Solution Manual for College Algebra, 10th Edition

Chapter 2 Graphs (f) Quadrant IV Section 2.1 1. 0 2. 5 − ( −3) = 8 = 8 3. 32 + 42 = 25 = 5 4. 112 + 602 = 121 + 3600 = 3721 = 612 Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle. 5. 1 bh 2 16. (a) Quadrant I (b) Quadrant III (c) Quadrant II (d) Quadrant I (e) y-axis (f) x-axis 6. true 7. x-coordinate or abscissa; y-coordinate or ordinate 8. quadrants 9. midpoint 10. False; the distance between two points is never negative. 11. False; points that lie in Quadrant IV will have a positive x-coordinate and a negative y-coordinate. The point ( −1, 4 ) lies in Quadrant II. 17. The points will be on a vertical line that is two units to the right of the y-axis.  x + x y + y2  12. True; M =  1 2 , 1 2   2 13. b 14. a 15. (a) Quadrant II (b) x-axis (c) Quadrant III (d) Quadrant I (e) y-axis 147 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 18. The points will be on a horizontal line that is three units above the x-axis. 28. d ( P1 , P2 ) = ( 6 − (− 4) )2 + ( 2 − (−3) )2 = 102 + 52 = 100 + 25 = 125 = 5 5 29. d ( P1 , P2 ) = (0 − a) 2 + (0 − b) 2 = ( − a ) 2 + ( −b ) 2 = a 2 + b 2 30. d ( P1 , P2 ) = (0 − a ) 2 + (0 − a) 2 = (−a)2 + (−a)2 = a 2 + a 2 = 2a 2 = a 19. d ( P1 , P2 ) = (2 − 0) 2 + (1 − 0) 2 = 22 + 12 = 4 + 1 = 5 31. A = (−2,5), B = (1,3), C = (−1, 0) d ( A, B ) = 20. d ( P1 , P2 ) = (−2 − 0) 2 + (1 − 0) 2 d ( B, C ) = 21. d ( P1 , P2 ) = (−2 − 1) 2 + (2 − 1) 2 = (−3) 2 + 12 = 9 + 1 = 10 (1 − (−2) )2 + (3 − 5)2 = 32 + (−2) 2 = 9 + 4 = 13 = (−2) 2 + 12 = 4 + 1 = 5 22. d ( P1 , P2 ) = 2 ( −1 − 1)2 + (0 − 3)2 = (−2) 2 + (−3)2 = 4 + 9 = 13 d ( A, C ) = ( 2 − (−1) )2 + (2 − 1)2 ( −1 − (−2) )2 + (0 − 5)2 = 12 + (−5) 2 = 1 + 25 = 26 = 32 + 12 = 9 + 1 = 10 23. d ( P1 , P2 ) = (5 − 3) 2 + ( 4 − ( −4 ) ) 2 2 = 22 + ( 8 ) = 4 + 64 = 68 = 2 17 24. d ( P1 , P2 ) = ( 2 − ( −1) ) + ( 4 − 0 )2 = ( 3)2 + 42 = 9 + 16 = 25. d ( P1 , P2 ) = ( 6 − (−3) )2 + (0 − 2)2 2 2 25 = 5 2 = 9 + (− 2) = 81 + 4 = 85 26. d ( P1 , P2 ) = ( 4 − 2 )2 + ( 2 − (−3) )2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 ( 13 ) + ( 13 ) = ( 26 ) 2 2 2 13 + 13 = 26 = 22 + 52 = 4 + 25 = 29 26 = 26 27. d ( P1 , P2 ) = (6 − 4) 2 + ( 4 − (−3) ) 2 = 22 + 7 2 = 4 + 49 = 53 The area of a triangle is A = problem, 148 Copyright © 2016 Pearson Education, Inc. 1 ⋅ bh . In this 2 Section 2.1: The Distance and Midpoint Formulas A = 1 ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] 2 1 = ⋅ 13 ⋅ 13 = 1 ⋅13 2 2 13 = 2 square units 32. A = (−2, 5), B = (12, 3), C = (10, − 11) d ( A, B ) = problem, 1 A = ⋅ [ d ( A, B ) ] ⋅ [ d ( B, C ) ] 2 1 = ⋅10 2 ⋅10 2 2 1 = ⋅100 ⋅ 2 = 100 square units 2 33. A = (− 5,3), B = (6, 0), C = (5,5) (12 − (−2) )2 + (3 − 5)2 d ( A, B) = = 142 + (−2) 2 ( 6 − (− 5) )2 + (0 − 3)2 = 196 + 4 = 200 = 112 + (− 3) 2 = 121 + 9 = 10 2 = 130 d ( B, C ) = d ( B, C ) = (10 − 12 )2 + (−11 − 3)2 ( 5 − 6 )2 + (5 − 0)2 = (−2) 2 + (−14) 2 = (−1) 2 + 52 = 1 + 25 = 4 + 196 = 200 = 26 d ( A, C ) = = 10 2 d ( A, C ) = (10 − (−2) )2 + (−11 − 5)2 ( 5 − (− 5) )2 + (5 − 3)2 = 102 + 22 = 100 + 4 = 122 + (−16) 2 = 104 = 144 + 256 = 400 = 2 26 = 20 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)] + [ d ( B, C )] = [ d ( A, C )] 2 2 (10 2 ) + (10 2 ) = ( 20 ) 2 2 2 2 200 + 200 = 400 400 = 400 1 The area of a triangle is A = bh . In this 2 [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 ( 104 ) + ( 26 ) = ( 130 ) 2 2 2 104 + 26 = 130 130 = 130 1 The area of a triangle is A = bh . In this 2 149 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs problem, 1 A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] 2 1 = ⋅ 29 ⋅ 2 29 2 1 = ⋅ 2 ⋅ 29 2 = 29 square units problem, 1 A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] 2 1 = ⋅ 104 ⋅ 26 2 1 = ⋅ 2 26 ⋅ 26 2 1 = ⋅ 2 ⋅ 26 2 = 26 square units 35. A = (4, −3), B = (0, −3), C = (4, 2) 34. A = (−6, 3), B = (3, −5), C = (−1, 5) ( 3 − (−6) )2 + (−5 − 3)2 d ( A, B) = = 92 + (−8) 2 = 81 + 64 = 145 2 ( −1 − 3) + (5 − (−5)) d ( B, C ) = 2 = (−4) 2 + 102 = 16 + 100 = 116 = 2 29 2 ( −1 − (− 6) ) + (5 − 3) d ( A, C ) = 2 2 = (− 4)2 + 02 = 16 + 0 = 16 =4 d ( B, C ) = ( 4 − 0 )2 + ( 2 − (−3) )2 = 42 + 52 = 16 + 25 = 41 d ( A, C ) = (4 − 4) 2 + ( 2 − (−3) ) 2 = 25 =5 = 29 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 ( 29 ) + ( 2 29 ) = ( 145 ) 2 2 = 02 + 52 = 0 + 25 2 = 5 + 2 = 25 + 4 2 d ( A, B ) = (0 − 4) 2 + ( −3 − (−3) ) 2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)]2 + [ d ( A, C )]2 = [ d ( B, C )]2 4 2 + 52 = 29 + 4 ⋅ 29 = 145 2 16 + 25 = 41 41 = 41 29 + 116 = 145 145 = 145 1 The area of a triangle is A = bh . In this 2 ( 41) The area of a triangle is A = 150 Copyright © 2016 Pearson Education, Inc. 1 bh . In this 2 Section 2.1: The Distance and Midpoint Formulas problem, 1 A = ⋅ [ d ( A, B) ] ⋅ [ d ( A, C ) ] 2 1 = ⋅4⋅5 2 = 10 square units The area of a triangle is A = 37. The coordinates of the midpoint are: x +x y +y  ( x, y ) =  1 2 , 1 2  2   2  3+5 −4+ 4 = , 2   2 8 0 = ,  2 2 = (4, 0) 2 = 02 + 42 = 0 + 16 = 16 =4 d ( B, C ) = ( 2 − 4 )2 + (1 − 1)2 38. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2 = (−2) 2 + 02 = 4 + 0 = 4 =2 d ( A, C ) = (2 − 4) 2 + (1 − (−3) ) 1 ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] 2 1 = ⋅4⋅2 2 = 4 square units A= 36. A = (4, −3), B = (4, 1), C = (2, 1) d ( A, B ) = (4 − 4) 2 + (1 − (−3) ) 1 bh . In this problem, 2 2 = (−2) 2 + 42 = 4 + 16 = 20 =2 5  −2 + 2 0 + 4  , = 2   2 0 4 = ,  2 2 = ( 0, 2 ) 39. The coordinates of the midpoint are: x +x y +y  ( x, y ) =  1 2 , 1 2  2   2  −3 + 6 2 + 0  = , 2   2 3 2 = ,  2 2 3  =  ,1 2  Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 ( 42 + 22 = 2 5 16 + 4 = 20 20 = 20 ) 2 40. The coordinates of the midpoint are: x +x y +y  ( x, y ) =  1 2 , 1 2  2   2  2 + 4 −3 + 2  = , 2   2  6 −1  = ,  2 2  1  =  3, −  2  151 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 41. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2  4 + 6 −3 + 1  , = 2   2  10 − 2  = ,   2 2  = (5, −1) 52 + b 2 = 132 25 + b 2 = 169 b 2 = 144 b = 12 Thus the coordinates will have an y value of −1 − 12 = −13 and −1 + 12 = 11 . So the points are ( 3,11) and ( 3, −13) . 42. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2  − 4 + 2 −3 + 2  , = 2   2  − 2 −1  ,  =  2 2  b. Consider points of the form ( 3, y ) that are a distance of 13 units from the point ( −2, −1) . d= ( x2 − x1 )2 + ( y2 − y1 )2 = ( 3 − (−2) )2 + ( −1 − y )2 = ( 5)2 + ( −1 − y )2 = 25 + 1 + 2 y + y 2 1  =  −1, −  2  y 2 + 2 y + 26 = 43. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2 a+0 b+0 = , 2   2 a b = ,   2 2 13 = y 2 + 2 y + 26 132 = ( y + 2 y + 26 ) 2 2 169 = y 2 + 2 y + 26 0 = y 2 + 2 y − 143 0 = ( y − 11)( y + 13) 44. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2 a+0 a+0 = , 2   2 a a = ,  2 2 45. The x coordinate would be 2 + 3 = 5 and the y coordinate would be 5 − 2 = 3 . Thus the new point would be ( 5,3) . 46. The new x coordinate would be −1 − 2 = −3 and the new y coordinate would be 6 + 4 = 10 . Thus the new point would be ( −3,10 ) 47. a. If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length. One of the legs of the triangle will be 2+3=5. Thus the other leg will be: y − 11 = 0 or y + 13 = 0 y = 11 y = −13 Thus, the points ( 3,11) and ( 3, −13) are a distance of 13 units from the point ( −2, −1) . 48. a. If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length. One of the legs of the triangle will be 2+6=8. Thus the other leg will be: 82 + b 2 = 17 2 64 + b 2 = 289 b 2 = 225 b = 15 Thus the coordinates will have an x value of 1 − 15 = −14 and 1 + 15 = 16 . So the points are ( −14, −6 ) and (16, −6 ) . 152 Copyright © 2016 Pearson Education, Inc. Section 2.1: The Distance and Midpoint Formulas b. Consider points of the form ( x, −6 ) that are x = 4 + 3 3 or x = 4 − 3 3 a distance of 17 units from the point (1, 2 ) . Thus, the points 4 + 3 3, 0 and 4 − 3 3, 0 are d= ( x2 − x1 )2 + ( y2 − y1 )2 = (1 − x )2 + ( 2 − ( −6 ) ) = x − 2 x + 1 + (8) 2 ( ) ( ) on the x-axis and a distance of 6 units from the point ( 4, −3) . 2 50. Points on the y-axis have an x-coordinate of 0. Thus, we consider points of the form ( 0, y ) that 2 = x 2 − 2 x + 1 + 64 are a distance of 6 units from the point ( 4, −3) . = x 2 − 2 x + 65 d= ( x2 − x1 )2 + ( y2 − y1 )2 17 = x 2 − 2 x + 65 = ( 4 − 0 ) 2 + ( −3 − y ) 2 ( x − 2x + 65 ) 17 2 = 2 2 = 42 + 9 + 6 y + y 2 = 16 + 9 + 6 y + y 2 289 = x 2 − 2 x + 65 0 = x 2 − 2 x − 224 0 = ( x + 14 )( x − 16 ) x + 14 = 0 or x − 16 = 0 x = −14 x = 16 Thus, the points ( −14, −6 ) and (16, −6 ) are a distance of 13 units from the point (1, 2 ) . 49. Points on the x-axis have a y-coordinate of 0. Thus, we consider points of the form ( x, 0 ) that are a distance of 6 units from the point ( 4, −3) . d= ( x2 − x1 )2 + ( y2 − y1 )2 = ( 4 − x )2 + ( −3 − 0 )2 = 16 − 8 x + x 2 + ( −3) 2 = 16 − 8 x + x 2 + 9 = x 2 − 8 x + 25 6 = x 2 − 8 x + 25 62 = ( x 2 − 8 x + 25 ) 2 2 36 = x − 8 x + 25 0 = x 2 − 8 x − 11 x= −(−8) ± (−8) 2 − 4(1)(−11) 2(1) 8 ± 64 + 44 8 ± 108 = 2 2 8±6 3 = = 4±3 3 2 = = y 2 + 6 y + 25 6= y 2 + 6 y + 25 62 = ( y + 6 y + 25 ) 2 2 36 = y 2 + 6 y + 25 0 = y 2 + 6 y − 11 y= ( −6) ± (6)2 − 4(1)( −11) 2(1) −6 ± 36 + 44 −6 ± 80 = 2 2 −6 ± 4 5 = = −3 ± 2 5 2 y = −3 + 2 5 or y = −3 − 2 5 = ( ) ( Thus, the points 0, −3 + 2 5 and 0, −3 − 2 5 ) are on the y-axis and a distance of 6 units from the point ( 4, −3) . 51. a. To shift 3 units left and 4 units down, we subtract 3 from the x-coordinate and subtract 4 from the y-coordinate. (2 − 3,5 − 4) = ( −1,1) b. To shift left 2 units and up 8 units, we subtract 2 from the x-coordinate and add 8 to the y-coordinate. ( 2 − 2,5 + 8) = ( 0,13) 153 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 52. Let the coordinates of point B be ( x, y ) . Using the midpoint formula, we can write  −1 + x 8 + y  ( 2,3) =  2 , 2  .   