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Solution Manual for A Graphical Approach to Precalculus with Limits, 7th Edition

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    S OLUTIONS M ANUAL D AVID A TWOOD Rochester Community and Technical College A G RAPHICAL A PPROACH TO P RECALCULUS WITH L IMITS A U NIT C IRCLE A PPROACH S EVENTH E DITION John Hornsby University of New Orleans Margaret L. Lial American River College Gary Rockswold Minnesota State University, Mankato

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    Contents Chapter 1 Linear Functions, Equations, and Inequalities 1 Chapter 2 Analysis of Graphs of Functions 69 Chapter 3 Quadratic Functions 153 Chapter 4 Polynomial Functions of Higher Degree 209 Chapter 5 Rational, Power, and Root Functions 279 Chapter 6 Inverse, Exponential, and Logarithmic Functions 371 Chapter 7 Systems and Matrices 451 Chapter 8 Conic Sections, Nonlinear Systems, and Parametric Equations 615 Chapter 9 The Unit Circle and the Functions of Trigonometry 683 Chapter 10 Trigonometric Identities and Equations 807 Chapter 11 Applications of Trigonometry and Vectors 893 Chapter 12 Further Topics in Algebra 977 Chapter 13 Limits, Derivatives, and Definite Integrals 1043 Chapter R Review: Basic Algebraic Concepts 1087 Appendices 1113

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    Section 1.1 1 Chapter 1: Linear Functions, Equations, and Inequalities 1.1: Real Numbers and the Rectangular Coordinate System 1. (a) The only natural number is 10. (b) The whole numbers are 0 and 10. (c) The integers are 12 6, (or 3), 0, 4    10. (d) The rational numbers are 12 5 6, (or 3), , 0, .31, .3, 4 8     and 10. (e) The irrational numbers are 3, 2 and 17.   (f) All of the numbers listed are real numbers. 2. (a) The natural numbers are 6 (or3),8, and 81(or 9). 2 (b) The whole numbers are 6 0, (or 3),8, and 81(or 9). 2 (c) The integers are 14 6 8, (or 2), 0, (or 3),8, and 81(or 9). 7 2    (d) The rational numbers are 14 6 8, (or 2), .245, (or 3),8, and 81(or 9). 7 2     (e) The only irrational number is 12. (f) All of the numbers listed are real numbers. 3. (a) There are no natural numbers listed. (b) There are no whole numbers listed. (c) The integers are 100(or 10) and 1.    (d) The rational numbers are 13 100 (or 10), , 1,5.23,9.14,3.14, 6     and 22 . 7 (e) There are no irrational numbers listed. (f) All of the numbers listed are real numbers. 4. (a) The natural numbers are 3, 18, and 56. (b) The whole numbers are 3, 18, and 56. (c) The integers are 49(or 7),3,18, and 56.   (d) The rational numbers are 49(or 7), .405, . 3,.1,3,18, and 56.     (e) The only irrational number is 6 .  (f) All of the numbers listed are real numbers. 5. The number 19,900,037,000,000 is a natural number, integer, rational number, and real number. 6. The number 700,000,000,000 is a natural number, integer, rational number, and real number. 7. The number 24  is an integer, rational, and real number.

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    2 Chapter 1 Linear Functions, Equations, and Inequalities 8. The number 17 is an integer, rational number, and real number 9. The number 71, 060  is an integer, rational number and real number. 10. The number 12.5  is a rational number and real number. 11. The number 7 2 is a real number. 12. The number  is a real number. 13. Natural numbers would be appropriate because population is only measured in positive whole numbers. 14. Natural numbers would be appropriate because distance on road signs is only given in positive whole numbers. 15. Rational numbers would be appropriate because shoes come in fraction sizes. 16. Rational numbers would be appropriate because gas is paid for in dollars and cents, a decimal part of a dollar. 17. Integers would be appropriate because temperature is given in positive and negative whole numbers. 18. Integers would be appropriate because golf scores are given in positive and negative whole numbers. 19. 20. 21. 22. 23. A rational number can be written as a fraction, , 0, p q q  where p and q are integers. An irrational number cannot be written in this way. 24. She should write 2 1.414213562.  Calculators give only approximations of irrational numbers. 25. The point 5 2, 7       is in Quadrant I. See Figure 25-34. 26. The point (1, 2) is in Quadrant I. See Figure 25-34. 27. The point ( 3, 2)  is in Quadrant II. See Figure 25-34. 28. The point ( 4,3)  is in Quadrant II. See Figure 25-34. 29. The point ( 5, 2)   is in Quadrant III. See Figure 25-34. 30. The point ( 2, 4)   is in Quadrant III. See Figure 25-34. 31. The point (2, 2)  is in Quadrant IV. See Figure 25-34.

