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Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Graphs, Equations, and Inequalities (f) Quadrant IV Section 1.1 3 32 42 25 112 602 121 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 bh 16 (a) (b) (c) (d) (e) (f) True x-coordinate; y-coordinate Quadrant I Quadrant III Quadrant II Quadrant I y-axis x-axis quadrants 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, lies in Quadrant II x x y y2 12 True; M , 17 The points will be on a vertical line that is two units to the right of the y-axis 13 d 14 c 15 (a) (b) (c) (d) (e) Quadrant II x-axis Quadrant III Quadrant I y-axis Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 18 The points will be on a horizontal line that is three units above the x-axis 26 X 90 X max 30 X scl 10 Y 50 Y max 70 Y scl 10 27 X 10 X max 110 X scl 10 Y 10 Y max 160 19 Y scl 10 1, ; Quadrant II 28 20 (3, 4); Quadrant I 21 (3, 1); Quadrant I 22 6, 4 ; Quadrant III 23 X 11 X max X scl Y 3 Y max Y scl 24 X 3 X max X scl 29 X 6 X max X scl Y 4 Y max Y scl 30 X 3 X max X scl Y 4 Y max Y scl 25 X 20 X max 110 X scl 10 Y 10 Y max 60 Y scl 10 Y 2 Y max Y scl X 30 X max 50 X scl 10 31 X 6 X max X scl Y 90 Y max 50 Y scl 10 Y 1 Y max Y scl Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; Introduction to Graphing Equations 32 X 9 X max X scl Y 12 Y max Y scl 42 d ( P1 , P2 ) 43 d ( P1 , P2 ) (6 4) (3) 22 2 49 53 44 d ( P1 , P2 ) ( 4) 2 (3) 2 102 52 100 25 125 5 Y Y max 10 Y scl 45 d ( P1 , P2 ) (0 a ) (0 b) a b 34 X 22 X max 10 X scl 46 d ( P1 , P2 ) (0 a) (0 a ) a a 2a a Y Y max Y scl 47 P1 1,3 ; P2 5,15 d P1 , P2 35 d ( P1 , P2 ) (2 0) (1 0) 12 15 32 2 12 2 16 144 160 10 36 d ( P1 , P2 ) ( 0)2 (1 0) 37 d ( P1 , P2 ) ( 1) (2 1) 10 (1) 2 (2 1)2 48 P1 8, 4 ; P2 2,3 d P1 , P2 10 8 4 10 2 2 100 49 39 d ( P1 , P2 ) (5 3) 4 22 2 1 2 149 49 P1 4, ; P2 4, 8 64 68 17 40 d ( P1 , P2 ) 3 16 25 4 d P1 , P2 4 8 2 8 2 14 2 64 196 41 d ( P1 , P2 ) 82 64 36 100 10 33 X X max X scl 38 d ( P1 , P2 ) 10 2 (3) 2 11 (5) 2 (9 3) 260 65 16 12 256 144 400 20 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 50 P1 0, ; P2 3, 8 d P1 , P2 problem, A d ( A, B) d ( B, C ) 13 13 13 13 square units 2 8 2 32 14 2 196 205 51 A ( 2,5), B (1,3), C (1, 0) d ( A, B ) 52 A ( 2, 5), B (12, 3), C (10, 11) 1 ( 2) 2 (3 5)2 d ( A, B) 32 ( 2) 142 ( 2) 13 d ( B, C ) 196 200 1 12 (0 3)2 10 ( 2) ( 3) d ( B, C ) 13 d ( A, C ) 12 ( 2) 2 (3 5)2 10 12 2 (11 3)2 ( 2) (14) 1 ( 2) 2 (0 5)2 196 200 12 ( 5)2 25 10 26 d ( A, C ) 10 ( 2) 2 (11 5)2 122 (16) 144 256 400 20 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 2 26 13 13 26 26 26 The area of a triangle is A bh In this Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; Introduction to Graphing Equations Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: d ( A, B)2 d ( B, C )2 d ( A, C )2 10 10 2 20 ( 5) 104 26 130 104 26 130 53 A ( 5,3), B (6, 0), C (5,5) d ( A, B) 200 200 400 400 400 The area of a triangle is A bh In this problem, A d ( A, B) d ( B, C ) 10 10 2 100 2 100 square units d ( A, C )2 d ( B, C )2 d ( A, B)2 (0 3) 130 130 The area of a triangle is A bh In this problem, A d ( A, C ) d ( B, C ) 104 26 26 26 26 26 square units 54 A ( 6, 3), B (3, 5), C (1, 5) d ( A, B) ( 6) 2 (5 3)2 112 ( 3) 121 92 ( 8) 81 64 130 145 d ( B, C ) 5 6 (5 0) d ( B, C ) 1 32 (5 (5))2 (1) 52 25 ( 4) 102 16 100 26 116 29 d ( A, C ) ( 5) (5 3) d ( A, C ) 1 ( 6) 2 (5 3)2 102 22 100 52 22 25 104 29 26 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs d ( A, C )2 d ( B, C )2 d ( A, B)2 29 116 145 29 116 145 145 145 The area of a triangle is A bh In this problem, A d ( A, C ) d ( B, C ) 29 116 29 29 29 29 square units Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: d ( A, B)2 d ( A, C )2 d ( B, C )2 52 55 A (4, 3), B (0, 3), C (4, 2) d ( A, B ) (0 4) 3 (3) 16 problem, A d ( A, B) d ( A, C ) 45 10 square units 4 2 (3) 2 42 52 16 25 41 02 52 25 25 5 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 41 41 The area of a triangle is A d ( A, C ) (4 4) (3) 16 25 41 ( 4)2 02 16 d ( B, C ) 41 bh In this Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; Introduction to Graphing Equations 56 A (4, 3), B (4, 1), C (2, 1) d ( A, B) (4 4) 1 (3) 02 42 16 16 57 The coordinates of the midpoint are: x x y y2 ( x, y ) , 4 d ( B, C ) 2 1 12 d ( A, B) d ( B, C ) 42 square units A 35 4 , 8 0 , 2 2 (4, 0) ( 2) 2 d ( A, C ) (2 4) 1 (3) ( 2) 42 16 20 2 58 The coordinates of the midpoint are: x x y y2 ( x, y ) , 2 , 0 4 , 2 2 0, 59 The coordinates of the midpoint are: x x y y2 ( x, y ) , 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 16 20 20 20 The area of a triangle is A bh In this problem, 5 11 3 , 6 6 , 2 2 3,3 60 The coordinates of the midpoint are: x x y y2 ( x, y ) , 10 3 , 12 , 2 6, Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 61 The coordinates of the midpoint x x y y2 are ( x, y ) , 66 y x3 x 03 13 1 13 00 1 1 1 (0, 0) and (1, –1) are on the graph of the equation 3 , 10 , 2 (5, 1) 67 y x 32 02 02 (3) 99 18 18 (0, 3) is on the graph of the equation 62 The coordinates of the midpoint are: x x y y2 ( x, y ) , 68 y x 23 3 , 1 , 2 13 03 1 82 00 11 (0, 1) and (–1, 0) are on the graph of the equation 69 x y 1 1, 2 63 The coordinates of the midpoint are: x x y y2 ( x, y ) , 2 ( 2)2 22 4 44 84 44 2, are on the graph of the equation 70 x y 02 12 22 02 22 12 44 44 54 (0, 1) and (2, 0) are on the graph of the equation 64 The coordinates of the midpoint are: x x y y2 ( x, y ) , 71 (–1, 0), (1, 0) 72 (0, 1) a0 a0 , a a , 2 2 73 , , , , (0, 1) 2 74 (–2, 0), (2, 0), (0, –3) 65 y x x 14 02 22 (0, 2) and a0 b0 , a b , 2 04 02 32 75 (1) 1 00 1 0 1 (0, 0) is on the graph of the equation 1, , 0, , 0, 2 76 (2, 0), (0, 2), (–2, 0), (0, –2) 77 (–4,0), (–1,0), (4, 0), (0, –3) 78 (–2, 0), (2, 0), (0, 3) Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; Introduction to Graphing Equations 79 y x 82 y 3x 80 y x 83 y x 81 y x 84 y x 9 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 85 y x 89 x y 36 90 x y 86 y x 91 y x 13 87 x y The x-intercept is x 6.5 and the y-intercept is y 13 88 x y 10 10 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.5: Circles b c x-intercepts: x 22 2 ( x 2)2 x 2 x 2 x 2 x or x 4 y-intercepts: y 22 c y2 x-intercepts: ( x 2) x 2 y2 y0 The intercepts are 4, and 0, 2 x2 x2 x 2 36 x y 12 y x2 y y x2 y y 2 x2 y 2 2 4 y a y-intercepts: (0 2) y Center: 0, ; Radius: r2 b No real solutions The intercepts are 2 , and , 2 y2 35 c 2 x 8x y x2 x y2 x2 x y a x 2 y 22 Center: 2, ; x-intercepts: x x2 x2 x0 y-intercepts: y Radius: r 2 y 2 b y2 y 2 y 22 y or y The intercepts are 0, and 0, 61 