms, adding logbM and logbN.
M = p6 N = r5 logb(MN) = logbM + logbN logb (p6r5) = logb p6 + logbr5
Practice 1. Write logb p + logbq2 as a single expression. Verify that the expression is in the form logbM + logbN. Identify M and N.
The expression is in the correct form. M= N=
Rewrite the expression as one logarithm, multiplying M and N.
logbM + logbN = logb(MN) logb p + logbq2 =
2. Write the expression logb(q3r2) in terms of logbq and logbr. 3. Write the expression logb (p5q2r) in terms of logb p and logbq and logbr. 4. Write the expression logb27 + logb6 as a single expression. 5. Write the expression logbx2 + logbx5 as a single expression.
Algebra 2
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Properties of Logarithms: Quotient Property The inverse of an exponential function is a logarithmic function. You write the inverse of the exponential function f(x) = bx as f −1(x) = logbx. In a logarithmic function, b is the base; b is positive and b 1. The expression logbx is called the base-b logarithm of x. Rules for Using the Quotient Property of Logarithms
To write a logarithm in expanded form: M 1. Verify that the logarithm is in the form logb__ . Identify M and N. N 2. Rewrite the logarithm as two logarithms, subtracting logbM − logbN M = logbM − logbN). (logb__ N To write a logarithm as a single expression: 1. Verify that the expression is in the form logbM − logbN. 2. Rewrite the expression as one expression, dividing M and N M (logbM − logbN = logb__ ). N
Example
p5 r
Write the expression logb__6 in terms of logb p and logbr.
p5
Step 1 Verify that the logarithm is in the
The expression logb__6 is in the correct form. r M = p5 N = r6
Step 2 Rewrite the logarithm in the form
logb__6 = logb p5 − logbr6
M . Identify M and N. form logb__ N
logbM − logbN.
p5 r
Practice 1. Write logb p − logbq2. Verify that the logarithm is in the form logbM − logbN.
The expression is in the correct form. M=
;N=
Identify M and N. Rewrite the logarithmic expression M. in the form logb__ N
logb p − logbq2 =
2. Rewrite logbp3 − logbr2. 5. 3. Rewrite logb __ 5 p
4. Write the expression logb27 − logb9 as a single expression.
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Properties of Logarithms: Power Property The inverse of an exponential function is a logarithmic function. You write the inverse of the exponential function f(x) = bx as f −1(x) = logbx. In a logarithmic function, b is the base; b is positive and b 1. The expression logbx is called the base-b logarithm of x. Rules for Using the Power Property of Logarithms
To write a logarithm in expanded form: 1. Verify that the logarithm is in the form logbMx. Identify M and x. 2. Rewrite the logarithm as a logarithm in which x becomes a coefficient (logbMx = x logbM). To rewrite a logarithm as a logarithm with an exponent: 1. Verify that the expression is in the form xlogbM. Identify x and M. 2. Rewrite the expression by moving x to an exponent (x logbM = logbMx).
Example Write the expression logba4 in expanded form. Step 1 Verify that the logarithm is in the form The expression logba4 is in the correct form.
logbMx. Identify M and x.
Step 2 Rewrite the logarithm as a logarithm
in which x becomes a coefficient (logbMx = x logbM).
M=a x=4 logbMx = xlogbM logba4 = 4logba
Practice 1. Write the expression 4logbz as a logarithm with an exponent. Verify that the logarithm is in the form xlogbM. Identify x and M.
The expression is in the correct form. x= M=
Rewrite the expression by moving x to an exponent (x logbM = logbMx).
x logbM = logbMx 4logbz =
2. Write the expression logbq3 in expanded form. 3. Write the expression logba3x in expanded form. 4. Write the expression 7logb2x as a logarithm with an exponent. 5. Write the expression x logb12 as a logarithm with an exponent.
Algebra 2
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Properties of Logarithms: Summary The inverse of an exponential function is a logarithmic function. You write the inverse of the exponential function f(x) = bx as f −1(x) = logbx. In a logarithmic function, b is the base; b is positive and b 1. The expression logbx is called the base-b logarithm of x. Properties of Logarithms
Product Property: logb(MN) = logbM + logbN M = log M – log N Quotient Property: logb__ b b N Power Property: logbMx = x logbM You may use more than one property in simplifying or expanding expressions. Examine each part of the expression for its general form. Apply the property rules for that general form.
Example Write the expression 4logba − logbz as a single logarithm. Step 1 Identify the general form of each
logarithm. Step 2 Apply the property rules for each
logarithm. Step 3 Identify the general form of each
logarithm. Step 4 Apply the property rules for each
logarithm.
4logba is in the form of xlogbM. logbz has no general form. 4logba − logbz logba4 − logbz logba4 − logbz is in the form logbM − logbN. 4
a logba4 − logbz = logb__ z
Practice 1. Expand the logarithm logb(5a5). Identify the general form of each logarithm.
The logarithm is in the form
Apply the property rules for each logarithm.
logb(5a5)
Identify the general form of each logarithm.
logba5 is in the form
Apply the property rules for each logarithm.
.
+ logba5 .
+ logba5 +
2. Write 4logba + logb4. 3. Write logba − 2logbc + 5logbm. a2 . 4. Expand logb __ 4
( )
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Solving Logarithmic Functions An equation that includes a logarithmic expression is a logarithmic equation. In some cases, you will need to use the properties of logarithms to solve a logarithmic equation. Rules for Solving Logarithmic Equations 1. If necessary, use the properties of logarithms to simplify the equation. 2. Rewrite the equation in exponential form: Remove the “log” by
writing an equivalent base-10 exponential expression. 3. Solve for x.
Example Solve. log(3x + 1) = 6 Step 1 If necessary, use the properties of
The equation is simplified.
logarithms to simplify the equation. Step 2 Rewrite the equation in exponential
form: Remove the “log” by writing an equivalent base-10 exponential expression. Step 3 Solve for x.
log(3x + 1) = 6 3x + 1 = 106
3x + 1 = 1,000,000 x = 333,333
Practice Solve.
1. log5 − log(2x) = −4 If necessary, use the properties of logarithms to simplify the equation.
The equation can be simplified. The left side is in the form logbM − logbN, so use the Property. log5 − log2x = −4 log
= 10
Rewrite the equation in exponential form: Remove the “log” by writing an equivalent base-10 exponential expression. Solve for x.
= –4
x=
2. 2logx = 1
4. logx + log4 = 2
3. log(3x) = 4
5. log(3x) + logx = 8
Algebra 2
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Solving a Natural Logarithmic Function The inverse of an exponential function with e as a base is called a natural logarithmic function. You can use the properties of logarithms to solve natural logarithmic functions. Rules for Solving a Natural Logarithmic Function 1. If necessary, use the properties of logarithms to simplify the equation. 2. Rewrite the equation in exponential form: Remove the “ln” by writing
an equivalent base-e exponential expression. 3. Solve for x.
Example Solve. ln(2x + 1)3 = 6 Step 1 If necessary, use the properties of
logarithms to simplify the equation.
Step 2 Rewrite the equation in exponential
Use the Power Property to simplify. Then divide by 3 to isolate the natural logarithm. ln(2x + 1)3 = 6 3ln(2x + 1) = 6 ln(2x + 1) = 2 2x + 1 = e2
form: Remove the “ln” by writing an equivalent base-e exponential expression. Step 3 Solve for x.
2x = e2 −1 e2 −1 x = ____ 2 = 3.19
Practice Solve.
1. ln(3x2) = 6 If necessary, use the properties of logarithms to simplify the equation.
Use the Power Property to simplify, then divide. ln(3x2) = 6 ln(3x) = 6 ln(3x) =
Rewrite the equation in exponential form: Remove the “ln” by writing an equivalent base-e exponential expression.
3x = e
Solve for x.
e x = ____ 3 =
——
2. ln(x + 1)4 = 2
4. 1 + lnx4 = 9
3. lnx5 = 25
5. 3ln(2x2) = 12
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Graphing Quadratic Functions The graph of a quadratic function is a U-shaped graph called a parabola. When you graph a quadratic function you find ordered pairs that satisfy the function. Rules for Graphing a Quadratic Function 1. Find the coordinates of the vertex. 2. Create an input/output table. 3. Select several other values for x.
Substitute the values for x into the equation. Solve for y. 4. Plot each ordered pair on the coordinate plane. Draw a parabola.
Example Graph. y = −2x2 + 1 Step 1 Find the coordinates of the vertex.
–b ____ –0 x-coordinate of vertex = __ 2a = 2(–2) = 0 y = −2x2 +1 = −2(0) 2 +1 = 1 The vertex is at (0, 1).
Step 2 Create an input/output table.
–2x2 + 1
x
Step 3 Select several other values for x.
Substitute the values for x into the equation. Solve for y.
y
2
–2(2)2
+1
–7
–2
–2(–2)2
+1
–7
1
–2(1)2
+1
–1
–1
–2(–1)2 + 1
–1
Step 4 Plot each ordered pair on the
coordinate plane. Draw a parabola connecting the points.
Practice Graph each function.
1. y = x2 + 2x + 3 Find the coordinates of the vertex.
b –000 _____ x-coordinate of vertex = – __ 2a = 2(000) = 2
y=
+ 2(
)+3=
The vertex is at Create an input/output table. Select several other values for x. Substitute the values for x into the equation. Solve for y.
.
x
x2 + 2x + 3
y
–2
(–2)2 + 2(–2) + 3
3
2
2
+ 2(
)+3
0
2
+ 2(
)+3
1
2
+ 2(
)+3
Plot each ordered pair on the coordinate plane. Draw a parabola connecting the points. 2. y = x2 + 2 3. y = 3x2 − 3x − 1 Algebra 2
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Properties of a Graph of a Quadratic Function The graph of a quadratic function is a U-shaped graph called a parabola. The vertex of a parabola is the lowest point of a parabola that opens up. In a parabola that opens down, the vertex is the highest point of the parabola. The line passing through that divides the parabola into two equal parts is the axis of symmetry.
axis of symmetry opens “up” vertex
Properties of the Graph of a Parabola
For quadratic equations in the form y = ax2 + bx + c : 1. If a is positive, the parabola opens up; if a is negative, the parabola opens down. –b 2. To find the x-coordinate of the vertex, use __ 2a ; plug the x-value into the equation to find the y-coordinate of the vertex. –b 3. To find the axis of symmetry, use x = __ 2a .
Example For the quadratic equation y = 2x2 − 4x − 3, tell whether the parabola opens up or down and find the coordinates of the vertex and the axis of symmetry. Step 1 If a is positive, the parabola opens up;
a is positive, the parabola opens up.
if a is negative, the parabola opens down. Step 2 To find the x-coordinate of the vertex, –b use __ 2a ; plug the x-value into the equation to find the y-coordinate of the vertex.
Step 3 To find the axis of symmetry, use –b x = __ 2a .
–(–4)
–b ____ The x-coordinate of the vertex: __ 2a = 2(2) = 1
y = 2x2 − 4x − 3 = 2(1)2 − 4(1) − 3 = −5 Vertex: (1, −5) –(–4)
–b ____ x = ___ 2a = 2(2) = 1
Practice For each quadratic function tell whether the parabola opens up or down and find the coordinates of the vertex and the axis of symmetry.
1. y = −x2 − 4x + 2 If a is positive, the parabola opens up; if a is negative, the parabola opens down.
a is
; the parabola
To find the x-coordinate of the vertex, –b use __ 2a ; plug the x-value into the equation to find the y-coordinate of the vertex.
The x-coordinate of the vertex:
opens
.
–(0000) –b ______ ___ 2a = 2(0000) =
y = −(
)2 − 4(
)+2=
Vertex: To find the axis of symmetry, use –b x = __ 2a .
–(0000)
–b ______ x = ___ 2a = 2(0000) =
2. y = 2x2
4. y = 3x2 − 6x + 1
3. y = −4x2 + 8x
5. y = −10x2 + 5x + 3
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Writing a Quadratic Function from Its Graph A quadratic function can be written in the form y = a(x − h)2 + k. This form is known as the vertex form of a quadratic function. Rules for Writing a Quadratic Function from Its Graph 1. Identify the vertex. The x-coordinate of the vertex is h, the y-coordinate is k. 2. Identify one other point on the graph. The x-coordinate of the point is x in
the vertex form, the y-coordinate is y in the vertex form. 3. Plug the values from Step 1 and Step 2 into the vertex form of a quadratic
function. Solve for a. 4. Write the function using the vertex for h and k from Step 1 and a from Step 3.
Example A parabola has a vertex at (3, −1). Another point on the graph is at (0, 8). Write the equation of the parabola in vertex form. Step 1 Identify the vertex. The x-coordinate
of the vertex is h, the y-coordinate of the vertex is k. Step 2 Identify one other point on the graph. The x-coordinate of the point is x in the vertex form, the y-coordinate is y in the vertex form. Step 3 Plug the values from Step 1 and Step 2 into the vertex form of a quadratic function. Solve for a. Step 4 Write the function using the vertex for h and k from Step 1 and a from Step 3.
Vertex: (3, −1); h = 3, k = −1
Other point: (0, 8) x = 0, y = 8 8 = a(0 − 3)2 + (−1) 1=a y = 1(x − 3)2 − 1 = (x − 3)2 − 1
Practice For each parabola, write an equation in vertex form.
1. Vertex: (1, −3); other point: (3, −5) Identify the vertex. The x-coordinate of the vertex is h, the y-coordinate of the vertex is k. Identify one other point on the graph. The x-coordinate of the point is x in the vertex form, the y-coordinate is y in the vertex form. Plug the values from Step 1 and Step 2 into the vertex form of a quadratic function. Solve for a. Write the function using the vertex for h and k from Step 1 and a from Step 3.
Vertex: (1, −3); h =
,k=
Other point: ( 3, −5 ) x=
,y= = a(
)2 +
−
a= y=
(x −
)2 +
2. Vertex: (1, −6); other point: (3, 0) 3. Vertex: (0, −3); other point: (3, 0) Algebra 2
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Quadratic Functions in Intercept Form A quadratic function can be written in the form y = a(x − p)(x − q). This form is known as the intercept form of a quadratic function. Rules for Writing a Quadratic Function in Intercept Form 1. Find one of the x-intercepts. The x-coordinate is p. 2. Find the other x-intercept. The x-coordinate of this intercept is q. 3. Identify the x- and y-coordinates of the vertex. These coordinates
are x and y in the intercept form equation. 4. Plug the values from Steps 1, 2, and 3 into the intercept form of a quadratic equation. Solve for a. 5. Write the equation using the values from Steps 1, 2, and 4.
Example A parabola has a vertex at (0, 2) and x-intercepts at (2, 0) and (−2, 0). Write the equation of the parabola in intercept form. Step 1 Find one of the x-intercepts. The Step 2 Step 3
Step 4
Step 5
x-intercept: (2, 0); p = 2
x-coordinate is p. Find the other x-intercept. The x-coordinate of this intercept is q. Identify the x- and y-coordinates of the vertex. These coordinates are x and y in the intercept form equation. Plug the values from Steps 1, 2, and 3 into the intercept form of a quadratic equation. Solve for a. Write the equation using the values from Steps 1, 2, and 4.
x-intercept: (−2, 0); q = −2 Vertex: (0, 2); x = 0, y = 2
2 = a(0 − 2)(0 − (−2)) – _12 = a y = a(x − p)(x − q) y = – _12 (x − 2)(x − (−2))
Practice Write the equation of each parabola in intercept form.
1. Vertex: (−1, −4); intercepts: (1, 0) and (−3, 0) Step 1 x-intercept: (1, 0); p = Step 2 x-intercept: (−3, 0); q = Step 3 Vertex: (−1, −4); x = ,y= Step 4 = a( − )( −
)
=a Step 5 y = a(x − p)(x − q)
y=
(x −
) (x −
)
2. Vertex: (−2, 3); intercepts: (−5, 0) and (1, 0) 3. Vertex: (1, −2); intercepts: (0, 0) and (2, 0)
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Solving Quadratic Equations Using Square Roots The following are examples of quadratic equations: 2x2 + 3x − 4 = 0
x2 + 4 = 0
4x2 − 2x = 0
When you solve a quadratic equation you are finding the points where the graph of the equation crosses the x-axis. In many quadratic equations, the graph crosses the x-axis at two locations. When solving a quadratic equation, look to see if there is an x-term (for example, 3x). If the equation does not have an x-term, then check to see if you can solve it using square roots. Rules for Solving a Quadratic Equation Using Square Roots 1. Isolate x2. 2. Find the square roots. Remember the solution of a square
root is both a positive number and a negative number.
Example Solve. 2x2 − 32 = 0 Step 1 Isolate x2.
2x2 − 32 + 32 = 0 + 32 2x2 = 32 2x2 ÷ 2 = 32 ÷ 2 x2 = 16 __
Step 2 Find the square roots. Remember
the solution of a square root is both a positive number and a negative number.
___
x2 = 16 → √ x2 = √ 16 x = 4, x = −4
Practice Solve.
1. 3x2 − 25 = 50 3x2 − 25
Isolate x2.
= 50
3x2 = 3x2 ÷
= 75 ÷
x2 = Find the square roots. Remember the solution of a square root is both a positive number and a negative number.
x2 = x=
__
→ √ x2 = √
___
,x=
2. x2 = 49
4. 2x2 − 2 = 6
3. x2 − 25 = 0
5. x2 + 15 = 115 Algebra 2
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Solving a Quadratic Equation by Completing the Square An equation such as x2 + 6x + 7 = 0 is not easy to solve. However, a method called completing the square is a way to solve a quadratic equation of this type. Rules for Solving a Quadratic Equation by Completing the Square 1. Identify the coefficient of the variable that is not squared. 2. Take _12 of this coefficient and square it. 3. Add the result to both sides of the equation. 4. Factor the expression on the left side; add the terms on the right side. 5. Take the square root of each side. 6. Isolate the variable.
Example Solve. x2 − 8x = 5 Step 1 Identify the coefficient of the variable
that is not squared. Step 2 Take _12 of this coefficient and square it. Step 3 Add the result to both sides of the
The coefficient of the variable that is not squared is −8. Half of −8 is −4; (−4)2 = 16. x2 − 8x + 16 = 5 + 16
equation. Step 4 Factor the expression on the left side;
add the terms on the right side. Step 5 Take the square root of each side. Step 6 Isolate the variable.
(x − 4)2 = 21 _______
___
___
= √ 21 → x – 4 = ±√ 21 √(x – 4)2 ___ x = 4 ± √ 21
Practice Solve.
1. x2 + 2x = 5 Identify the coefficient of the variable that is not squared.
The coefficient in front of the variable that is
Take _12 of this coefficient and square it.
Half of
Add the result to both sides of the equation.
x2 + 2x +
Factor the expression on the left side; add the terms on the right side.
not squared is
. is
2
;(
)2 =
.
=5+ =
_________
_______
Take the square root of each side.