This leads to two equations we can solve. −1 + x 8+ y =3 =2 2 2 −1 + x = 4 8+ y = 6 y = −2 x=5 Point B has coordinates ( 5, −2 ) . and 1 = x2 = (− 4) 2 + (− 1) 2 = 16 + 1 = 17 = (− 4) 2 + 22 = 16 + 4 = 20 = 2 5 d ( A, F ) = (2 − 0) 2 + (5 − 0) 2 = 22 + 52 = 4 + 25 y + y2 y= 1 2 6 + y2 4= 2 8 = 6 + y2 2 = y2 56. Let P1 = (0, 0), P2 = (0, 4), P = ( x, y ) d ( P1 , P2 ) = (0 − 0) 2 + (4 − 0) 2 = 16 = 4 d ( P1 , P ) = ( x − 0) 2 + ( y − 0) 2 = x2 + y 2 = 4 → x 2 + y 2 = 16 d ( P2 , P ) = ( x − 0) 2 + ( y − 4) 2 Thus, P2 = (1, 2) . = x 2 + ( y − 4) 2 = 4  x + x y + y2  54. M = ( x, y ) =  1 2 , 1 . 2   2 P2 = ( x2 , y2 ) = (7, −2) and ( x, y ) = (5, −4) , so x +x x= 1 2 2 x1 + 7 5= 2 10 = x1 + 7 and 3 = x1 ( 2 − 6 )2 + (2 − 0)2 d ( B, E ) = = 29  x + x y + y2  53. M = ( x, y ) =  1 2 , 1 . 2   2 P1 = ( x1 , y1 ) = (−3, 6) and ( x, y ) = (−1, 4) , so x +x x= 1 2 2 −3 + x2 −1 = 2 −2 = −3 + x2 ( 0 − 4 )2 + (3 − 4)2 d (C , D) = y + y2 y= 1 2 y1 + (−2) −4 = 2 −8 = y1 + (−2) → x 2 + ( y − 4) 2 = 16 Therefore, y2 = ( y − 4) 2 y 2 = y 2 − 8 y + 16 8 y = 16 y=2 which gives x 2 + 22 = 16 x 2 = 12 −6 = y1 Thus, P1 = (3, −6) . 0+6 0+0 , 55. The midpoint of AB is: D =  2   2 = ( 3, 0 ) 0+4 0+4 The midpoint of AC is: E =  , 2   2 = ( 2, 2 ) x = ±2 3 Two triangles are possible. The third vertex is ( − 2 3, 2 ) or ( 2 3, 2) . 57. Let P1 = ( 0, 0 ) , P2 = ( 0, s ) , P3 = ( s, 0 ) , and P4 = ( s, s ) . 6+4 0+4 , The midpoint of BC is: F =  2   2 = ( 5, 2 ) 154 Copyright © 2016 Pearson Education, Inc. Section 2.1: The Distance and Midpoint Formulas y (0, s) (s , s ) (0, 0) (s, 0) x The points P1 and P4 are endpoints of one diagonal and the points P2 and P3 are the endpoints of the other diagonal. 0+s 0+s   s s  M 1,4 =  , = ,  2  2 2  2 0+s s+0  s s  M 2,3 =  , = ,  2  2 2  2 The midpoints of the diagonals are the same. Therefore, the diagonals of a square intersect at their midpoints. 0+a 0+0  a  P4 = M P1P2 =  ,  =  , 0 2  2   2   a 3 a   3a 3 a   P5 = M P2 P3 =  a + 2 0 + 2  =  ,  , 4     4 2  2     a 3a  0+ 0+  2, 2  =a, 3a P6 = M P1P3 =    2  2   4 4      2   3a a   3 a d ( P4 , P5 ) =  −  +  − 0   4 2  4  2 a  3a =   +    4   4  58. Let P1 = ( 0, 0 ) , P2 = ( a, 0 ) , and a 3a P3 =  ,  . To show that these vertices 2 2  form an equilateral triangle, we need to show that the distance between any pair of points is the same constant value. d ( P1 , P2 ) = ( x2 − x1 )2 + ( y2 − y1 )2 = ( a − 0 )2 + ( 0 − 0 )2 = d ( P2 , P3 ) = 2  a   3a =  − a  +  − 0  2 2     = d ( P1 , P3 ) = 2 2 4a = a2 = a 4 2 2 a 3a + = 4 4 2  a   3a =  − 0  +  − 0  2   2  = 2 2 a a 2 3a 2 + = 16 16 2 2 3a  3a a   3 a d ( P5 , P6 ) =  −  +  −  4   4 4  4 2 2 a =   + 02 2 a a2 = 4 2 Since the sides are the same length, the triangle is equilateral. = 2 a a 2 3a 2 + = 16 16 2 2  a  3a =  −  +    4   4  2 ( x2 − x1 ) + ( y2 − y1 ) 2 2  a a  3a d ( P4 , P6 ) =  −  +  − 0  4 2  4  a2 = a ( x2 − x1 )2 + ( y2 − y1 )2 2 = 2 a 2 3a 2 4a 2 + = = a2 = a 4 4 4 Since all three distances have the same constant value, the triangle is an equilateral triangle. Now find the midpoints: = 155 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 59. d ( P1 , P2 ) = (− 4 − 2) 2 + (1 − 1) 2 61. d ( P1 , P2 ) = ( 0 − (− 2) )2 + ( 7 − (−1) )2 = (− 6) 2 + 02 = 22 + 82 = 4 + 64 = 68 = 36 =6 = 2 17 d ( P2 , P3 ) = 2 ( − 4 − (− 4) ) + (−3 − 1) 2 = 0 + (− 4) = 32 + (− 5) 2 = 9 + 25 2 = 34 = 16 =4 ( 3 − (−2) )2 + ( 2 − (−1) )2 d ( P1 , P3 ) = = 52 + 32 = 25 + 9 d ( P1 , P3 ) = (− 4 − 2) 2 + (−3 − 1) 2 = 34 Since d ( P2 , P3 ) = d ( P1 , P3 ) , the triangle is isosceles. = (− 6) 2 + (− 4) 2 = 36 + 16 = 52 Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , 2 = 2 13 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , 2 2 2 the triangle is a right triangle. 60. d ( P1 , P2 ) = ( 3 − 0 )2 + (2 − 7)2 d ( P2 , P3 ) = 2 ( − 4 − 7 )2 + ( 0 − 2 )2 = 7 2 + (− 2) 2 = (−11) 2 + (− 2) 2 = 49 + 4 = 121 + 4 = 125 =5 5 = 53 d ( P2 , P3 ) = ( 4 − 6 )2 + (−5 − 2)2 d ( P2 , P3 ) = = 100 = 10 = 4 + 49 = 53 d ( P1 , P3 ) = 2 ( 4 − (−1) ) + (−5 − 4)2 2 = 5 + (− 9) ( 4 − 7 )2 + ( 6 − 2 )2 = (−3) 2 + 42 = 9 + 16 2 = 25 =5 = 25 + 81 = 106 Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , 2 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , 2 ( 4 − (− 4) )2 + (6 − 0)2 = 82 + 62 = 64 + 36 = (− 2)2 + (− 7) 2 d ( P1 , P3 ) = 2 the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle. 62. d ( P1 , P2 ) = ( 6 − (−1) )2 + (2 − 4)2 2 2 2 the triangle is a right triangle. the triangle is a right triangle. Since d ( P1 , P2 ) = d ( P2 , P3 ) , the triangle is isosceles. Therefore, the triangle is an isosceles right triangle. 156 Copyright © 2016 Pearson Education, Inc. 2 2 Section 2.1: The Distance and Midpoint Formulas 63. Using the Pythagorean Theorem: 902 + 902 = d 2 First: (60, 0), Second: (60, 60) Third: (0, 60) 66. a. 8100 + 8100 = d 2 16200 = d y 2 (0,60) (60,60) d = 16200 = 90 2 ≈ 127.28 feet 90 90 x d (0,0) b. Using the distance formula: 90 90 (60,0) d = (180 − 60) 2 + (20 − 60) 2 = 1202 + (− 40) 2 = 16000 64. Using the Pythagorean Theorem: 602 + 602 = d 2 = 40 10 ≈ 126.49 feet 3600 + 3600 = d 2 → 7200 = d 2 Using the distance formula: c. d = 7200 = 60 2 ≈ 84.85 feet d = (220 − 0) 2 + (220 − 60)2 = 2202 + 1602 = 74000 60 60 = 20 185 ≈ 272.03 feet d 60 60 65. a. First: (90, 0), Second: (90, 90), Third: (0, 90) Y (0,90) 67. The Focus heading east moves a distance 30t after t hours. The truck heading south moves a distance 40t after t hours. Their distance apart after t hours is: d = (30t ) 2 + (40t ) 2 = 900t 2 + 1600t 2 (90,90) = 2500t 2 = 50t miles 30t X (0,0) 40t (90,0) d b. Using the distance formula: d = (310 − 90) 2 + (15 − 90) 2 = 2202 + (−75)2 = 54025 = 5 2161 ≈ 232.43 feet c. 68. 15 miles 5280 ft 1 hr ⋅ ⋅ = 22 ft/sec 1 hr 1 mile 3600 sec d = 1002 + ( 22t ) Using the distance formula: d = (300 − 0) 2 + (300 − 90)2 2 = 10000 + 484t 2 feet = 3002 + 2102 = 134100 = 30 149 ≈ 366.20 feet 157 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs pair ( 2013, 23624) . The midpoint is 22t ( year, $) = 2003 +2 2013 , 18660 +2 23624 d 100 = 4016 42284 , 2 2 = ( 2008, 21142) 69. a. The shortest side is between P1 = (2.6, 1.5) and P2 = (2.7, 1.7) . The estimate for the desired intersection point is:  x1 + x2 y1 + y2   2.6 + 2.7 1.5 + 1.7  ,   2 , 2 = 2 2     5.3 3.2   , =   2 2  = ( 2.65, 1.6 ) b. Using the distance formula: Using the midpoint, we estimate the poverty level in 2008 to be $21,142. This is lower than the actual value. 72. Answers will vary. 73. To find the domain, we know the denominator cannot be zero. 2x − 5 = 0 2x = 5 5 x= 2 d = (2.65 − 1.4) 2 + (1.6 − 1.3) 2 So the domain is all real numbers not equal to { = (1.25) 2 + (0.3) 2 or x | x ≠ = 1.5625 + 0.09 74. = 1.6525 ≈ 1.285 units 70. Let P1 = (2007, 345) and P2 = (2013, 466) . The midpoint is: x +x y +y ( x, y ) = 1 2 2 , 1 2 2 = 2007 + 2013 345 + 466 , 2 2 = 4020 811 , 2 2 = ( 2010, 405.5) The estimate for 2010 is $405.5 billion. The estimate net sales of Wal-Mart Stores, Inc. in 2010 is $0.5 billion off from the reported value of $405 billion. 71. For 2003 we have the ordered pair (2003,18660) and for 2013 we have the ordered } 5 . 2 3 x 2 − 7 x − 20 = 0 (3x + 5)( x − 4) = 0 (3x + 5) = 0 or ( x − 4) = 0 5 x = − or x = 4 3 5 So the solution set is: − ,4 3 75. (7 + 3i )(1 − 2i) = 7 − 14i + 3i − 6i 2 = 7 − 11i − 6( −1) = 7 − 11i + 6 = 13 − 11i 76. 5( x − 3) + 2 x ≥ 6(2 x − 3) − 7 5 x − 15 + 2 x ≥ 12 x − 18 − 7 7 x − 15 ≥ 12 x − 25 −5 x ≥ −10 x≤2 158 Copyright © 2016 Pearson Education, Inc. 5 2 Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry Section 2.2 15. y 2 = x 2 + 9 32 = 02 + 9 02 = 32 + 9 02 = (−3) 2 + 9 9=9 0 ≠ 18 0 ≠ 18 The point (0, 3) is on the graph of the equation. 1. 2 ( x + 3) − 1 = −7 2 ( x + 3) = −6 x + 3 = −3 16. y 3 = x + 1 x = −6 The solution set is {−6} . 23 = 1 + 1 03 = −1 + 1 13 = 0 + 1 8≠2 0=0 1=1 The points (0, 1) and (–1, 0) are on the graph of the equation. 2. x 2 − 9 = 0 x2 = 9 x = ± 9 = ±3 The solution set is {−3,3} . 17. x 2 + y 2 = 4 3. intercepts (− 2) 2 + 22 = 4 4=4 8≠4 (0, 2) and 4. y = 0 ( 2) +( 2) = 4 2 2 ( 2, 2 ) are on the graph of the 4=4 equation. 5. y-axis 18. x 2 + 4 y 2 = 4 6. 4 7. 02 + 22 = 4 02 + 4 ⋅12 = 4 22 + 4 ⋅ 02 = 4 22 + 4 ( 12 ) = 4 4=4 4=4 5≠4 The points (0, 1) and (2, 0) are on the graph of the equation. 2 ( −3, 4 ) 8. True 9. False; the y-coordinate of a point at which the graph crosses or touches the x-axis is always 0. The x-coordinate of such a point is an x-intercept. 10. False; a graph can be symmetric with respect to both coordinate axes (in such cases it will also be symmetric with respect to the origin). For example: x 2 + y 2 = 1 19. y = x + 2 x-intercept: 0= x+2 −2 = x y-intercept: y = 0+2 y=2 The intercepts are ( −2, 0 ) and ( 0, 2 ) . 11. d 12. c 13. y = x 4 − x 4 = (2) 4 − 2 0 = 04 − 0 1 = 14 − 1 0=0 1≠ 0 4 ≠ 16 − 2 The point (0, 0) is on the graph of the equation. 3 14. y = x − 2 x 3 3 3 0 = 0 −2 0 1=1 −2 1 −1 = 1 − 2 1 0=0 1 ≠ −1 −1 = −1 The points (0, 0) and (1, –1) are on the graph of the equation. 20. y = x − 6 x-intercept: 0 = x−6 6=x 159 Copyright © 2016 Pearson Education, Inc. y-intercept: y = 0−6 y = −6 Chapter 2: Graphs The intercepts are ( 6, 0 ) and ( 0, −6 ) . 23. y = x 2 − 1 x-intercepts: 0 = x2 − 1 x2 = 1 y-intercept: y = 02 − 1 y = −1 x = ±1 The intercepts are ( −1, 0 ) , (1, 0 ) , and ( 0, −1) . 21. y = 2 x + 8 x-intercept: y-intercept: 0 = 2x + 8 y = 2 (0) + 8 2 x = −8 y =8 x = −4 The intercepts are ( −4, 0 ) and ( 0,8 ) . 24. y = x 2 − 9 x-intercepts: 0 = x2 − 9 x2 = 9 y-intercept: y = 02 − 9 y = −9 x = ±3 The intercepts are ( −3, 0 ) , ( 3, 0 ) , and ( 0, −9 ) . 22. y = 3 x − 9 x-intercept: y-intercept: 0 = 3x − 9 y = 3(0) − 9 3x = 9 y = −9 x=3 The intercepts are ( 3, 0 ) and ( 0, −9 ) . 