Solution Manual for Precalculus: Graphs and Models, A Right Triangle Approach, 6th Edition

INSTRUCTOR’S SOLUTIONS MANUAL JUDITH A. PENNA A LGEBRA & T RIGONOMETRY G RAPHS AND M ODELS S IXTH E DITION P RECALCULUS G RAPHS AND M ODELS , A R IGHT T RIANGLE A PPROACH S IXTH E DITION Marvin L. Bittinger Indiana University Purdue University Indianapolis Judith A. Beecher David J. Ellenbogen Community College of Vermont Judith A. Penna Boston Columbus Indianapolis New York San Francisco Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2017, 2013, 2009 Pearson Education, Inc. Publishing as Pearson, 501 Boylston Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-438340-8 ISBN-10: 0-13-438340-0 www.pearsonhighered.com Contents Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . 333 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . 387 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . 429 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . 483 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 583 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 667 Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 709 Just-in-Time Review 1. Real Numbers √ 2 1 8 1. Rational numbers: , 6, −2.45, 18.4, −11, 3 27, 5 , − , 3 6 7 √ 0, 16 √ √ √ 3. −5 = − 25, and − 26 is to the left of√− 25, or −5, on the number line. Thus it is true that − 26 −25. 16 5 25 16 25 4 =− and − = − ; − is to the right of − , 5 20 4 20 20 20 5 4 so it is true that − > − . 5 4 1 2 2. Rational numbers but not integers: , −2.45, 18.4, 5 , 3 6 8 − 7 √ √ √ √ 3. Irrational numbers: 3, 6 26, 7.151551555 . . . , − 35, 5 3 6. − (Although there is a pattern in 7.151551555 . . . , there is no repeating block of digits.) √ √ 4. Integers: 6, −11, 3 27, 0, 16 √ √ 5. Whole numbers: 6, 3 27, 0, 16 4. Absolute Value 6. Real numbers: All of them 3. |4.7| = 4.7 (|a| = a, if a ≥ 0.) 2 2 (|a| = −a, if a < 0.) 4. − = 3 3 2. Properties of Real Numbers 1. | − 98| = 98 (|a| = −a, if a < 0.) 2. |0| = 0 (|a| = a, if a ≥ 0.) 5. | − 7 − 13| = | − 20| = 20, or 1. −24 + 24 = 0 illustrates the additive inverse property. |13 − (−7)| = |13 + 7| = |20| = 20 2. 7(xy) = (7x)y illustrates the associative property of multiplication. 6. |2 − 14.6| = | − 12.6| = 12.6, or 3. 9(r − s) = 9r − 9s illustrates a distributive property. 7. | − 39 − (−28)| = | − 39 + 28| = | − 11| = 11, or 4. 11 + z = z + 11 illustrates the commutative property of addition. | − 28 − (−39)| = | − 28 + 39| = |11| = 11 3 15 6 15 21 21 , or 8. − − = − − = − = 4 8 8 8 8 8 15 − − 3 = 15 + 6 = 21 = 21 8 4 8 8 8 8 5. −20 · 1 = −20 illustrates the multiplicative identity property. 6. 5(x + y) = (x + y)5 illustrates the commutative property of multiplication. 7. q + 0 = q illustrates the additive identity property. 8. 75 · 1 = 1 illustrates the multiplicative inverse property. 75 9. (x+y)+w = x+(y+w) illustrates the associative property of addition. 10. 8(a + b) = 8a + 8b illustrates a distributive property. |14.6 − 2| = |12.6| = 12.6 5. Operations with Real Numbers 1. 8 − (−11) = 8 + 11 = 19 1 3·1 3 1 1 1 3 = = · =1· = 2. − · − 10 3 10 · 3 3 10 10 10 3. 15 ÷ (−3) = −5 4. −4 − (−1) = −4 + 1 = −3 3. Order on the Number Line 5. 7 · (−50) = −350 1. 