√(000000)2 = √(00000) →
Isolate the variable.
x=
±√
___
2. x2 − 4x = −3
4. x2 + 2x = 5
3. x2 − 2x = 8
5. x2 + 4x = −1
______
= ±√ (0000)
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Quadratic Formula When you solve a quadratic equation, you are finding the points where the graph of the equation crosses the x-axis. The graph of a quadratic equation is U-shaped and may cross the x-axis at two points. Therefore, in finding the solution, you are finding the x-coordinate. The y-coordinates are always 0. One way to solve a quadratic equation is to use the quadratic formula. To use the quadratic formula, your equation must be in the form of ax2 + bx + c = 0. _______
± √ b2 – 4ac __________ The quadratic formula is x = –b . 2a
Example Use the quadratic formula to solve the following quadratic equation. x2 + 5x − 50 = 0 Step 1 Identify a, b, and c.
a = 1, b = 5, c = −50
Step 2 Plug the values for a, b, and c into the
± √ b2 – 4ac __________ x = –b 2a
_______
quadratic formula.
___________
–5 ± √ 52–4(1)(–50) x = ______________ 2(1) ________
Step 3 Solve.
± √ 25 + 200 ___________ x = –5 2 x = 5 or x = −10
Practice Solve using the quadratic formula.
1. x2 + 3x – 4 = 0 Identify a, b, and c.
,b=
a=
,c=
_______
Plug the values for a, b, and c into the quadratic formula.
2
± √ b – 4ac __________ x = –b 2a
_______________
–______ ± √ _____2 – 4______ x = ______________________ 2______
x=
Solve.
x=
or x =
2. x2 + 15x + 26 = 0 3. x2 – 6x − 72 = 0 4. 2x2 − 10x + 12 = 0 5. 3x2 − 12x − 15 = 0
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Solving a Quadratic Equation by Factoring You have learned to factor an equation in the form ax2 + bx + c. By applying this method of factoring and by using the Zero-Product Property, you can solve a quadratic equation. Rules for Solving a Quadratic Equation by Factoring 1. Be sure the equation is in the form ax2 + bx + c = 0. Set up a
FOIL table to help factor the equation. 2. Use the FOIL table to identify the numbers in each binomial.
Write the factored form of the original equation. 3. Set each binomial equal to 0 and solve for the two values of x.
Example Solve. 2x2 + 5x + 2 = 0 Step 1 Be sure the equation is in the
form ax2 + bx + c = 0. Set up a FOIL table to help factor the equation.
2x2
+ 5x
+2
F
O
+
I
=
?
L
2×1
2×1 2×2
+ +
2×1 1×1
= =
4 5
1×2
Step 2 Use the FOIL table to identify the
The outer terms are 2 and 2; the inner terms numbers in each binomial. Write the are 1 and 1. factored form of the original equation. 2x2 + 5x + 2 = (2x + 1)(x + 2) = 0 Step 3 Set each binomial equal to 0 and solve 2x + 1 = 0 x+2=0 1 _ for the two values of x. x = −2 x = −2
Practice Solve. 2x2 5x 2 1. 2x2 − 5x + 2 = 0 Be sure the equation is in the F O + I = ? L form ax2 + bx + c = 0. Set up 2×1 2× + × 1 = −4 a FOIL table to help factor the 2× + × 1 = −5 equation. Use the FOIL table to identify the The outer terms are and ; the numbers in each binomial. Write the inner terms are and . 2 factored form of the original equation. )( )=0 2x – 5x + 2 = ( Set each binomial equal to 0 and solve =0 =0 for the two values of x. x= x=
2. 6x2 − 23x + 7 = 0 3. 2x2 + x − 3 = 0 4. 3x2 − 7x − 6 = 0
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Using the Discriminant The solution to a quadratic equation is where the graph of the equation crosses the x-axis. The graph of a quadratic equation can cross the x-axis at two points, one point, or no point. You can find out the number of solutions a quadratic equation has by using the discriminant. The discriminant is the b2 − 4ac part of the quadratic formula. Rules for using the Discriminant 1. Identify a, b, and c in a quadratic equation. 2. Plug the numbers for a, b, and c into b2 − 4ac. 3. Solve. If the result is positive: There are two solutions.
If the result is 0: There is one solution. If the result is negative: There are no solutions.
Example Find the number of solutions for 3x2 – 5x − 1 = 0. Step 1 Identify a, b, and c in the quadratic
equation. Step 2 Plug the numbers for a, b, and c into
b2
3x2 − 5x − 1 = 0 a = 3, b = −5, c = −1 b2 − 4ac = (−5)2 − 4(3)(−1)
− 4ac.
Step 3 Solve.
25 + 12 = 37 The result is positive, so there are two solutions for the equation.
Practice Find the number of solutions for the following equations.
1. x2 + 3x + 7 = 0 Identify a, b, and c in the quadratic equation.
x2 + 3x + 7 = 0 a = 1, b = ,c=
Plug the numbers for a, b, and c into b2 − 4ac.
b2 − 4ac =
Solve.
−
2
− 4(1)
=
Since the result is
, there are
solutions. 2. x2 + 2x + 1 = 0 3. 2x2 − 7x + 4 = 0 4. x2 − 5 = 0 5. 3x2 − 9x + 12 = 0 Algebra 2
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Methods for Solving Quadratic Functions The chart below summarizes the methods you can use to solve a quadratic function. What to look for
Method(s) to use
ax2 = c
Finding square roots
Discriminant is a positive square number.
Factor Quadratic formula Completing the square
Discriminant is a positive non-square number.
Quadratic formula Completing the square
Discriminant is 0.
Factor Quadratic formula Completing the square
Discriminant is negative.
Quadratic formula Completing the square
Example Solve the equation. x2 − 12x + 5 = 0
The equation is not in the form ax2 = c. = c; if not, find the form The discriminant: discriminant. b2 − 4ac = (−12)2 − 4(1)(5) = 124 Step 2 Identify the nature of the discriminant; The discriminant is a positive non-square number; use the quadratic formula or determine a method for solving the completing the square. quadratic equation. Use completing the square. b 2 –12 2 _ ___ Step 3 Solve. = 36 2 = 2 x2 − 12x = −5 → x2 − 12x + 36 = −5 + 36 ___ (x − 6)2 = 31, so x = 6 ± √ 31
Step 1 Determine if the equation is in the
ax2
() ( )
Practice Solve each quadratic equation.
1. x2 − 2x − 15 = 0 Step 1 The equation is not in the form ax2 = c.
The discriminant: b2 − 4ac = (
)2 − 4 (
)(
)=
Step 2 The discriminant is
Factor, use_______ –b ± √
b2
– 4ac
. _________________
–___ ± √
.
(000)2
– 4(000)(000) 2(000)
Step 3 x = __________ = _____________________ 2a
x=
or
2. 3x2 − 243 = 0
4. −3x2 + 14x − 8 = 0
3. x2 + 6x + 8 = 0
5. 3x2 + 23x − 40 = 0
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Writing an Equation of an Ellipse The equation of an ellipse is based upon the orientation of the ellipse along the major axis. Major axis: Horizontal
Major axis: Vertical
Vertices: (±a, 0); co-vertices: (0, ±b)
Vertices: (0, ±a); co-vertices: (±b, 0)
y2 x2 __ Standard form of the equation: __ + 2=1 2 a b
2
y2 a
x __ Standard form of the equation: __ 2 + 2 =1 b
Rules for Writing the Equation of an Ellipse 1. If the y-coordinate of the vertex does not equal 0, then
the major axis is vertical. If the x-coordinate of the vertex does not equal 0, then the major axis is horizontal. 2. Identify a and b. If the major axis is horizontal, a is the x-coordinate of the vertex, b is the y-coordinate of a co-vertex. If the major axis is vertical, a is the y-coordinate of a vertex, and b is the x-coordinate of a co-vertex. 3. Choose the equation of the ellipse based on the major axis. Plug a and b into the equation.
Example Write the equation of an ellipse with a vertex at (4, 0) and a co-vertex at (0, −2). Step 1 Identify the orientation of the major
Step 2 Identify a and b, the vertex and
Vertex: (4, 0); co-vertex: (0, −2) The x-coordinate of the vertex does not equal 0; the major axis is horizontal. a = 4, b = −2
co-vertex. Step 3 Choose the equation of the ellipse based on the major axis. Plug a and b into the equation.
y2 x2 __ __ + =1 a2 b 2 2 y y2 x2 ____ x2 + __ __ __ + = 1 → 4 =1 16 42 (–2)2
axis.
Practice Write the equation of an ellipse with the given vertex and co-vertex.
1. vertex: (0, −6); co-vertex: (3, 0) Identify the orientation of the major axis.
Identify a and b, the vertex and co-vertex. Choose the equation of the ellipse based on the major axis. Plug a and b into the equation.
Vertex: (0, −6); co-vertex: (3, 0) The
-coordinate of the vertex does not
equal 0; the axis is a= ,b=
.
2
y x2 __ __ + =1 b2 a2 y2 y2 x2 x2 + _____ _____ + _____2 = 1 → _____ =1 2 ____ ____ ____ ____
2. Vertex: (0, −5); co-vertex: (−1, 0) 3. Vertex: (−4, 0); co-vertex: (0, 3) Algebra 2
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Foci of an Ellipse The foci of an ellipse are always located along the major axis, c units from the center. Rules for Finding the Foci of an Ellipse 1. Identify the larger denominator in the equation of the ellipse. 2. If the larger denominator is that of the y2 term, the major axis is vertical; if
the larger denominator is that of the x2 term, the major axis is horizontal. 3. Use the equation c2 = a2 − b2; the larger denominator is a2, the smaller denominator is b2. 4. Solve for c. The + and − values of c are the x-coordinates of the foci of an ellipse with a horizontal major axis, or they are the y-coordinates of the foci of an ellipse with a vertical major axis.
Example
2
y2
x Find the foci of the ellipse with the equation. __ + ___ =1 9 25 Step 1 Identify the larger denominator in the 25 is the larger denominator, 9 is the smaller
equation of the ellipse. Step 2 Identify the orientation of the major axis. Step 3 Use the equation c2 = a2 − b2; the larger denominator is a2, the smaller denominator is b2. Step 4 Solve for c, the coordinates of the foci.
denominator. The larger denominator is that of the y2 term; the major axis is vertical. a2 = 25, b2 = 9 c2 = a2 − b2 = 25 − 9 = 16 c = ±4 The major axis is vertical; foci are at (0, 4) and (0, −4).
Practice Find the foci of each ellipse. 2
y x2 __ 1. ___ 100 + 36 = 1 Identify the larger denominator in the equation of the ellipse.
Identify the orientation of the major axis. Use the equation c2 = a2 − b2; the larger denominator is a2, the smaller denominator is b2.
is the larger denominator; is the smaller denominator. The larger denominator is that of the x2 term; the major axis is a2 = b2 = c2 = a2 − b2 =
Solve for c, the coordinates of the foci. y2
=
and
.
y2
___ 2. ___ 144 + 225 = 1
x2 __ 4. __ 36 + 81 = 1
y x2 __ 3. __ 36 + 16 = 1
y x2 ___ 5. __ 64 + 100 = 1
2
−
c= Foci are at
x2
.
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Standard Deviation You can look at how the data in a data set are spread out. This is known as looking at measures of variation. One measure of variation is the standard deviation—a measure of how each value in the data set varies from the mean. The Greek letter σ represents standard deviation. Rules for Finding Standard Deviation 1. Find the mean of the data set, (x). 2. Find the differences between each value and the mean.
Square each difference, (x − x)2. 3. Find the sum of the squares in Step 2; divide by the total number of items in the data set. 4. Take the square root of the result.
Example Find the standard deviation of the data in the following data set: 24, 30, 29, 21, 22. (24 + 30 + 29 + 21 + 22) ÷ 5 = 25.2 Step 1 Find the mean of the data set, (x). Step 2 Find the differences between each x x (x − x)2
value and the mean. Square each difference, (x − x)2.
24 30 29 21 22
25.2 25.2 25.2 25.2 25.2
1.44 23.04 14.44 17.64 10.24
Step 3 Find the sum of the squares in Step 2
Sum of squares = 66.8 and divide by the total number of items. 66.8 ÷ 5 = 13.36 _____ Step 4 Take the square root of the result. √13.36 = 3.66
Practice Find the standard deviation of the data.
1. 53, 47, 39, 33, 40 Find the mean of the data set, (x). Find the differences between each value and the mean. Square each difference, (x − x)2.
(53 + 47 + 39 + 33 + 40) ÷ 5 = x 53 47 39 33 40
x
Find the sum of the squares in Step 2 and divide by the total number of items.
Sum of squares =
Take the square root of the result.
√00000 =
_____
2. 427, 466, 372, 299, 381
(x − x)2
÷5=
3. 54, 59, 35, 41, 60 Algebra 2
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Margin of Error The margin of error is used to indicate the expected variability in the data. For example, a margin of error of ±5% means that the data are likely within 5% of the population. As you will see, the greater the population, the smaller the margin of error. Rules for Using the Margin of Error
To find margin of error: 1. Identify the population sample size. 1__ 2. Plug the sample number into the formula Margin of error = ±___ , where √n n is the sample size. Multiply the result by 100 to get the percentage. To find the population size: 1. Identify the margin of error. Change the percentage to a decimal. 1__ 2. Plug the margin of error into the formula Margin of error = ±___ . Solve √n for n, the population size.
Example Find the margin of error for a survey with a sample size of 2,000. Step 1 Identify the population sample size.
The sample size (n) is 2,000.
Step 2 Plug the sample number into the
1__ Margin of error = ±___ , √n 1 _____ _____ = ±0.0224 × 100 = ±2.24% ± √2,000
1__ , formula Margin of error = ±___ √n
where n is the sample size. Multiply the result by 100 to get the percentage.
Practice 1. A survey has a margin of error of ±2.5%. What is the sample size? Identify the margin of error; change to a decimal. Plug the margin of error into the 1__ . formula Margin of error = ±___ √n Solve for n, the population size.
Margin of error = ±2.5% =
__ 1 √ n = ± ______ _______
1__ = ±___ √n
=
Square each side: __
(√ n )2 =
2
n= For the population size, find the margin of error.
2. Population: 1,000
3. Population: 3,000
For the margin of error, find the population size.
4. Margin of error = ±1%
5. Margin of error = ±4%
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Binomial Theorem Expanding a binomial (raising it to a power) is easily done using a pattern and Pascal’s Triangle. Rules for Expanding a Binomial 1. The first term in the binomial is a, the second term is b. The first expanded term is
the value for a raised to the power of the binomial times b raised to power 0. 2. Each next expanded term is a raised to a power one less than the preceding term
times b raised to a power one greater than the preceding term, until a is raised to the power 0 and b is raised to the power of the original binomial. 3. Find the row in Pascal’s Triangle with a value for the second number equal to the power of the original binomial. The numbers along that row are the values for the coefficients in the expanded binomial.
Example Expand. (x + 2)3 Step 1 The first term in the binomial is a, the
a = x; b = 2; power (exponent) = 3 second term is b. The first expanded term is the value for (x + 2)3 = x3(2)0 + . . . a raised to the power of the binomial times b raised to power 0. Step 2 Each next expanded term is a raised (x + 2)3 = x3 + 2x2 + x1(2)2 + x0(2)3 or to a power one less than the preceding (x + 2)3 = x3 + 2x2 + 4x + 8 term times b raised to a power one greater than the preceding term. Step 3 Find the row in Pascal’s Triangle with
The numbers in the row with 3 as the second value are 1, 3, 3, 1. (x + 2)3 = (1)x3 + (3)2x2 + (3)4x + (1)8 or (x + 2)3 = x3 + 6x2 + 12x + 8
a value for the second number equal to the power of the original binomial. The numbers along that row are the values for the coefficients in the expanded binomial.
Practice 1. Expand. (x + 3)4 Step 1 a =
;b=
; power (exponent) =
(x + 3)4 =
4
Step 2 (x + 3)4 =
4
0+
...
+
3
+
2
+
+
3
+
2
+
+
Step 3 The numbers are 1, 4, 6, 4, 1.
(x + 3)4 = 2. (x − 4)3
4
+
3. (x + 3)5
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Writing a System of Equations as a Matrix You can represent a system of equations as a matrix equation. 2x + 2y = 6 4x – y = 10
2
2
x
4
–1
y
Coefficient Matrix
=
6 10
Variable Matrix
Constant Matrix
Rules for Writing a System as a Matrix Equation 1. Identify the coefficient of the x-terms in each equation.
This is the first column of the coefficient matrix. 2. Identify the coefficient of the y-terms in each equation.
This is the second column of the coefficient matrix. 3. Write the variable matrix, x on the top, y on the bottom.
Write the constant matrix.
Example Write the following system as a matrix equation. −x + 3y = 4; 4x − 5y = 24 Step 1 Identify the coefficient of the x-terms
The coefficients of the x-terms are −1 and 4. in each equation. This is the first column of the coefficient matrix. Step 2 Identify the coefficient of the y-terms The coefficients of the y-terms are 3 and −5. in each equation. This is the second column of the coefficient matrix. Step 3 Write the variable matrix, x on the top, –1 x 4 3 = y on the bottom. Write the constant y 24 4 –5 matrix.
Practice Write each system as a matrix equation.
1. 3y = 1; 2x + 4y = 8 Identify the coefficient of the x-terms in each equation. This is the first column of the coefficient matrix. Identify the coefficient of the y-terms in each equation. This is the second column of the coefficient matrix. Write the variable matrix, x on the top, y on the bottom. Write the constant matrix. 2. 4x – 4y = 8; x = 6 3. –2x – 4y + 7 = 10; –x + 2y + 2 = 12
There is no x-term in the first equation, so the coefficient of x is 0. The coefficient of the x-term in the second equation is The y-coefficients are and
. .
=
= =
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Using Matrices to Solve a System of Two Equations A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions to all the equations. You can use a matrix equation to solve systems of equations. Rules for Using Matrices to Solve a System of Two Equations 1. Write the system as a matrix equation. 2. Find the inverse of the coefficient matrix. 3. Multiply the inverse of the coefficient matrix by the constant
matrix to solve for x and y.
Example Solve. 3x + 2y = −6; −4x − 3y = 10 Step 1 Write the system as a matrix equation.
3
2
x
–4
–3
y
Step 2 Find the inverse of the coefficient 1 _____ ad – bc
matrix.
=
–6 10
d
–b
–c
a
–3 1 _______________ 3 × (–3) – (2 × (–4)) 4
Step 3 Multiply the inverse of the coefficient
matrix by the constant matrix to solve for x and y.
x y
=
–2
=
3
3
2
–6
–4
–3
10
=
3
2
–4
–3
2 –6
Practice Solve.
1. 5x − 3y = 5; 4x − 2y = 10 x
Write the system as a matrix equation.
=
y
Find the inverse of the coefficient matrix.