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 37 Center at (0, 0); containing point (–2, 3) r 2 13 Equation: ( x 0)2 ( y 0) 13 43 Center at (–1, 3); tangent to the line y = This means that the circle contains the point (–1, 2), so the radius is r = Equation: ( x 1) ( y 3) (1) ( x 1) ( y 3) x y 13 38 Center at (1, 0); containing point (–3, 2) r 3 12 2 16 20 Equation: ( x 1) ( y 0) 20 ( x 4) ( y 2) ( x 1) y 20 45 (c); Center: 1, 2 ; Radius = 39 Center at (2, 3); tangent to the x-axis r 3 Equation: ( x 2) ( y 3) 32 ( x 2) ( y 3) 40 Center at (–3, 1); tangent to the y-axis r 3 Equation: ( x 3) ( y 1) 32 41 Endpoints of a diameter are (1, 4) and (–3, 2) The center is at the midpoint of that diameter: (3) Center: , 1,3 2 Radius: r (1 (1)) (4 3) Equation: ( x (1)) ( y 3) 5 ( x 1) ( y 3) 42 Endpoints of a diameter are (4, 3) and (0, 1) The center is at the midpoint of that diameter: 1 Center: , 2, 2 Radius: r (4 2) (3 2) Equation: ( x 2) ( y 2) 5 46 (d) ; Center: 3,3 ; Radius = 47 (b) ; Center: 1, ; Radius = 48 (a) ; Center: 3,3 ; Radius = 49 Let the upper-right corner of the square be the point x, y The circle and the square are both ( x 3) ( y 1) 44 Center at (4, –2); tangent to the line x = This means that the circle contains the point (1, –2), so the radius is r = Equation: ( x 4) ( y 2) (3) 2 ( x 2) ( y 2) centered about the origin Because of symmetry, we have that x y at the upper-right corner of the square Therefore, we get x2 y x2 x2 2x2 9 x2 2 The length of one side of the square is 2x Thus, the area is x 2 A s 2 18 square units 50 The area of the shaded region is the area of the circle, less the area of the square Let the upperright corner of the square be the point x, y The circle and the square are both centered about the origin Because of symmetry, we have that x y at the upper-right corner of the square Therefore, we get 62 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.5: Circles x y 36 x (mx b) r 54 a x x 36 x m x 2bmx b r 2 x 36 (1 m ) x 2bmx b r There is one solution if and only if the discriminant is zero (2bm) 4(1 m )(b r ) x 18 x3 The length of one side of the square is 2x Thus, the area of the square is 4b m 4b 4r 4b m 4m r 72 square 4b 4r 4m r units From the equation of the circle, we have r The area of the circle is b2 r m2 r r 36 square units r (1 m ) b Therefore, the area of the shaded region is A 36 72 square units b 51 The diameter of the Ferris wheel was 250 feet, so the radius was 125 feet The maximum height was 264 feet, so the center was at a height of 264 125 139 feet above the ground Since the center of the wheel is on the y-axis, it is the point (0, 139) Thus, an equation for the wheel is: 2bm bm bmr mr b 2(1 m ) b b2 2 r mr y m b b x x 2 y 139 2 1252 x y 139 15, 625 52 The diameter of the wheel is 520 feet, so the radius is 260 feet The maximum height is 550 feet, so the center of the wheel is at a height of 550 260 290 feet above the ground Since the center of the wheel is on the y-axis, it is the point (0, 290) Thus, an equation for the wheel is: c x 0 y 290 2602 x y 290 67, 600 Using the quadratic formula, the result from part (a), and knowing that the discriminant is zero, we get: (1 m ) x 2bmx b r m2 r m2 r b2 r b b b b The slope of the tangent line is m The slope of the line joining the point of tangency and the center is: r2 0 b