25. y = − x 2 + 4 x-intercepts: y-intercepts: 2 0 = −x + 4 y = − ( 0) + 4 x2 = 4 y=4 2 x = ±2 160 Copyright © 2016 Pearson Education, Inc. Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry The intercepts are ( −2, 0 ) , ( 2, 0 ) , and ( 0, 4 ) . 28. 5 x + 2 y = 10 x-intercepts: 5 x + 2 ( 0 ) = 10 y-intercept: 5 ( 0 ) + 2 y = 10 5 x = 10 2 y = 10 x=2 y=5 The intercepts are ( 2, 0 ) and ( 0,5 ) . 26. y = − x 2 + 1 x-intercepts: y-intercept: 2 0 = −x +1 y = − (0) + 1 x2 = 1 y =1 2 x = ±1 The intercepts are ( −1, 0 ) , (1, 0 ) , and ( 0,1) . 29. 9 x 2 + 4 y = 36 x-intercepts: y-intercept: 2 9 x + 4 ( 0 ) = 36 9 ( 0 ) + 4 y = 36 9 x 2 = 36 4 y = 36 y=9 2 2 x =4 x = ±2 The intercepts are ( −2, 0 ) , ( 2, 0 ) , and ( 0,9 ) . 27. 2 x + 3 y = 6 x-intercepts: 2x + 3(0) = 6 2x = 6 x=3 y-intercept: 2 (0) + 3 y = 6 3y = 6 y=2 The intercepts are ( 3, 0 ) and ( 0, 2 ) . 30. 4 x 2 + y = 4 x-intercepts: y-intercept: 2 4x + 0 = 4 4 ( 0) + y = 4 4 x2 = 4 y=4 2 2 x =1 x = ±1 161 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs The intercepts are ( −1, 0 ) , (1, 0 ) , and ( 0, 4 ) . 34. 35. y 5 31. (c) = (−5, 2) (a) = (5, 2) −5 5 (5, −2) (b) = (−5, −2) −5 36. 32. 37. 33. 38. 162 Copyright © 2016 Pearson Education, Inc. Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry 39. 49. a. x-intercept: [ −2,1] , y-intercept 0 b. Not symmetric to x-axis, y-axis, or origin. 50. a. x-intercept: [ −1, 2] , y-intercept 0 b. Not symmetric to x-axis, y-axis, or origin. 51. a. Intercepts: none b. Symmetric with respect to the origin. 40. 52. a. Intercepts: none b. Symmetric with respect to the x-axis. 53. 41. a. Intercepts: ( −1, 0 ) and (1, 0 ) b. Symmetric with respect to the x-axis, y-axis, and the origin. 42. a. 54. Intercepts: ( 0,1) b. Not symmetric to the x-axis, the y-axis, nor the origin 43. a. ( ) ( Intercepts: − π2 , 0 , ( 0,1) , and π2 , 0 ) b. Symmetric with respect to the y-axis. 44. a. 55. Intercepts: ( −2, 0 ) , ( 0, −3) , and ( 2, 0 ) b. Symmetric with respect to the y-axis. 45. a. Intercepts: ( 0, 0 ) b. Symmetric with respect to the x-axis. 46. a. Intercepts: ( −2, 0 ) , ( 0, 2 ) , ( 0, −2 ) , and ( 2, 0 ) 56. b. Symmetric with respect to the x-axis, y-axis, and the origin. 47. a. Intercepts: ( −2, 0 ) , ( 0, 0 ) , and ( 2, 0 ) b. Symmetric with respect to the origin. 48. a. Intercepts: ( −4, 0 ) , ( 0, 0 ) , and ( 4, 0 ) b. Symmetric with respect to the origin. 163 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs Test y-axis symmetry: Let x = − x 57. y 2 = x + 4 x-intercepts: 02 = x + 4 −4 = x y-intercepts: y2 = 0 + 4 y2 = 4 y = ±2 The intercepts are ( −4, 0 ) , ( 0, −2 ) and ( 0, 2 ) . Test x-axis symmetry: Let y = − y ( − y )2 = x + 4 y 2 = x + 4 same Test y-axis symmetry: Let x = − x y 2 = − x + 4 different Test origin symmetry: Let x = − x and y = − y . ( − y )2 = − x + 4 Test origin symmetry: Let x = − x and y = − y − y = 3 −x = −3 x y = 3 x same Therefore, the graph will have origin symmetry. 60. y = 5 x x-intercepts: y-intercepts: 3 y=50 =0 0= x 0=x The only intercept is ( 0, 0 ) . Test x-axis symmetry: Let y = − y − y = 5 x different y 2 = − x + 4 different Therefore, the graph will have x-axis symmetry. 58. y 2 = x + 9 x-intercepts: (0) 2 = − x + 9 0 = −x + 9 x=9 y = 3 − x = − 3 x different Test y-axis symmetry: Let x = − x y = 5 − x = − 5 x different Test origin symmetry: Let x = − x and y = − y y-intercepts: y2 = 0 + 9 y2 = 9 y = ±3 The intercepts are ( −9, 0 ) , ( 0, −3) and ( 0,3) . Test x-axis symmetry: Let y = − y 2 (− y) = x + 9 y 2 = x + 9 same − y = 5 −x = −5 x y = 5 x same Therefore, the graph will have origin symmetry. 61. x 2 + y − 9 = 0 x-intercepts: x2 − 9 = 0 x2 = 9 y-intercepts: 02 + y − 9 = 0 y=9 y = − x + 9 different x = ±3 The intercepts are ( −3, 0 ) , ( 3, 0 ) , and ( 0,9 ) . Test origin symmetry: Let x = − x and y = − y . Test x-axis symmetry: Let y = − y ( − y )2 = − x + 9 x 2 − y − 9 = 0 different Test y-axis symmetry: Let x = − x 2 y 2 = − x + 9 different Test y-axis symmetry: Let x = − x Therefore, the graph will have x-axis symmetry. 59. y = 3 x x-intercepts: y-intercepts: 3 y=30 =0 0= x 0=x The only intercept is ( 0, 0 ) . ( − x )2 + y − 9 = 0 x 2 + y − 9 = 0 same Test origin symmetry: Let x = − x and y = − y ( − x )2 − y − 9 = 0 x 2 − y − 9 = 0 different Therefore, the graph will have y-axis symmetry. Test x-axis symmetry: Let y = − y − y = 3 x different 164 Copyright © 2016 Pearson Education, Inc. Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry 62. x 2 − y − 4 = 0 x-intercepts: y-intercept: x2 − 0 − 4 = 0 02 − y − 4 = 0 −y = 4 x2 = 4 y = −4 x = ±2 The intercepts are ( −2, 0 ) , ( 2, 0 ) , and ( 0, −4 ) . Test x-axis symmetry: Let y = − y x2 − ( − y ) − 4 = 0 64. 4 x 2 + y 2 = 4 x-intercepts: y-intercepts: 2 4 x 2 + 02 = 4 4 ( 0) + y2 = 4 4 x2 = 4 y2 = 4 y = ±2 x2 = 1 x = ±1 The intercepts are ( −1, 0 ) , (1, 0 ) , ( 0, −2 ) , and ( 0, 2 ) . x 2 + y − 4 = 0 different Test y-axis symmetry: Let x = − x ( − x )2 − y − 4 = 0 Test x-axis symmetry: Let y = − y 2 4 x2 + ( − y ) = 4 4 x 2 + y 2 = 4 same x 2 − y − 4 = 0 same Test origin symmetry: Let x = − x and y = − y Test y-axis symmetry: Let x = − x 2 ( − x )2 − ( − y ) − 4 = 0 4 ( −x) + y2 = 4 Therefore, the graph will have y-axis symmetry. Test origin symmetry: Let x = − x and y = − y 4 x 2 + y 2 = 4 same x 2 + y − 4 = 0 different 2 y-intercepts: 2 2 9 x 2 + 4 ( 0 ) = 36 9 ( 0 ) + 4 y 2 = 36 9 x = 36 x2 = 4 x = ±2 4 y = 36 y2 = 9 y = ±3 2 2 The intercepts are ( −2, 0 ) , ( 2, 0 ) , ( 0, −3) , and ( 0,3) . Test x-axis symmetry: Let y = − y 2 9 x 2 + 4 ( − y ) = 36 9 x 2 + 4 y 2 = 36 same Test y-axis symmetry: Let x = − x 2 9 ( − x ) + 4 y = 36 2 4 x 2 + y 2 = 4 same Therefore, the graph will have x-axis, y-axis, and origin symmetry. 65. y = x 3 − 27 x-intercepts: 0 = x3 − 27 y-intercepts: y = 03 − 27 y = −27 x3 = 27 x=3 The intercepts are ( 3, 0 ) and ( 0, −27 ) . Test x-axis symmetry: Let y = − y − y = x3 − 27 different Test y-axis symmetry: Let x = − x 9 x 2 + 4 y 2 = 36 same 3 Test origin symmetry: Let x = − x and y = − y 2 2 4(−x) + (− y) = 4 63. 9 x 2 + 4 y 2 = 36 x-intercepts: 2 9 ( − x ) + 4 ( − y ) = 36 y = ( − x ) − 27 y = − x3 − 27 different Test origin symmetry: Let x = − x and y = − y 9 x 2 + 4 y 2 = 36 same 3 Therefore, the graph will have x-axis, y-axis, and origin symmetry. − y = ( − x ) − 27 y = x3 + 27 different Therefore, the graph has none of the indicated symmetries. 165 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 66. y = x 4 − 1 x-intercepts: 0 = x4 − 1 Test x-axis symmetry: Let y = − y y-intercepts: y = 04 − 1 y = −1 x4 = 1 x = ±1 The intercepts are ( −1, 0 ) , (1, 0 ) , and ( 0, −1) . Test x-axis symmetry: Let y = − y − y = x 4 − 1 different Test y-axis symmetry: Let x = − x 4 y = (−x) −1 y = x 4 − 1 same Test origin symmetry: Let x = − x and y = − y 4 − y = (−x) −1 − y = x 4 − 1 different Therefore, the graph will have y-axis symmetry. 67. y = x 2 − 3x − 4 x-intercepts: 0 = x 2 − 3x − 4 0 = ( x − 4 )( x + 1) y-intercepts: y = 02 − 3 ( 0 ) − 4 y = −4 x = 4 or x = −1 The intercepts are ( 4, 0 ) , ( −1, 0 ) , and ( 0, −4 ) . Test x-axis symmetry: Let y = − y − y = x 2 − 3 x − 4 different Test y-axis symmetry: Let x = − x 2 − y = x 2 + 4 different Test y-axis symmetry: Let x = − x 2 y = (−x) + 4 y = x 2 + 4 same Test origin symmetry: Let x = − x and y = − y 2 − y = (−x) + 4 − y = x 2 + 4 different Therefore, the graph will have y-axis symmetry. 69. y = 3x 2 x +9 x-intercepts: y-intercepts: 3(0) 0 3x y= 2 = =0 0= 2 0 +9 9 x +9 3x = 0 x=0 The only intercept is ( 0, 0 ) . Test x-axis symmetry: Let y = − y 3x −y = different x +9 Test y-axis symmetry: Let x = − x 3( −x ) y= ( − x )2 + 9 y=− 2 3x x2 + 9 different y = ( −x) − 3( −x) − 4 Test origin symmetry: Let x = − x and y = − y y = x 2 + 3 x − 4 different −y = Test origin symmetry: Let x = − x and y = − y 2 − y = ( −x ) − 3( − x ) − 4 3( −x) ( − x )2 + 9 −y = − − y = x 2 + 3x − 4 different Therefore, the graph has none of the indicated symmetries. y= 3x 2 x +9 3x x2 + 9 Therefore, the graph has origin symmetry. 2 68. y = x + 4 x-intercepts: 0 = x2 + 4 same y-intercepts: y = 02 + 4 y=4 x 2 = −4 no real solution The only intercept is ( 0, 4 ) . 166 Copyright © 2016 Pearson Education, Inc. Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry x2 − 4 2x x-intercepts: x2 − 4 0= 2x 2 x −4 = 0 Test y-axis symmetry: Let x = − x 70. y = 3 y-intercepts: 02 − 4 −4 y= = 2 (0) 0 undefined 2 x =4 x = ±2 The intercepts are ( −2, 0 ) and ( 2, 0 ) . Test x-axis symmetry: Let y = − y x2 − 4 −y = different 2x Test y-axis symmetry: Let x = − x ( − x )2 − 4 y= 2(−x) y=− x −4 different 2x (−x) − 4 2(−x) x2 − 4 −2 x x2 − 4 y= same 2x Therefore, the graph has origin symmetry. −y = − x3 x2 − 9 x-intercepts: − x3 0= 2 x −9 3 −x = 0 y-intercepts: −03 0 y= 2 = =0 0 − 9 −9 Test x-axis symmetry: Let y = − y y= − x3 x2 − 9 x3 x2 − 9 ( − x )2 − 9 x3 Test origin symmetry: Let x = − x and y = − y −y = −y = y= − (−x) 3 ( − x )2 − 9 x3 x2 − 9 − x3 same x2 − 9 Therefore, the graph has origin symmetry. x4 + 1 0= y-intercepts: 04 + 1 1 y= = 5 0 2 ( 0) x4 + 1 2 x5 undefined x 4 = −1 no real solution There are no intercepts for the graph of this equation. Test x-axis symmetry: Let y = − y −y = x4 + 1 different 2 x5 Test y-axis symmetry: Let x = − x y= y= ( − x )4 + 1 5 2(−x) x4 + 1 −2 x5 −y = −y = different different x2 − 9 different Test origin symmetry: Let x = − x and y = − y x=0 The only intercept is ( 0, 0 ) . −y = − (−x) 2 x5 x-intercepts: 2 71. y = y= 72. y = 2 Test origin symmetry: Let x = − x and y = − y −y = y= y= ( − x )4 + 1 5 2(−x) x4 + 1 −2 x5 x4 + 1 2 x5 same Therefore, the graph has origin symmetry. 