1 _____ ad – bc
d
–b
–c
a
1 _____________________ ___________–___________
Multiply the inverse of the coefficient matrix by the constant matrix to solve for x and y. 2. −7x − 5y = 9; −3x − 2y = 4
x y
=
=
=
3. 5x + 3y = 7; 2x + y = 11 Algebra 2
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Cramer’s Rule Linear systems can also be solved using determinants and a pattern called Cramer’s Rule. Rules for Using Cramer’s Rule 1. Find the determinant (D) using the x- and y-coefficients. 2. Find the Dx determinant using the constants in the place of the x-coefficients. 3. Find the Dy determinant using the constants in the place of the y-coefficients. Dy
D
x 4. Find the coordinates of the solution using the formulas x = __ and y = __ . D D
Example Solve. 3x + 2y = −6; −4x − 3y = 10 Step 1 Find the determinant (D) using
a = 3, b = 2, c = −4, d = −3
the x- and y-coefficients. D= Step 2 Find the Dx determinant using
b
c
d
= ad – bc = 3(−3) − (2)(−4) = −1
m = −6; n = 10
the constants in the place of the x-coefficients.
Dx=
Step 3 Find the Dy determinant using
Dy =
the constants in the place of the y-coefficients.
Step 4 Find the coordinates of the solution
Dy
D
| | | | | | a
m
b
n
d
a
m
c
n
= md – bn = −6(−3) − (2)(10) = −2 = an – mc = 3(10) − (−6)(−4) = 6 Dy
D
6 x –2 __ __ = __ x = __ –1 = 2; y = D = –1 = −6 D
x using the formulas x = __ and y = __ . The solution is (2, –6). D D
Practice Solve.
1. 5x − 3y = 5; 4x − 2y = 10 Step 1 a = 5, b = −3, c = 4, d = −2
D=
| | | | | | a
b
c
d
= ad – bc = (5)(−2) −
=
Step 2 m = 5, n =
Dx=
Step 3 Dy =
D
m
b
n
d
a
m
c
n
x Step 4 x = __ = D
= md – bn =
−
=
= an – mc =
−
=
=
2. –7x – 5y = 9; –3x – 2y = 4
Dy
; y = __ = D
= 3. 5x + 3y = 7; 2x + y = 11
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Simplifying Radical Expressions by Removing Perfect Squares A radical expression contains a number or expression under a square root sign. The following expressions are radial expressions. ____
√121
___
_____
√4x2
2√ x + 5
You can simplify a radical expression by finding and removing perfect squares. Rules for Simplifying Radical Expressions by Removing Perfect Squares 1. Look at the number or expression under the square root sign. Find two
factors, one of which is a perfect square. 2. Rewrite the radical expression as the product of the two square root factors. 3. Place the square root of the perfect square factor outside the square root
symbol, leaving the non-perfect-square factor(s) inside.
Example ___ Simplify. √ 50
Step 1 Look at the number or expression
The factors of 50 are: 1 and 50, 2 and 25, under the square root sign. Find two 5 and 10; 25 is a perfect square; so use the factors, one of which is a perfect square. factors 2 and 25.
Step 2 Rewrite the radical expression as the
___
product of the two square root factors. Step 3 Place the square root of the perfect
___
__
√50 = √25 × √2 __
__
5 × √ 2 = 5√ 2
square factor outside the square root symbol, leaving the non-perfect-square factor(s) inside.
Practice Simplify. ___
1. √ 27
Look at the number or expression under the square root sign. Find two factors, one of which is a perfect square.
The factors of 27 are 1 and 3 and 3 and ___
; .
_______
is a perfect square, so use __
____
Rewrite the radical expression as the product of the two square root factors.
√27 = √3 × 000 = √3 × √000
Place the square root of the perfect square factor outside the square root symbol, leaving the non-perfect-square factor(s) inside.
√3 ×
____
__
=
___
2. √ 500
5. √ 75
3. √ 80
6. √ 120
4. √ 48
7. √ 162
___ ___
;
____ ____
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Simplifying Radical Expressions with Variables A____ radical_____ expression contains a number or expression under a square root sign, such as ___ √121 , 2√x + 5 , or √4x2 . You can simplify a radical expression by finding and removing perfect squares. Rules for Simplifying Radical Expressions with Variables 1. Find two factors (including variables) of the expression under the square
root symbol, one of which is a perfect square. 2. Rewrite the radical expression as the product of the two square root factors. 3. Place the square root of the perfect square factor outside the square root
symbol, leaving the non-perfect-square factor(s) inside.
Example _____ Simplify. √ 27a5
Step 1 Find two factors (including variables)
Look for a factor that is a perfect square; of the expression under the square root group the perfect square with the variable symbol; one of the factors must be a raised to an even power. perfect square. The factors of 27a5 are 3a and 9a4.
Step 2 Rewrite the radical expression as the
√
____
_______
27a5
9a4
=√
___
___
× 3a = √ 9a4 × √ 3a
product of the two square root factors. Step 3 Place the square root of the perfect
___
3a2√ 3a
square factor outside the square root symbol, leaving the non-perfect-square factor(s) inside.
Practice Simplify.
____
1.
√32x7 Find two factors (including variables) of the expression under the square root symbol; one of the factors must be a perfect square.
Look for a factor that is a perfect square; group the perfect square with the variable raised to an even power. The factors of 32x7 are ____
Rewrite the radical expression as the product of the two square root factors.
√32x7 = √
3.
√____ √48x5
3
___
× √ 2x
___
____
____
2.
× 2x = √
and 2x.
________
√2x
Place the square root of the perfect square factor outside the square root symbol, leaving the non-perfect-square factor(s) inside. 50x2
____________
4. 5.
63x6 √_____ √180x9
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Adding Radical Expressions Adding radical expressions is not unlike adding variable expressions. When you do so, you combine like terms. When adding radical expressions, you must have like radicals. Like radicals have the same index and the same radicand. Rules for Adding Radical Expressions 1. Simplify the radicals so you have like radicals. 2. Use the Distributive Property to separate the
numbers in front of each radical from the radical. 3. Add.
Example ____
___
Add. 3√ 12x + 2√ 3x
Step 1 Simplify the radicals so you have
like radicals. Step 2 Use the Distributive Property to
____
______
___
___
3√___ 12x = 3√ 4 × 3x = 2 × 3√ 3x = 6√ 3x 2√ 3x is in simplest form. ___
___
___
6√ 3x + 2√ 3x = (6 + 2)(√ 3x )
separate the numbers in front of each radical from the radical. ___
Step 3 Add.
___
(6 + 2)(√ 3x ) = 8√ 3x
Practice Add.
____
___
1. 5√ 32x + 3√ 2x
____
Simplify the radicals so you have like radicals.
5√ 32x =
=
___
3√ 2x is
.
___
+ 3√ 2x
Use the Distributive Property to separate the numbers in front of each radical from the radical.
___
=(
)(√ 2x ) ___
(
Add. ___
___
____
____
____
____
___
____
)(√ 2x ) =
___
√2x
2. 4√ 18y + 3√ 2y
3. 3√ 16x + 6√ 25x 4. 4√ 32x + 2√ 98x
5. 5√ 8x3 + 3√ 18x3
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Subtracting Radical Expressions Subtracting radical expressions is not unlike subtracting variable expressions. When you do so, you subtract like terms. When subtracting radical expressions, you must have like radicals. Like radicals have the same index and the same radicand. Rules for Subtracting Radical Expressions 1. Simplify the radicals so you have like radicals. 2. Use the Distributive Property to separate the
numbers in front of each radical from the radical. 3. Subtract.
Example
____
___
Subtract. 3√ 27x − 2√ 3x
Step 1 Simplify the radicals so you have
like radicals. Step 2 Use the Distributive Property to
____
___
___
___
___
___
9√ 3x − 2√ 3x = (9 − 2)(√ 3x )
separate the numbers in front of each radical from the radical. Step 3 Subtract.
______
27x = 3√ 9 × 3x = 3 × 3√ 3x = 9√ 3x 3√___ 2√ 3x is in simplest form.
___
___
(9 − 2)(√ 3x ) = 7√ 3x
Practice Subtract. ____
___
1. 5√ 18x − 3√ 2x
Simplify the radicals so you have like radicals.
____
5√ 18x = 5 3√ 2x is
___
(
___
___
____
____
___
____
___
− 3√ 2x = (
Use the Distributive Property to separate the numbers in front of each radical from the radical. Subtract.
=
___
)(√ 2x ) =
___
.
)(√ 2x ) ___
√2x
2. 4√ 18y − 3√ 2y
3. 3√ 16x − 6√ 25x
4. 5√ 8x3 − 3√ 18x3 ____
____
5. 2√ 98x − 4√ 32x
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Multiplying Radical Expressions Multiplying radical expressions is different from adding or subtracting a radical expression. When multiplying radical expressions, you must be sure the radicals have the same index. Unlike adding or subtracting radical expressions, you do not need to have like radicals. Rules for Multiplying Radical Expressions 1. Make sure that each term is using the same root.
Place each term under the same root symbol separated by the multiplication symbol. 2. Multiply. 3. Simplify.
Example ____
____
Multiply. √ 5x4 × √ 4x3
___
___
Step 1 Make sure that each term is using the
________
√5x4 × √4x3 = √5x4 × 4x3
same root. Place each term under the same root symbol separated by the multiplication symbol.
________
____
5x4 × 4x3 = √ 20x7 √____ _________ ___ √20x7 = √4 × x6 × 5x = 2x3√5x
Step 2 Multiply. Step 3 Simplify.
Practice Multiply.
1.
3
____
3
____
√25xy × √5xy2
Make sure that each term is using the same root. Place each term under the same root symbol separated by the multiplication symbol.
3
____
3
_______
3
√25xy × √
____
5xy2
3
__________________
= √ 0000000 0×00000002 3
√0000000 3
Multiply. Simplify. ___
______
___
___
2. 2√ 3x2 × √ 14x3y2 ____
5. 3√ 5y2 × 2√ 4y5
3.
√4y5 × √20y2
6. 4√ 2x2 × √ 6xy3
4.
√5x3 × √20xy5
7. 5√ xy6 × 2√ 2x6y
___
___
_____
3
___
___
3
____ ____
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Dividing Radical Expressions Dividing radical expressions is different from adding or subtracting a radical expression. When dividing radical expressions, you must be sure the radicals have the same index. Unlike adding or subtracting radical expressions, you do not need to have like radicals. Rules for Dividing Radical Expressions 1. Make sure that each radical is to the same root.
Place each term under the same root symbol. 2. Divide the terms under the symbol. 3. Simplify.
Example _____
4 √12x ___ Divide. _____
√3x
Step 1 Make sure that each radical is to the
same root. Place each term under the same root symbol. Step 2 Divide the terms under the symbol. Step 3 Simplify.
____
____
4 √12x 12x4 _____ ___ = ____ 3x √3x
√
____
√
___
12x4 = 4x3 ____ √_________ 3x ___
__
√4x3 = √4 × x2 × x = 2x√x
Practice Divide._____ 4 √–81x ___ 1. ______
√9x
Make sure that each radical is to the same root. Place each term under the same root symbol. Divide the terms under the symbol. Simplify. 3
_____
_____
4 √–81x 00000 ______ ___ = ____ 00000 9x √
√
______
√000000 _________ √000000000 =
____
√0000
_____
5 √270x ____ 2. ______ 3 √10x
____ 8 √48x ___ 3. _____ √3x3 3
______ 2 5
y √332x ___ 4. ______
√2xy
______ 6 7
y √56x ___ 5. ______
√7xy
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Rationalizing the Denominator of a Radical Expression There are times when working with radical expressions that you will have a radical in the denominator of a fraction. In these cases you may need to rationalize the denominator (rewrite the denominator so it has no radicals). Rules for Rationalizing the Denominator of a Radical Expression 1. Find the conjugate of the denominator. 2. Multiply the numerator and denominator of the fraction by the conjugate. 3. Simplify.
Example
__
3 – √2 __ . Rationalize the denominator of ______ 5 + √2
__
Step 1 Find the conjugate of the denominator. The denominator is __ 5 + √2 ;
its conjugate is 5 – √ 2 . __
Step 2 Multiply the numerator and
denominator of the fraction by the conjugate.
__
__
__
__
15 – 3√ 2 – 5√__2 + (√ 2 )2 3 – √ 2__ _____ _____ × 5 – √__2 = _________________ 5 + √2 5 – √2 52 – (√ 2 )2 __
__
__
__
__
15 – 3√ 2 – 5√__2 + (√ 2 )2 _______ 17 – 8√ 2 _________________ = 1725– –8√22 = _______ 23 52 – (√ 2 )2
Step 3 Simplify.
Practice Rationalize the denominator. ___ 5______ – √ 11 ___ 1. 2 + √ 11
Find the conjugate of the denominator.
___
The denominator is 2 + √ 11 . The ___ conjugate is
Multiply the numerator and denominator of the fraction by the conjugate.
.
5 – √ 11 ______ ___ × 2 + √ 11
= =
Simplify. = __
4 – √ 7__ 2. _____ 9 + √7
__
–5 + √__3 3. ______ 1 – √3
___
5 – √ 10 ___ 4. ______ 2 + √ 10 __
1 – √ 3__ 5. _____ 2 + √3
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Square Root of a Negative Real Number: Imaginary Numbers So far you have worked with a group of numbers called real numbers. These numbers included integers, whole numbers, natural numbers, rational numbers, and irrational numbers.____ In your work with real numbers, you did not consider the value of a number such as √ –16 . The imaginary number,___ i, is defined as a number whose square equals −1. Therefore, i2 = −1 and √ –1 = i. Rules for Finding the Square Root of a Negative Number 1. Factor the number under the square root sign so one of the factors is −1. ___ 2. Apply the multiplication property of roots so that you have –1 as one of your factors. √ ___ 3. Simplify. Remember √ –1 = i. Simplify the other roots.
Example ___ Simplify. √ –9
Step 1 Factor the number under the square
__
______
√9 = √–1 × 9
root sign so one of the factors is −1. Step 2 Apply the multiplication___ property of
roots so that you have √ –1 as one of your factors. ___
Step 3 Simplify. Remember √ –1 = i. Simplify
______
___
__
√–1 × 9 = √–1 × √9 ___
__
√–1 × √9 = i × 3 = 3i
the other roots.
Practice Simplify. ____
1. √ –12
____
___________
Factor the number under the square root sign so one of the factors is −1.
√–12 = √–1 ×
Apply the multiplication___ property of roots so that you have √ –1 as one of your factors.
√–1 ×
___
Simplify. Remember √ –1 = i. Simplify the other roots.
___________ 0 0=
0
___
0
√–1 ×
___
√–1 ×
=i×
=
i
____
2. √ –16
____
3. √ –25
____
4. √ –27
____
5. √ –40
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Complex Numbers Imaginary numbers and real numbers make up the set of complex numbers. Complex numbers are written in the form: a
+
bi
Real part
Imaginary part
Rules for Simplifying Complex Numbers 1. Identify the real part of the number. Identify what will become the imaginary part. 2. Simplify the imaginary part of the complex number. 3. Rewrite the complex number in the form a + bi.
Example
____
Rewrite the complex number. √ –81 + 3 in the form a + bi Step 1 Identify the real part of the number.
Identify what will become the imaginary part. Step 2 Simplify the imaginary part of the
complex number. Step 3 Rewrite the complex number in the
____
√–81 = imaginary part 3 = real part ____
___
___
–81 = √ –1 × √ 81 √____ √–81 = 9i 3 + 9i
form a + bi.
Practice Rewrite each complex number in the form a + bi. ____
1. √ –18 − 2
Identify the real part of the number. Identify what will become the imaginary part.
= imaginary part = real part ____
___
Simplify the imaginary part of the complex number.
√–18 = √–1 ×
Rewrite the complex number in the form a + bi.
+
___
____
√–18 =
____
2. √ –8 + 4
5. √ –25 − 10
3. √ –24 + 6
6. −√ –36 − 3
4. √ –49 + 7
7. 2√ –32 + 4
____ ____
____
____
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Adding Complex Numbers Imaginary numbers and real numbers make up the set of complex numbers. Complex numbers are written in the form: a
+
bi
Imaginary part
Real part
You can apply what you know about operations with real numbers to any questions with complex numbers. Rules for Adding Complex Numbers 1. For each complex number (in the form a + bi) identify the real
part and the imaginary part. 2. Group the real parts together; group the imaginary parts together. 3. Simplify. Express the sum in terms of a + bi.
Example Add. (3 + 4i) + (−2 + 6i) Step 1 For each complex number (in the
form a + bi) identify the real part and the imaginary part. Step 2 Group the real parts together; group
Real parts = +3 and −2 Imaginary parts = +4i and +6i (3 − 2) + (4i + 6i)
the imaginary parts together. Step 3 Simplify. Express the sum in terms
1 + 10i
of a + bi.
Practice Add.
1. (−5 − 4i) + (3 + 6i) For each complex number (in the form a + bi) identify the real part and the imaginary part.
Real parts = −5 and
Group the real parts together; group the imaginary parts together.
(−5 +
Simplify. Express the sum in terms of a + bi.
Imaginary parts = −4i and ) + (−4i +
)
+
2. (4 + 4i) + (3 − i)
5. (8 + 6i) + (8 − 6i)
3. (−7 + 2i) + (6 − 6i)
6. (12 − 3i) + (−9 + i)
4. (12 − 3i) + (−12 − 6i)
7. (4 + √ –16 ) + (2 + √ –25 )
____
____
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Subtracting Complex Numbers You can apply what you know about operations with real numbers to any operation with complex numbers. Rules for Subtracting Complex Numbers 1. For the complex number to the right of the minus sign, change the sign in front
of the real part of the number complex and in front of the imaginary part. Change the minus sign (between the two complex numbers) to a plus. 2. For each complex number, identify the real part and the imaginary part. 3. Group the real parts together; group the imaginary parts together. Separate the real part from the imaginary part with a plus sign. 4. Simplify. Express the difference in terms of a + bi.
Example Subtract. (3 + 4i) – (–4 + 2i) Step 1 For the complex number to the right of
the minus sign, change the sign in front of the real part of the number complex and in front of the imaginary part. Change the minus sign (between the two complex numbers) to a plus. Step 2 For each complex number, identify the
real part and the imaginary part. Step 3 Group the real parts together; group
(3 + 4i) − (−4 + 2i) = (3 + 4i) + (4 − 2i)
Real parts: 3 and 4 Imaginary parts: 4i and −2i (3 + 4) + (4i − 2i)
the imaginary parts together. Separate the real part from the imaginary part with a plus sign. Step 4 Simplify. Express the difference in
(3 + 4) + (4i − 2i) = 7 + 2i
terms of a + bi.
Practice Subtract.
1. (9 + 4i) − (2 + 5i) Step 1 (9 + 4i) − (2 + 5i)
9 + 4i + Step 2 Real parts: 9 and Step 3 (9
) + (4i
; imaginary parts: 4i and )
Step 4
2. (10 + 2i) − (4 + i)
4. 5 − (2 + 6i)
3. (5 − 3i) − (−3 − 2i)
5. (6 + 3i) − 3i Algebra 2
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Multiplying Complex Numbers You can apply what you know about operations with real numbers to any operation with complex numbers. Rules for Multiplying Complex Numbers
For two imaginary numbers: 1. Multiply the whole numbers; multiply i by i, if applicable. 2. Remember, i × i = i2 = −1. For two complex numbers: 1. Use the FOIL method. 2. Remember, i × i = i2 = −1. 3. Simplify by combining like terms.