r b 1 m mr b mr b Therefore, the tangent line is perpendicular to the line containing the center of the circle and the point of tangency 53 x y x y 4091 x x y y 4091 x x y y 4091 55 x y Center: (0, 0) x 12 y 2 4096 The circle representing Earth has center 1, 2 Slope from center to 1, 2 and radius = 4096 64 So the radius of the satellite’s orbit is 64 0.6 64.6 units The equation of the orbit is 2 0 2 2 1 Slope of the tangent line is x 12 y 2 64.6 2 x y x y 4168.16 63 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ is 1 2 Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs Equation of the tangent line is: y2 x 1 2 y2 x 4 4y 8 x slope 2k h x 4y 2 x 4y 56 x y x y ( x x 4) ( y y 9) 4 ( x 2) ( y 3) Center: (2, –3) Slope from center to 3, 2 is 2 (3) 2 2 3 1 Slope of the tangent line is: 2 Equation of the tangent line: ( x 3) y 2 3 y2 3 x 4 y 12 x The slope from (h, k ) to (3, –1) is 1 k 3h 2 2k h x y 11 12 57 Let (h, k ) be the center of the circle x 2y 2y x y x2 The slope of the tangent line is The slope from (h, k ) to (0, 2) is –2 h 2k Solve the two equations in h and k : k 2(1 2k ) k 4k 3k k 0 h 2(0) The center of the circle is (1, 0) 58 Find the centers of the two circles: x2 y2 4x y ( x x 4) ( y y 9) ( x 2) ( y 3) Center: 2, 3 x2 y x y ( x x 9) ( y y 4) ( x 3) ( y 2)2 Center: 3, 2 Find the slope of the line containing the centers: (3) m 3 Find the equation of the line containing the centers: y ( x 2) 5 y 15 x x y 13 x y 13 2k 2 0h k 2h The other tangent line is y x , and it has 64 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 59 Consider the following diagram: (2,2) c slope y x 1, 1 , 2,3 a distance C 2 r b 6 2 r 2 2 3r The radius is units long 61 (b), (c), (e) and (g) We need h, k and 0, on the graph c 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 2 b 16 x 3 y 2 2 y 1 4 x 2 3 distance 4 42 c Thus, h, k 3, and r 4 midpoint , 8 4 , 4, 2 2 slope y 4 12 , undefined x 44 d Undefined slope means the points lie on a vertical line, in this case x Chapter Review 2, 1 , 3, 1 a distance 2 1 (1) 2 25 25 0, , 4, distance b 2 2 16 20 2 2 1 (1) midpoint , 1 2 1 , , 1 2 2 2 65 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 144 144 12 standard form x h y k r x 3 y slope 4, 4 , 4,8 a 2 1 midpoint , 1 , ,1 2 d When x increases by units, y decreases by units 62 (b), (e) and (g) We need h , k , and h r a 2 12 1 16 25 6 2 r 04 02 2 midpoint , , 2,1 2 2 d When x increases by units, y increases by unit Therefore, the path of the center of the circle has the equation y 60 b Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs c slope y 1 (1) 0 x 2 d Slope = means the points lie on a horizontal line, in this case y 1 4, , 0, , 2, , 0, 2 , 0, , 0, y x 15 y x x-intercepts: x2 x 3 x 3 y-intercept: y 02 y 9 x or x 3 The intercepts are 3, , 3, , and 0, 9 The intercepts are 0,15 , 3.87, , and 3.87, x y 3 y 2 x y x2 x-intercepts: 2x 30 2x x3 x y 16 y-intercept: 0 y 3 y y 2 The intercepts are 3, and 0, 2 y x 16 y x2 x-intercepts: x 16 x 16 x 4 The intercepts are 4, , 4, , and 0,8 66 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ y-intercept: 02 y 16 y 16 y 8 Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises The intercepts are (2, 0), (2, 0), (0, 4), and (0, 2) Test x-axis symmetry: Let y y y x 3x2 y x x different Test y-axis symmetry: Let x x