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    Section 1.1 3 32. The point (3, 3)  is in Quadrant IV. See Figure 25-34. 33. The point (3, 0) is located on the x -axis, therefore is not in a quadrant. See Figure 25-34. 34. The point ( 2, 0)  is located on the x -axis, therefore is not in a quadrant. See Figure 25-34. Figure 25-34 35. If 0, then either 0 and 0 Quadrant I, or 0 and 0 Quadrant III. xy x y x y        36. If 0, then either 0 and < 0 Quadrant IV, or 0 and 0 Quadrant II. xy x y x y       37. If 0, then either 0 and < 0 Quadrant IV, or 0 and 0 Quadrant II. x x y x y y       38. If 0, then either 0 and > 0 Quadrant I, or 0 and 0 Quadrant III. x x y x y y       39. Any point of the form (0, ) b is located on the y -axis. 40. Any point of the form ( , 0) a is located on the x -axis. 41. [ 5,5]by[ 25,25]   42. [ 25, 25]by[ 5,5]   43. [ 60, 60]by[ 100,100]   44. [ 100,100]by[ 60,60]   45. [ 500,300]by[ 300,500]   46. [ 300,300]by[ 375,150]   47. See Figure 47. 48. See Figure 48. 49. See Figure 49. 50. See Figure 50.

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    4 Chapter 1 Linear Functions, Equations, and Inequalities [-10,10] by [-10,10] [-40,40] by [-30,30] [-5,10] by [-5,10] [-3.5,3.5] by [-4,10] Xscl = 1 Yscl = 1 Xscl = 5 Yscl = 5 Xscl= 3 Yscl = 3 Xscl = 1 Yscl= 1 Figure 47 Figure 48 Figure 49 Figure 50 51. See Figure 51. 52. See Figure 52. [-100,100] by [-50,50] [-4.7,4.7] by [-3.1,3.1] Xscl = 20 Yscl = 25 Xscl = 1 Yscl = 1 Figure 51 Figure 52 53. There are no tick marks, which is a result of setting Xscl and Yscl to 0. 54. The axes appear thicker because the tick marks are so close together. The problem can be fixed by using larger values for Xscl and Yscl such as Xscl = Yscl =10. 55. 58 7.615773106 7.616   56. 97 9.848857802 9.849   57. 3 33 3.20753433 3.208   58. 3 91 4.497941445 4.498   59. 4 86 3.045261646 3.045   60. 4 123 3.330245713 3.330   61. 1/ 2 19 4.35889844 4.359   62. 1/3 29 3.072316826 3.072   63. 1.5 46 311.9871792 311.987   64. 2.75 23 5555.863268 5555.863   65. (5.6 3.1) / (8.9 1.3) .25    66. (34 25) / 23 2.57   67. ( ^ 3 1) 5.66    68. 2 3 ( 2.1 6 ) 3.24    69. 2 3(5.9) 2(5.9) 6 98.63   