Example Multiply. (5 + 7i)(−2 + 6i) Step 1 Use the FOIL method.
(5 + 7i)(−2 + 6i) = −10 + 30i + (−14i) + 42i2
Step 2 Remember, i × i = i2 = −1.
−10 + (30i + (−14i)) + 42(−1)
Step 3 Simplify by combining like terms.
−10 + (30i + (−14i)) − 42 = −10 + 16i − 42 = −52 + 16i
Practice Multiply.
1. (3 + 6i)(4 − 8i) Use the FOIL method.
Remember, i × i = i2 = −1.
Simplify by combining like terms.
(3 + 6i)(4 − 8i) = 12 − 24i +
+
12 − 24i +
+
12 − 24i +
+
12 +
+
=
=
2. (5 + 6i)(3 − 4i) 3. (3 + 2i)(5 + 3i) 4. (7 + 3i)(4 − 2i) 5. (9 + 4i)(3 + 4i)
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Dividing Complex Numbers You can apply what you know about operations with real numbers to any operation with complex numbers. Rules for Dividing Complex Numbers 1. Multiply numerator and denominator by the complex conjugate
of the denominator. Use the same complex number, but with the opposite operation sign between the real and imaginary parts. 2. Follow the rules for multiplying two complex numbers. 3. Simplify. Express the quotient in a + bi form.
Example + 2i Solve. 2_____ 3–i
Step 1 Multiply numerator and denominator
by the complex conjugate of the denominator. Use the same complex number, but with the opposite operation sign between the real and imaginary parts. Step 2 Follow the rules for multiplying two complex numbers. Step 3 Simplify. Express the quotient in a + bi form.
The complex conjugate of 3 − i is 3 + i. 2 + 2i ____ i _____ × 33 + 3–i +i
2 + 2i ____ 3 + i ___________ 6 + 2i + 6i + 2i2 _____ × = 3–i 3+i 9 – 3i + 3i – i2 6 + 2i + 6i + 2i2 _____ + 8i __ 8i _ 4 2 _ 4 ___________ = 4 10 = 10 + __ 10 = 5 + 5 i 9 – 3i + 3i – i2
Practice Solve.
5–i 1. _____ –2 + 4i
Multiply numerator and denominator by the complex conjugate of the denominator. Use the same complex number, but with the opposite operation sign between the real and imaginary parts. Follow the rules for multiplying two complex numbers. Simplify. Express the quotient in a + bi form.
The complex conjugate of −2 + 4i is
.
5–i _____ × –2 + 4i
5–i _____ × –2 + 4i
= =
=
–
=
–
2. (2 + 4i) ÷ (6 + i) 3. (3 + 3i) ÷ (2 − i) 4. (−2 + i) ÷ (3 − 2i) 5. (2 + 5i) ÷ (8 + i) Algebra 2
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Absolute Value and Complex Numbers You can apply what you know about real numbers, including absolute value, to complex numbers. The absolute value of a complex number is its distance from the origin on the complex number plane. Rules for Finding the Absolute Value of a Complex Number 1. Write the complex number in the form | a + bi | . 2. Put the real number and the coefficient of the imaginary number ______
into the formula √ a2 + b2 . 3. Simplify. The absolute value is always a positive number.
Example Find |4 − 3i|. Step 1 Write the complex number in
the form | a + bi | . Step 2 Put the real number and the
coefficient of the imaginary______ number into the formula √ a2 + b2 . Step 3 Simplify. The absolute value is
always a positive number.
| 4 − 3i | _________
| 4 − 3i | = √42 + (–3)2 _________
______
___
√42 + (–3)2 = √16 + 9 = √25 = 5
Practice Find the absolute value.
1. | 4 − 6i | Write the complex number in the form | a + bi | .
| 4 − 6i |
Put the real number and the coefficient of the imaginary______ number into the formula √ a2 + b2 .
| 4 − 6i | = √
______________
0+
2
Simplify. The absolute value is always a positive number. 2. | 2 + 4i | 3. | 5 − 2i | 4. | 7 − 4i | 5. | 4 − 2i | 6. | 5i + 12 | 7. | 8i + 2 |
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Finding a Complex Solution to a Simple Quadratic Equation You can use the concept of imaginary numbers to help find the solution to certain quadratic equations. The solution of some quadratic equations is a complex number. Rules for Finding Complex Solutions to Simple Quadratic Equations 1. Isolate the term with the x2 variable on one side of the equation. 2. Simplify the equation so you have x2 on one side. 3. Find the square root of each side. The result will be x = ±
Example Solve. 5x2 + 125 = 0 Step 1 Isolate the term with the x2 variable
on one side of the equation. Step 2 Simplify the equation so you have x2
on one side.
5x2 + 125 − 125 = 0 − 125 5x2 = −125 5x2 ÷ 5 = −125 ÷ 5 x2 = −25 __
Step 3 Find the square root of each side. The
result will be x = ± complex number.
____
_______
√x2 = √–25 = √–1 × 25 x = ±i × 5 = ±5i
Practice Solve.
1. 4x2 + 32 = 0 Isolate the term with the x2 variable on one side of the equation. Simplify the equation so you have x2 on one side.
4x2 + 32 4x2 = 4x2 ÷
=
÷
x2 = __
Find the square root of each side. The result will be x = ± complex number.
=0
______
______
√x2 = √000000 = √000000 x=±
2. x2 + 16 = 0 3. 3x2 + 48 = 0 4. x2 + 10 = −90 5. x2 + 5 = −22
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Finding a Complex Solution to a Quadratic Equation You can apply what you know about operations with real numbers to any operation with complex numbers. You can use the concept of imaginary numbers to help find the solution to certain quadratic equations. The solution of some quadratic equations is a complex number. Rules for Finding Complex Solutions to Quadratic Equations 1. Write the quadratic equation in standard form (ax2 + bx + c = 0). 2. Determine the values for a, b, and c. 3. Plug a, b, and c into the quadratic formula. 4. If the number under the square root sign is negative, apply the
concept of imaginary numbers to simplify.
Example Solve. x2 = −4x − 29 Step 1 Write the quadratic equation in
standard form
(ax2
x2 + 4x + 29 = 0
+ bx + c = 0).
Step 2 Determine the values for a, b, and c. Step 3 Plug a, b, and c into the quadratic
formula. Step 4 If the number under the square root
sign is negative, apply the concept of imaginary numbers to simplify.
a = 1, b =_______ 4, c = 29
___________
–4 ± √ 42 – 4(1)(29) –b ± √ b2 – 4ac ______________ x = __________ = 2a 2(1) _____
–4 ± √ –100 –4 ± 10i = ______ = –2 ± 5i x = _________ 2 2
Practice Solve.
1. x2 = 2x − 26 Write the quadratic equation in standard form (ax2 + bx + c = 0).
x2 − 2x + 26 = 0
Determine the values for a, b, and c.
a=
Plug a, b, and c into the quadratic formula.
,b=
_______
,c=
b2
–b ± √ – 4ac = x = __________ 2a _____ 0
If the number under the square root sign is negative, apply the concept of imaginary numbers to simplify.
000 ± √ 0000 000 ± 000 0 = _________ = x = __________ 000 000
2. 6x2 = −4x − 8
4. 4x2 = −16x − 24
3. 2x2 − 4x = −10
5. –2x2 = −10x + 14
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Solving Cubic Equations: Finding x-Intercepts A cubic function is a function written in the form f(x) = ax3 + bx2 + cx + d. As you can see, in a cubic function, one of the terms has a variable with an exponent of 3. To solve a cubic function, you find the points where the graph of the function crosses the x-axis. To find the x-intercepts, you write the cubic function in intercept form (f(x) = a(x − p)(x − q)(x − r)). Rules for Solving a Cubic Function 1. Write the cubic function in intercept form. 2. Set each factor equal to 0. 3. Solve each factor for x.
Example Solve. f(x) = 5(x − 3)(x − 7)(x − 9) Step 1 Write the cubic function in intercept
form.
5(x − 3)(x − 7)(x − 9) = 0 Divide both sides by 5. (x − 3)(x − 7)(x − 9) = 0
Step 2 Set each factor equal to 0.
x − 3 = 0; x − 7 = 0; x − 9 = 0
Step 3 Solve each factor for x.
x−3=0→x=3 x−7=0→x=7 x−9=0→x=9
Practice Solve.
1. f(x) = (2x + 4)(x − 5)(x − 6) Write the cubic function in intercept form.
(2x + 4)(x − 5)(x − 6) = 0 Factor the first term. (x − 5)(x − 6) = 0 Divide both sides by
.
(x − 5)(x − 6) = 0 Set each factor equal to 0.
= 0;
Solve each factor for x.
=0→x=
= 0;
=0
=0→x= =0→x= 2. f(x) = (x − 2)(x − 4)(x − 5) 3. f(x) = 4(x + 9)(x + 3)(x − 3) 4. f(x) = 2(x − 3)(x + 10)(x − 7) 5. f(x) = (2x + 4)(3x − 6)(5x + 10) Algebra 2
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Factoring Cubic Equations: Sum of Cubes A cubic function is a function written in the form f(x) = ax3 + bx2 + cx + d. As you can see, in a cubic function, one of the terms has a variable with an exponent of 3. You may recall that certain quadratic functions were the sums of squares. Similarly, certain cubic functions are the sums of cubes. Functions of this type are factored following a special factoring pattern. Rules for Using a Sum of Cubes to Factor a Cubic Function 1. Verify that the function is in the form (a3 + b3). Identify a3 and b3. 2. Find the cube roots of a3 and b3. 3. Rewrite the expression by first writing the term (a + b). 4. Rewrite the expression by writing the term (a2 − ab + b2). 5. Multiply the terms in Step 3 and Step 4.
Example Factor. x3 + 64 Step 1 Verify that the function is in the form
(a3
+
b3). Identify a3
and
Step 2 Find the cube roots of
a3
b 3. and
b3.
Step 3 Rewrite the expression by first writing
x3 + 64 a3 = x3; b3 = 64 3
__
3
___
a = √ x3 = x; b = √ 64 = 4 (a + b) = (x + 4)
the term (a + b). Step 4 Rewrite the expression by writing the
(a2 − ab + b2) = (x2 − 4x + 16)
term (a2 − ab + b2). Step 5 Multiply the terms in Step 3 and Step 4. x3 + 64 = (x + 4)(x2 − 4x + 16)
Practice Factor.
1. 8x3 + 125 Verify that the function is in the form (a3 + b3). Identify a3 and b3.
8x3 + 125 a3 = 8x3; b3 = ____
___
√8x3 =
Find the cube roots of a3 and b3.
a=
Rewrite the expression by first writing the term (a + b).
(a + b) = (
Rewrite the expression by writing the term (a2 − ab + b2).
(a2 − ab + b2) = (
Multiply the terms in Step 3 and Step 4.
8x3 + 125 =
2. x3 + 8
4. 27x3 + 125
3. x3 + 27
5. 64x3 + 8
;b=
____
_____
√00000 =
) )
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Factoring Cubic Equations: Difference of Cubes A cubic function is a function written in the form f(x) = ax3 + bx2 + cx + d. As you can see, in a cubic function, one of the terms has a variable with an exponent of 3. You may recall that certain quadratic functions were the differences of squares. Similarly, certain cubic functions are the differences of cubes. Functions of this type are factored following a special factoring pattern. Rules for Using a Difference of Cubes to Factor a Cubic Function 1. Verify that the function is in the form (a3 − b3). Identify a3 and b3. 2. Find the cube roots of a3 and b3. 3. Rewrite the expression by first writing the term (a − b). 4. Rewrite the expression by writing the term (a2 + ab + b2). 5. Multiply the terms in Step 3 and Step 4.
Example Factor. x3 − 125 Step 1 Verify that the function is in the form
(a3
−
b3). Identify a3
and
Step 2 Find the cube roots of
a3
b 3. and
b3.
Step 3 Rewrite the expression by first writing
x3 − 125 a3 = x3; b3 = 125 3
__
3
____
a = √ x3 = x; b = √ 125 = 5 (a − b) = (x – 5)
the term (a − b). Step 4 Rewrite the expression by writing the
(a2 + ab + b2) = (x2 + 5x + 25)
term (a2 + ab + b2). Step 5 Multiply the terms in Step 3 and Step 4. x3 − 125 = (x – 5)(x2 + 5x + 25)
Practice Factor.
1. 8x3 − 64 Verify that the function is in the form (a3 − b3). Identify a3 and b3.
8x3 − 64 a3 = 3
; b3 = _____
Find the cube roots of a3 and b3.
a = √ 00000 =
Rewrite the expression by first writing the term (a − b).
(a − b) = (
Rewrite the expression by writing the term (a2 + ab + b2).
(a2 + ab + b2) = (
Multiply the terms in Step 3 and Step 4.
8x3 – 64 =
2. x3 − 8
4. 64x3 − 125
3. x3 − 27
5. 125x3 − 8
;b=
= ) )
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Writing Cubic Equations in Intercept Form To solve a cubic function you find the points where the graph of the function crosses the x-axis. To find the x-intercepts you write the cubic function in intercept form (f(x) = a(x − p)(x − q)(x − r)). If you know the x-intercepts and one other point, you can use the intercept form of a cubic function to write a cubic function. Rules for Writing a Cubic Function in Intercept Form 1. Identify the x-intercepts of the function. One intercept becomes
p, one q, and one r. 2. Plug p, q, and r into the intercept form of a cubic function,
f(x) = a(x − p)(x − q)(x − r). 3. To find a use the other point. Substitute the x-coordinate for each occurrence of x; substitute the y-coordinate for f(x). Solve for a. 4. Rewrite the function in Step 2 to include the value of a from Step 3.
Example A cubic function has x-intercepts of −1, 2, and 4 and passes through (0, 16). Step 1 Identify the x-intercepts of the
x-intercepts: −1, 2, and 4 p = −1, q = 2, r = 4
function. One intercept becomes p, one q, and one r. Step 2 Plug p, q, and r into the intercept
f(x) = a(x + 1)(x − 2)(x − 4)
form of a cubic function, f(x) = a(x − p)(x − q)(x − r). Step 3 To find a use the other point.
Use the point (0, 16). 16 = a(0 + 1)(0 − 2)(0 − 4) 2=a
Substitute the x-coordinate for each occurrence of x; substitute the y-coordinate for f(x). Solve for a. Step 4 Rewrite the function in Step 2 to
f(x) = 2(x + 1)(x − 2)(x − 4)
include the value of a from Step 3.
Practice Write a cubic equation.
1. x-intercepts: −2, −1, 2; passes through (1, 18) Step 1 x-intercepts: −2, −1, 2; p = −2, q = Step 2 f(x) = a(x +
)(x
)(x
,r= )
Step 3 Use point (1, 18).
=a =a Step 4 f(x) =
(x +
)(x
)(x
)
2. x-intercepts: 1, 2, 3; passes through (0, 6) 3. x-intercepts: 1, 3, 6; passes through (2, 4) Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Evaluating Polynomials Using Synthetic Substitution A process called synthetic substitution is an easy way to evaluate a polynomial for any given value of x. Rules for Synthetic Substitution 1. Write the coefficients of the polynomial in a row along with the value of x. 2. Bring down the first coefficient. Multiply the value of x by the value of the first
coefficient. Place the product under the second coefficient. 3. Add the second coefficient and the answer to Step 2. 4. Multiply the sum from Step 3 by the value for x. Place the product under the third coefficient. 5. Repeat the process of multiplication and addition until you reach the last coefficient.
Example Evaluate 5x3 + x2 − 6x + 1 for x = −2. Step 1 Write the coefficients of the polynomial Step 2
Step 3 Step 4
Step 5
in a row along with the value of x. Bring down the first coefficient. Multiply the value of x by the value of the first coefficient. Place the product under the second coefficient. Add the second coefficient and the answer to Step 2. Multiply the sum from Step 3 by the value for x. Place the product under the third coefficient. Repeat the process of multiplication and addition until you reach the last coefficient.
–2]
5
–2]
5
1
–6
1
1 –6 1 –10 18 –24
5 –9 12 –23
The answer is −23.
Practice Evaluate each polynomial.
1. 4x3 + 2x2 + x − 1 for x = 3 Step 1 3] 4 2 1 –1 Steps 2–4 3] 4 2 1
Step 5
2.
x2
4 The answer is
–1
.
− 2x + 6 for x = −3
3. 8x3 − 4x2 + 2x − 1 for x = 2
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Multiplying Polynomials When you multiply polynomials you apply the rules you learned for multiplying whole numbers. However, in polynomial multiplication you treat each term as a digit. Rules for Polynomial Multiplication 1. Place the polynomials in vertical format. 2. Multiply the right-most of the lower polynomial
terms by each term in the upper polynomial. 3. Repeat with the next term in the lower polynomial. 4. Combine like terms.
Example Multiply. (−3x2 + x − 4) × (6x + 5) Step 1 Place the polynomials in vertical
format. Step 2 Multiply the right-most of the lower
polynomial terms by each term in the upper polynomial. Step 3 Repeat with the next term in the lower polynomial. Step 4 Combine like terms.
–3x2 + x – 4 × 6x + 5 –15x2 + 5x – 20 –18x3 + 6x2 – 24x –18x3 – 9x2 – 19x – 20
Practice Multiply.
1. (7x2 + 6x + 5) × (x − 4) Place the polynomials in vertical format.
7x2 + 6x + 5 × x –4
Multiply the right-most of the lower polynomial terms by each term in the upper polynomial. Repeat with the next term in the lower polynomial. Combine like terms. 2. (4x2 − 5x + 8) × (2x – 3) 3. (x2 − 5x − 1) × (x + 1) 4. (−x2 + 8x − 9) × (x2 − x) 5. (10x2 + x + 6) × (−3x3 + x2)
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Dividing Polynomials Using Synthetic Division You can use the principles of synthetic substitution to divide polynomials. Rules for Dividing Polynomials Using Synthetic Division 1. Write the coefficients in a row. Reverse the sign of the constant
term in the divisor and place it to the left of the list. 2. Perform a series of multiplications and additions as you would for
synthetic substitution. 3. In the result the value farthest to the right is the remainder. The other values are the coefficients for the other terms, from the constant on the right to the term with the highest power on the left.
Example Divide. (6x3 − 5x2 + 9) ÷ (x + 1) Step 1 Write the coefficients in a row. Reverse
–1]
6
Step 2 Perform a series of multiplications and –1]
6
–5
0
9
the sign of the constant term in the divisor and place it to the left of the list. additions as you would for synthetic substitution.
–5 0 9 –6 11 –11 6 –11 11 –2
Step 3 In the result the value farthest to the
The remainder is −2. right is the remainder. The other values The quotient is 6x2 − 11x + 11. are the coefficients for the other terms, from the constant on the right to the term with the highest power on the left.
Practice Divide.
1. (x2 + 6x + 10) ÷ (x + 2) Write the coefficients in a row. Reverse the sign of the constant term in the divisor and place it to the left of the list.
]
1
Perform a series of multiplications and additions as you would for synthetic substitution.