y x 3x 4 y x 3x Test origin symmetry: Let x x and y y 10 x y x-axis: x y y x 3 x 4 2 x y different origin: x y 13 y x3 x x-axis: y x3 x y x3 x different 2 x y different Therefore, the graph of the equation has x-axis symmetry y-axis: y x x y x3 x different 11 x y 16 origin: y x x x-axis: x y 16 y x3 x x y 16 same y x3 x same Therefore, the equation of the graph has origin symmetry y-axis: x y 16 x y 16 same 14 x x y y origin: x y 16 2 x-axis: x x y y x y 16 same Therefore, the graph of the equation has x-axis, y-axis, and origin symmetry x x y y different y-axis: x x y y 12 y x x x-intercepts: x 3x different Therefore, the graph will have y-axis symmetry y-axis: x y 2 y x 3x2 y x 3x2 x y same same x x y y different x2 x2 y-intercepts: y (0) 3(0) 4 origin: x 2 x y 2 y x x y y different Therefore, the graph of the equation has none of the three symmetries x2 x2 x 2 67 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 15 x y x3 x, y 2 y 2 8 y 0 y 2 2, 8 0, 2,8 3 The solution set is 1.14,1.64 18 Slope = –2; containing (3,–1) y y1 m( x x1 ) y (1) 2( x 3) y 2x y 2x x y or y x 16 x3 x Use the Zero option from the CALC menu 19 Slope = 0; containing the point (–5, 4) y y1 m( x x1 ) y 0( x (5)) y4 y4 The solution set is 2.49, 0.66,1.83 17 x x Use the Intersect option on the CALC menu 68 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 20 Slope undefined; containing (–3, 4) This is a vertical line x 3 No slope-intercept form y ( 3, 4) 22 y-intercept = –2; containing (5,–3) Points are (5,–3) and (0,–2) (3) 1 m 05 5 y mx b y x2 x y 10 or y x 5 (3, 0) 5 x 5 21 x-intercept = or the point (2, 0); Containing the point (4, –5) 5 5 m 42 2 y y1 m x x1 x 2 y x5 y0 5 x y 10 or y x 23 Containing the points (3,–4) and (2, 1) ( 4) m 5 23 1 y y1 m x x1 y ( 4) 5 x 3 y 5 x 15 y 5 x 11 x y 11 or y 5 x 11 69 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 24 Parallel to x y ; 26 ( x h) ( y k ) r 2 ; containing (–5,3) y y1 m( x x1 ) x 2 y 32 42 Slope x 2 y 32 16 2 10 ( x ( 5)) y x 3 19 y x 3 19 x y 19 or y x 3 y 3 27 ( x h) ( y k ) r x 1 y 2 2 12 x 12 y 2 28 x y x y x2 x y y ( x x 1) ( y y 4) ( x 1) ( y 2) ( x 1) ( y 2) 32 Center: (1,–2) Radius = x 02 x x2 2x x 25 Perpendicular to 3x y ; Containing the point (–2, 4) y y1 m( x x1 ) Slope of perpendicular 2 2 2 1 4 20 1 22 1 02 y y y ( x ( 2)) y4 x 3 10 y x 3 10 x y 10 or y x 3 y2 y y 4 42 1 4 1 4 2 2 Intercepts: 4 32 1 5, 0 , 1 0, 2 2 70 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 5, , 0, 2 2 , and Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 30 Given the points A ( 2, 0), B ( 4, 4), C (8, 5) a Find the distance between each pair of points d A, B ( ( 2)) (4 0) 16 20 d B, C (8 ( 4)) (5 4) 29 3x y x 12 y 144 145 x y 2x y d A, C (8 ( 2)) (5 0) x2 x y2 y 100 25 125 5 ( x x 1) ( y y 4) ( x 1) ( y 2) ( x 1) ( y 2) Center: (1,–2) Radius = 5 20 125 145 20 125 145 145 145 The Pythagorean Theorem is satisfied, so this is a right triangle b Find the slopes: 40 mAB 2 ( 2) 54 mBC 12 3x x 12 3x x 3x x x or x mAC y 12 y y 12 y 50 10 1 Since the product of the slopes is –1, the sides of the triangle are perpendicular and the triangle is a right triangle mAB mAC y y 4 y or y 4 Intercepts: 0, , 2, , and 0, 4 1 1; 62 1 1 slope of BC 86 Since lines AB and BC are parallel, and have a point in common (B), they are the same line