9 is to the right of −9 on the number line, so it is false that 9 < −9. 2. −10 is to the left of −1 on the number line, so it is true that −10 ≤ −1. 6. −0.5 − 5 = −0.5 + (−5) = −5.5 7. −3 + 27 = 24 8. −400 ÷ −40 = 10 9. 4.2 · (−3) = −12.6 c 2017 Pearson Education, Inc. Copyright 2 Just-in-Time Review 10. −13 − (−33) = −13 + 33 = 20 w−4 z9 a−m bn = 4 Using −n = m −9 z w b a 2 2 z z 4. = 2 Raising a quotient to a power y y 3. 11. −60 + 45 = −15 1 2 3 4 1 1 2 = + − =− 12. − = + − 2 3 2 3 6 6 6 13. −24 ÷ 3 = −8 5. 1000 = 1 Using a0 = 1, a = 0 14. −6 + (−16) = −22 5 1 8 1·8 1·2 /·4 4 1 = = =− · − = 15. − ÷ − 2 8 2 5 2·5 /·5 2 5 6. a5 = a5−(−3) = a5+3 = a8 a−3 7. (2xy 3 )(−3x−5 y) = 2(−3)x · x−5 · y 3 · y Using the quotient rule = −6×1+(−5) y 3+1 = −6x−4 y 4 , or − 6. Interval Notation 8. x−4 · x−7 = x−4+(−7) = x−11 , or 1. This is a closed interval, so we use brackets. Interval notation is [−5, 5]. 9. (mn)−6 = m−6 n−6 , or 10. (t−5 )4 = t−5·4 = t−20 , or 3. This interval is of unlimited extent in the negative direction, and the endpoint −2 is included. Interval notation is (−∞, −2]. 8. Scientific Notation 5. {x|7 7}. This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7, ∞). 6. The endpoints −2 and 2 are not included in the interval, so we use parentheses. Interval notation is (−2, 2). 7. The endpoints −4 and 5 are not included in the interval, so we use parentheses. Interval notation is (−4, 5). 8. The interval is of unlimited extent in the positive direction, and the endpoint 1.7 is included. Internal notation is [1.7, ∞). 9. The endpoint −5 is not included in the interval, so we use a parenthesis before −5. The endpoint −2 is included in the interval, so we use a bracket after −2. Interval notation is (−5, −2]. 10. This interval is of unlimited extent in the negative direc√ tion, and the√endpoint 5 is not included. Interval notation is (−∞, 5). 1 x11 1 m6 n 6 2. This is a half-open interval. We use a parenthesis on the left and a bracket on the right. Interval notation is (−3, −1]. 4. This interval is of unlimited extent in the positive direction, and the endpoint 3.8 is not included. Interval notation is (3.8, ∞). 6y 4 x4 1 t20 1. Convert 18,500,000 to scientific notation. We want the decimal point to be positioned between the 1 and the 8, so we move it 7 places to the left. Since 18,500,000 is greater than 10, the exponent must be positive. 18, 500, 000 = 1.85 × 107 2. Convert 0.000786 to scientific notation. We want the decimal point to be positioned between the 7 and the 8, so we move it 4 places to the right. Since 0.000786 is between 0 and 1, the exponent must be negative. 0.000786 = 7.86 × 10−4 3. Convert 0.0000000023 to scientific notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 9 places to the right. Since 0.0000000023 is between 0 and 1, the exponent must be negative. 0.0000000023 = 2.3 × 10−9 4. Convert 8,927,000,000 to scientific notation. We want the decimal point to be positioned between the 8 and the 9, so we move it 9 places to the left. Since 8,927,000,000 is greater than 10, the exponent must be positive. 8, 927, 000, 000 = 8.927 × 109 7. Integers as Exponents 5. Convert 4.3 × 10−8 to decimal notation. 1. 3−6 = 2. 