]
1
In the result the value farthest to the right is the remainder. The other values are the coefficients for the other terms, from the constant on the right to the term with the highest power on the left.
6
6
10
10 –8
1 The remainder is The quotient is
. .
2. (4x3 + 6x2 − 13x + 10) ÷ (x + 4) 3. (6x2 + 7x + 5) ÷ (x − 3) Algebra 2
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End Behavior of a Polynomial Function As x takes on large positive and negative values, a polynomial function can behave in one of several ways. You describe this end behavior using infinity notation. Sign of the leading coefficient
Degree of the function
End behavior
Positive
Odd
Up to the right Down to the left
As x → +⬁, f(x) → As x → −⬁, f(x) →
. .
Positive
Even
Up to the right Up to the left
As x → +⬁, f(x) → As x → −⬁, f(x) →
. .
Negative
Odd
Down to the right Up to the left
As x → +⬁, f(x) → As x → −⬁, f(x) →
. .
Negative
Even
Down to the right Down to the left
As x → +⬁, f(x) → As x → −⬁, f(x) →
. .
End behavior using ⴥ
Rules for Describing the End Behavior of a Rational Function 1. Determine the sign (positive or negative) of the leading coefficient. 2. Determine the degree of the polynomial; determine if the degree is odd or even. 3. Use the chart above to describe the end behavior of the function.
Example Describe the end behavior of the function f(x) = 5x4 + x − 1. Step 1 Determine the sign (positive or
negative) of the leading coefficient. Step 2 Determine the degree of the polynomial; determine if the degree is odd or even. Step 3 Use the chart above to describe the end behavior of the function.
The leading coefficient is 5; its sign is positive. The degree of the polynomial is 4; the degree is even. The coefficient is positive; the degree is even. As x → +⬁, f(x) → +⬁. As x → −⬁, f(x) → +⬁.
Practice 1. Describe the end behavior of the function f(x) = −7x2 + 3x + 2. Step 1 The leading coefficient is ; its sign is Step 2 The degree of the polynomial is ; the degree is Step 3 The coefficient is ; the degree is
2. f(x) =
As x → +⬁, f(x) →
.
As x → −⬁, f(x) →
.
−3x2
. . .
+ 2x + 1
3. f(x) = 10x10 + x3 − 3 Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Inverse Variation When you have a set of data in which one variable increases while the other decreases and the product of the variables is constant, you have an inverse variation. Rules for an Inverse Variation. 1. Examine the data. Does one variable increase while the other one decreases? 2. Is the product xy a constant value? 3. If the answers to both 1 and 2 are “yes,” the data show an inverse variation.
Example Do the data in the table represent an inverse variation?
x
2
4
6
8
y
12
6
4
3
Step 1 Examine the data. Does one variable
Yes, as the value for x increases, the value for y increase while the other one decreases? decreases.
Step 2 Is the product xy a constant value?
Yes, the product of xy is the same value, 24. For example, 2 × 12 = 24; 4 × 6 = 24.
Step 3 If the answers to both 1 and 2 are “yes,” The answers to both are yes. The data show
the data show an inverse variation.
an inverse variation.
Practice Do the data in the table below represent an inverse variation?
1.
x
9
6
3
1
y
4
6
12
36
Examine the data. Does one variable increase while the other one decreases?
As the value for x decreases, the value for y
Is the product xy a constant value?
The product xy
. a constant. For
example, 9 × 4 = If the answers to both 1 and 2 are “yes,” the data show an inverse variation.
;6×6=
The answer to question 1 is answer to question 2 is
. , the
; the data
show an inverse variation. 2.
3.
x
2
4
6
8
10
y
4
8
12
16
20
x
4
12
20
30
y
15
5
3
2
4.
x
18
12
9
3
y
4
6
8
24
Algebra 2
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Simplifying Rational Expressions The following expressions are examples of rational expressions. 4 _ x
5 ____ x+3
2+ 2 x_____ y2 – 1
x ____ y–2
As you can see, a rational function has a variable in the denominator. You may have noticed that a rational expression looks like a fraction. Like a fraction, a rational expression is in simplest form if the numerator and denominator have no common factors other than 1. Rules for Simplifying a Rational Expression 1. Factor the numerator and denominator. 2. Divide out common factors. 3. Simplify.
Example
10x2 Simplify. ____ 4 15x
2
(5x )(2) 10x2 ________ Step 1 Factor the numerator and denominator. ____ 4 = 2 2
Find two or more factors for each expression. Step 2 Divide out common factors.
A factor in the numerator cancels out the same factor in the denominator. Step 3 Simplify.
15x
(5x )(3x )
(5x2)(2) 10x2 ________ ____ = 2 2 4 15x (5x )(3x )
2 ___ 3x2
Practice Simplify. 18x3 1. _____ 3x + 6
Factor the numerator and denominator.
18x3 _____ 3x + 6 =
Divide out common factors. Simplify. + 15 ______ 2. 6x 18 5x4 3. ____ 7
35x + 12 ______ 4. 4x 2x2 + 12 ______ 5. 3x 2x + 8 + 5x ______ 6. 20 10 + 5x Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Multiplying Rational Expressions When multiplying a fraction you multiply the numerators and multiply the denominators. You then simplify the product as needed. 3 1____ ×3 _ 3 1 _ _ 2 × 4 = 2×4 = 8
When multiplying rational expressions you follow the same rules. Rules for Multiplying Rational Expressions 1. Multiply the numerators. 2. Multiply the denominators. 3. Express the resulting rational expression in simplest form.
Example 5 4 __ Multiply. __ 2 × 3 x
x
Step 1 Multiply the numerators. Step 2 Multiply the denominators. Step 3 Express the resulting rational
expression in simplest form.
4×5 20 4 __ __ × 5 = ______ = ______ x2 x3 (x2)(x3) (x2)(x3) 20 20 ______ = __ (x2)(x3) x5 20 __ is in simplest form. x5
Practice Multiply. x2 5x __ 1. ____ x–1 × 4 Multiply the numerators.
Multiply the denominators. Express the resulting rational expression in simplest form.
(x2)(5x) x2 5x _______ 0000 ____ __ _______ x – 1 × 4 = (x – 1)(4) = (x – 1)(4) 0000 _______ = (x – 1)(4)
in simplest form.
3x2 ___ 6 2. ___ 2 × 4 5x
+ 4 _____ 2x2 3. x____ x × 2x – 3 –1 _____ 4. _23 × 4x 4 x
3x x____ +1 5. ____ x–2 × 3
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Dividing Rational Expressions When dividing fractions you flip the second fraction in the expression and then follow the rules for multiplying fractions. 3 1____ ×3 _ 3 1 _ 2 _ 1 _ _ 2 ÷ 3 = 2 × 2 = 2×2 = 4
When multiplying rational expressions you follow the same rules. Rules for Dividing Rational Expressions 1. Flip the second expression and change the division sign to a multiplication sign. 2. Multiply the numerators. 3. Multiply the denominators. 4. Express the resulting rational expression in simplest form.
Example –3 2 Divide. x____ ÷ _____ x+1 x2 Step 1 Flip the second expression and
change the division sign to a multiplication sign. Step 2 Multiply the numerators.
Leave the numerator in factored form. Step 3 Multiply the denominators. Step 4 Express the resulting rational
expression in simplest form.
x – 3 ____ x – 3 ____ ____ ÷ x +2 1 = ____ × x +2 1 x2 x2
(x – 3)(x + 1) x – 3 ____ ____ × x +2 1 = __________ 2 x (x2)(2)
(x – 3)(x + 1) __________ (x – 3)(x + 1) __________ = (x2)(2) 2x2 (x – 3)(x + 1) __________ is in simplest form. 2x2
Practice Divide
x – 2 __ 2 1. ____ x + 2 ÷ x2 Flip the second expression and change the division sign to a multiplication sign.
Multiply the numerators.
x – 2 __ x–2 2 ____ ____ x + 2 ÷ x2 = x + 2 ×
x–2 ____ x+2 ×
=
= Multiply the denominators. Express the resulting rational expression in simplest form.
= The resulting rational expression simplest form.
3x3 ____ –5x4 ÷ 2. ___ 2 4
– 1 5x +6 _____ 4. x____ x – 3 ÷ x2
+ 2 x____ –1 3. x____ x+3 ÷ 2
+ 6 ____ 8x2 _____ 5. 2x 4x ÷ x + 1
in
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Rational Functions—Finding Vertical Asymptotes p(x)
A rational function is a function that has the form f(x) = ___ when p(x) and q(x) are q(x) polynomials and q(x) ⫽ 0. The following are examples of rational functions. +4 f(x) = x____ x–3
+5 f(x) = x____ x+2
x–3 f(x) = _______ 2
x +x–6
An asymptote is a line that a graph approaches but never reaches. Rules for Finding the Vertical Asymptote of a Rational Function 1. Factor the numerator and denominator. Cancel out common factors. 2. Set the denominator as an equation equal to 0. 3. Solve the equation.
Example
x2 – 1 Find the vertical asymptote of the graph of f(x) = ___________ . 2 2x + 5x – 12
Step 1 Factor the numerator and
denominator. Cancel out common factors. Step 2 Set the denominator as an equation
equal to 0.
(x – 1)(x + 1) x2 – 1 _________ = ___________ 2 2x + 5x – 12 (2x – 3)(x + 4)
There are no common factors. (2x − 3)(x + 4) = 0 2x − 3 = 0 x+4=0 x = _32
Step 3 Solve the equation.
x = −4
The vertical asymptotes are x = _32 and x = −4.
Practice Find the vertical asymptotes. –1 _____ 1. f(x) = 2x 2 x –4
Factor the numerator and denominator. Cancel out common factors. Set the denominator as an equation equal to 0.
2x – 1 ___________ 2x – 1 _____ = x2 – 4 (00000)(00000)
There are no common factors. (
)=0
=0 x=
Solve the equation.
)(
=0 x=
The vertical asymptotes are x = and x =
.
2 2. f(x) = _________ 2
x + 3x – 28 2 ____ 3. f(x) = 2x 2 x –1 3 + 5x2 4. f(x) = x______ x2 – 9 Algebra 2
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End Behavior of Rational Functions: m < n p(x)
A rational function is a function that has the form f(x) = ___ when p(x) and q(x) are q(x) polynomials and q(x) ⫽ 0. As x takes on large positive or negative values, a rational function can behave in one of several ways. You describe this end behavior using infinity notation. Rules for Describing the End Behavior of a Rational Function When m < n 1. Find the degree of the polynomial in the numerator. The degree is m. 2. Find the degree of the polynomial in the denominator. The degree is n. 3. If m < n, then as x → +⬁, f(x) → 0 and as x → −⬁, f(x) → 0.
Example 3x Describe the end behavior of the function f(x) = __________ . 2 x + 3x – 28
Step 1 Find the degree of the polynomial in
The degree of 3x is 1; m = 1.
the numerator. The degree is m. Step 2 Find the degree of the polynomial in
The degree of x2 + 3x − 28 is 2; n = 2.
the denominator. The degree is n. Step 3 If m < n, then as x → +⬁, f(x) → 0
and as x → −⬁, f(x) → 0.
1 < 2, so m < n. As x → +⬁, f(x) → 0. As x → −⬁, f(x) → 0.
Practice Describe the end behavior. 2
8x – 8 1. f(x) = ________ 3
x – 2x – 3
Find the degree of the polynomial in the numerator. The degree is m.
The degree of 8x2 − 8 is
Find the degree of the polynomial in the denominator. The degree is n.
The degree of x3 − 2x − 3 is
If m < n, then as x → +⬁, f(x) → 0 and as x → −⬁, f(x) → 0.
<
;m=
, so m
;n=
. .
n.
As x → +⬁, f(x) →
.
As x → −⬁, f(x) →
.
x–5 2. f(x) = _________ 3 2
x – 5x + 4x 2 _____ 3. f(x) = x3 – 9 x + 5x _________ 4. f(x) = 2 x – 1 2x + 5x – 12 2 ________ 5. f(x) = 3 4x 2 2x + x – 2 Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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End Behavior of Rational Functions: m = n As x takes on large positive or negative values, a rational function can behave in one of several ways. You describe this end behavior using infinity notation. Rules for Describing the End Behavior of a Rational Function When m = n 1. Find the degree m of the polynomial in the numerator. Find the degree n
of the polynomial in the denominator. 2. If m = n, compare the leading coefficient of the polynomial in the
numerator (a) to that of the denominator (b). Find _ab . 3. The end behavior is as x → ±⬁, f(x) → _a , and the graph has a horizontal b asymptote of y = _ab .
Example
x2 – 2x + 12 Describe the end behavior of the function f(x) = ___________ . 2 2x + 3x + 11
Step 1 Find the degree m of the polynomial
in the numerator. Find the degree n of the polynomial in the denominator. Step 2 If m = n, compare the leading
coefficient of the polynomial in the numerator (a) to that of the denominator (b). Find _ab . Step 3 The end behavior is as x → ±⬁,
f(x) → _ab , and the graph has a horizontal asymptote of y = _ab .
The degree of x2 −2x +12 is 2; m = 2. The degree of 2x2 + 3x + 11 is 2; n = 2. 2 = 2, so m = n. a = 1; b = 2 a _ _ =1 b 2 As x → ±⬁, f(x) → _12 . Horizontal asymptote: y = _12
Practice Divide.
2
12x 1. f(x) = ________ 2
4x + x – 1
Find the degree m of the polynomial in the numerator. Find the degree n of the polynomial in the denominator. If m = n, compare the leading coefficient of the polynomial in the numerator (a) to that of the denominator (b). Find _ab . The end behavior is as x → ±⬁, f(x) → _ab , and the graph has a horizontal asymptote of y = _ab .
The degree of 12x2 is
;m=
The degree of 4x2 + x − 1 is = a=
, so m ;b=
As x → ±⬁, f(x) →
. ;n=
.
n. ; _ab =
=
.
Horizontal asymptote: y =
3
x + 5x 2. f(x) = 7______ 3 2x – 8 2_____ 3. f(x) = xx+– 51
Algebra 2
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End Behavior of Rational Functions: m > n As x takes on large positive and negative values, a rational function can behave in one of several ways. You describe this end behavior using infinity notation. Rules for Describing the End Behavior of a Rational Function When m > n 1. Find the degree m of the polynomial in the numerator. Find the degree n
of the polynomial in the denominator. 2. If m > n, compare the leading coefficient of the polynomial in the
numerator (a) to that of the denominator (b). Find _ab xm–n. 3. Use what you know about the end behavior of a polynomial function to describe the end behavior of the rational function.
Example
6 + 11 _______ Describe the end behavior of the function f(x) = 5x . 4
x +1
Step 1 Find the degree m of the polynomial
in the numerator. Find the degree n of the polynomial in the denominator. Step 2 If m > n, compare the leading coefficient of the polynomial in the numerator (a) to that of the denominator (b). Find _ab xm–n. Step 3 Use what you know about the end behavior of a polynomial function to describe the end behavior of the rational function.
The degree of 5x6 + 11 is 6; m = 6. The degree of x4 + 1 is 4; n = 4. 6 > 4, so m > n. a = 5; b = 1; _ab xm–n = _51 x6–4 = 5x2 The coefficient is positive, the degree is even. As x → +⬁, f(x) → +⬁. As x → −⬁, f(x) → +⬁.
Practice
2
+ 3x + 2 __________ 1. Describe the end behavior of the function f(x) = –6x . 3x – 4 The degree of −6x2 + 3x + 2 is Find the degree m of the polynomial in the numerator. Find the degree n m= . of the polynomial in the denominator. The degree of 3x − 4 is ;n=
If m > n, compare the leading coefficient of the polynomial in the numerator (a) to that of the denominator (b). Find _ab xm–n. Use what you know about the end behavior of a polynomial function to describe the end behavior of the rational function.
> a=
, so m ;b=
;
.
n. ; _ab xm–n =
x
= The coefficient is
, the
degree is
.
As x → +⬁, f(x) →
.
As x → −⬁, f(x) →
.
2x4 2. f(x) = _______ 2 4x + 10 7 + x2 + 1 _________ 3. f(x) = –x x4 – x Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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End Behavior of Rational Functions: Using All Three Conditions As x takes on large positive or negative values, a rational function can behave in one of several ways. You describe this end behavior using infinity notation. Rules for Describing the End Behavior of a Rational Function 1. Find the degree of the polynomial in the numerator. The degree is m. 2. Find the degree of the polynomial in the denominator. The degree is n. 3. Compare m and n. Use the comparison to identify the end behavior
case. Apply the rules of the identified case.
Example
6x2 Describe the end behavior of the function f(x) = _____ . 4 x –1
Step 1 Find the degree of the polynomial in
The degree of 6x2 is 2; m = 2.
the numerator. The degree is m. Step 2 Find the degree of the polynomial in
The degree of x4 − 1 is 4; n = 4.
the denominator. The degree is n. Step 3 Compare m and n. Use the comparison 2 < 4, so m < n.
to identify the end behavior case. Apply Use the behavior for when m < n. the rules of the identified case. As x → +⬁, f(x) → 0. As x → −⬁, f(x) → 0.
Practice Describe the end behavior of each function. 4
+ 7x + 2 1. f(x) = x________ x+4
Find the degree of the polynomial in the numerator. The degree is m.
The degree of x4 + 7x + 2 is
Find the degree of the polynomial in the denominator. The degree is n.
The degree of x + 4 is
Compare m and n. Use the comparison to identify the end behavior case. Apply the rules of the identified case.
>
;m= ;n=
, so m
. .
n.
Use the behavior for when m > n. a m–n _ x = b
x
=
As x → +⬁, f(x) →
.
As x → −⬁, f(x) →
.
4
+2 _____ 2. f(x) = 2x 4 x –1 –2x4 ____ 3. f(x) = 2 x –3
Algebra 2
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Recursive Formulas: Arithmetic Sequences In a recursion you repeatedly apply the same operation to a term to get to the next term. A recursion will help you write a formula to find a particular term in a sequence. A recursive formula has two parts: a starting value (t1) and a recursion equation that you can use to find each term of the sequence. Rules for Writing a Recursive Formula for an Arithmetic Sequence 1. Identify the starting value (t1). 2. Identify the common difference (d) between each pair of terms. 3. Substitute the value for the common difference into the recursive
equation tn = tn−1 + d. 4. State the recursive formula for the sequence.
Example Write a recursive formula for the sequence. −4, 1, 6, 11 . . . Step 1 Identify the starting value (t1).
The starting value of t1 is −4.
Step 2 Identify the common difference (d)
−4
between each pair of terms. Step 3 Substitute the value for the common
difference into the recursive equation tn = tn−1 + d. Step 4 State the recursive formula for the
sequence.
1
6
11
5 5 5 The common difference (d) is +5. tn = tn−1 + d tn = tn−1 + 5 t1 = −4 tn = tn−1 + 5
Practice Write a recursive formula for each sequence.
1. 7, 7.25, 7.50, 7.75 . . . Identify the starting value (t1).
The starting value (t1) is
Identify the common difference (d) between each pair of terms.