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    Section 1.1 5 70. 2 ^ 3 5 3 9.66      71. 2 2 (4 6) (7 1) ) 8.25     72.   2 2 ( 1 3 ) ( 5 3) 8.25        73. ( 1) / (1 ) .72      74. 3 ( 4.3E5 3.7E2) 76.65   75. 3 2 / (1 5) 2.82    76. 1 4.5 / (3 2) 1.84     77. 2 2 2 2 2 2 2 2 8 15 64 225 289 17 a b c c c c c             78. 2 2 2 2 2 2 2 2 7 24 49 576 625 25 a b c c c c c             79. 2 2 2 2 2 2 2 2 13 85 169 7225 7056 84 a b c b b b b             80. 2 2 2 2 2 2 2 2 14 50 196 2500 2304 48 a b c b b b b             81. 2 2 2 2 2 2 2 2 5 8 25 64 89 89 a b c c c c c             82. 2 2 2 2 2 2 2 2 9 10 81 100 181 181 a b c c c c c             83. 2 2 2 2 2 2 2 2 ( 13) ( 29) 13 29 16 4 a b c a a a a             84. 2 2 2 2 2 2 2 2 ( 7) ( 11) 7 11 4 2 a b c a a a a             85. (a) 2 2 2 2 (2 ( 4)) (5 3) (6) (2) 36 4 40 2 10 d            (b) 4 2 3 5 2 8 , , ( 1, 4) 2 2 2 2 M                     86. (a) 2 2 2 2 (2 ( 3)) (1 4) (5) ( 5) 25 25 50 5 2 d             (b) 3 2 4 ( 1) 1 3 1 3 , , , 2 2 2 2 2 2 M                           87. (a) 2 2 2 2 (6 ( 7)) ( 2 4) (13) ( 6) 169 36 205 d             (b) 7 6 4 ( 2) 1 2 1 , , ,1 2 2 2 2 2 M                            88. (a) 2 2 2 2 (1 ( 3)) (4 ( 3)) (4) (7) 16 49 65 d            (b) 3 1 3 4 2 1 1 , , 1, 2 2 2 2 2 M                            89. (a) 2 2 2 2 (2 5) (11 7) ( 3) (4) 9 16 25 5 d            (b) 5 2 7 11 7 18 7 , , ,9 2 2 2 2 2 M                        90. (a) 2 2 2 2 (4 ( 2)) ( 3 5) (6) ( 8) 36 16 100 10 d             

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    6 Chapter 1 Linear Functions, Equations, and Inequalities (b) 2 4 5 ( 3) 2 2 , , (1,1) 2 2 2 2 M                    91. (a) 2 2 2 2 ( 3 ( 8)) (( 5) ( 2)) (5) ( 3) 25 9 34 d               (b) 8 ( 3) 2 ( 5) 11 7 11 7 , , , 2 2 2 2 2 2 M                                92. (a) 2 2 2 2 (6 ( 6)) (5 ( 10)) (12) (15) 144 225 369 3 41 d             (b) 6 6 10 5 0 5 5 , , 0, 2 2 2 2 2 M                            93. (a) 2 2 2 2 (6.2 9.2) (7.4 3.4) ( 3) (4) 9 16 25 5 d            (b) 9.2 6.2 3.4 7.4 15.4 10.8 , , (7.7,5.4) 2 2 2 2 M                  94. (a) 2 2 2 2 (3.9 8.9) (13.6 1.6) ( 5) (12) 25 144 169 13 d            (b) 8.9 3.9 1.6 13.6 12.8 15.2 , , (6.4, 7.6) 2 2 2 2 M                  95. (a) 2 2 2 2 2 2 2 (6 13 ) ( ( 23 )) ( 7 ) (24 ) 49 576 625 25 d x x x x x x x x x x             (b) 13 6 23 19 22 19 , , , 11 2 2 2 2 2 x x x x x x M x x                           96. (a) 2 2 2 2 2 2 2 (20 12 ) (12 ( 3 )) (8 ) (15 ) 64 225 289 17 d y y y y y y y y y y            (b) 12 20 3 12 32 9 9 , , 16 , 2 2 2 2 2 y y y y y y M y y                         97. Using the midpoint formula we get: 2 2 2 2 2 7 4 7 , (8,5) 8 7 16 9 2 2 2 x y x x x                         and 2 2 2 4 5 4 10 14. 2 y y y          Therefore the coordinates are: (19,14). Q 98. Using the midpoint formula we get: 2 2 2 2 13 5 13 , ( 2, 4) 2 13 4 2 2 2 x y x x                     2 2 2 2 5 17 and 4 5 8 13. 2 y x y y             Therefore the coordinates are: ( 17, 13). Q   99. Using the midpoint formula we get: 2 2 2 5.64 8.21 5.64 , ( 4.04,1.60) 4.04 2 2 2 x y x                2 5.64 8.08 x     2 2 2 2 8.21 13.72 and 1.60 8.21 3.20 5.01. 2 y x y y           Therefore the coordinates are: ( 13.72, 5.01). Q  