1 36 Using a−m = 1 = (0.2)5 (0.2)−5 1 am Using a−m = The exponent is negative, so the number is between 0 and 1. We move the decimal point 8 places to the left. 1 am 4.3 × 10−8 = 0.000000043 c 2017 Pearson Education, Inc. Copyright Just-in-Time Review 3 6. Convert 5.17 × 106 to decimal notation. 6. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 5.17 × 106 = 5, 170, 000 7. Convert 6.203 × 1011 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 11 places to the right. 6.203 × 1011 = 620, 300, 000, 000 8. Convert 2.94 × 10−5 to scientific notation. The exponent is negative, so the number is between 0 and 1. We move the decimal point 5 places to the left. 2.94 × 10−5 = 0.0000294 4(8 − 6)2 − 4 · 3 + 2 · 8 31 + 190 2 4·2 −4·3+2·8 Calculating in the = 3+1 numerator and in the denominator 4·4−4·3+2·8 = 4 16 − 12 + 16 = 4 4 + 16 = 4 20 = 4 =5 7. 64 ÷ [(−4) ÷ (−2)] = 64 ÷ 2 = 32 8. 9. Order of Operations 6[9 − (3 − 2)] + 4(2 − 3) = 6[9 − 1] + 4(2 − 3) 1. 3 + 18 ÷ 6 − 3 = 3 + 3 − 3 = 6−3=3 2. 3. = 6 · 8 + 4(−1) Dividing = 48 − 4 Adding and subtracting = 44 = 5 · 3 + 8 · 32 + 4(6 − 2) = 5 · 3 + 8 · 32 + 4 · 4 Working inside parentheses = 5·3+8·9+4·4 Evaluating 32 10. Introduction to Polynomials = 15 + 72 + 16 Multiplying = 87 + 16 Adding in order = 103 from left to right 5[3 − 8 · 32 + 4 · 6 − 2] 1. 5 − x6 The term of highest degree is −x6 , so the degree of the polynomial is 6. 2. x2 y 5 − x7 y + 4 = 5[3 − 8 · 9 + 4 · 6 − 2] The degree of x2 y 5 is 2 + 5, or 7; the degree of −x7 y is 7 + 1, or 8; the degree of 4 is 0 (4 = 4×0 ). Thus the degree of the polynomial is 8. = 5[3 − 72 + 24 − 2] = 5[−69 + 24 − 2] = 5[−45 − 2] 3. 2a4 − 3 + a2 = 5[−47] The term of highest degree is 2a4 , so the degree of the polynomial is 4. = −235 4. 16 ÷ 4 · 4 ÷ 2 · 256 = 4 · 4 ÷ 2 · 256 = 16 ÷ 2 · 256 4. −41 = −41×0 , so the degree of the polynomial is 0. Multiplying and dividing in order from left to right 5. 4x − x3 + 0.1×8 − 2×5 The term of highest degree is 0.1×8 , so the degree of the polynomial is 8. = 8 · 256 6. x − 3 has two terms. It is a binomial. = 2048 5. 26 · 2−3 ÷ 210 ÷ 2−8 = 2 ÷2 3 −7 =2 10 ÷2 −8 ÷2 −8 7. 14y 5 has one term. It is a monomial. 1 8. 2y − y 2 + 8 has three terms. It is a trinomial. 4 =2 11. Add and Subtract Polynomials 1. (8y − 1) − (3 − y) = (8y − 1) + (−3 + y) = (8 + 1)y + (−1 − 3) = 9y − 4 c 2017 Pearson Education, Inc. Copyright 4 Just-in-Time Review 2. (3×2 − 2x − x3 + 2) − (5×2 − 8x − x3 + 4) 2. = (3×2 − 2x − x3 + 2) + (−5×2 + 8x + x3 − 4) (5x − 3)2 = (5x)2 − 2 · 5x · 3 + 32 = (3 − 5)x + (−2 + 8)x + (−1 + 1)x + (2 − 4) 2 [(A − B)2 = A2 − 2AB + B 2 ] 3 = −2×2 + 6x − 2 3. = 25×2 − 30x + 9 (2x + 3y + z − 7) + (4x − 2y − z + 8)+ 3. (−3x + y − 2z − 4) (2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 = (2 + 4 − 3)x + (3 − 2 + 1)y + (1 − 1 − 2)z+ [(A+B)2 = A2 +2AB +B 2 ] (−7 + 8 − 4) = 4×2 + 12xy + 9y 2 = 3x + 2y − 2z − 3 4. 4. (3ab − 4a b − 2ab + 6)+ 2 2 (a − 5b)2 = a2 − 2 · a · 5b + (5b)2 (−ab − 5a b + 8ab + 4) 2 2 [(A−B)2 = A2 −2AB +B 2 ] = (3 − 1)ab + (−4 − 5)a b + (−2 + 8)ab + (6 + 4) 2 2 = 2ab − 9a b + 6ab + 10 2 5. = a2 − 10ab + 25b2 2 5. (5×2 + 4xy − 3y 2 + 2) − (9×2 − 4xy + 2y 2 − 1) = n2 − 62 = (5x + 4xy − 3y + 2) + (−9x + 4xy − 2y + 1) 2 2 2 2 = (5 − 9)x2 + (4 + 4)xy + (−3 − 2)y 2 + (2 + 1) = −4×2 + 8xy − 5y 2 + 3 (n + 6)(n − 6) [(A + B)(A − B) = A2 − B 2 ] = n − 36 2 6. (3y + 4)(3y − 4) = (3y)2 − 42 [(A + B)(A − B) = A2 − B 2 ] = 9y 2 − 16 12. Multiply Polynomials 1. 