167 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 73. y = x3 77. If the point ( a, 4 ) is on the graph of y = x 2 + 3 x , then we have 4 = a 2 + 3a 0 = a 2 + 3a − 4 0 = ( a + 4 )( a − 1) a + 4 = 0 or a − 1 = 0 a = −4 a =1 Thus, a = −4 or a = 1 . 78. If the point ( a, −5 ) is on the graph of 74. x = y 2 y = x 2 + 6 x , then we have −5 = a 2 + 6a 0 = a 2 + 6a + 5 0 = ( a + 5 )( a + 1) a + 5 = 0 or a + 1 = 0 a = −5 a = −1 Thus, a = −5 or a = −1 . 79. For a graph with origin symmetry, if the point ( a, b ) is on the graph, then so is the point 75. y = x ( −a, −b ) . Since the point (1, 2 ) is on the graph of an equation with origin symmetry, the point ( −1, −2 ) must also be on the graph. 80. For a graph with y-axis symmetry, if the point ( a, b ) is on the graph, then so is the point ( −a, b ) . Since 6 is an x-intercept in this case, the point ( 6, 0 ) is on the graph of the equation. Due to the y-axis symmetry, the point ( −6, 0 ) must 76. y = 1 x also be on the graph. Therefore, −6 is another xintercept. 81. For a graph with origin symmetry, if the point ( a, b ) is on the graph, then so is the point ( −a, −b ) . Since −4 is an x-intercept in this case, the point ( −4, 0 ) is on the graph of the equation. Due to the origin symmetry, the point ( 4, 0 ) must also be on the graph. Therefore, 4 is another x-intercept. 82. For a graph with x-axis symmetry, if the point ( a, b ) is on the graph, then so is the point ( a, −b ) . Since 2 is a y-intercept in this case, the 168 Copyright © 2016 Pearson Education, Inc. Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry point ( 0, 2 ) is on the graph of the equation. Due Test origin symmetry: Let x = − x and y = − y (( − x ) + ( − y ) − ( − x )) = ( − x ) + ( − y ) to the x-axis symmetry, the point ( 0, −2 ) must 2 also be on the graph. Therefore, −2 is another yintercept. 83. a. ( x + y − x) = x + y 2 2 2 2 2 2 ( x − x) = x 2 2 2 84. a. 2 2 x 4 − 2 x3 = 0 x3 ( x − 2 ) = 0 x3 = 0 or x−2 = 0 x=0 x=2 (( 0) + y − 0) = ( 0) + y 2 2 (y ) = y 2 2 2 4 2 y =y 2 different 16 y 2 = 120 x − 225 x-intercepts: 16 y 2 = 120 ( 0 ) − 225 2 16 ( 0 ) = 120 x − 225 0 = 120 x − 225 −120 x = −225 −225 15 x= = −120 8 2  15  The only intercept is  , 0  . 8  4 y − y2 = 0 ( 2 y-intercepts: y-intercepts: 2 2 16 y 2 = −225 225 y2 = − 16 no real solution x 4 − 2 x3 + x 2 = x 2 2 2 2 Thus, the graph will have x-axis symmetry. ( x + ( 0) − x ) = x + ( 0) 2 2 ( x + y + x) = x + y 2 x-intercepts: 2 2 2 b. Test x-axis symmetry: Let y = − y ) 2 16 ( − y ) = 120 x − 225 y2 y2 −1 = 0 y 2 = 0 or y2 −1 = 0 16 y 2 = 120 x − 225 same y=0 y2 = 1 y = ±1 Test y-axis symmetry: Let x = − x 16 y 2 = 120 ( − x ) − 225 The intercepts are ( 0, 0 ) , ( 2, 0 ) , ( 0, −1) , 16 y 2 = −120 x − 225 different and ( 0,1) . Test origin symmetry: Let x = − x and y = − y 2 16 ( − y ) = 120 ( − x ) − 225 b. Test x-axis symmetry: Let y = − y ( x + (− y) − x) = x + (− y) 2 2 2 2 ( x + y − x) = x + y 2 2 2 2 2 16 y 2 = −120 x − 225 different 2 Thus, the graph will have x-axis symmetry. same 85. a. Test y-axis symmetry: Let x = − x (( − x ) + y − ( − x )) = ( − x ) + y 2 2 2 2 ( x + y + x) = x + y 2 2 2 2 2 2 different 169 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 88. Answers will vary 89. Answers will vary 90. Answers will vary. Case 1: Graph has x-axis and y-axis symmetry, show origin symmetry. ( x, y ) on graph → ( x, − y ) on graph (from x-axis symmetry) ( x, − y ) on graph → ( − x, − y ) on graph ( from y-axis symmetry ) Since the point ( − x, − y ) is also on the graph, the graph has origin symmetry. Case 2: Graph has x-axis and origin symmetry, show y-axis symmetry. ( x, y ) on graph → ( x, − y ) on graph b. Since x 2 = x for all x , the graphs of y = x 2 and y = x are the same. c. For y = ( x ) , the domain of the variable 2 x is x ≥ 0 ; for y = x , the domain of the variable x is all real numbers. Thus, ( x ) = x only for x ≥ 0. 2 d. For y = x 2 , the range of the variable y is y ≥ 0 ; for y = x , the range of the variable y is all real numbers. Also, if x ≥ 0 . Otherwise, x 2 = x only x2 = − x . 86. Answers will vary. A complete graph presents enough of the graph to the viewer so they can “see” the rest of the graph as an obvious continuation of what is shown. 87. Answers will vary. One example: y x ( from x-axis symmetry ) ( x, − y ) on graph → ( − x, y ) on graph ( from origin symmetry ) Since the point ( − x, y ) is also on the graph, the graph has y-axis symmetry. Case 3: Graph has y-axis and origin symmetry, show x-axis symmetry. ( x, y ) on graph → ( − x, y ) on graph ( from y-axis symmetry ) ( − x, y ) on graph → ( x, − y ) on graph ( from origin symmetry ) Since the point ( x, − y ) is also on the graph, the graph has x-axis symmetry. 91. Answers may vary. The graph must contain the points ( −2,5 ) , ( −1,3) , and ( 0, 2 ) . For the graph to be symmetric about the y-axis, the graph must also contain the points ( 2,5 ) and (1,3) (note that (0, 2) is on the y-axis). For the graph to also be symmetric with respect to the x-axis, the graph must also contain the points ( −2, −5 ) , ( −1, −3) , ( 0, −2 ) , ( 2, −5 ) , and (1, −3) . Recall that a graph with two of the symmetries (x-axis, y-axis, origin) will necessarily have the third. Therefore, if the original graph with y-axis symmetry also has x170 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines axis symmetry, then it will also have origin symmetry. 92. 6 + ( −2) 4 1 = = 6 − ( − 2) 8 2 93. 3x 2 − 30 x + 75 = 3( x 2 − 10 x + 25) = 3( x − 5)( x − 5) = 3( x − 5) 2 6. m1 = m2 ; y-intercepts; m1 ⋅ m2 = −1 7. 2 8. − 9. False; perpendicular lines have slopes that are opposite-reciprocals of each other. 10. d 94. −196 = ( −1)(196) = 14i 11. c 95. x2 − 8x + 4 = 0 12. b x 2 − 8 x = −4 x 2 − 8 x + 16 = −4 + 16 1 2 13. a. ( x − 4)2 = 12 Slope = b. If x increases by 2 units, y will increase by 1 unit. x − 4 = ± 12 x = 4 ± 12 = 4±2 3 1− 0 1 = 2−0 2 14. a. Slope = 1− 0 1 =− −2−0 2 b. If x increases by 2 units, y will decrease by 1 unit. Section 2.3 15. a. 1. undefined; 0 2. 3; 2 x-intercept: 2 x + 3(0) = 6 2x = 6 x=3 y-intercept: 2(0) + 3 y = 6 3y = 6 y=2 Slope = b. If x increases by 3 units, y will decrease by 1 unit. 16. a. Slope = 2 −1 1 = 2 − (−1) 3 b. If x increases by 3 units, y will increase by 1 unit. 3. True 4. False; the slope is 3 . 2 2 y = 3x + 5 3 5 y = x+ 2 2 1− 2 1 =− 1 − (− 2) 3 17. Slope = y2 − y1 0 − 3 3 = =− x2 − x1 4 − 2 2 ? 5. True; 2 (1) + ( 2 ) = 4 ? 2 + 2=4 4 = 4 True 171 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 18. Slope = y2 − y1 4 − 2 2 = = = −2 x2 − x1 3 − 4 −1 22. Slope = y2 − y1 2−2 0 = = =0 x2 − x1 −5 − 4 −9 19. Slope = y2 − y1 1− 3 −2 1 = = =− x2 − x1 2 − (− 2) 4 2 23. Slope = y2 − y1 −2−2 −4 = = undefined. x2 − x1 −1 − (−1) 0 20. Slope = y2 − y1 3 −1 2 = = x2 − x1 2 − (− 1) 3 24. Slope = y2 − y1 2 − 0 2 = = undefined. x2 − x1 2 − 2 0 21. Slope = y2 − y1 −1 − (−1) 0 = = =0 x2 − x1 2 − (−3) 5 25. P = (1, 2 ) ; m = 3 ; y − 2 = 3( x − 1) 172 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines 26. P = ( 2,1) ; m = 4 ; y − 1 = 4( x − 2) 30. P = ( 2, −4 ) ; m = 0 ; y = −4 3 3 27. P = ( 2, 4 ) ; m = − ; y − 4 = − ( x − 2) 4 4 31. P = ( 0,3) ; slope undefined ; x = 0 (note: the line is the y-axis) 28. P = (1,3) ; m = − 2 2 ; y − 3 = − ( x − 1) 5 5 29. P = ( −1, 3) ; m = 0 ; y − 3 = 0 32. P = ( −2, 0 ) ; slope undefined x = −2 4 ; point: (1, 2 ) 1 If x increases by 1 unit, then y increases by 4 units. Answers will vary. Three possible points are: x = 1 + 1 = 2 and y = 2 + 4 = 6 33. Slope = 4 = ( 2, 6 ) x = 2 + 1 = 3 and y = 6 + 4 = 10 ( 3,10 ) x = 3 + 1 = 4 and y = 10 + 4 = 14 ( 4,14 ) 173 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 2 ; point: ( −2,3) 1 If x increases by 1 unit, then y increases by 2 units. Answers will vary. Three possible points are: x = −2 + 1 = −1 and y = 3 + 2 = 5 34. Slope = 2 = ( −1,5) ( −1, −5 ) x = −1 + 1 = 0 and y = −5 − 2 = −7 ( 0, −7 ) x = 0 + 1 = 1 and y = −7 − 2 = −9 (1, −9 ) x = −1 + 1 = 0 and y = 5 + 2 = 7 ( 0, 7 ) −1 ; point: ( 4,1) 1 If x increases by 1 unit, then y decreases by 1 unit. Answers will vary. Three possible points are: x = 4 + 1 = 5 and y = 1 − 1 = 0 38. Slope = −1 = x = 0 + 1 = 1 and y = 7 + 2 = 9 (1,9 ) 3 −3 = ; point: ( 2, −4 ) 2 2 If x increases by 2 units, then y decreases by 3 units. Answers will vary. Three possible points are: x = 2 + 2 = 4 and y = −4 − 3 = −7 35. Slope = − ( 4, −7 ) ( 5, 0 ) x = 5 + 1 = 6 and y = 0 − 1 = −1 ( 6, −1) x = 6 + 1 = 7 and y = −1 − 1 = −2 x = 4 + 2 = 6 and y = −7 − 3 = −10 ( 6, −10 ) x = 6 + 2 = 8 and y = −10 − 3 = −13 (8, −13) 4 ; point: ( −3, 2 ) 3 If x increases by 3 units, then y increases by 4 units. Answers will vary. Three possible points are: x = −3 + 3 = 0 and y = 2 + 4 = 6 36. Slope = ( 0, 6 ) x = 0 + 3 = 3 and y = 6 + 4 = 10 ( 3,10 ) x = 3 + 3 = 6 and y = 10 + 4 = 14 ( 6,14 ) −2 ; point: ( −2, −3) 1 If x increases by 1 unit, then y decreases by 2 units. 