7
7.25
7.50
The common difference (d) is Substitute the value for the common difference into the recursive equation tn = tn−1 + d. State the recursive formula for the sequence.
. 7.75
.
tn = tn−1 + d tn = tn−1 + t1 = tn = tn−1 +
2. 21, 17, 13, 9, . . .
4. −3, −8, −13, −18, . . .
3. 0, 6, 12, 18, . . .
5. 70, 62, 54, 46, . . .
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Recursive Formulas: Geometric Sequences In a recursion you repeatedly apply the same operation to a term to get to the next term. A recursion will help you write a formula to find a particular term in a sequence. A recursive formula has two parts: a starting value (t1) and a recursion equation that you can use to find each term of the sequence. Rules for Writing a Recursive Formula for a Geometric Sequence 1. Identify the starting value (t1). 2. Identify the common ratio (r) between each pair of terms. 3. Substitute the value for the common ratio into the recursive equation tn = r(tn−1). 4. State the recursive formula for the sequence.
Example Write a recursive formula for the sequence. 575, 115, 23, 4.6, . . . Step 1 Identify the starting value (t1).
The starting value (t1) is 575.
Step 2 Identify the common ratio (r)
575
between each pair of terms.
115 1 _ 5
23 1 _ 5
4.6 1 _ 5
The common ratio (r) is _15. Step 3 Substitute the value for the common
ratio into the recursive equation tn = r(tn−1). Step 4 State the recursive formula for the
sequence.
tn = r(tn−1) tn = _15 (tn−1) t1 = 575 tn = _15 (tn−1)
Practice Write a recursive formula for each geometric sequence.
1. −12, 48, −192, 768, . . . Identify the starting value (t1).
The starting value (t1) is
Identify the common ratio (r) between each pair of terms.
−12
48
−192
. 768
The common ratio (r) is Substitute the value for the common ratio into the recursive equation tn = r(tn−1).
tn = r(tn−1)
State the recursive formula for the sequence.
t1 =
tn =
tn =
.
(tn−1)
(tn−1)
2. 2, 3, 4.5, 6.75, . . .
16 4. 6, 4, _83, __ 9,...
3. 64, 32, 16, 8, . . .
5. 675, 67.5, 6.75, 0.675, . . . Algebra 2
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Finding the Sum of a Finite Arithmetic Series An arithmetic series is a series whose terms form an arithmetic sequence. A finite arithmetic series is one in which the terms can be counted individually. A finite arithmetic series has a known beginning term and a known end term. Rules for Finding the Sum of a Finite Arithmetic Series 1. Determine the number of terms in the series. Identify the first and last values for x. 2. Substitute each value of x into the formula for the series to find the value of the
first term and the last term.
n 3. Use the formula Sn = __ 2 (a1 + an), where n is the number of terms, a1 is the value of
the first term, and an is the value of the last term.
Example Find the sum of the series whose terms are given by the formula 2x + 1, where x = 1 for the first term and x = 10 for the last term. Step 1 Determine the number of terms in the There are 10 terms in the series, so n = 10.
series. Identify the first and last values for x. Step 2 Substitute each value of x into the formula for the series to find the value of the first term and the last term.
The first term has x = 1, the last term has x = 10. Formula: 2x + 1 First term: 2(1) + 1 = 3 Last term: 2(10) + 1 = 21 n n __ Step 3 Use the formula Sn = __ (a + a ), where S n n = 2 (a1 + an) 2 1 n is the number of terms, a1 is the 10 Sn = __ 2 (3 + 21) = 120 value of the first term, and an is the value of the last term.
Practice Find the sum of each series.
1. Formula: 2x + 3; first term: x = 1; last term: x = 6 There are terms. Determine the number of terms in the series. Identify the first and last values The first term has x = , the last term has for x. x= . Formula: 2x + 3 Substitute each value of x into the formula for the series to find the value First term: 2( )+3= of the first term and the last term. Last term: 2( )+3= Use the formula Sn = __n2 (a1 + an), where n is the number of terms, a1 is the value of the first term, and an is the value of the last term.
000 Sn = ___ 2 (
+
)=
2. Formula: −2x − 2; first term: x = 1; last term: x = 10 3. Formula: 2x2; first term: x = 1; last term: x = 10 8x 4. Formula: __ 3 ; first term: x = 2; last term: x = 10 Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Finding the Sum of a Finite Geometric Series A geometric series is a series whose terms form a geometric sequence. A finite geometric series is one in which the terms can be counted individually. A finite geometric series has a known beginning term and a known end term. Rules for Finding the Sum of a Finite Geometric Series 1. Identify the number of terms, n, in the series. 2. Identify the first term in the series, which is a. 3. Determine the common ratio in the series, which is r. a(1 – rn) 4. Plug the values in Steps 1, 2, and 3 into the formula Sn = ______ 1–r .
Example Find the sum of the series. −1, 6, −36, 216 Step 1 Identify the number of terms, n, in
There are 4 terms in the series.
the series. Step 2 Identify the first term in the series,
The first term is −1.
which is a. Step 3 Determine the common ratio in the
The common ratio is −6.
series, which is r. Step 4 Plug the values in Steps 1, 2, and 3 a(1 – rn)
into the formula Sn = ______ 1–r .
a(1 – rn)
–1(1 – (–6)4)
__________ Sn = ______ 1 – r = 1 – (–6) =185
Practice Find the sum of each series.
1. 5, 15, 45, 135, 405 Identify the number of terms, n, in the series.
There are
Identify the first term in the series, which is a.
The first term is
Determine the common ratio in the series, which is r.
The common ratio is
Plug the values in Steps 1, 2, and 3 a(1 – rn) into the formula Sn = ______ 1–r .
___________ = Sn = ______ 1–r = 1 –000
a(1 – rn)
terms in the series. . .
000(1 –000000)
2. 1, 2, 4, 8, 16, 32 3. 10, 15, 22.5, 33.75 4. 1, 0.5, 0.25, 0.125, 0.0625 5. 1,024, 512, 256, 128, 64 Algebra 2
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Finding the Sum of an Infinite Geometric Series An infinite geometric series is one that has no end. In other words, the series continues and does not have an end term. You indicate an infinite series with ellipsis points (. . .). For some geometric series you can find the sum of the series. You can find the sum of a geometric series when the absolute value of the common ratio is less than 1. Rules for Finding the Sum of an Infinite Geometric Series 1. Find the common ratio, r. The absolute value of the ratio
must be less than 1 (| r | < 1).
2. Identify the first term, a. a 3. Use the formula S = ____ 1 – r.
Example Find the sum of the series. 1, 0.5, 0.25, 0.125 . . . Step 1 Find the common ratio, r. The
The common ratio is 0.5.
absolute value of the ratio must be less than 1 (| r | < 1). Step 2 Identify the first term, a.
The first term, a, is 1.
a Step 3 Use the formula S = ____ 1 – r.
a 1 _____ S = ____ 1 – r = 1 – 0.5 = 2
Practice Find the sum of each infinite geometric series.
1. 1, 0.25, 0.0625, 0.015625 . . . Find the common ratio, r. The absolute value of the ratio must be less than 1 (| r | < 1).
The common ratio is
Identify the first term, a.
The first term is
a Use the formula S = ____ 1–r .
a 000 ______ S = ____ 1 – r = 1 –00000 =
.
.
2. 243, 81, 27, 9 . . . 3. 100, 10, 1, 0.1, 0.01 . . . 8 4. 1, _23 , _49 , __ 27 . . . 5. 3, 1, _13 , _19 . . .
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Function Operations—Adding Functions Just as you can add, subtract, multiply, and divide real numbers, so too you can add, subtract, multiply, and divide functions. Rules for Adding Functions 1. Identify each function. 2. Rewrite the two functions as one function
with a plus sign between the two functions. 3. Identify like terms. 4. Simplify by combining like terms.
Example Find (f + g)(x) when f(x) = 2x + 6 and g(x) = x − 10. Step 1 Identify each function.
f(x) = 2x + 6 g(x) = x − 10
Step 2 Rewrite the two functions as one
(f + g)(x) = f(x) + g(x) = (2x + 6) + (x − 10)
function with a plus sign between the two functions. Step 3 Identify like terms.
Like terms: 2x and x 6 and −10
Step 4 Simplify by combining like terms.
(f + g)(x) = 2x + x + 6 − 10 = 3x − 4
Practice Find (f + g)(x).
1. f(x) = 6x2 + 2x; g(x) = 3x − 1 Identify each function.
f(x) = g(x) =
Rewrite the two functions as one function with a plus sign between the two functions.
(f + g)(x) = f(x) + g(x)
Identify like terms.
Like terms:
Simplify by combining like terms.
(f + g)(x) =
=
+
= 2. f(x) = −4x − 4; g(x) = 2x − 7 3. f(x) = x2 + 5; g(x) = 3x2 − 2 4. f(x) = −2x3 − 5; g(x) = x2 − 4x + 10 5. f(x) = −4x + 7; g(x) = 3x − 5 Algebra 2
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Function Operations—Subtracting Functions Just as you can add, subtract, multiply, and divide real numbers, so too you can add, subtract, multiply, and divide functions. Rules for Subtracting Functions 1. Identify each function. 2. Rewrite the two functions as one function with a minus sign between the
two functions. Be sure to enclose the second function with parentheses. 3. Change the minus sign between the two functions to a plus sign and
reverse the signs in front of each term in the second function. 4. Simplify by combining like terms.
Example Find (f − g)(x) when f(x) = 2x + 6 and g(x) = x − 10. Step 1 Identify each function.
f(x) = 2x + 6 g(x) = x − 10
Step 2 Rewrite the two functions as one
(f − g)(x) = f(x) − g(x) (f − g)(x) = (2x + 6) − (x − 10)
function with a minus sign between the two functions. Be sure to enclose the second function with parentheses. Step 3 Change the minus sign between the
(f − g)(x) = (2x + 6) + (−x + 10)
two functions to a plus sign and reverse the signs in front of each term in the second function. Step 4 Simplify by combining like terms.
(f − g)(x) = x + 16
Practice Subtract.
1. Find (g − f )(x) when f(x) = 6x2 + 2x and g(x) = 3x − 1. f(x) = Identify each function. g(x) = Rewrite the two functions as one function with a minus sign between the two functions. Be sure to enclose the second function with parentheses.
(g − f )(x) = g(x) − f(x)
Change the minus sign between the two functions to a plus sign and reverse the signs in front of each term in the second function.
(g − f )(x) =
Simplify by combining like terms.
(g − f )(x) =
(g − f)(x) =
−
2. Find (f − g)(x) when f(x) = −4x − 4 and g(x) = 2x − 7. 3. Find (g − f)(x) when f(x) = −2x3 − 5 and g(x) = x2 − 4x + 10. Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Function Operations—Multiplying Functions Just as you can add, subtract, multiply, and divide real numbers, so too you can add, subtract, multiply, and divide functions. Rules for Multiplying Functions 1. Identify each function. 2. Rewrite the two functions as one function with a multiplication sign between the
two functions. 3. Multiply the two functions as you would multiply two polynomials. In other words,
each term in the first function is multiplied by each term in the second function. 4. Simplify by combining like terms.
Example Find (f × g)(x) when f(x) = 2x + 6 and g(x) = x − 10. Step 1 Identify each function.
f(x) = 2x + 6 g(x) = x − 10
Step 2 Rewrite the two functions as one
(f × g)(x) = f(x) × g(x) = (2x + 6) × (x − 10)
function with a multiplication sign between the two functions. Step 3 Multiply the two functions as you
(2x + 6) × (x − 10) = 2x2 − 20x + 6x − 60
would multiply two polynomials. In other words, each term in the first function is multiplied by each term in the second function. Step 4 Simplify by combining like terms.
2x2 − 14x − 60
Practice Find (f × g)(x).
1. f(x) = x2 − 1; g(x) = x4 + 2 Identify each function.
f(x) = g(x) =
Rewrite the two functions as one function with a multiplication sign between the two functions.
(f × g)(x) = f(x) × g(x) = (x2 − 1) × (
Multiply the two functions as you would multiply two polynomials. In other words, each term in the first function is multiplied by each term in the second function.
(x2 − 1) × (
)
)
=
Simplify by combining like terms. 2. f(x) = −4x − 4; g(x) = 2x − 7 3. f(x) = 3x − 1; g(x) = 6x2 + 2x Algebra 2
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Function Operations—Dividing Functions Just as you can add, subtract, multiply, and divide real numbers, so too you can add, subtract, multiply, and divide functions. Rules for Dividing Functions 1. Identify each function. 2. Rewrite the two functions as a fraction. 3. Factor the numerator and the denominator. 4. Simplify by canceling common factors in
the numerator and denominator.
Example f 2 4 Find __ g (x) when f(x) = x + 1 and g(x) = x − 1.
Step 1 Identify each function.
f(x) = x2 + 1 g(x) = x4 − 1
f f(x) x2 + 1 Step 2 Rewrite the two functions as a fraction. _g (x) = ___ = _____ 4 g(x)
Step 3 Factor the numerator and the
denominator.
(x4
x –1
− 1) factors to (x2 − 1)(x2 + 1).
x2 + 1 ___________ x2 + 1 _____ = x4 – 1 (x2 – 1)(x2 + 1)
x2 + 1 is a factor found in the numerator and the denominator. Step 4 Simplify by canceling common factors
in the numerator and denominator.
x2 + 1 1 ___________ = ____ 2 (x – 1)(x2 + 1) x2 – 1
Practice Divide.
f
1. Find _g(x) when f(x) = 3x2 + 2x − 8 and g(x) = x + 2. f(x) = 3x2 + 2x − 8 Identify each function. g(x) = Rewrite the two functions as a fraction.
f(x) _f (x) = ___ = g g(x)
Factor the numerator and the denominator.
3x2 + 2x − 8 factors to
and
.
3x2 + 2x – 8 _________ 00000 =
Simplify by canceling common factors in the numerator and denominator.
=
f
2. Find _g(x) when f(x) = 2x2 and g(x) = 3x2 − 2x. g
3. Find _f (x) when f(x) = 2x + 5 and g(x) = 6x2 + 11x − 10. Algebra 2 Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com
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Composition of Functions In the composition of functions, the output from one function forms the input to a second function. The composition of function f with function g is written using a “º” between the two functions, as in f ºg. It is also written as (f ºg)(x) or f(g(x)) and reads “f of g of x.” Rules for Evaluating the Composition of Functions 1. Identify the two functions. Identify the value of the variable. 2. Identify the inner function. Plug the value of the variable
into the inner function. Evaluate. 3. Use the answer from Step 2 as the value of the variable for
the outer function. Evaluate the outer function.
Example Find(f ºg)(4) when f(x) = x + 2 and g(x) = 2x. Step 1 Identify the two functions. Identify
the value of the variable. Step 2 Identify the inner function. Plug the
value of the variable into the inner function. Evaluate. Step 3 Use the answer from Step 2 as the
value of the variable for the outer function. Evaluate the outer function.
f(x) = x + 2 g(x) = 2x x=4 (f ºg)(x) = f(g(x)); g(x) is the inner function. g(x) = 2x = 2(4) = 8 f(x) is the outer function; use 8 as the value for x. f(x) = x + 2 = 8 + 2 = 10
Practice Find each composition of functions.
1. Find (g ºf )(−2) when f(x) = 3x + 8 and g(x) = 2x2 − 12. Step 1 f(x) =
g(x) = x= Step 2 (g ºf )(x) = g(f(x)), so
is the inner function. =
Step 3
is the outer function; use =
= as the value of x. =
2. Find (f ºg)(3). f(x) = 2x2; g(x) = x + 2 3. Find (f ºg)(−4). f(x) = _3x ; g(x) = x2 − 1 4. Find (g ºf )(5). f(x) = 2x − 5; g(x) = 2x + 2
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Inverse of a Function The inverse of a function f is written as f −1. In the inverse of a function, the range becomes the domain, and the domain becomes the range. Keep in mind that the inverse of a function may not be a function. However, in the examples that follow assume that the inverse of each function is a function. Rules for Finding the Inverse of a Function 1. Write the function as an equation using y in the place of f(x). 2. Exchange x for y and y for x. 3. Solve for y.
Example Find f −1 for the function f(x) = 20 − 6x. Step 1 Write the function as an equation using f(x) = 20 − 6x
y in the place of f(x).
y = 20 − 6x
Step 2 Exchange x for y and y for x.
y = 20 − 6x → x = 20 − 6y
Step 3 Solve for y.
x = 20 − 6y x + 6y = 20 6y = 20 − x 20 – x y = _____ 6
Practice Find f −1.
1. f(x) = x2 + 2 Write the function as an equation using y in the place of f(x).
f(x) = x2 + 2
Exchange x for y and y for x.
y = x2 + 2 →
Solve for y.
y = x2 + 2
=
=
+2
+2
=y 2. f(x) = 4x − 8 3. f(x) = 2x2 + 4 4. f(x) = x + 6 5. f(x) = −x2 + 2
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Finding Trigonometric Ratios As you know, the sides of a right triangle exhibit a special relationship expressed with the Pythagorean theorem. The sides of a right triangle exhibit other special properties: The ratios of pairs of sides of a right triangle are called trigonometric ratios. A hypotenuse
leg adjacent to A C
B
leg opposite to A
There are 3 basic trigonometric ratios—sine, cosine, and tangent. These ratios are based on the lengths of two sides of a right triangle. length of the leg opposite A opposite = sin A = ________ sine of A = ____________________ length of the hypotenuse hypotenuse length of the leg adjacent to A adjacent cosine of A = _____________________ = cos A = ________ length of the hypotenuse hypotenuse length of the leg opposite A opposite tangent of A = _____________________ = tan A = ______ length of the leg adjacent to A adjacent
Example Find sin A, cos A, and tan A.
5
3
A
4
length of the leg opposite A
opposite
Step 1 Find the sine of A.
sin A = ____________________ = ________ = _3 length of the hypotenuse hypotenuse 5
Step 2 Find the cosine of A.
cos A = ______________________ = ________ = _4 length of the hypotenuse hypotenuse 5
length of the leg adjacent to A
length of the leg opposite A length of the leg adjacent to A
adjacent
opposite adjacent
Step 3 Find the tangent of A. tan A = _____________________ = ______ = _34
Practice
B
1. Find sin A, cos A, and tan A.
5 C
13 12
A
Find the sine of A.
sin A =
=
Find the cosine of A.
cos A =
=
Find the tangent of A.
tan A =
=
2. Use the triangle above to find sin B, cos B, and tan B. 3. Use the triangle to the right to find sin X, cos X, and tan X.
Y 6
10
4. Use the triangle to the right to find sin Y, cos Y, and tan Y. 8
X
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Law of Sines You can use the law of sines for either of the following situations: 1. If you know the measure of any two angles and the measure of any side, you can find the measure of an unknown side. 2. If you know the measure of two sides and the measure of one of the angles opposite either of the known sides, you can find the measure of an unknown angle. Rules for Using the Law of Sines 1. Label (or relabel) the triangle as ABC. Identify the given
side(s) and angle measure(s). Identify the unknown. 2. Identify a side and the angle opposite that side. 3. Create a proportion. Let each ratio be the sine of the angle
over the measure of the side, including the unknown. 4. Plug the values into the proportion. Solve for the unknown.