14. Factor Polynomials; The FOIL Method (3a2 )(−7a4 ) = [3(−7)](a2 · a4 ) = −21a6 2. 3. 4. 1. 3x + 18 = 3 · x + 3 · 6 = 3(x + 6) (y − 3)(y + 5) Using FOIL = y 2 + 2y − 15 Collecting like terms = (3x − 1)(x2 + 6) = x2 + 3x + 6x + 18 Using FOIL = x2 + 9x + 18 Collecting like terms t3 + 6t2 − 2t − 12 = (t + 6)(t2 − 2) Using FOIL Collecting like terms (2x + 3y)(2x + y) 2 = 4x + 2xy + 6xy + 3y = 4x + 8xy + 3y 4. = t2 (t + 6) − 2(t + 6) (2a + 3)(a + 5) 2 3×3 − x2 + 18x − 6 = x2 (3x − 1) + 6(3x − 1) = 2a2 + 13a + 15 6. 3. (x + 6)(x + 3) = 2a2 + 10a + 3a + 15 5. 2. 2z 3 − 8z 2 = 2z 2 · z − 2z 2 · 4 = 2z 2 (z − 4) = y 2 + 5y − 3y − 15 2 Using FOIL 2 5. w2 − 7w + 10 We look for two numbers with a product of 10 and a sum of −7. By trial, we determine that they are −5 and −2. w2 − 7w + 10 = (w − 5)(w − 2) 6. t2 + 8t + 15 (11t − 1)(3t + 4) We look for two numbers with a product of 15 and a sum of 8. By trial, we determine that they are 3 and 5. = 33t2 + 44t − 3t − 4 Using FOIL = 33t2 + 41t − 4 t2 + 8t + 15 = (t + 3)(t + 5) 7. 2n2 − 20n − 48 = 2(n2 − 10n − 24) 13. Special Products of Binomials 1. Now factor n2 − 10n − 24. We look for two numbers with a product of −24 and a sum of −10. By trial, we determine that they are 2 and −12. Then n2 − 10n − 24 = (n + 2)(n − 12). We must include the common factor, 2, to have a factorization of the original trinomial. (x + 3)2 = x2 + 2 · x · 3 + 32 [(A + B)2 = A2 + 2AB + B 2 ] 2n2 − 20n − 48 = 2(n + 2)(n − 12) 2 = x + 6x + 9 c 2017 Pearson Education, Inc. Copyright Just-in-Time Review 5 8. y 4 − 9y 3 + 14y 2 = y 2 (y 2 − 9y + 14) Now factor y − 9y + 14. Look for two numbers with a product of 14 and a sum of −9. The numbers are −2 and −7. Then y 2 − 9y + 14 = (y − 2)(y − 7). We must include the common factor, y 2 , in order to have a factorization of the original trinomial. 2 4. Split the middle term using the numbers found in step (3): −6x = −12x + 6x 5. Factor by grouping. 8×2 − 6x − 9 = 8×2 − 12x + 6x − 9 = 4x(2x − 3) + 3(2x − 3) y 4 − 9y 3 + 14y 2 = y 2 (y − 2)(y − 7) = (2x − 3)(4x + 3) 9. 2n2 + 9n − 56 1. There is no common factor other than 1 or −1. 1. Factor out the largest common factor, 2. 2. The factorization must be of the form (2n+ )(n+ ). 10t2 + 4t − 6 = 2(5t2 + 2t − 3) 3. Factor the constant term, −56. The possibilities are −1 · 56, 1(−56), −2 · 28, 2(−28), −4 · 16, 4(−16), −7 · 8, and 7(−8). The factors can be written in the opposite order as well: 56(−1), −56 · 1, 28(−2), −28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7. 4. Find a pair of factors for which the sum of the outer and the inner products is the middle term, 9n. By trial, we determine that the factorization is (2n − 7)(n + 8). Now factor 5t2 + 2t − 3. 2. Multiply the leading coefficient and the constant: 5(−3) = −15. 3. Try to factor −15 so that the sum of the factors is the coefficient of the middle term, 2. The factors we want are 5 and −3. 4. Split the middle term using the numbers found in step (3): 2t = 5t − 3t. 10. 2y 2 + y − 6 1. There is no common factor other than 1 or −1. 2. The factorization (2y+ )(y+ ). 