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    Section 1.1 7 100. Using the midpoint formula we get: 2 2 2 2 10.32 8.55 10.32 , (1.55, 2.75) 1.55 10.32 3.10 2 2 2 x y x x                     2 2 2 2 8.55 13.42. 2.75 8.55 5.50 14.05. 2 y x y y            Therefore the coordinates are: ( 13.42, 13.05). Q   101. 2011 2015 36.53 67.39 4026 103.92 , , (2013,51.96); 2 2 2 2 M                  the revenue was about $51.96 billion. 102. 2006 2012 7505 3335 4018 10840 , , (2009,5420); 2 2 2 2 M                  the revenue was about $5420 million. The result is quite a bit higher than the actual figure. 103. In 2012 , 2011 2013 22,350 23,550 4024 45,900 , , (2012, 22,950); 2 2 2 2 M                  poverty level was approximately $22,950. In 2014, 2013 2015 23,350 24, 250 4028 47,800 , , 2 2 2 2 M                  (2014, 23,900); poverty level was approximately $23,900. 104. For 2017, 2016 2018 7194 7500 4034 14, 694 , , (2017 , 7347); 2 2 2 2 M                  enrollment was 7347 thousand. For 2019, 2018 2020 7500 7706 4038 15, 206 , , (2019, 7603); 2 2 2 2 M                  Enrollment was about 7603 thousand. 105. (a) From (0, 0) to (3, 4): 2 2 2 2 1 (3 0) (4 0) (3) (4) 9 16 25 5. d           From (3,4) to (7, 1): 2 2 2 2 2 (7 3) (1 4) (4) ( 3) 16 9 25 5. d            From (0, 0) to (7, 1): 2 2 2 2 3 (7 0) (1 0) (7) (1) 49 1 50 5 2. d           Since d 1 = d 2 , the triangle is isosceles. (b) From (−1, −1) to (2, 3): 2 2 2 2 1 (2 ( 1)) (3 ( 1)) (3) (4) 9 16 25 5. d             From (2, 3) to (−4, 3): 2 2 2 2 2 ( 4 2) (3 3) ( 6) (0) 36 0 36 6. d             From (−4, 3) to (−1, −1): 2 2 2 2 3 ( 1 ( 4)) ( 1 3) (3) ( 4) 9 16 25 5. d               Since 1 2 , d d  the triangle is not equilateral. (c) From (−1, 0) to (1, 0): 2 2 2 2 1 (1 ( 1)) (0 0) (2) (0) 4 0 4 2. d            From (-1, 0) to   0, 3 : 2 2 2 2 2 ( 1 0) (0 3) ( 1) ( 3) 1 3 4 2. d             

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    8 Chapter 1 Linear Functions, Equations, and Inequalities From (1, 0) to   0, 3 : 2 2 2 2 3 (1 0) (0 3) (1) ( 3) 1 3 4 2. d           Since 1 2 3 , d d d   the triangle is equilateral and isosceles. (d) From (−3, 3) to (-1, 3): 2 2 2 2 1 ( 3 ( 1)) (3 3) ( 2) (0) 4 0 4 2. d              From (-3, 3) to (−2, 5): 2 2 2 2 2 ( 3 ( 2)) (3 5) ( 1) ( 2) 1 4 5. d              From (−1, 3) to (−2, 5): 2 2 2 2 3 ( 1 ( 2)) (3 5) (1) ( 2) 1 4 5. d             Since 2 3 , d d  the triangle is not isosceles. 106. Let 1 d represent the distance between P and M and let 2 d represent the distance between M and Q . 2 2 2 2 1 2 1 2 1 1 2 1 1 2 1 1 1 2 2 2 2 2 2 x x y y x x x y y y d x y                                      2 2 2 2 1 2 1 2 1 1 2 1 2 ( ) ( ) 1 ( ) ( ) 4 4 2 x x y y d x x y y         2 2 2 2 1 2 1 2 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 x x y y x x x y y y d x y                                      2 2 2 2 2 1 2 1 2 2 1 2 1 ( ) ( ) 1 ( ) ( ) 4 4 2 x x y y d x x y y         Since 2 2 1 2 2 1 ( ) ( ) x x x x    and 2 2 1 2 2 1 ( ) ( ) y y y y    , the distances are the same. Since 1 2 , d d  the sum 2 2 2 2 1 2 2 2 1 2 1 2 1 2 1 1 2 2 ( ) ( ) ( ) ( ) . 2 d d d x x y y x x y y                 That is, the sum is equal to the distance between P and Q. 1.2: Introduction to Relations and Functions 1. The interval is ( 1, 4).  2. The interval is [ 3, ).   3. The interval is ( , 0).  4. The interval is (3, 8). 5. The interval is   1, 2 . ( 5, 4].   6. The interval is 7. ( 4,3) { | 4 3} x x      8. [2, 7) { | 2 7} x x    9. ( , 1] { | 1} x x     