Example Find AC.
20
B 40˚ 20˚
A
C
AB = 20, B = 40°, C = 20° ABC. Identify the given side(s) and AC is the unknown. angle measure(s). Identify the unknown.
Step 1 Label (or relabel) the triangle as
Step 2 Identify a side and the angle opposite
that side. Step 3 Create a proportion. Let each ratio be
the sine of the angle over the measure of the side, including the unknown. Step 4 Plug the values into the proportion.
Solve for the unknown.
Practice
AC is opposite B. AB is opposite C. sin B ____ ____ = sin C0 AC AB AC sin 40° _____ _____ = sin2020°; AC = 37.59 AC AC
A 25
15
1. Find mC.
120˚
B
C
Step 1 AB = 15, AC = 25, B = Step 2 AC is opposite
; mC is the unknown. is opposite C.
;
C sin 000 ____ _____ = sin 0000 AC AC sin C sin 000 ____ Step 3 _____ 25 = 000 ; mC =
2. Find mB.
3. Find AB.
B
B
16 A
80˚ 12
40˚ C
A
45˚ 25
C
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Law of Cosines You can use the law of cosines for either of the following situations: 1. If you know the measure of two sides and the angle between the sides, you can find the measure of an unknown side. 2. If you know the measure of all three sides, you can find the measure of an unknown angle. Rules for Using the Law of Cosines 1. Label (or relabel) the triangle as ABC. Identify the given sides
and the angle measure, if any. Identify the unknown. 2. If a side is unknown, use the form of the law of cosines with the unknown side on one side. If an angle is unknown, use the form of the law of cosines that contains the unknown angle. a2 = b2 + c2 − 2bc cos A b2 = a2 + c2 − 2ac cos B c2 = a2 + b2 − 2bc cos A 3. Plug the values into the chosen formula. Solve for the unknown. A
Example Find b.
b
c = 10
105˚ a = 12
B
C
Step 1 Label (or relabel) the triangle as
a = 12; c = 10; B = 105° b is the unknown.
Step 2 If a side is unknown, use the form of
Since b is unknown, use b2 = a2 + c2 − 2ac cos B.
ABC. Identify the given sides and angle measure. Identify the unknown. the law of cosines with the unknown side on one side. If an angle is unknown, use the form of the law of cosines that contains the unknown angle.
b2 = 122 + 102 − 2(12)(10)cos 105° b = 17.50
Step 3 Plug the values into the chosen
formula. Solve for the unknown. A
Practice
b = 14
c = 16
1. Find mA.
B
C
a = 13
Step 1 a = 13; b =
; A is the unknown.
;c=
Step 2 Since A is unknown, use Step 3 132 =
2
2
+
− 2(
. )(
)cos
A = 2. Find c.
A c B
3. Find mB.
A
b = 10 25˚ a = 21
b = 18
c=8 C
B
a = 13
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Answer Key PAGE 1 Properties of Real
PAGE 6 Graphing Absolute
Numbers Complete the table. Summary of Properties of Real Numbers Property Using Symbols
Commutative
Associative
Summary
Sample: You can add or multiply any two numbers in any order. Sample: When you add or multiply three or more numbers, you can regroup without changing the result.
Distributive
PAGE 12 Writing the Inverse of
Value Inequalities
Sample: When a sum is multiplied by a number, you can distribute the number to each part of the sum.
Practice 1. Sample: The order of the
numbers inside the parentheses had changed. Commutative Property 2. Commutative 3. Associative 4. 3x + (2y + 5) 5. 2x + 5x PAGE 2 Absolute Value Practice 1. 10
–10 –| 10| = –10 true –| –10| = –10 true 2. 3, –3 5. –13, 13 3. 0 6. –5 4. 7 7. –6 PAGE 3 Absolute Value
Equations Practice 1. 5, | 2(5) – 1| = 9; | 9| = 9
–4, | 2(–4) – 1| = 9; | –9| = 9 2. –4, 12 4. –7, 7 3. –18, 12 5. –11, 9
PAGE 4 Compound Inequalities Practice 1. x < 2, x > 4
x < 2 or x > 4 2. x < –2 or x > 4 3. x < –3 or x > 1 PAGE 5 Absolute Value
Inequalities Practice 1. 2x + 4 < 12
2x + 4 > –12 2x + 4 < 12 → x < 4 2x + 4 > –12 → x > –8 –8 < x < 4 2. x 6 or x 0 4. x > 4 or x < 2 3. –15 x 5 5. x > 1 or x < –3
a Matrix
Practice 1. x + 5 –3 x –2 or x –8
Practice 1. a = –2, b = –5, c = 1, d = 3
ad – bc = (–2)(3) – (–5)(1) = –1
–8
0
–2
1 __ –1
2. x > 1 or x < –9 0
–9
3. –17 x 3
–17
0
3
2, 2, B22 = 7 2. 2 × 2 matrix, Z21 = –2 3. 3 × 4 matrix, –10 = Z23
2.
6 14
7 0
3.
3 –13
9 23
4.
7 5
8
–3 2
4.
3.
–8 3
5 –2
5.
2 –10
–4 –7
3.
7 –1
–16 8
4.
15 –6
12 5
(6 × 2) – (–5 × (–2))
X= X=
–12
2 2
5 6
2.5 3 1 1
2.5 3
1(8) + 2.5(3) 1(8) + 3(3) –2 2
8 3 15.5 17
=
3. X =
18 29
–45 30
–6 –27 33
–48 6 –18
20 32
–48 –8
48 0
24 –12
10 41
Range
3 5 8
The relation is a function. 3. is not
PAGE 15 Types of Functions Practice 1. no, no, yes, rational 2. quadratic 5. rational 3. linear 6. exponential 4. rational 7. linear
3×5
PAGE 16 Direct Variation Practice 1. increases, is not, does, do not 2. yes 3. yes
–20 15
PAGE 17 Slope-Intercept Form Practice 1. m = slope = –3
–6 54
3×1+7×3
–2 –1 0 2
2. is
18 –2
6×5
55 –25
Functions Practice 1. Domain
1
PAGE 11 Matrix Multiplication Practice 1. 2 × 1 + (–4) × 3 2 × 9 + (–4) × 2 =
3.
1 1
2. X =
–5 × 5
2.
2 5
Equation Practice 1 1. _______________
7 –10
PAGE 10 Scalar Multiplication Practice 1. 11 × 5 –9 × 5 –4 × 5
–10 24
1 2
PAGE 13 Solving a Matrix
5 7
–1 – (–2)
2.
4.
0.3 –0.2 –0.7 0.8
PAGE 14 Relations and
–4 – 8
3.
–5 2
2 –1
=
PAGE 9 Matrix Subtraction Practice 1. 3 – 6 3 – (–2) = –3 5
2.
–3 1
=
2.
PAGE 8 Matrix Addition Practice 1. –5 + 11 8 + (–1) = 6 –3 + (–7)
5 –2
1
PAGE 7 Introduction to Matrices Practice 1. 2, 4, 2 × 4
3+5
3 –1
3×9+7×2
2. 3. 4. 5. 6.
b = y-intercept = 7 slope = _13 , y-int. = –3 slope = 1, y-int. = 1 slope = _34 , y-int. = –13 slope = – _23 , y-int. = 3 y = –5x + 3 7. y = _12 x – 5
70 79 28 48
23 34
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PAGE 18 Point-Slope Form I Practice 1. 6
The x-coordinate is –3; the y-coordinate is –1. y – (–1) = 6(x – (–3)) or y + 1 = 6(x + 3) 2. y – 1 = – _12 (x – 7) 3. y – (–3) = 2(x – (–3)) or y + 3 = 2(x + 3) 4. y – (–5) = _23 (x – 4) or y + 5 = _23(x – 4) 5. y – 3 = –3(x + 1)
log22
1.3424 x = _____ 1.204 = 1.12 2. x = 1.76 4. x = –3.32 3. x = 0.66 5. x = –3.94
–2, –2 y – (–2) = –1(x – (–2)) or y + 2 = –(x + 2) 2. y – 2 = _12 (x – 2) or y – 1 = _12 x 3. y – 4 = –(x – (–6)) or y + 5 = –(x – 3) 4. y – 6 = 3(x – 2) or y = 3x 5. y – 2 = 1(x – 5) or y+4=x+1 6. y = _23 (x – 6) or y + 2 = _23 (x – 3)
PAGE 20 Linear Parametric
Equations Practice 1. Let t = –2, –1, 0, 2 y = 2t – 2 y = 2(–2) – 2 = –6 y = 2(–1) – 2 = –4 y = 2(0) – 2 = –2 y = 2(2) – 2 = 2
(–8, –6), (–4, –4), (0, –2), (8, 2)
PAGE 21 Exponential Growth
and Decay Practice 1. a = 500, x = 20 years, r = 6.6%,
b = 1 – 0.066 = 0.934 y = (500)(0.934)20 = 127.62 2. 4,062.67 3. 672.40 4. 25,003.39 PAGE 22 Writing an Exponential
base
1 2−5 = __ 25 1 1 1 ____________ __ __ 5 = 2 × 2 × 2 × 2 × 2 = 32 2
2. _18 3. _19
1 4. – ___ 125 1 5. __ 81
1 6. __ 2 a
4 7. __ 3 x
PAGE 25 Rational Exponents Practice 1. 2nd
–322 = 1,024 55 th _____ √1,024 = 4 2. 9 4. 4 3. –125 5. 1,000 Exponents Practice 1. divided, Subtract, x2–(–11) = x13 2. x30 3. x 4. 9y6
1 5. __ 9
x 1 6. ___ x4y4 1 7. __ x2
PAGE 27 Doubling Time Practice 1. b = 1 + 0.055 = 1.055
y = 450(1.055)x y = 900 900 = 450(1.055)x 2 = 1.055x x = 12.95 2. 7.27 3. 31.15 PAGE 28 The Number e Practice 1. g = original amount = 50 g
Function Practice 1. 1 = ab3 1 1 ÷ b3 = ab3 ÷ b3 → __ 3 =a b
b4 1 4 __ 1 2 = __ 3 b = 3 =b →b=2 b
1 1 1 __ _ a = __ 3 = 3 = 8 b
PAGE 24 Negative Exponents Practice exponent 1. 2−5
PAGE 26 Properties of Rational
2. (0, −2), (−1, −1), (−2, 0), (−4, 2) 3. (1, −1), (2, 0), (3, 1), (5, 3)
2
y = _18(2)x 2. y = 2(2)x
Practice 1. log42x = log22
x = ____ 2log4
0 – (–2)
(b )
Equation 2xlog4 = log22 2xlog4 ÷ 2log4 = log22 ÷ 2log4
PAGE 19 Point-Slope Form II Practice –4 – (–2) –2 1. Slope = _______ = ___ 2 = –1
t x = 4t –2 x = 4(–2) = –8 –1 x = 4(–1) = –4 0 x = 4(0) = 0 2 x = 4(2) = 8
PAGE 23 Solving an Exponential
3. y = _12 (8)x
t = time = 30 seconds A = 50e–0.1386(30) e–0.1386(30) = 0.01564 A = (50)(0.01564) = 0.782 g 2. $3,364.06 3. 71.34 g
PAGE 29 Logarithmic Functions Practice 1. log432 = x
32 = 4x 2, 2 32 = (22)x 25 = (22)x = 22x 5 = 2x 5 _ 2=x
2. 3 3. 2
4. 3 5. _12
PAGE 30 Properties of
Logarithms: Product Property Practice 1. M = p, N = q2, logb(pq2) 2. logbq3 + logbr2 3. logb p5 + logbq2 + logbr 4. logb162 5. logbx7 PAGE 31 Properties of
Logarithms: Quotient Property Practice p 1. M = p, N = q2, logb__2 q
p3 r
2. logb__2 3. logb5 – logb p5 4. logb3 PAGE 32 Properties of
Logarithms: Power Property Practice 1. x = 4, M = z, logbz4 2. 3logbq 4. logb(2x)7 3. 3xlogba 5. logb12x PAGE 33 Properties of
Logarithms: Summary Practice 1. logb(MN)
logb5 logbMx logb5, logb5 + 5logba 2. logb(4a4) am5 3. logb___ c2 4. 2logba – logb4
PAGE 34 Solving Logarithmic
Functions Practice 1. Quotient
5 log __ 2x = –4 5 __ –4 2x = 10 x = 25,000 2. x = 3.16 4. x = 25 3. x = 3,333.33 5. x = 5,773.50
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PAGE 39 Quadratic Functions in
Logarithmic Function Practice 1. 2ln(3x) = 6, ln(3x) = 3
Intercept Form Practice 1. p = 1, q = –3
e3
3x = e3 x = __ 3 = 6.70 2. x = 0.65 4. x = 7.39 3. x = 148.41 5. x = 5.22 PAGE 36 Graphing Quadratic
Functions Practice –2 1. x-coordinate of vertex = ____
x2 + 2x + 3 (–2)2 + 2(–2) + 3 22 + 2(2) + 3 02 + 2(0) + 3 12 + 2(1) + 3
x –2 2 0 1
y 3 11 3 6
2. (0, 2), (–2, 6), (2, 6), (1, 3), (–1, 3) 3. (0.5, –1.75), (1, –1), (–2, 17),
(2, 5), (0, –1)
PAGE 40 Solving Quadratic
Equations Using Square Roots Practice 1. 3x2 – 25 + 25 = 50 + 25, 3x2 = 75 __3, x2 ___ 3x2 ÷ 3 = 75 ÷ = 25 x2 = 25 → √ x2 = √ 25 x = 5, x = –5 2. x = 7, x = –7 4. x = 2, x = –2 3. x = 5, x = –5 5. x = 10, x = –10
PAGE 41 Solving a Quadratic
PAGE 37 Properties of a
Graph of a Quadratic Function –(–4)
x-coord. of vertex = ____ = –2 2(–1) y = –(–2)2 – 4(–2) + 2 = 6 vertex: (–2, 6) –(–4) axis of symmetry: x = ____ = –2 2(–1) 2. up, (0, 0), x = 0 3. down, (1, 4), x = 1 4. up, (1, –2), x = 1 29 1 _ 5. down, _14 , __ 8 ,x=4
(
)
PAGE 38 Writing a Quadratic
Function from Its Graph Practice 1. h = 1, k = –3
x = 3, y = –5 –5 = a(3 – 1)2 + (–3) a = – _12 y = – _12 (x – 1)2 + (–3)
( )
2. y = _32 (x – 1)2 + (–6) 3. y = _13 x2 – 3
PAGE 44 Using the Discriminant Practice 1. b = 3, c = 7
b2 – 4ac = 32 – 4(1)(7) 9 – 28 = –19 negative, no 2. result: 0; one solution 3. result: positive (17); two solutions 4. result: positive (20); two solutions 5. result: negative (–63); no solutions PAGE 45 Methods for Solving
Quadratic Functions
Practice 1. b2 – 4ac = (–2)2 – 4(1)(–15) = 64
a positive square number the quadratic formula, or complete_______________ the square
Equation by Completing the Square Practice 1. 2
Practice 1. negative, down
3. (2x + 3)(x − 1) = 0; x = – _32 , 1
4. (x − 3)(3x + 2) = 0; x = 3, – _23
x = –1, y = –4 –4 = a(–1 – 1)(–1 – (–3)) 1=a y = 1(x – 1)(x – (–3)) 2. y = – _13 (x + 5)(x – 1) 3. y = 2x(x – 2)
2(1)
= –1 y = (–1)2 + 2(–1) + 3 = 2 vertex: (–1, 2)
2. (3x − 1)(2x – 7) =0; x = _13 , _72
–(–2) ± (–2)2 – 4(1)(–15)
Half of 2 is 1; 12 = 1 x2 + 2x + 1 = 5 + 1 (x_______ + 1)2 = 6 __ __ √(x + 1)2 = √__6 → x + 1 = ± √6 x = (–1) ±__ √ 6 2. x = 2 ± √__ 1 = 3 or 1 3. x = 1 ± √ 9__= 4 or –2 4. x = –1 ± √__ 6 5. x = –2 ± √ 3 PAGE 42 Quadratic Formula Practice 3, c = –4 1. a = 1, b =___________ 2
–3 ± √ 3 – 4(1)(–4) x = ______________ 2(1) ______
± √ 9 + 16 _________ x = –3 2 x = –4 or x = 1 2. x = –13 or x = –2 3. x = –6 or x = 12 4. x = –1 or x = 6 5. x = –1 or x = 5
PAGE 43 Solving a Quadratic
Equation by Factoring Practice 1. 2x2 F O 2 × 1 2 × –1 2 × –2
–5x + + +
I –2 × 1 –1 × 1
2 = L = −4 –1 × –2 = −5
outer terms: 2 and –2 inner terms: 1 and –1 (2x − 1)(x − 2) = 0 2x − 1 = 0, x − 2 = 0 x = _12, x = 2
√ x = ___________________ 2(1) = –3 or 5 2. x = ±9 4. x = _46 or 4 3. x = –4 or –2 5. x = –5 or – _83 PAGE 46 Writing an Equation of
an Ellipse
Practice 1. y, vertical
a = –6, b = 3 y2 y2 x2 __ x2 ____ __ __ 2 = 1 → 9 + 36 = 1 2 +
(–6) 3 y2 x2 __ __ 2. 1 + 25 = 1
2
y x2 __ 3. __ 16 + 9 = 1
PAGE 47 Foci of an Ellipse Practice 1. 100, 36
horizontal a2 = 100, b2 = 36 c2 = 100 – 36 = 64, c = ±8 foci: (8, 0) and (–8, 0) 2. (0, 9) __ and (0, –9) __ 3. (2√ 5 , __ 0) and (–2√ 5 , __0) 4. (0, 3√ 5 ) and (0, –3√ 5 ) 5. (0, 6) and (0, –6) PAGE 48 Standard Deviation Practice 1. mean: 42.4
x x (x − x)2 53 42.4 112.36 47 42.4 21.16 39 42.4 11.56 33 42.4 88.36 40 42.4 5.76 sum: 239.2 239.2 _____÷ 5 = 47.84 √47.84 = 6.92 2. = 56.26 3. = 10.03
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PAGE 49 Margin of Error Practice 1. 0.025
PAGE 54 Simplifying Radical
Expressions by Removing Perfect Squares
0.025 __ 1 √ n = ±____ 0.025 = 40 __ (√ n )2 = 402, n = 1,600 2. ±3.16% 3. ±1.83% 4. 10,000 5. 625
(x + 3)4 = x4(3)0 + . . . (x + 3)4 = x4 + 3x3 + 9x2 + 27x + 81 (x + 3)4 = x4 + 12x3 + 54x2 + 108x + 81 2. x3 − 12x2 + 48x − 64 3. x5 + 15x4 + 90x3 + 270x2 + 405x + 243 PAGE 51 Writing a System of
Equations as a Matrix Practice 1. 2 3 4
x y
=
1 8
2.
4 1
–4 0
x y
=
8 6
3.