2. 10t + 4t − 6 2 must be of the form 3. Factor the constant term, −6. The possibilities are −1 · 6, 1(−6), −2 · 3, and 2(−3). The factors can be written in the opposite order as well: 6(−1), −6 · 1, 3(−2) and −3 · 2. 4. Find a pair of factors for which the sum of the outer and the inner products is the middle term, y. By trial, we determine that the factorization is (2y − 3)(y + 2). 5. Factor by grouping. 5t2 + 2t − 3 = 5t2 + 5t − 3t − 3 = 5t(t + 1) − 3(t + 1) = (t + 1)(5t − 3) Include the largest common factor in the final factorization. 10t2 + 4t − 6 = 2(t + 1)(5t − 3) 3. 18a2 − 51a + 15 1. Factor out the largest common factor, 3. 18a2 − 51a + 15 = 3(6a2 − 17a + 5) Now factor 6a2 − 17a + 5. 11. b2 − 6bt + 5t2 We look for two numbers with a product of 5 and a sum of −6. By trial, we determine that they are −1 and −5. b2 − 6bt + 5t2 = (b − t)(b − 5t) 2. Multiply the leading coefficient and the constant: 6(5) = 30. 3. Try to factor 30 so that the sum of the factors is the coefficient of the middle term, −17. The factors we want are −2 and −15. 12. x4 − 7×2 − 30 = (x2 )2 − 7×2 − 30 We look for two numbers with a product of −30 and a sum of −7. By trial, we determine that they are 3 and −10. x4 − 7×2 − 30 = (x2 + 3)(x2 − 10) 4. Split the middle term using the numbers found in step (3): −17a = −2a − 15a. 5. Factor by grouping. 15. Factoring Polynomials; The ac-Method 6a2 − 17a + 5 = 6a2 − 2a − 15a + 5 = 2a(3a − 1) − 5(3a − 1) = (3a − 1)(2a − 5) 1. 8×2 − 6x − 9 1. There is no common factor other than 1 or −1. 2. Multiply the leading coefficient and the constant: 8(−9) = −72. Include the largest common factor in the final factorization. 18a2 − 51a + 15 = 3(3a − 1)(2a − 5) 3. Try to factor −72 so that the sum of the factors is the coefficient of the middle term, −6. The factors we want are −12 and 6. c 2017 Pearson Education, Inc. Copyright 6 Just-in-Time Review 5. 16. Special Factorizations 1. z 2 − 81 = z 2 − 92 = (z + 9)(z − 9) 6. 7pq 4 − 7py 4 = 7p(q 4 − y 4 ) = 7p(q 2 + y 2 )(q + y)(q − y) 7. 3m − 7 = −13 + m Subtracting m 2m = −6 Adding 7 m = −3 Dividing by 2 2(x + 7) = 5x + 14 2x + 14 = 5x + 14 5. 9z 2 − 12z + 4 = (3z)2 − 2 · 3z · 2 + 22 = (3z − 2)2 −3x + 14 = 14 Subtracting 5x a + 24a + 144a −3x = 0 Subtracting 14 = a(a2 + 24a + 144) x=0 3 6. 2 = a(a2 + 2 · a · 12 + 122 ) = a(a + 12)2 The solution is 0. 8. x3 + 64 = x3 + 43 5y − (2y − 10) = 25 5y − 2y + 10 = 25 = (x + 4)(x2 − 4x + 16) 3y + 10 = 25 m − 216 = m − 6 3 3 3 = (m − 6)(m + 6m + 36) 2 9. Dividing by 12 The solution is −3. x2 + 12x + 36 = x2 + 2 · x · 6 + 62 = (x + 6)2 8. Adding 1 2m − 7 = −13 = 7p(q 2 + y 2 )(q 2 − y 2 ) 7. Adding 5y 12y = 24 y=2 = 7p[(q 2 )2 − (y 2 )2 ] 4. 12y − 1 = 23 The solution is 2. 2. 16×2 − 9 = (4x)2 − 32 = (4x + 3)(4x − 3) 3. 7y − 1 = 23 − 5y 3y = 15 Subtracting 10 y=5 Dividing by 3 The solution is 5. 3a − 24a = 3a2 (a3 − 8) 5 Collecting like terms 2 = 3a2 (a3 − 23 ) = 3a2 (a − 2)(a2 + 2a + 4) 10. 18. Inequality-Solving Principles t6 + 1 = (t2 )3 + 13 = (t2 + 1)(t4 − t2 + 1) 1. p ≥ −125 2. 7t = 70 t = 10 x 6 3 3 x < − ·6 2 Dividing by 7 The solution is 10. 2. Subtracting 25 The solution set is [−125, ∞). 17. Equation-Solving Principles 1. p + 25 ≥ −100 9x − 1 < 17 9x < 18 Adding 1 x<2 Dividing by 9 The solution set is (−∞, 2). 