37. Slope = −2 = Answers will vary. Three possible points are: x = −2 + 1 = −1 and y = −3 − 2 = −5 ( 7, −2 ) 39. (0, 0) and (2, 1) are points on the line. 1− 0 1 Slope = = 2−0 2 y -intercept is 0; using y = mx + b : 1 y = x+0 2 2y = x 0 = x − 2y 1 x − 2 y = 0 or y = x 2 40. (0, 0) and (–2, 1) are points on the line. 1− 0 1 1 Slope = = =− −2−0 −2 2 y -intercept is 0; using y = mx + b : 1 y = − x+0 2 2 y = −x x + 2y = 0 1 x + 2 y = 0 or y = − x 2 174 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines 41. (–1, 3) and (1, 1) are points on the line. 1− 3 −2 Slope = = = −1 1 − (−1) 2 Using y − y1 = m( x − x1 ) y − 1 = −1( x − 1) y −1 = −x +1 y = −x + 2 x + y = 2 or y = − x + 2 42. (–1, 1) and (2, 2) are points on the line. 2 −1 1 Slope = = 2 − (−1) 3 Using y − y1 = m( x − x1 ) 46. y − y1 = m( x − x1 ), m = 1 y − 1 = 1( x − (−1)) y −1 = x +1 y = x+2 x − y = − 2 or y = x + 2 47. Slope = 3; containing (–2, 3) y − y1 = m( x − x1 ) y − 3 = 3( x − (− 2)) y − 3 = 3x + 6 y = 3x + 9 3x − y = − 9 or y = 3 x + 9 48. Slope = 2; containing the point (4, –3) y − y1 = m( x − x1 ) 1 ( x − (−1) ) 3 1 y − 1 = ( x + 1) 3 1 1 y −1 = x + 3 3 1 4 y = x+ 3 3 y −1 = y − (−3) = 2( x − 4) y + 3 = 2x − 8 y = 2 x − 11 2 x − y = 11 or y = 2 x − 11 1 4 x − 3 y = − 4 or y = x + 3 3 2 49. Slope = − ; containing (1, –1) 3 y − y1 = m( x − x1 ) 2 ( x − 1) 3 2 2 y +1 = − x + 3 3 2 1 y = − x− 3 3 43. y − y1 = m( x − x1 ), m = 2 y − 3 = 2( x − 3) y − 3 = 2x − 6 y = 2x − 3 2 x − y = 3 or y = 2 x − 3 y − (−1) = − 44. y − y1 = m( x − x1 ), m = −1 y − 2 = −1( x − 1) y − 2 = −x +1 y = −x + 3 x + y = 3 or y = − x + 3 2 x + 3 y = −1 or y = − 45. y − y1 = m( x − x1 ), m = − 1 2 1 y − 2 = − ( x − 1) 2 1 1 y−2 = − x+ 2 2 1 5 y = − x+ 2 2 2 1 x− 3 3 1 50. Slope = ; containing the point (3, 1) 2 y − y1 = m( x − x1 ) 1 ( x − 3) 2 1 3 y −1 = x − 2 2 1 1 y = x− 2 2 y −1 = x − 2 y = 1 or y = 1 5 x + 2 y = 5 or y = − x + 2 2 175 Copyright © 2016 Pearson Education, Inc. 1 1 x− 2 2 Chapter 2: Graphs 51. Containing (1, 3) and (–1, 2) 2 − 3 −1 1 m= = = −1 − 1 − 2 2 56. x-intercept = –4; y-intercept = 4 Points are (–4, 0) and (0, 4) 4−0 4 m= = =1 0 − (− 4) 4 y = mx + b y = 1x + 4 y = x+4 x − y = − 4 or y = x + 4 y − y1 = m( x − x1 ) 1 ( x − 1) 2 1 1 y −3 = x− 2 2 1 5 y = x+ 2 2 y −3 = x − 2 y = − 5 or y = 57. Slope undefined; containing the point (2, 4) This is a vertical line. x=2 No slope-intercept form. 1 5 x+ 2 2 52. Containing the points (–3, 4) and (2, 5) 5−4 1 m= = 2 − (−3) 5 y − y1 = m( x − x1 ) 1 y − 5 = ( x − 2) 5 1 2 y −5 = x − 5 5 1 23 y = x+ 5 5 x − 5 y = − 23 or y = 58. Slope undefined; containing the point (3, 8) This is a vertical line. x=3 No slope-intercept form. 59. Horizontal lines have slope m = 0 and take the form y = b . Therefore, the horizontal line passing through the point ( −3, 2 ) is y = 2 . 60. Vertical lines have an undefined slope and take the form x = a . Therefore, the vertical line passing through the point ( 4, −5 ) is x = 4 . 1 23 x+ 5 5 53. Slope = –3; y-intercept =3 y = mx + b y = −3 x + 3 3x + y = 3 or y = −3x + 3 54. Slope = –2; y-intercept = –2 y = mx + b y = − 2 x + (− 2) 2 x + y = − 2 or y = − 2 x − 2 55. x-intercept = 2; y-intercept = –1 Points are (2,0) and (0,–1) −1 − 0 −1 1 m= = = 0−2 −2 2 y = mx + b 1 y = x −1 2 1 x − 2 y = 2 or y = x − 1 2 61. Parallel to y = 2 x ; Slope = 2 Containing (–1, 2) y − y1 = m( x − x1 ) y − 2 = 2( x − (−1)) y − 2 = 2x + 2 → y = 2x + 4 2 x − y = − 4 or y = 2 x + 4 62. Parallel to y = −3 x ; Slope = –3; Containing the point (–1, 2) y − y1 = m( x − x1 ) y − 2 = −3( x − (− 1)) y − 2 = −3 x − 3 → y = −3x − 1 3x + y = −1 or y = −3 x − 1 176 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines 63. Parallel to 2 x − y = − 2 ; Slope = 2 Containing the point (0, 0) y − y1 = m( x − x1 ) y − 0 = 2( x − 0) y = 2x 2 x − y = 0 or y = 2 x 69. Perpendicular to 2 x + y = 2 ; Containing the point (–3, 0) 1 Slope of perpendicular = 2 y − y1 = m( x − x1 ) 1 1 3 ( x − (−3)) → y = x + 2 2 2 1 3 x − 2 y = − 3 or y = x + 2 2 y−0 = 64. Parallel to x − 2 y = − 5 ; 1 Slope = ; Containing the point ( 0, 0 ) 2 y − y1 = m( x − x1 ) 1 1 ( x − 0) → y = x 2 2 1 x − 2 y = 0 or y = x 2 y−0 = 65. Parallel to x = 5 ; Containing (4,2) This is a vertical line. x = 4 No slope-intercept form. 66. Parallel to y = 5 ; Containing the point (4, 2) This is a horizontal line. Slope = 0 y=2 1 x + 4; Containing (1, –2) 2 Slope of perpendicular = –2 y − y1 = m( x − x1 ) 67. Perpendicular to y = 70. Perpendicular to x − 2 y = −5 ; Containing the point (0, 4) Slope of perpendicular = –2 y = mx + b y = −2 x + 4 2 x + y = 4 or y = −2 x + 4 71. Perpendicular to x = 8 ; Containing (3, 4) Slope of perpendicular = 0 (horizontal line) y=4 72. Perpendicular to y = 8 ; Containing the point (3, 4) Slope of perpendicular is undefined (vertical line). x = 3 No slope-intercept form. 73. y = 2 x + 3 ; Slope = 2; y-intercept = 3 y − (− 2) = − 2( x − 1) y + 2 = − 2x + 2 → y = − 2x 2 x + y = 0 or y = − 2 x 68. Perpendicular to y = 2 x − 3 ; Containing the point (1, –2) 1 Slope of perpendicular = − 2 y − y1 = m( x − x1 ) 74. y = −3 x + 4 ; Slope = –3; y-intercept = 4 1 y − (− 2) = − ( x − 1) 2 1 1 1 3 y+2= − x+ → y = − x− 2 2 2 2 1 3 x + 2 y = −3 or y = − x − 2 2 177 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 75. 76. 1 y = x −1 ; y = 2x − 2 2 Slope = 2; y-intercept = –2 1 79. x + 2 y = 4 ; 2 y = − x + 4 → y = − x + 2 2 1 Slope = − ; y-intercept = 2 2 1 1 x+ y = 2; y = − x+2 3 3 1 Slope = − ; y-intercept = 2 3 77. y = 80. − x + 3 y = 6 ; 3 y = x + 6 → y = Slope = 1 x+2 3 1 ; y-intercept = 2 3 1 1 x + 2 ; Slope = ; y-intercept = 2 2 2 81. 2 x − 3 y = 6 ; −3 y = − 2 x + 6 → y = Slope = 78. y = 2 x + 2 ; y-intercept = –2 3 1 1 ; Slope = 2; y -intercept = 2 2 178 Copyright © 2016 Pearson Education, Inc. 2 x−2 3 Section 2.3: Lines 3 82. 3 x + 2 y = 6 ; 2 y = − 3 x + 6 → y = − x + 3 2 3 Slope = − ; y-intercept = 3 2 86. y = −1 ; Slope = 0; y-intercept = –1 87. y = 5 ; Slope = 0; y-intercept = 5 83. x + y = 1 ; y = − x + 1 Slope = –1; y-intercept = 1 88. x = 2 ; Slope is undefined y-intercept – none 84. x − y = 2 ; y = x − 2 Slope = 1; y-intercept = –2 89. y − x = 0 ; y = x Slope = 1; y-intercept = 0 85. x = − 4 ; Slope is undefined y-intercept – none 179 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 90. x + y = 0 ; y = − x Slope = –1; y-intercept = 0 93. a. x-intercept: 2 x + 3 ( 0 ) = 6 2x = 6 x=3 The point ( 3, 0 ) is on the graph. y-intercept: 2 ( 0 ) + 3 y = 6 3y = 6 y=2 The point ( 0, 2 ) is on the graph. y b. 91. 2 y − 3 x = 0 ; 2 y = 3x → y = 5 3 x 2 (0, 2) 3 Slope = ; y-intercept = 0 2 (3, 0) 5 −5 x −5 94. a. x-intercept: 3x − 2 ( 0 ) = 6 3x = 6 x=2 The point ( 2, 0 ) is on the graph. y-intercept: 3 ( 0 ) − 2 y = 6 3 92. 3x + 2 y = 0 ; 2 y = −3 x → y = − x 2 3 Slope = − ; y-intercept = 0 2 −2 y = 6 y = −3 The point ( 0, −3) is on the graph. y b. 5 (2, 0) 5 −5 (0, −3) −5 180 Copyright © 2016 Pearson Education, Inc. x Section 2.3: Lines 95. a. x-intercept: −4 x + 5 ( 0 ) = 40 97. a. x-intercept: 7 x + 2 ( 0 ) = 21 −4 x = 40 7 x = 21 x = −10 The point ( −10, 0 ) is on the graph. x=3 The point ( 3, 0 ) is on the graph. y-intercept: −4 ( 0 ) + 5 y = 40 y-intercept: 7 ( 0 ) + 2 y = 21 5 y = 40 y =8 2 y = 21 y= The point ( 0,8 ) is on the graph. 21 2  21  The point  0,  is on the graph.  2 b. b. 96. a. x-intercept: 6 x − 4 ( 0 ) = 24 6 x = 24 x=4 The point ( 4, 0 ) is on the graph. 98. a. x-intercept: 5 x + 3 ( 0 ) = 18 5 x = 18 x= y-intercept: 6 ( 0 ) − 4 y = 24 −4 y = 24 18 5  18  The point  , 0  is on the graph. 5  y = −6 The point ( 0, −6 ) is on the graph. y-intercept: 5 ( 0 ) + 3 y = 18 3 y = 18 b. y=6 The point ( 0, 6 ) is on the graph. b. 181 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 99. a. 1 1 x + (0) = 1 2 3 1 x =1 2 x=2 The point ( 2, 0 ) is on the graph. x-intercept: y-intercept: 1 1 ( 0) + y = 1 2 3 1 y =1 3 y=3 101. a. x-intercept: 0.2 x − 0.5 ( 0 ) = 1 0.2 x = 1 x=5 The point ( 5, 0 ) is on the graph. y-intercept: 0.2 ( 0 ) − 0.5 y = 1 −0.5 y = 1 y = −2 The point ( 0, −2 ) is on the graph. b. The point ( 0,3) is on the graph. b. 102. a. x-intercept: −0.3 x + 0.4 ( 0 ) = 1.2 −0.3x = 1.2 100. a. x = −4 The point ( −4, 0 ) is on the graph. 2 x-intercept: x − ( 0 ) = 4 3 x=4 The point ( 4, 0 ) is on the graph. 2 y=4 3 2 − y=4 3 y = −6 y-intercept: −0.3 ( 0 ) + 0.4 y = 1.2 0.4 y = 1.2 y=3 y-intercept: ( 0 ) − The point ( 0,3) is on the graph. b. The point ( 0, −6 ) is on the graph. b. 103. The equation of the x-axis is y = 0 . (The slope is 0 and the y-intercept is 0.) 104. The equation of the y-axis is x = 0 . (The slope is undefined.) 105. The slopes are the same but the y-intercepts are different. Therefore, the two lines are parallel. 182 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines 106. The slopes are opposite-reciprocals. That is, their product is −1 . Therefore, the lines are perpendicular. 107. The slopes are different and their product does not equal −1 . Therefore, the lines are neither parallel nor perpendicular. 108. The slopes are different and their product does not equal −1 (in fact, the signs are the same so the product is positive). Therefore, the lines are neither parallel nor perpendicular. 109. Intercepts: ( 0, 2 ) and ( −2, 0 ) . Thus, slope = 1. y = x + 2 or x − y = − 2 110. Intercepts: ( 0,1) and (1, 0 ) . Thus, slope = –1. y = − x + 1 or x + y = 1 115. P1 = ( −1, 0 ) , P2 = ( 2,3) , P3 = (1, −2 ) , P4 = ( 4,1) m12 = m34 = 3−0 3 1− 3 = = 1 ; m24 = = −1 ; 2 − ( −1) 3 4−2 1 − ( −2 ) 4 −1 = 3 −2 − 0 = 1 ; m13 = = −1 3 1 − ( −1) Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is −1 ). Therefore, the vertices are for a rectangle. 116. P1 = ( 0, 0 ) , P2 = (1,3) , P3 = ( 4, 2 ) , P4 = ( 3, −1) 3−0 2−3 1 =− ; = 3 ; m23 = 4 −1 3 1− 0 −1 − 2 −1 − 0 1 m34 = = 3 ; m14 = =− 3− 4 3−0 3 m12 = d12 = (1 − 0 )2 + ( 3 − 0 )2 = 1 + 9 = 10 1 111. Intercepts: ( 3, 0 ) and ( 0,1) . Thus, slope = − . 3 1 y = − x + 1 or x + 3 y = 3 3 d 23 = ( 4 − 1)2 + ( 2 − 3)2 = 9 + 1 = 10 d34 = ( 3 − 4 )2 + ( −1 − 2 )2 = 1 + 9 = 10 d14 = ( 3 − 0 )2 + ( −1 − 0 )2 = 9 + 1 = 10 112. Intercepts: ( 0, −1) and ( −2, 0 ) . Thus, Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is −1 ). In addition, the length of all four sides is the same. Therefore, the vertices are for a square. 1 slope = − . 2 1 y = − x − 1 or x + 2 y = − 2 2 5−3 2 2 = =− −2 − 1 −3 3 3−0 3 = P2 = (1,3) , P3 = ( −1, 0 ) : m2 = 1 − ( −1) 2 113. P1 = ( −2,5 ) , P2 = (1,3) : m1 = Since m1 ⋅ m2 = −1 , the line segments P1 P2 and P2 P3 are perpendicular. Thus, the points P1 , P2 , and P3 are vertices of a right triangle. 