–2 –1
–4 2
x y
=
3 10
square, _____ so use __ 3 and __ 9 3 × 9 = √ 3__× √ 9 √__ √3 ×__ 3 = 3√3 __ 2. 10√ 5. 5√___ 3 __5 3. 4√__ 5 6. 2√__ 30 4. 4√ 3 7. 9√ 2 PAGE 55 Simplifying Radical
Expressions with Variables Practice 6 ________ 1. 16x
____ __ 4. 3x3√___ 7
Practice 1. 5
x y
–3 –2
4
5 10
=
1 ________________ (5 × (–2)) – (4 × (–3))
= x y
2.
–2 1
=
–2 –4
3 5
5. 6x4√ 5x
PAGE 56 Adding Radical
Expressions Practice _______ ___ 16 × 2x = 20√ 2x 1. 5√___
–1 –2
Practice ______ ___ 9 × 2x = 15√ 2x 1. 5√___
Expressions 1.5 2.5
3.
5 10
=
Practice __________
10 15
√ 2 3 √125x __ y
n = 10 Dx = (5)(−2) − (10)(−3) = 20 Dy = (5)(10) − (4)(5) = 30 20 30 __ x = __ 2 = 10; y = 2 = 15 2. (–2, 1) 3. (26, –41)
5y √ x2____
___
PAGE 62 Complex Numbers Practice ____ 1. √ –18
–2____ ___ ___ –18 = √ –1__× √ 18 √____ √–18 = 3i__√2 −2 + 3i__ √2 2. 4 + 2i √__ 2 5. –10 + 5i 3. 6 + 2i √ 6 6. −3 − 6i__ 4. 7 + 7i 7. 4 + 8i √ 2 Numbers Practice 1. 3, 6i
(−5 + 3) + (−4i + 6i) –2 + 2i 2. 7 + 3i 5. 16 3. –1 – 4i 6. 3 − 2i 4. –9i 7. 6 + 9i
Practice 1. (−2 – 5i)
3
2. 2x2y √ 42x ___ 3 3. 4y √ 5y __ 4. 10x2y2√ y
___
Numbers
3
PAGE 53 Cramer’s Rule Practice 1. D = (5)(−2) − (−3)(4) = 2
Practice _______ –1 × 12 1. √_______
PAGE 64 Subtracting Complex
3
1. _______ 25xy × 5xy2
26 –41
Negative Real Number: Imaginary Numbers
PAGE 63 Adding Complex
3√ 2x___is in simplest form. ___ ___ 15√ 2x – 3√___ 2x = (15 ___ – 3)(√ 2x ) (15___– 3)(√ 2x ) = 12√ 2x ___ 2. 9√ 2y __ 4. x √ 2x ___ 3. –18√ x 5. –2√ 2x PAGE 58 Multiplying Radical
–1 1.5 –2 2.5
___
7√ 10 – 20 4. _______ 6 __ 5. 5 – 3√ 3
–1 × 12___= √ –1 × _____ √___ √12 __ √–1 × √12 = i × √4__× 3 = 2i √3 2. 4i 4. 3i √___ 3 3. 5i 5. 2i √ 10
Expressions Solve a System of Two Equations
43 – 13√ 7 2. _______ 74 __ 3. 1 + 2√ 3
___
PAGE 57 Subtracting Radical
PAGE 52 Using Matrices to
___
– √ 11 5 – √ 11 ______ ______ ___ ___ × 2 2 + √ 11 ___2 – √ 11 ___ ___ 10 – 5√ 11 – 2√___ 11 + (√ 11 )2 = ___________________ 2 (√ 11 )2 ___ 2 – ___ ___ ___ 10 – 5√ 11 – 2√___ 11 + (√ 11 )2 21 – 7√ 11 _______ ___________________ = 2 2 –7 ___ 2 – (√ 11 )
PAGE 61 Square Root of a
√16x6 ×___2x = √16x6 × √2x
= 4x__3√ 2x 2. 5x √ 2___ 3. 4x2√ 3x
Practice ___ 11 1. 2 – √___
= √ 11 –__3
3√ 2x___is in simplest form. ___ ___ 20√ 2x + 3√___ 2x = (20 ___ + 3)(√ 2x ) (20 ___ + 3)(√ 2x ) = 23√___ 2x 2. 15√__ 2y 4. 30√ 2x___ 3. 42√ x 5. 19x √ 2x
y-coefficients: 3 and 4 0 2
Denominator of a Radical Expression
Practice 1. 1 and 27; 3 and 9; 9 is a perfect
PAGE 50 Binomial Theorem Practice 1. a = x; b = 3; power = 4
PAGE 60 Rationalizing the
___
5. 12y3√___ 5y 3 6. 4xy √ 12 ___ 7. 10x3y3√ 2xy
PAGE 59 Dividing Radical
–2, –5i (9 − 2) + (4i − 5i) 7–i 2. 6 + i 4. 3 – 6i 3. 8 – i 5. 6
Expressions Practice _____ –81x4 1. ____ 9x
√
____
–9x3 √______________
√9 ×__ (–1) × x2 × x 3
2. 3x √ x __ 3. 4x2√ x
___
=___ 3x √ –x 4. 2y √ 2xy___ 5. 2x2y3√ 2x 3
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PAGE 65 Multiplying Complex
PAGE 70 Solving Cubic
Numbers Practice 1. 12 − 24i + 24i + (−48i2)
12 − 24i + 24i + (−48)(−1) 12 − 24i + 24i + 48 12 + 0i + 48 = 60 2. (39 − 2i) 4. (34 − 2i) 3. (9 + 19i) 5. (11 + 48i) PAGE 66 Dividing Complex
Numbers
Practice 1. 2(x + 2); 2; (x + 2)
x + 2 = 0; x – 5 = 0; x – 6 = 0 x + 2 = 0 → x = –2 x – 5 = 0 →x = 5 x – 6 = 0 →x = 6 2. x = 2, 4, 5 4. x = 3, –10, 7 3. x = –9, –3, 3 5. x = –2, 2 Equations: Sum of Cubes
1. –2 – 4i –2 – 4i 5–i ______ × ______ –2 – 4i –2 + 4i –2 – 4i 5–i ______ × ______ –2 – 4i –2 + 4i –10 – 20i + 2i + 4i2 = ______________ 4 + 8i – 8i – 16i2 – 18i –10 – 20i + 2i + 4i2 _______ ______________ = –14 = 20 4 + 8i – 8i – 16i2 9 18i __ –7 __ –14 ___ ___ 20 – 20 = 10 – 10 i –8 __ i 16 __ __ 2. 37 + 22 4. __ 37 i 13 – 13 38 21 __ 3. _35 + _95 i 5. __ 65 + 65 i
PAGE 67 Absolute Value and
Practice 1. 125 3 ___
Practice _________ 42 + (–6)2 1. √_________
_______
2 = √ 16 + 36 = 42 + (–6)___ √___
√52__ = 2√13 __ 2√___ 5 5. 2√ 5 29 6. 13 ___ √___ 7. 2√ 17 √65
PAGE 68 Finding a Complex
Solution to a Simple Quadratic Equation Practice 1. 4x2 + 32 − 32 = 0 − 32; 4x2= –32 2 ÷ 4 = −32 ÷ 4; x2 = –8 4x___ ______ –8 √ = √–1 __ × 8 x = ±2i √ 2 2. x = ±4i 4. x = ±10i __ 3. x = ±4i 5. x = ±3i √ 3
PAGE 69 Finding a Complex
Solution to a Quadratic Equation Practice –2, c = 26 1. a = 1, b = _____________ –(–2) ± √ (–2)2 – 4(1)(26) __________________ _____ 2(1) 2________ ± √ –100 2_____ ± 10i 2 = 1 ±__5i 2 ___= i √ 11 1 ____ _ 2. − 3 ± 3 4. −2 ± i__√ 2 i √3 3. 1 ± 2i 5. _52 ± ___ 2
3
____
a = √ 8x3 = 2x; b = √ 125 = 5 2x + 5 4x2 − 10x + 25 (2x + 5)(4x2 − 10x + 25) 2. (x + 2)(x2 − 2x + 4) 3. (x + 3)(x2 − 3x + 9) 4. (3x + 5)(9x2 − 15x + 25) 5. (4x + 2)(16x2 − 8x + 4)
PAGE 72 Factoring Cubic
Equations: Difference of Cubes
Complex Numbers
Practice 3 3 1. a3 =3 8x ___; b = 64
3
Polynomials Practice
– 28x2 –24x – 20 7x3 + 6x2 + 5x 7x3 – 22x2 – 19x – 20 2. 8x3 − 22x2 + 31x − 24 3. x3 − 4x2 − 6x − 1 4. −x4 + 9x3 − 17x2 + 9x 5. −30x5 + 7x4 − 17x3 + 6x2 1.
PAGE 71 Factoring Cubic
Practice
2. 3. 4.
PAGE 75 Multiplying
Equations: Finding x-Intercepts
___
a = √ 8x3 = 2x; = √ 64 = 4 2x – 4 4x2 + 8x + 16 (2x – 4)(4x2 + 8x + 16) 2. (x – 2)(x2 + 2x + 4) 3. (x – 3)(x2 + 3x + 9) 4. (4x – 5)(16x2 + 20x + 25) 5. (5x – 2)(25x2 + 10x + 4)
PAGE 73 Writing Cubic
Equations in Intercept Form Practice 1. q = –1, r = 2
f(x) = a(x + 2)(x + 1)(x −2) 18 = a(1 + 2)(1 + 1)(1 − 2) –3 = a f(x) = −3(x + 2)(x + 1)(x − 2) 2. f(x) = –(x – 1)(x – 2)(x − 3) 3. f(x) = (x – 1)(x – 3)(x − 6) PAGE 74 Evaluating Polynomials
PAGE 76 Dividing Polynomials
Using Synthetic Division Practice 1. –2
–2] 1
6 10 –2 –8 1 4 2 remainder: 2; quotient: (x + 4) 2. 4x2 − 10x + 27, R −98 3. 6x + 25, R 80 PAGE 77 End Behavior of a
Polynomial Function Complete the table. Sign of the leading coefficient
Degree of the function
Positive
Odd
Positive
Even
Negative
Odd
Negative
Even
End behavior using
ⴥ
As x → +, f(x) → + As x → –, f(x) → – As x → +, f(x) → + As x → –, f(x) → + As x → +, f(x) → – As x → –, f(x) → + As x → +, f(x) → – As x → –, f(x) → –
Practice 1. –7, negative
2, even negative, even −, − 2. As x → +, f(x) → −, as x → −, f(x) → −. 3. As x → +, f(x) → +, as x → −, f(x) → +. PAGE 78 Inverse Variation Practice 1. increases
is, 36, 36 yes, yes, do 2. no 3. yes 4. yes
Using Synthetic Substitution Practice 1. 3] 4
2 12 4 14 answer: 128 2. 21 3.
1 42 43
–1 129 128
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PAGE 79 Simplifying Rational
PAGE 83 End Behavior of
Expressions
Rational Functions: m
Practice (3)(6x3) 1. ______ 3(x + 2) (3)(6x3) ______ 3(x + 2) 6x3 ____ x+2 +5 _____ 2. 2x 6 1 3. ___ 7x3 2x +6 4. _____ x2
Practice 1. 2, 2
5. _32 +x 6. 4____ 2+x
PAGE 80 Multiplying Rational
Expressions Practice 5x3 1. _______
2. 3. 4. 5.
(x – 1)(4) 5x3 5x3 _______ = _____ (x – 1)(4) 4x – 4 5x3 _____ 4x – 4 is in simplest form. 9 ___ 5x2 2 + 8x 2x(x + 4) 2x ______ _______ 2x – 3 or 2x – 3 8x –2 _____ 3x4 2+x x(x + 1) x_____ ______ x – 2 or x – 2
PAGE 84 End Behavior of
Rational Functions: m=n Practice 1. 2, 2
PAGE 81 Dividing Rational
Expressions Practice x2 x – 2 __ 1. ____ x+2 × 2 – 2)(x2) _______ x2 (x x3 – 2x2 x – 2 __ ________ ____ x + 2 × 2 = (x + 2)(2) = (x + 2)(2) 3 – 2x2 x3 – 2x2 _______ = x______ 2x + 4 (x + 2)(2)
is –6 2. __ 5x 2x + 4 __________ 3. (x + 3)(x – 1)
3, 3 2 < 3, so m < n. f(x) → 0 f(x) → 0 2. As x → +, f(x) → 0; as x → −, f(x) → 0. 3. As x → +, f(x) → 0; as x → −, f(x) → 0. 4. As x → +, f(x) → 0; as x → −, f(x) → 0. 5. As x → +, f(x) → 0; as x → −, f(x) → 0.
x3 – x 2 4. ___________ (x – 3)(5x + 6) (x + 3)(x + 1) 5. __________ 16x3
PAGE 82 Rational Functions—
Finding Vertical Asymptotes Practice 2x – 1 1. __________
(x + 2)(x – 2)
(x + 2)(x − 2) = 0 (x + 2) = 0 (x − 2) = 0 x = –2 x=2 vertical asymptotes: x = –2 and x=2 2. x = –7 and x = 4 3. x = –1 and x = 1 4. x = –3 and x = 3
2, 2 2 = 2, so m = n. 12 a = 12; b = 4; _ab = __ 4 =3 f(x) → 3 horizontal asymptote: y → 3 2. As x → ±, f(x) → _72 ; horizontal asymptote: y = _72. 3. As x → ±, f(x) → 2; horizontal asymptote: y = 2. PAGE 85 End Behavior of
Rational Functions: m>n Practice 1. 2, 2
1, 1 2 > 1, so m > n. –6 2–1 a = –6; b = 3: __ = –2x 3x negative, odd As x → +, f(x) → −. As x → −, f(x) → +. 2. As x → +, f(x) → +; as x → −, f(x) → +. 3. As x → +, f(x) → −; as x → −, f(x) → +. PAGE 86 End Behavior of
Rational Functions: Using All Three Conditions Practice 1. 4, 4
PAGE 87 Recursive Formulas:
Arithmetic Sequences Practice 1. 7
0.25, 0.25, 0.25 d = 0.25 tn = tn−1 + 0.25 t1 = 7, tn = tn−1 + 0.25 2. t1 = 21, tn = tn−1 – 4 3. t1 = 0, tn = tn−1 + 6 4. t1 = –3, tn = tn−1 – 5 5. t1 = 70, tn = tn−1 – 8 PAGE 88 Recursive Formulas:
Geometric Sequences Practice 1. –12
–4, –4, –4 r = –4 tn = −4(tn−1) t1 = −12 tn = −4(tn−1) 2. t1 = 2, tn = 1.5(tn−1) 3. t1 = 64, tn = _12 (tn−1) 4. t1 = 6, tn = _23 (tn−1) 5. t1 = 675, tn = 0.1(tn−1) PAGE 89 Finding the Sum of a
Finite Arithmetic Series Practice 1. 6, 1, 6
2(1) + 3 = 5 2(6) + 3 = 15 Sn = _62(5 + 15) = 60 2. Sn = −130 3. Sn = 1,010 4. Sn = 144 PAGE 90 Finding the Sum of a
Finite Geometric Series Practice 1. 5, 5, 3
5(1 – 35)
Sn = _______ 1 – 3 = 605 2. Sn = 63 4. Sn = 1.9375 3. Sn = 81.25 5. Sn = 1,984 PAGE 91 Finding the Sum of
an Infinite Geometric Series Practice 1. 0.25
1 1 S = ______ 1 – 0.25 = 1.33 2. S = 364.5 4. S = 3 3. S = 111.11 5. S = 4.5
1, 1 4 > 1, so m > n. 1 4–1 _ = x3 1x As x → +, f(x) → +. As x → −, f(x) → −. 2. As x → ±, f(x) → 2. 3. As x → +, f(x) → −; as x → −, f(x) → −.
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PAGE 92 Function Operations—
Adding Functions Practice 1. f(x) = 6x2 + 2x
g(x) = 3x − 1 (6x2 + 2x) + (3x − 1) 6x2, 2x and 3x, –1 6x2 + 2x + 3x − 1 = 6x2 + 5x − 1 2. (f + g)(x) = −2x − 11 3. (f + g)(x) = 4x2 + 3 4. (f + g)(x) = −2x3 + x2 − 4x + 5 5. (f + g)(x) = −x + 2 PAGE 93 Function Operations—
Subtracting Functions Practice 1. f(x) = 6x2 + 2x
g(x) = 3x − 1 (3x − 1) – (6x2 + 2x) (3x − 1) + (–6x2 – 2x) –6x2 + x − 1 2. (f – g)(x) = −6x + 3 3. (g – f)(x) = 2x3 + x2 – 4x + 15 PAGE 94 Function Operations—
Multiplying Functions Practice 1. f(x) = x2 – 1
g(x) = x4 + 2 (x2 – 1) × (x4 + 2) (x2 – 1) × (x4 + 2) = x6 + 2x2 – x4 – 2 6 x – x4 + 2x2 – 2 2. (f × g)(x) = −8x2 + 20x + 28 3. (f × g)(x) = 18x3 – 2x
PAGE 95 Function Operations—
PAGE 98 Finding Trigonometric
Dividing Functions Practice 1. g(x) = x + 2
Ratios Practice 1. sin A =
2 + 2x – 8 3x _________ x+2
length of the leg opposite A __ 5 ___________________ = 13 length of the hypotenuse
(3x – 4) and (x + 2)
cos A = length of the leg adjacent to A __ _____________________ = 12 13 length of the hypotenuse
2 + 2x – 8 (3x – 4)(x + 2) 3x _________ = ___________ x+2 x+2 (3x – 4)(x + 2) ___________ = 3x –4 x+2 f 2x _____ _ 2. g (x) = 3x – 2 g 3. _(x) = 3x − 2 f
tan A = length of the leg opposite A 5 _____________________ = __ length of the leg adjacent to A 12 5 __ 12 12 __ 2. __ 13 , 13 , 5 3 3 _ 4 _ _ 3. 5 , 5 , 4 4. _45 , _35 , _43
PAGE 96 Composition of
Functions Practice 1. f(x) = 3x + 8
g(x) = 2x2 – 12 x = –2 f(x), f(x) = 3x + 8 = 3(−2) + 8 = 2 g(x), 2 g(x) = 2x2 − 12 = 2(2)2 − 12 = −4 2. 50 3. 5 4. 12 PAGE 97 Inverse of a Function Practice 1. x = y2 + 2
x = y2 + 2 x –____ 2 = y2 ±√ x – 2 = y +8 2. y = x____ 4 ____
4. y = x − 6
−4 3. y = ± x____ 2
5. y = ±√ 2 – x
√
PAGE 99 Law of Sines Practice 1. 120°
B, AB C sin B ____ ____ = sin AB AC
AC AB sin C sin 120° ____ ______ = 15 ; mC = 31.31° 25
2. 47.61°
3. 27.50
PAGE 100 Law of Cosines Practice 1. b = 14, c = 16
a2 = b2 + c2 – 2bccos A 132 = 142 + 162 − 2(14)(16)cos A A = 50.82° 2. 12.66 3. 115.94°
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