4. −x − 16 ≥ 40 −x ≥ 56 6x − 15 = 45 6x = 60 Adding 15 x = 10 Dividing by 6 The solution is 10. x ≤ −56 Adding 6 Multiplying by −1 and reversing the inequality symbol The solution set is (−∞, −56]. c 2017 Pearson Education, Inc. Copyright Just-in-Time Review 5. 1 y−6 < 3 3 1 y<9 3 y 0 and q 0, then −p < 0 so both coordinates of the point (q, −p) are negative and (q, −p) is in the third quadrant. (x − 2)2 + [y − (−7)]2 = 36 (x − 2)2 + (y + 7)2 = 36 124. Let the point be (x, 0). We set the distance from (−4, −3) to (x, 0) equal to the distance from (−1, 5) to (x, 0) and solve for x. (−4 − x)2 + (−3 − 0)2 = (−1 − x)2 + (5 − 0)2 √ √ 16 + 8x + x2 + 9 = 1 + 2x + x2 + 25 √ √ x2 + 8x + 25 = x2 + 2x + 26 x2 + 8x + 25 = x2 + 2x + 26 Squaring both sides 8x + 25 = 2x + 26 The point is c 2017 Pearson Education, Inc. Copyright 1 ,0 . 6 6x = 1 1 x= 6 26 Chapter 1: Graphs, Functions, and Models 125. Let (0, y) be the required point. We set the distance from (−2, 0) to (0, y) equal to the distance from (4, 6) to (0, y) and solve for y. [0 − (−2)]2 + (y − 0)2 = (0 − 4)2 + (y − 6)2 4 + y 2 = 16 + y 2 − 12y + 36 129. 4 + y 2 = 16 + y 2 − 12y + 36 Squaring both sides −48 = −12y 4=y The point is (0, 4). 130. x2 + y 2 = 1 02 + (−1)2 ? 1 1 1 TRUE 126. We first find the distance between each pair of points. For (−1, −3) and (−4, −9): d1 = [−1 − (−4)]2 + [−3 − (−9)]2 √ √ = 32 + 62 = 9 + 36 √ √ = 45 = 3 5 x2 + y 2 = 1 2 √ 2 3 1 + − ? 1 2 2 3 1 + 4 4 1 1 TRUE √ 3 1 ,− lies on the unit circle. 2 2 (0, −1) lies on the unit circle. 131. x2 + y 2 = 1 √ 2 √ 2 2 2 + ? 1 − 2 2 2 2 + 4 4 1 1 TRUE √ √ 2 2 , lies on the unit circle. − 2 2 132. x2 + y 2 = 1 2 √ 2 1 3 + − ? 1 2 2 1 3 + 4 4 1 1 TRUE √ 3 1 ,− lies on the unit circle. 2 2 For (−1, −3) and (2, 3): d2 = (−1 − 2)2 + (−3 − 3)2 √ = (−3)2 + (−6)2 = 9 + 36 √ √ = 45 = 3 5 For (−4, −9) and (2, 3): d3 = (−4 − 2)2 + (−9 − 3)2 √ = (−6)2 + (−12)2 = 36 + 144 √ √ = 180 = 6 5 Since d1 + d2 = d3 , the points are collinear. 127. a) When the circle is positioned on a coordinate system as shown in the text, the center lies on the y-axis and is equidistant from (−4, 0) and (0, 2). Let (0, y) be the coordinates of the center. (−4−0)2 +(0−y)2 = (0−0)2 +(2−y)2 42 + y 2 = (2 − y)2 133. See the answer section in the text. 16 + y 2 = 4 − 4y + y 2 12 = −4y Exercise Set 1.2 −3 = y The center of the circle is (0, −3). b) Use the point (−4, 0) and the center (0, −3) to find the radius. (−4 − 0)2 + [0 − (−3)]2 = r2 25 = r 2 5=r The radius is 5 ft. b h , by the midpoint formula. 2 2 By the distance formula, each of the distances √ from P to b2 + h 2 . (0, h), from P to (0, 0), and from P to (b, 0) is 2 128. The coordinates of P are 1. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 2. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 3. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 4. This correspondence is not a function, because there is a member of the domain (1) that corresponds to more than one member of the range (4 and 6). 5. This correspondence is not a function, because there is a member of the domain (m) that corresponds to more than one member of the range (A and B). c 2017 Pearson Education, Inc. Copyright