114. P1 = (1, −1) , P2 = ( 4,1) , P3 = ( 2, 2 ) , P4 = ( 5, 4 ) 1 − ( −1) 2 4 −1 = ; m24 = = 3; 4 −1 3 5−4 2 − ( −1) 4−2 2 =3 m34 = = ; m13 = 5−2 3 2 −1 Each pair of opposite sides are parallel (same slope) and adjacent sides are not perpendicular. Therefore, the vertices are for a parallelogram. m12 = 117. Let x = number of miles driven, and let C = cost in dollars. Total cost = (cost per mile)(number of miles) + fixed cost C = 0.60 x + 39 When x = 110, C = ( 0.60)(110) + 39 = $105.00 . When x = 230, C = ( 0.60)( 230) + 39 = $177.00 . 118. Let x = number of pairs of jeans manufactured, and let C = cost in dollars. Total cost = (cost per pair)(number of pairs) + fixed cost C = 8 x + 500 When x = 400, C = ( 8 )( 400 ) + 500 = $3700 . When x = 740, C = ( 8 )( 740 ) + 500 = $6420 . 119. Let x = number of miles driven annually, and let C = cost in dollars. Total cost = (approx cost per mile)(number of miles) + fixed cost C = 0.17 x + 4462 183 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 120. Let x = profit in dollars, and let S = salary in dollars. Weekly salary = (% share of profit)(profit) + weekly pay S = 0.05 x + 375 121. a. C = 0.0821x + 15.37 ; 0 ≤ x ≤ 800 b. c. For 200 kWh, C = 0.0821(200) + 15.37 = $31.79 d. For 500 kWh, C = 0.0821(500) + 15.37 = $56.42 e. For each usage increase of 1 kWh, the monthly charge increases by $0.0821 (that is, 8.21 cents). 122. a. 123. (°C , ° F ) = (0, 32); (°C , ° F ) = (100, 212) 212 − 32 180 9 = = 100 − 0 100 5 9 ° F − 32 = (°C − 0) 5 9 ° F − 32 = (°C ) 5 5 °C = (° F − 32) 9 If ° F = 70 , then 5 5 °C = (70 − 32) = (38) 9 9 °C ≈ 21.1° slope = 124. a. b. C = 0.0907 x + 7.24 ; 0 ≤ x ≤ 1000 b. 125. a. K =º C + 273 5 º C = (º F − 32) 9 5 K = (° F − 32) + 273 9 5 160 + 273 K = ºF − 9 9 5 2297 K = ºF + 9 9 The y-intercept is (0, 30), so b = 30. Since the ramp drops 2 inches for every 25 inches −2 2 of run, the slope is m = = − . Thus, 25 25 2 the equation is y = − x + 30 . 25 b. Let y = 0. c. 0=− For 200 kWh, C = 0.0907 ( 200) + 7.24 = $25.38 d. For 500 kWh, C = 0.0907 (500) + 7.24 = $52.59 e. For each usage increase of 1 kWh, the monthly charge increases by $0.0907 (that is, 9.07 cents). 2 x + 30 25 2 x = 30 25 25  2  25  x  = ( 30 ) 2  25  2 x = 375 The x-intercept is (375, 0). This means that the ramp meets the floor 375 inches (or 31.25 feet) from the base of the platform. 184 Copyright © 2016 Pearson Education, Inc. Section 2.3: Lines c. No. From part (b), the run is 31.25 feet which exceeds the required maximum of 30 feet. d. First, design requirements state that the maximum slope is a drop of 1 inch for each 1 12 inches of run. This means m ≤ . 12 Second, the run is restricted to be no more than 30 feet = 360 inches. For a rise of 30 inches, this means the minimum slope is 30 1 1 . That is, m ≥ . Thus, the = 12 360 12 1 . The only possible slope is m = 12 diagram indicates that the slope is negative. Therefore, the only slope that can be used to obtain the 30-inch rise and still meet design 1 requirements is m = − . In words, for 12 every 12 inches of run, the ramp must drop exactly 1 inch. 126. a. The year 2000 corresponds to x = 0, and the year 2012 corresponds to x = 12. Therefore, the points (0, 20.6) and (12, 9.3) are on the line. Thus, 9.3 − 20.6 11.3 m= =− = −0.942 . The y12 − 0 12 intercept is 20.6, so b = 20.6 and the equation is y = −0.942 x + 20.6 0 = −0.942 x + 20.6 0.942 x = 20.6 x = 21.9 y-intercept: y = −0.942 ( 0) + 20.6 = 20.6 b. x-intercept: The intercepts are (21.9, 0) and (0, 20.6). c. The y-intercept represents the percentage of twelfth graders in 2000 who had reported daily use of cigarettes. The x-intercept represents the number of years after 2000 when 0% of twelfth graders will have reported daily use of cigarettes. d. The year 2025 corresponds to x = 25. y = −0.942 ( 25) + 20.6 = −2.95 This prediction is not reasonable. 127. a. ( x2 , A2 ) = (200, 000, 60, 000) 60, 000 − 40, 000 200, 000 − 100, 000 20, 000 1 = = 100, 000 5 1 A − 40, 000 = ( x − 100, 000 ) 5 1 A − 40, 000 = x − 20, 000 5 1 A = x + 20, 000 5 slope = b. If x = 300,000, then 1 A = ( 300, 000 ) + 20, 000 = $80, 000 5 c. Each additional box sold requires an additional $0.20 in advertising. 128. Find the slope of the line containing ( a, b ) and ( b, a ) : a−b = −1 b−a The slope of the line y = x is 1. slope = Since −1 ⋅1 = −1 , the line containing the points (a, b) and (b, a) is perpendicular to the line y = x. The midpoint of (a, b) and (b, a) is  a+b b+a  M = , . 2   2 Since the coordinates are the same, the midpoint lies on the line y = x . Note: a+b b+a = 2 2 129. 2x − y = C Graph the lines: 2x − y = − 4 2x − y = 0 2x − y = 2 All the lines have the same slope, 2. The lines Let x = number of boxes to be sold, and A = money, in dollars, spent on advertising. We have the points ( x1 , A1 ) = (100, 000, 40, 000); 185 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs are parallel. slope 1 and y-intercept (0, −1). Thus, the lines are parallel with positive slopes. One line has a positive y-intercept and the other with a negative y-intercept. 134. (d) The equation y − 2 x = 2 has slope 2 and yintercept (0, 2). The equation x + 2 y = −1 has 1 1  and y-intercept  0, −  . The lines 2 2   1 are perpendicular since 2  −  = −1 . One line  2 has a positive y-intercept and the other with a negative y-intercept. slope − 130. Refer to Figure 47. length of OA = d ( O, A ) = 1 + m12 length of OB = d ( O, B ) = 1 + m2 2 135 – 137. Answers will vary. 138. No, the equation of a vertical line cannot be written in slope-intercept form because the slope is undefined. length of AB = d ( A, B ) = m1 − m2 Now consider the equation ( 1+ m ) + ( 1+ m ) = (m − m ) 2 1 2 2 2 2 1 2 2 If this equation is valid, then ΔAOB is a right triangle with right angle at vertex O. ( 1 + m12 ) ( 2 + 1 + m2 2 ) = (m − m ) 2 1 2 2 1 + m12 + 1 + m2 2 = m12 − 2m1m2 + m2 2 2 + m12 + m2 2 = m12 − 2m1m2 + m2 2 139. No, a line does not need to have both an xintercept and a y-intercept. Vertical and horizontal lines have only one intercept (unless they are a coordinate axis). Every line must have at least one intercept. 140. Two lines with equal slopes and equal y-intercepts are coinciding lines (i.e. the same). 2 + m12 + m2 2 = m12 − 2 ( −1) + m2 2 141. Two lines that have the same x-intercept and yintercept (assuming the x-intercept is not 0) are the same line since a line is uniquely defined by two distinct points. 2 + m12 + m2 2 = m12 + 2 + m2 2 0=0 Therefore, by the converse of the Pythagorean Theorem, ΔAOB is a right triangle with right angle at vertex O. Thus Line 1 is perpendicular to Line 2. 142. No. Two lines with the same slope and different xintercepts are distinct parallel lines and have no points in common. Assume Line 1 has equation y = mx + b1 and Line 2 has equation y = mx + b2 , But we are assuming that m1m2 = −1 , so we have 131. (b), (c), (e) and (g) The line has positive slope and positive y-intercept. 132. (a), (c), and (g) The line has negative slope and positive y-intercept. b Line 1 has x-intercept − 1 and y-intercept b1 . m b2 and y-intercept b2 . Line 2 has x-intercept − m Assume also that Line 1 and Line 2 have unequal x-intercepts. If the lines have the same y-intercept, then b1 = b2 . 133. (c) The equation x − y = −2 has slope 1 and yintercept (0, 2). The equation x − y = 1 has 186 Copyright © 2016 Pearson Education, Inc. Section 2.4: Circles b b b b b1 = b2  1 = 2  − 1 = − 2 m m m m b1 b2 But − = −  Line 1 and Line 2 have the m m same x-intercept, which contradicts the original assumption that the lines have unequal x-intercepts. Therefore, Line 1 and Line 2 cannot have the same y-intercept. 143. Yes. Two distinct lines with the same y-intercept, but different slopes, can have the same x-intercept if the x-intercept is x = 0 . Assume Line 1 has equation y = m1 x + b and Line 2 has equation y = m2 x + b , Line 1 has x-intercept − b and y-intercept b . m1 146. = x4 y5 = −2 x2 x 4 y5 y3 −2 1 x 2 y8 x 2 y8 1 2 = x 4 y16 147. h 2 = a 2 + b 2 = 82 + 152 = 16 + 225 = 289 h = 289 = 17 148. b and y-intercept b . m2 Assume also that Line 1 and Line 2 have unequal slopes, that is m1 ≠ m2 . If the lines have the same x-intercept, then b b − =− . m1 m2 b b =− m1 m2 − m2 b = −m1b −m2b + m1b = 0 −2 = Line 2 has x-intercept − ( x − 3)2 + 25 = 49 ( x − 3)2 = 24 x − 3 = ± 24 x − 3 = ±2 6 x = 3± 2 6 { 2 x − 5 + 7 < 10 2x − 5 < 3 −3 < 2 x − 5 < 3 2 < 2x < 8 But − m2 b + m1b = 0  b ( m1 − m2 ) = 0 b=0 or m1 − m2 = 0  m1 = m2 Since we are assuming that m1 ≠ m2 , the only way that the two lines can have the same x-intercept is if b = 0. 1< x < 4 The solution set is: { x | 1 < x 0 and ( 0, 0 ) on the graph. 62. (b), (e) and (g) We need h r . 63. Answers will vary. 64. The student has the correct radius, but the signs of the coordinates of the center are incorrect. The student needs to write the equation in the standard form ( x − h ) + ( y − k ) = r 2 . 2 Center: ( 2, −3) 2 ( x + 3) + ( y − 2 ) = 16 2 x2 + y 2 + 6 x + 4 y + 9 = 0 2 ( x − ( −3)) + ( y − 2) = 4 2 ( x 2 + 6 x + 9) + ( y 2 + 4 y + 4) = − 9 + 9 + 4 ( x + 3) 2 + ( y + 2)2 = 4 2 2 Thus, ( h, k ) = ( −3, 2 ) and r = 4 . Center: ( −3, −2 ) Find the slope of the line containing the centers: − 2 − (−3) 1 m= =− −3 − 2 5 Find the equation of the line containing the centers: 1 y + 3 = − ( x − 2) 5 5 y + 15 = − x + 2 x + 5 y = −13 x + 5 y + 13 = 0 59. Consider the following diagram: C = 2π r 65. A = π r 2 = π (13) 2 = 169π cm 2 C = 2π r = 2π (13) = 26π cm 66. (3 x − 2)( x 2 − 2 x + 3) = 3 x 3 − 6 x 2 + 9 x − 2 x 2 + 4 x − 6 = 3 x3 − 8 x 2 + 13 x − 6 67. 