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    Section 1.2 9 10. (3, ) { | 3} x x    11.   2 6 x x    12.   0 8 x x   13.   4 x x   14.   3 x x  15. A parenthesis is used if the symbol is , , , or      or . A square bracket is used if the symbol is or .   16. No real number is both greater than 7  and less than 10.  Part (d) should be written 10 7. x     17. See Figure 17 18. See Figure 18 19. See Figure 19 Figure 17 Figure 18 Figure 19 20. See Figure 20 21. See Figure 21 22. See Figure 22 Figure 20 Figure 21 Figure 22 23. See Figure 23 24. See Figure 24

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    10 Chapter 1 Linear Functions, Equations, and Inequalities Figure 23 Figure 24 25. The relation is a function. Domain:   5,3, 4, 7 Range:   1, 2,9, 6 . 26. The relation is a function. Domain:   8,5,9,3 , Range:   0, 4,3,8 . 27. The relation is a function. Domain:   1, 2,3 , Range:   6 . 28. The relation is a function. Domain:   10, 20, 30 ,    Range:   5 . 29. The relation is not a function. Domain:   4,3, 2 ,  Range:   1, 5,3, 7 .  30. The relation is not a function. Domain:   0,1 , Range:   5,3, 4 .  31. The relation is a function. Domain:   11,12,13,14 , Range:   6, 7 .   32. The relation is not a function. Domain:   1 , Range:   12,13,14,15 . 33. The relation is a function. Domain:   0,1, 2,3, 4 , Range:   2, 3, 5, 6, 7 . 34. The relation is a function. Domain: 1 1 1 1 1, , , , , 2 4 8 16       Range:   0, 1, 2, 3, 4 .     35. The relation is a function. Domain:   , ,   Range:   , .   36. The relation is a function. Domain:   , ,   Range:   , 4 .  37. The relation is not a function. Domain:   4, 4 ,  Range:   3,3 .  38. The relation is a function. Domain:   2, 2 ,  Range:   0, 4 . 39. The relation is a function. Domain:   2, ,  Range:   0, .  40. The relation is a function. Domain:   , ,   Range:   1, .  41. The relation is not a function. Domain:   9, ,   Range:   , .   42. The relation is a function. Domain:   , ,   Range:   , .   43. The relation is a function. Domain:   5, 2, 1, .5, 0,1.75,3.5 ,     Range:   1, 2,3,3.5, 4,5.75, 7.5 .  44. The relation is a function. Domain:   2, 1, 0,5,9,10,13 ,   Range:   5, 0, 3,12, 60, 77,140 .  45. The relation is a function. Domain:{2,3,5,11,17} Range:   1, 7, 20 .

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    Section 1.2 11 46. The relation is not a function. Domain:   1, 2,3,5 , Range: {10,15,19,27} 47. From the diagram, ( 2) 2. f   48. From the diagram, (5) 12. f  49. From the diagram, (11) 7. f  50. From the diagram, (5) 1. f  51. f(1) is undefined since 1 is not in the domain of the function. 52. f(10) is undefined since 10 is not in the domain of the function. 53. ( 2) 3( 2) 4 6 4 10 f          54. ( 5) 5( 5) 6 25 6 19 f          55. 2 (1) 2(1) (1) 3 2 1 3 4 f        56. 2 (2) 3(2) 2(2) 5 12 4 5 11 f        57. 2 (4) (4) (4) 2 16 4 2 10 f           58. 2 (3) (3) (3) 6 9 3 6 18 f           59. (9) 5 f  60. (12) 4 f   61. 3 ( 2) ( 2) 12 8 12 4 2 f          62. 2 3 3 3 (2) (2) (2) 6 4 2 6 8 2 f         63. 3 2 3 (8) 8 64 4 f    64. 2 3 3 ( 8) ( 8) 64 4 f      65. Given that ( ) 5 , f x x  then ( ) 5 , ( 1) 5( 1) 5 5, and (3 ) 5(3 ) 15 f a a f b b b f x x x         66. Given that ( ) 5, f x x   then ( ) 5, ( 1) 1 5 4, and (3 ) 3 5 f a a f b b b f x x           67. Given that ( ) 2 5, f x x   then ( ) 2 5, ( 1) 2( 1) 5 2 2 5 2 3, and f a a f b b b b            (3 ) 2(3 ) 5 6 5 f x x x     68. Given that 2 ( ) , f x x  then 2 2 2 ( ) , ( 1) ( 1) ( 1)( 1) 2 1, and f a a f b b b b b b           2 2 (3 ) (3 ) 9 f x x x   69. Given that 2 ( ) 1 , f x x   then 2 2 2 2 ( ) 1 , ( 1) 1 ( 1) 1 ( 2 1) 2 , and f a a f b b b b b b              2 2 (3 ) 1 (3 ) 1 9 f x x x     70. (a) Given that 2 ( ) 2 4, f x x   then 2 ( ) 2 4 f a a   (b) Given that 2 ( ) 2 4, f x x   then 2 2 ( 1) 2( 1) 4 2( 2 1) 4 f b b b b          2 2 2 2 2 4 2 2 6 b b b b      