2 x 2 + 3x − 1 = x + 1 2 x 2 + 3x − 1 = ( x + 1) 2 2 x 2 + 3x − 1 = x 2 + 2 x + 1 (2,2) x2 + x − 2 = 0 ( x + 2)( x − 1) = 0 x = −2 or x = 1 We need to check each possible solution: 198 Copyright © 2016 Pearson Education, Inc. Section 2.5: Variation Check x = −2 5. y = kx 2 = 10k 2 1 k= = 10 5 1 y= x 5 2 2( −2) + 3( −2) − 1 = ( −2) + 1 2(4) − 6 − 1 = −1 no Check x = 1 6. v = kt 16 = 2k 8=k v = 8t 2 2(1) + 3(1) − 1 = (1) + 1 2 + 3 −1 = 2 4=2 yes The solution is {1} 68. Let t represent the time it takes to do the job together. Part of job done Time to do job in one minute 1 Aaron 22 22 1 Elizabeth 28 28 1 Together t t 1 1 1 + = 22 28 t 14t + 11t = 308 25t = 308 t = 12.32 Working together, the job can be done in 12.32 minutes. Section 2.5 1. y = kx 2. False. If y varies directly with x, then y = kx, where k is a constant. 7. A = kx 2 4π = k (2) 2 4π = 4k π =k A = π x2 8. V = kx3 36π = k (3)3 36π = 27 k 36π 4 k= = π 27 3 4 3 V = πx 3 k d2 k 10 = 2 5 k 10 = 25 k = 250 250 F= 2 d 9. F = 10. y = 3. b 4. c 4= k x k 9 k 3 k = 12 12 y= x 4= 199 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs ( ) 5 = k (3 + 4 ) 11. z = k x 2 + y 2 2 ( 16 k 4= 2 k =8 ) d2 ( ) 2 = k (9 + 4 ) 16. z 3 = k x 2 + y 2 1= k ( )( ) 3 T = 3 x d2 2 2 8 = k ( 97 ) kd 2 8 97 8 2 z3 = x + y2 97 k= x ( ) ( k 42 9 16k 24 = 3  3 9 k = 24   =  16  2 17. V = 19. A = 2 x ( ) 1 = k (2 + 3 ) 14. z = k x3 + y 2 ) 4π 3 r 3 18. c 2 = a 2 + b 2 9d 2 3 8a 3 T2 = 2 18 = k (18 ) M = 42 k (8) 4= ( )( ) 18 = k ( 8 ) ( 3 ) 3 ( ) k 23 22 = 12. T = k 3 x d 2 24 = d2 2 5 = k (25) 5 1 k= = 25 5 1 z = x2 + y2 5 13. M = ka3 15. T 2 = 1 bh 2 20. p = 2 ( l + w ) 2 1 = k (17 )  mM  21. F = 6.67 × 10−11  2   d  1 17 1 3 z= x + y2 17 22. T = ( k= ( ) 23. ) 2π 32 l p = kB 6.49 = k (1000 ) 0.00649 = k Therefore we have the linear equation p = 0.00649 B . If B = 145000 , then p = 0.00649 (145000 ) = $941.05 . 200 Copyright © 2016 Pearson Education, Inc. Section 2.5: Variation If R = 576, then 12, 288 576 = l 576l = 12, 288 12, 288 64 inches l= = 576 3 p = kB 24. 8.99 = k (1000 ) 0.00899 = k Therefore we have the linear equation p = 0.00899 B . If B = 175000 , then p = 0.00899 (175000 ) = $1573.25 . 25. s = kt 16 = k (1) 47.40 = k (12 ) 3.95 = k Therefore, we have the linear equation R = 3.95 g . 2 k = 16 Therefore, we have equation s = 16t 2 . If g = 10.5 , then R = ( 3.95 )(10.5 ) ≈ $41.48 . 2 If t = 3 seconds, then s = 16 ( 3) = 144 feet. C = kA 30. 23.75 = k ( 5 ) If s = 64 feet, then 64 = 16t 2 t2 = 4 t = ±2 Time must be positive, so we disregard t = −2. It takes 2 seconds to fall 64 feet. 26. R = kg 29. 2 v = kt 4.75 = k Therefore, we have the linear equation C = 4.75 A. If A = 3.5 , then C = ( 4.75 )( 3.5 ) = $16.63 . D = 156 , p = 2.75 ; a. 64 = k ( 2 ) k 2.75 k = 429 429 . So, D = p 156 = k = 32 Therefore, we have the linear equation v = 32t. If t = 3 seconds, then v = 32 ( 3) = 96 ft/sec. 27. E = kW 3 = k ( 20 ) k= Therefore, we have the linear equation E = If W = 15, then E = 28. D= b. 3 20 3 W. 20 3 (15 ) = 2.25 . 20 k l k 256 = 48 k = 12, 288 k p 31. D = 32. t = 429 = 143 bags of candy 3 k s a. R= t = 40 , s = 30 ; k 40 = 30 k = 1200 So, we have the equation t = b. Therefore, we have the equation R = 12, 288 . l t= 1200 = 30 minutes 40 k P V = 600, P = 150 ; 33. V = 201 Copyright © 2016 Pearson Education, Inc. 1200 . s Chapter 2: Graphs k 150 k = 90, 000 600 = 39. I = So, we have the equation V = If P = 200 , then V = 90, 000 P 0.075 = 90, 000 = 450 cm3 . 200 If d = 5, then I = k and k = 240 . 8 240 . So, we have the equation i = R 240 If R = 10, then i = = 24 amperes . 10 If i = 30, R = 8 , then 30 = k 39602 1 Av 2 . 880 If A = 47.125 and v = 36.5 , then 1 F= ( 47.125)( 36.5)2 ≈ 71.34 pounds. 880 and k = 1,960, 200, 000 41. 45 = (0.06)(125)d 3 45 = 7.5d 3 2 6 = d3 k d = 3 6 ≈ 1.82 inches 39602 k = 862, 488, 000 So, we have the equation W = 862, 488, 000 . d2 If d =3965, then 862, 488, 000 W= ≈ 54.86 pounds. 39652 37. V = π r 2 h π 3 h = ksd 3 36 = k (75)(2)3 36 = 600k 0.06 = k So, we have the equation h = 0.06sd 3 . If h = 45 and s = 125, then k 55 = = 0.012 foot-candles. So, we have the equation F = 1,960, 200, 000 . So, we have the equation W = d At the top of Mt. McKinley, we have d = 3960 + 3.8 = 3963.8 , so 1,960, 200, 000 W= ≈ 124.76 pounds. ( 3963.8)2 d 52 11 = k (20)(22) 2 11 = 9860k 11 1 k= = 9680 880 k d2 If W = 125, d = 3960 then 125 = 0.3 0.3 . d2 40. F = kAv 2 35. W = 38. V = k and k = 0.3 . 22 So, we have the equation I = k 34. i = R 36. W = k d2 If I = 0.075, d = 2 , then 42. kT P k (300) 100 = 15 100 = 20k 5=k V= So, we have the equation V = r 2h 202 Copyright © 2016 Pearson Education, Inc. 5T . P Section 2.5: Variation If V = 80 and T = 310, then 5(310) 80 = P 80 P = 1550 1550 P= = 19.375 atmospheres 80 43. 46. 2 K = 0.5 ( 25 )(15 ) = 2812.5 Joules 44. 375wt 2 . l If l = 10, w = 6, and t = 2, then K = kmv 2 1250 = k (25)(10) 2 1250 = 2500k k = 0.5 So, we have the equation K = 0.5mv 2 . If m = 25 and v = 15, then R= 1.24 = So, we have the equation S = S= 51. 3×3 + 25 x 2 − 12 x − 100 = (3 x3 + 25 x 2 ) − (12 x + 100) 2 d k ( 432 ) = x 2 (3 x + 25) − 4(3 x + 25) = ( x 2 − 4)(3 x + 25) = ( x − 2)( x + 2)(3 x + 25) 2 52. S= 5 x−2 5 x−2 + 2 = + x + 3 x + 7 x + 12 x + 3 ( x + 3)( x + 4) x−2 5( x + 4) = + ( x + 3)( x + 4) ( x + 3)( x + 4) 5( x + 4) + ( x − 2) = ( x + 3)( x + 4) 5 x + 20 + x − 2 = ( x + 3)( x + 4) 6 x + 18 = ( x + 3)( x + 4) 6( x + 3) 6 = = ( x + 3)( x + 4) ( x + 4) 0.6 = k 4 2 = 25 So, we have the equation R = 1.24l 27 d . 2 If R = 1.44 and d = 3, then 1.24l 1.44 = 27(3) 2 1.24l 1.44 = 243 349.92 = 1.24l 349.92 ≈ 282.2 feet l= 1.24 45. 375(6)(2) 2 = 900 pounds. 10 47 – 50. Answers will vary. kl (4) 1.24 = 27 k 1.24 k= 27 kwt 2 l k (4)(2)2 750 = 8 750 = 2k 375 = k S= kpd t k (25)(5) 100 = 0.75 75 = 125k 3 53. 0.6 pd . t If p = 40, d = 8, and t = 0.50, then So, we have the equation S = S= 0.6(40)(8) = 384 psi. 0.50 2 5 3 = 1 3 4 2 25 8 125 54. The term needed to rationalize the denominator is 7 + 2 . 203 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs Chapter 2 Review Exercises 4. y = x 2 + 4 1. P1 = ( 0, 0 ) and P2 = ( 4, 2 ) a. d ( P1 , P2 ) = (−1, 5) = 16 + 4 = 20 = 2 5 b. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2 0+4 0+2  4 2 , =  =  ,  = ( 2,1) 2  2 2  2 c. Δy 2 − 0 2 1 = = = slope = Δx 4 − 0 4 2 d. For each run of 2, there is a rise of 1. 2. P1 = (1, −1) and P2 = ( −2,3) a. d ( P1 , P2 ) = ( −2 − 1)2 + ( 3 − ( −1) ) 2 = 9 + 16 = 25 = 5 b. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2  1 + ( −2 ) −1 + 3  = ,  2   2  −1 2   1  =  ,  =  − ,1  2 2  2  c. slope = 3. P1 = ( 4, −4 ) and P2 = ( 4,8 ) −1 5 x 5. x-intercepts: −4, 0, 2 ; y-intercepts: −2, 0, 2 Intercepts: (−4, 0), (0, 0), (2, 0), (0, −2), (0, 2) 6. 2 x = 3 y 2 x-intercepts: y-intercepts: 2 2 x = 3(0) 2(0) = 3 y 2 2x = 0 0 = y2 y=0 x=0 The only intercept is (0, 0). Test x-axis symmetry: Let y = − y 2 x = 3(− y ) 2 2 x = 3 y 2 same Test y-axis symmetry: Let x = − x 2(− x) = 3 y 2 7. x 2 +4 y 2 =16 x-intercepts: ( 4 − 4 )2 + (8 − ( −4 ) ) 2 = 0 + 144 = 144 = 12 b. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2  4 + 4 −4 + 8   8 4  , =  =  ,  = ( 4, 2 ) 2  2 2  2 c. (1, 5) (0, 4) 2(− x) = 3(− y ) 2 −2 x = 3 y 2 different Therefore, the graph will have x-axis symmetry. Δy 3 − ( −1) 4 4 = = =− Δx −2 − 1 −3 3 d ( P1 , P2 ) = −5 (2, 8) −2 x = 3 y 2 different Test origin symmetry: Let x = − x and y = − y . d. For each run of 3, there is a rise of −4. a. 9 (−2, 8) ( 4 − 0 )2 + ( 2 − 0 )2 y Δy 8 − ( −4 ) 12 slope = = = , undefined Δx 4−4 0 2 y-intercepts: x +4 ( 0 ) =16 ( 0 )2 +4 y 2 =16 x 2 = 16 4 y 2 = 16 x = ±4 y2 = 4 y = ±2 2 The intercepts are (−4, 0), (4, 0), (0, −2), and (0, 2). Test x-axis symmetry: Let y = − y 2 x 2 + 4 ( − y ) =16 x 2 + 4 y 2 =16 same d. An undefined slope means the points lie on a vertical line. There is no change in x. 204 Copyright © 2016 Pearson Education, Inc. Chapter 2 Review Exercises Test y-axis symmetry: Let x = − x 2 Test y-axis symmetry: Let x = − x y = ( − x )3 − ( − x ) ( − x ) + 4 y =16 2 y = − x3 + x x 2 + 4 y 2 =16 same Test origin symmetry: Let x = − x and y = − y . Test origin symmetry: Let x = − x and y = − y . − y = ( − x )3 − ( − x ) − y = − x3 + x y = x3 − x same ( − x )2 + 4 ( − y )2 =16 x 2 +4 y 2 =16 same Therefore, the graph will have x-axis, y-axis, and origin symmetry. x-intercepts: 0 = x 4 +2 x 2 +1 ( Therefore, the graph will have origin symmetry. 10. x 2 + x + y 2 + 2 y = 0 8. y = x 4 +2 x 2 +1 )( x-intercepts: x 2 + x + (0) 2 + 2(0) = 0 x2 + x = 0 x( x + 1) = 0 x = 0, x = −1 y-intercepts: y = (0) 4 +2(0) 2 +1 =1 ) 0 = x2 + 1 x2 + 1 2 x +1 = 0 y-intercepts: (0) 2 + 0 + y 2 + 2 y = 0 y2 + 2 y = 0 y ( y + 2) = 0 y = 0, y = −2 The intercepts are (−1, 0), (0, 0), and (0, −2). x 2 = −1 no real solutions The only intercept is (0, 1). Test x-axis symmetry: Let y = − y − y = x4 + 2 x2 + 1 y = − x 4 − 2 x 2 − 1 different Test x-axis symmetry: Let y = − y x 2 + x + (− y ) 2 + 2(− y ) = 0 Test y-axis symmetry: Let x = − x 4 x 2 + x + y 2 − 2 y = 0 different Test y-axis symmetry: Let x = − x (− x) 2 + (− x) + y 2 + 2 y = 0 x 2 − x + y 2 + 2 y = 0 different Test origin symmetry: Let x = − x and y = − y . 2 y = (−x) + 2(−x) +1 y = x4 + 2 x2 + 1 same Test origin symmetry: Let x = − x and y = − y . 4 2 − y = (−x) + 2(−x) +1 − y = x4 + 2 x2 + 1 y = − x4 − 2 x2 − 1 (− x) 2 + (− x) + (− y ) 2 + 2(− y ) = 0 different x 2 − x + y 2 − 2 y = 0 different The graph has none of the indicated symmetries. Therefore, the graph will have y-axis symmetry. 11. 9. y = x3 − x x-intercepts: 0 = x3 − x ( 2 different ) 0 = x x −1 ( x − ( −2 ) ) + ( y − 3)2 = 42 2 y-intercepts: y = (0)3 − 0 =0 0 = x ( x + 1)( x − 1) ( x + 2 )2 + ( y − 3)2 = 16 12. x = 0, x = −1, x = 1 ( x − h) 2 + ( y − k ) 2 = r 2 ( x − h) 2 + ( y − k ) 2 = r 2 ( x − ( −1) ) + ( y − ( −2 ) ) = 12 2 The intercepts are (−1, 0), (0, 0), and (1, 0). Test x-axis symmetry: Let y = − y 2 ( x + 1)2 + ( y + 2 )2 = 1 − y = x3 − x y = − x 3 + x different 205 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 2 x-intercepts: ( x − 1) + ( 0 + 2 ) = 32 2 ( x − 1)2 + 4 = 9 ( x − 1)2 = 5 2 13. x 2 + ( y − 1) = 4 x 2 + ( y − 1) = 22 Center: (0,1); Radius = 2 2 x −1 = ± 5 x = 1± 5 y-intercepts: ( 0 − 1) + ( y + 2 ) = 32 2 2 1 + ( y + 2) = 9 2 ( y + 2 )2 = 8 y+2= ± 8 2 x-intercepts: x 2 + ( 0 − 1) = 4 y + 2 = ±2 2 x2 + 1 = 4 ( 2 y-intercepts: 0 + ( y − 1) = 4 y − 1 = ±2 x2 + y2 − 2 x + 4 y = 0 y = 1± 2 y = 3 or y = −1 x2 − 2x + y2 + 4 y = 0 ) ( 3, 0) , ( 0, −1) , The intercepts are − 3, 0 , 2 3x 2 + 3 y 2 − 6 x + 12 y = 0 15. ( y − 1) 2 = 4 ( ) ( 0, −2 − 2 2 ) , and ( 0, −2 + 2 2 ) . x=± 3 and ( 0, 3) . ) ( The intercepts are 1 − 5, 0 , 1 + 5, 0 , x2 = 3 2 y = −2 ± 2 2 ( x − 2 x + 1) + ( y + 4 y + 4) = 1 + 4 ( x − 1) + ( y + 2 ) = ( 5 ) 2 2 2 2 Center: (1, –2) Radius = 2 5 2 x + y − 2x + 4 y − 4 = 0 14. x2 − 2 x + y 2 + 4 y = 4 ( x − 2 x + 1) + ( y + 4 y + 4 ) = 4 + 1 + 4 2 2 ( x − 1)2 + ( y + 2 )2 = 32 Center: (1, –2) Radius = 3 2 2 x-intercepts: ( x − 1) + ( 0 + 2 ) = ( 5) 2 ( x − 1)2 + 4 = 5 ( x − 1)2 = 1 x − 1 = ±1 x = 1±1 x = 2 or x = 0 206 Copyright © 2016 Pearson Education, Inc. Chapter 2 Review Exercises 2 2 y-intercepts: ( 0 − 1) + ( y + 2 ) = ( 5) 2 2 1 + ( y + 2) = 5 ( y + 2 )2 = 4 y + 2 = ±2 y = −2 ± 2 y = 0 or y = −4 The intercepts are ( 0, 0 ) , ( 2, 0 ) , and ( 0, −4 ) . 20. Parallel to 2 x − 3 y = −4 2x − 3y = − 4 −3 y = −2 x − 4 −3 y −2 x − 4 = −3 −3 2 4 y = x+ 3 3 2 Slope = ; containing (–5,3) 3 y − y1 = m ( x − x1 ) 16. Slope = –2; containing (3,–1) y − y1 = m ( x − x1 ) 2 ( x − (−5) ) 3 2 y − 3 = ( x + 5) 3 2 10 y −3 = x+ 3 3 2 19 y = x+ or 2 x − 3 y = −19 3 3 y −3 = y − (−1) = −2 ( x − 3) y + 1 = −2 x + 6 y = −2 x + 5 or 2 x + y = 5 17. vertical; containing (–3,4) Vertical lines have equations of the form x = a, where a is the x-intercept. Now, a vertical line containing the point (–3, 4) must have an x-intercept of –3, so the equation of the line is x = −3. The equation does not have a slopeintercept form. 18. y-intercept = –2; containing (5,–3) Points are (5,–3) and (0,–2) 1 1 − 2 − (−3) m= = =− 0−5 −5 5 y = mx + b 21. Perpendicular to x + y = 2 x+ y = 2 y = −x + 2 The slope of this line is −1 , so the slope of a line perpendicular to it is 1. Slope = 1; containing (4,–3) y − y1 = m( x − x1 ) y − (−3) = 1( x − 4) y+3= x−4 1 y = − x − 2 or x + 5 y = −10 5 y = x − 7 or x − y = 7 19. Containing the points (3,–4) and (2, 1) 1 − (−4) 5 m= = = −5 2−3 −1 y − y1 = m ( x − x1 ) y − (− 4) = −5 ( x − 3) y + 4 = −5 x + 15 y = −5 x + 11 or 5 x + y = 11 207 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 22. 4 x − 5 y = − 20 −5 y = −4 x − 20 4 y = x+4 5 4 slope = ; y-intercept = 4 5 24. 2 x − 3 y = 12 x-intercept: 2 x − 3(0) = 12 2 x = 12 x=6 y-intercept: 2(0) − 3 y = 12 −3 y = 12 y = −4 The intercepts are ( 6, 0 ) and ( 0, −4 ) . x-intercept: Let y = 0. 4 x − 5(0) = − 20 4 x = − 20 x = −5 25. 23. 1 1 1 x− y = − 2 3 6 1 1 1 − y =− x− 3 2 6 3 1 y = x+ 2 2 3 1 slope = ; y -intercept = 2 2 x-intercept: Let y = 0. 1 1 1 x − (0) = − 2 3 6 1 1 x=− 2 6 1 x=− 3 1 1 x+ y = 2 2 3 x-intercept: 1 1 x + (0) = 2 2 3 1 x=2 2 x=4 y-intercept: 1 1 (0) + y = 2 2 3 1 y=2 3 y=6 The intercepts are ( 4, 0 ) and ( 0, 6 ) . 208 Copyright © 2016 Pearson Education, Inc. Chapter 2 Review Exercises 30. Given the points A = (− 2, 0), B = (− 4, 4), and C = (8, 5). 26. y = x3 a. Find the distance between each pair of points. d ( A, B ) = (− 4 − (− 2)) 2 + (4 − 0) 2 = 4 + 16 = 20 = 2 5 d ( B, C ) = (8 − (− 4)) 2 + (5 − 4) 2 = 144 + 1 = 145 d ( A, C ) = (8 − (− 2)) 2 + (5 − 0) 2 = 100 + 25 27. y = x = 125 = 5 5 2 2  d ( A, B )  +  d ( A, C )  =  d ( B, C )  ( 20 ) + ( 125 ) = ( 145 ) 2 2 2 2 20 + 125 = 145 145 = 145 The Pythagorean Theorem is satisfied, so this is a right triangle. 28. slope = 2 , containing the point (1,2) 3 b. Find the slopes: mAB = 4−0 4 = = −2 − 4 − (− 2) − 2 mBC = 5−4 1 = 8 − ( − 4 ) 12 mAC = 5−0 5 1 = = 8 − ( − 2 ) 10 2 1 = −1 , the sides AB 2 and AC are perpendicular and the triangle is a right triangle. Since mAB ⋅ mAC = − 2 ⋅ 29. Find the distance between each pair of points. d A, B = (1 − 3) 2 + (1 − 4) 2 = 4 + 9 = 13 d B,C = (− 2 − 1) 2 + (3 − 1) 2 = 9 + 4 = 13 d A,C = (− 2 − 3) 2 + (3 − 4) 2 = 25 + 1 = 26 Since AB = BC, triangle ABC is isosceles. 31. Endpoints of the diameter are (–3, 2) and (5,–6). The center is at the midpoint of the diameter:  −3 + 5 2 + ( − 6 )  , Center:   = (1, − 2 ) 2  2  Radius: r = (1 − (−3)) 2 + (− 2 − 2) 2 = 16 + 16 = 32 = 4 2 209 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs Chapter 2 Test 2 2 ( Equation: ( x − 1) + ( y + 2 ) = 4 2 ) 2 1. d ( P1 , P2 ) = ( x − 1)2 + ( y + 2 )2 = 32 = 62 + ( −4 ) = 52 = 2 13 2. The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  1 2 , 1 2   2 p = kB  −1 + 5 3 + (−1)  , = 2   2 4 2 = ,  2 2 = ( 2, 1) 854 = k (130, 000 ) k= 854 427 = 130, 000 65, 000 Therefore, we have the equation p = 427 B. 65, 000 If B = 165, 000 , then 427 p= (165, 000 ) = $1083.92 . 65, 000 34. w = k d2 200 = 2 = 36 + 16 1− 5 = −1 6−2 −1 − 5 slope of AC = = −1 8−2 Therefore, the points lie on a line. 32. slope of AB = 33. ( 5 − (−1) )2 + ( −1 − 3)2 3. a. m= y2 − y1 −1 − 3 −4 2 = = =− x2 − x1 5 − (−1) 6 3 b. If x increases by 3 units, y will decrease by 2 units. 4. y = x 2 − 9 k 39602 ( ) k = ( 200 ) 39602 = 3,136,320, 000 Therefore, we have the equation 3,136,320, 000 w= . d2 If d = 3960 + 1 = 3961 miles, then 3,136,320, 000 w= ≈ 199.9 pounds. 39612 35. H = ksd 135 = k (7.5)(40) 135 = 300k k = 0.45 So, we have the equation H = 0.45sd . If s = 12 and d = 35, then H = 0.45 (12)(35) = 189 BTU 5. y 2 = x y 5 (1, 1) (4, 2) (9, 3) y2 = x 10 x (0, 0) (1,−1) −5 210 Copyright © 2016 Pearson Education, Inc. (4,−2) (9,−3) Chapter 2 Test 6. x 2 + y = 9 x-intercepts: x2 + 0 = 9 9. y-intercept: (0) 2 + y = 9 y=9 x2 = 9 x = ±3 The intercepts are ( −3, 0 ) , ( 3, 0 ) , and ( 0, 9 ) . x2 + y2 + 4 x − 2 y − 4 = 0 x2 + 4 x + y 2 − 2 y = 4 ( x 2 + 4 x + 4) + ( y 2 − 2 y + 1) = 4 + 4 + 1 ( x + 2) 2 + ( y − 1) 2 = 32 Center: (–2, 1); Radius = 3 y 5 Test x-axis symmetry: Let y = − y x2 + ( − y ) = 9 (−2, 1) x 2 − y = 9 different Test y-axis symmetry: Let x = − x 5 x −5 2 (−x) + y = 9 x 2 + y = 9 same −5 Test origin symmetry: Let x = − x and y = − y ( − x )2 + ( − y ) = 9 x 2 − y = 9 different Therefore, the graph will have y-axis symmetry. 7. Slope = −2 ; containing (3, −4) 10. 2 x + 3 y = 6 3 y = −2 x + 6 2 y = − x+2 3 Parallel line Any line parallel to 2 x + 3 y = 6 has slope y − y1 = m( x − x1 ) y − (−4) = −2( x − 3) 2 m = − . The line contains (1, −1) : 3 y − y1 = m( x − x1 ) y + 4 = −2 x + 6 y = −2 x + 2 2 y − (−1) = − ( x − 1) 3 2 2 y +1 = − x + 3 3 2 1 y = − x− 3 3 Perpendicular line Any line perpendicular to 2 x + 3 y = 6 has slope 8. ( x − h) 2 + ( y − k ) 2 = r 2 3 . The line contains (0, 3) : 2 y − y1 = m( x − x1 ) m= ( x − 4 )2 + ( y − (−3) )2 = 52 ( x − 4 )2 + ( y + 3)2 = 25 General form: ( x − 4 )2 + ( y + 3)2 = 25 x 2 − 8 x + 16 + y 2 + 6 y + 9 = 25 x2 + y 2 − 8x + 6 y = 0 3 ( x − 0) 2 3 y −3 = x 2 3 y = x+3 2 y −3 = 211 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 11. Let R = the resistance, l = length, and r = radius. l Then R = k ⋅ 2 . Now, R = 10 ohms, when r l = 50 feet and r = 6 × 10−3 inch, so 50 10 = k ⋅ 2 6 × 10−3 ( 4. x 2 − 2 x − 2 = 0 x= 2± 4+8 2 2 ± 12 = 2 2±2 3 = 2 = 1± 3 −3 2 −6 50 Therefore, we have the equation l R = 7.2 × 10−6 2 . r ( ) ) ( { 5. x 2 + 2 x + 5 = 0 ) x= −2 ± 22 − 4 (1)( 5 ) 2 (1) −2 ± 4 − 20 2 −2 ± −16 = 2 No real solutions = Chapter 2 Cumulative Review 1. 3x − 5 = 0 3x = 5 5 x= 3 6. 2 2x = 8 x 2 − x − 12 = 0 ( x − 4 )( x + 3) = 0 x = 4 or x = −3 2 x2 − 5x − 3 = 0 ( 2 x + 1)( x − 3) = 0 x=− 1 or x = 3 2  1  The solution set is − ,3 .  2  2 2x +1 = 9 x=4 Check: The solution set is {−3, 4} . 3. 2x +1 = 3 ( 2x + 1) = 3 5  The solution set is   . 3 2. } The solution set is 1 − 3, 1 + 3 . If l = 100 feet and r = 7 × 10−3 inch, then 100 ≈ 14.69 ohms. R = 7.2 ×10−6 2 7 × 10−3 ( ( −2 )2 − 4 (1)( −2 ) 2 (1) = ) ( 6 ×10 ) = 7.2 ×10 k = 10 ⋅ − ( −2 ) ± 2(4) + 1 = 3? 9 = 3? 3 = 3 True The solution set is {4} . 7. x −2 =1 x − 2 = 1 or x − 2 = −1 x=3 x =1 The solution set is {1,3} . 212 Copyright © 2016 Pearson Education, Inc. Chapter 2 Cumulative Review 12. −1 < x + 4 < 5 −5 < x < 1 { x − 5 < x 3 2 + x 3 x 1 { x x 1} or ( −∞, − 5) ∪ (1, ∞ ) 4 + 8 2 + 8 − 8 − 8 2 = 2? { 4 = 2 True } The solution set is −2 − 2 2, −2 + 2 2 . 15. d ( P, Q ) = 9. x 2 = −9 = x = ± −9 x = ±3i The solution set is {−3i,3i} . − ( −2 ) ± ( −5 )2 + ( 5)2 = 50 = 5 2  −1 + 4 3 + ( −2 )   3 1  Midpoint =  , = ,  2  2  2 2 ( −2 )2 − 4 (1)( 5 ) 2 ± 4 − 20 = 2 (1) 2 2 ± −16 2 ± 4i = = = 1 ± 2i 2 2 The solution set is {1 − 2i, 1 + 2i} . 11. 2 x − 3 ≤ 7 2 x ≤ 10 x≤5 { x x ≤ 5} or ( −∞,5] 2 = 25 + 25 10. x 2 − 2 x + 5 = 0 x= ( −1 − 4 )2 + ( 3 − ( −2 ) ) 16. y = x3 − 3x + 1 a. ( −2, −1) : ( −2 )3 − ( 3)( −2 ) + 1 = −8 + 6 + 1 = −1 ( −2, −1) is on the graph. b. ( 2,3) : ( 2 )3 − ( 3)( 2 ) + 1 = 8 − 6 + 1 = 3 ( 2,3) is on the graph. c. ( 3,1) : ( 3)3 − ( 3)( 3) + 1 = 27 − 9 + 1 = 19 ≠ 1 ( 3,1) is not on the graph. 213 Copyright © 2016 Pearson Education, Inc. Chapter 2: Graphs 17. y = x3 20. x2 + y2 − 4x + 8 y − 5 = 0 x2 − 4 x + y 2 + 8 y = 5 ( x 2 − 4 x + 4) + ( y 2 + 8 y + 16) = 5 + 4 + 16 ( x − 2) 2 + ( y + 4) 2 = 25 ( x − 2) 2 + ( y + 4) 2 = 52 Center: (2,–4); Radius = 5 18. The points (–1,4) and (2,–2) are on the line. −2 − 4 −6 Slope = = = −2 2 − (−1) 3 y − y1 = m( x − x1 ) y − 4 = −2 ( x − ( −1) ) y − 4 = −2 ( x + 1) y = −2 x − 2 + 4 y = −2 x + 2 19. Perpendicular to y = 2 x + 1 ; Contains ( 3,5 ) 1 Slope of perpendicular = − 2 y − y1 = m( x − x1 ) Chapter 2 Project Internet Based Project 1 y − 5 = − ( x − 3) 2 1 3 y −5 = − x + 2 2 1 13 y = − x+ 2 2 214 Copyright © 2016 Pearson Education, Inc.