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    12 Chapter 1 Linear Functions, Equations, and Inequalities (c) Given that 2 ( ) 2 4, f x x   then 2 2 2 (3 ) 2(3 ) 4 2(9 ) 4 18 4 f x x x x       71. Since ( 2) 3, f   the point ( 2,3)  lies on the graph of ƒ. 72. Since (3) 9.7, f   the point (3, 9.7)  lies on the graph of ƒ. 73. Since the point (7,8) lies on the graph of ƒ, (7) 8. f  74. Since the point ( 3, 2)  lies on the graph of ƒ, ( 3) 2. f   75. From the graph: (a) ( 2) 0, f   (b) (0) 4, f  (c) (1) 2, f  and (d) (4) 4. f  76. From the graph: (a) ( 2) 5, f   (b) (0) 0, f  (c) (1) 2, f  and (d) (4) 4. f  77. From the graph: (a) ( 2) f  is undefined , (b) (0) 2, f   (c) (1) 0, f  and (d) (4) 2. f  78. From the graph: (a) ( 2) 3, f   (b) (0) 3, f  (c) (1) 3, f  and (d) (4) f is undefined. 79. (a)           Apple, 216 , Alphabet, 90 , Google, 89 , Microsoft, 85 A  , The U.S. total revenue in 2011 for Apple was $216,000,000 dollars. (b) See Figure 81. (c)   Apple, Alphabet, Google, Microsoft D  ,   216, 90, 89, 85 R  80. (a)           0,1.0 , 2, 2.0 , 7,5.5 , 12,11.0 T  (b) See Figure 80 (c)     0, 2, 7, 12 ; 1.0, 2.0, 5.5, 11.0 D R   Figure 79 Figure 80 81. (a) See Figure 81. (b) (2000) 12.8 f  In 2000 there were 12,800 radio stations on the air. (c) Domain:   1990, 2000, 2005, 2012, 2015 , Range:   10.8,12.8,13.5,15.1,16.4 . 82. (a) See Figure 82. (b) (20012) 2366 f  In 2012, there were $2366 billion spent on personal health care. (c)   : 2010, 2011, 2012, 2013, 2014, 2015 D   R : 2195, 2273, 2366, 2436, 2563, 2717

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    Section 1.3 13 Figure 81 Figure 82 Reviewing Basic Concepts (Sections 1.1 and 1.2) 1. See Figure 1. 2. The distance is 2 2 (6 ( 4)) ( 2 5) 100 49 149. d          The midpoint is 4 6 5 2 3 , 1, . 2 2 2 M                  3. 3 5 1.168 ( 3 1)     4. 2 2 (12 ( 4)) ( 3 27) 256 900 1156 34 d           5. Using Pythagorean Theorem, 2 2 2 2 2 2 2 11 61 61 11 3600 60 b b b b          inches. 6. The set   2 5 x x    is the interval ( 2,5].  The set   4 x x  is the interval [4, ).  7. The relation is not a function because it does not pass the vertical line test. Domain:   2, 2 ,  Range:   3,3 .  8. See Figure 8. 9. Given ( ) 3 4 f x x   then ( 5) 3 4( 5) 23 and ( 4) 3 4( 4) 3 4 16 4 13 f f a a a a                10. From the graph, (2) 3 f  and ( 1) 3. f    Figure 1 Figure 8 1.3: Linear Functions 1. The graph is shown in Figure 1. (a) x -intercept: 4 (b) y -intercept: 4  (c) Domain: ( , )   (d) Range: ( , )  

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