Therefore, A , B , and C are collinear 31 slope of AB 71 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs 32 A circle with center 1, has equation x 1 y 2 r 35 slope = 2 , containing the point (1,2) point A 1,5 1 1 r 2 r r 13 point B 2, 1 r 2 r r 13 point C 3,5 3 1 r 2 r r 13 Therefore the points A, B and C lie on a circle with center point 1, and radius = 13 33 Endpoints of the diameter are (–3, 2) and (5,–6) The center is at the midpoint of the diameter: 3 ( 6) Center: , 1, 36 Answers will vary 37 a x is a vertical line passing through the origin, that is, x is the equation of the y-axis b y is a horizontal line passing through the origin, that is, y is the equation of the x-axis c x y implies that y x Thus, the graph of x y is the line passing through the origin with slope = –1 d xy implies that either x , y , or both are x is the y-axis and y is the x-axis Thus, the graph of xy is a graph consisting of the coordinate axes e x y implies that both x and Radius: r (1 (3)) ( 2) 16 16 32 ( x 1) ( y 2) Equation: 2 x x y y 32 x y x y 27 34 Using the distance formula on the points (–3, 2) and (5, y) yields d (5 (3)) ( y 2) 64 ( y 2) Now set d 10 and solve for y y Thus, the graph of x y is a graph consisting of a single point - the origin 10 64 ( y 2) 102 64 ( y 2) 100 64 ( y 2) 2 36 ( y 2) 6 y so y y and y 6 y 4 Chapter Test a d x2 x1 2 y2 y1 2 2 3 82 36 64 100 10 72 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 2 Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Test x x y y2 M , 2 3 , 2 b 2 2 , 2 2 1,1 2 x y 21 7 y 2 x 21 y x3 x y 72 x The intercepts are 0, 5 , 2.24, , and 2.24, x, y 7 5 7, 5 3 0, 3 7 y 72 y y 1 7, 1 x3 x x Since this equation has on one side, we will use the Zero option from the CALC menu It will be important to carefully select the window settings so as not to miss any solutions The intercepts are 0, 3 and 10.5, The solutions to the equation are 1 , 0.5, and x x Since this equation has on one side, we will use the Zero option from the CALC menu y x x y x2 3 y 3 1 y 1 4 y 5 y 1 4 y 3 2 x, y 3, 1, 4 0, 5 1, 4 3, The solutions, rounded to two decimal places, are 2.50 and 2.50 73 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter 1: Graphs x3 x x x Since there are nonzero expressions on both sides of the equation, we will use the Intersect option from the CALC menu Enter the left side of the equation in Y1 and the right side in Y2 Test x-axis symmetry: Let y y x2 y x y different Test y-axis symmetry: Let x x x 2 y x y same Test origin symmetry: Let x x and y y x 2 y x y different Therefore, the graph will have y-axis symmetry 10 Slope = 2 ; containing (3, 4) y y1 m( x x1 ) y (4) 2( x 3) The solutions, rounded to two decimal places, are 2.46 , 0.24 , and 1.70 a m y 2 x y 2 x y2 y1 1 4 x2 x1 (1) b If x increases by units, y will decrease by units y x y y2 x x 2 y (3) 2 52 x 2 y 32 25 x x y x-intercepts: x2 x2 11 ( x h) ( y k ) r General form: x 2 y 32 25 y-intercept: (0) y y9 x x 16 y y 25 x2 y 8x y x 3 The intercepts are 3, , 3, , and 0,9 74 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Project 12 Chapter Project x2 y2 x y x2 x y y ( x x 4) ( y y 1) ( x 2) ( y 1) 32 Internet-based Project Center: (–2, 1); Radius = y x 13 x y y 2 x y x2 Parallel line Any line parallel to x y has slope m The line contains (1, 1) : y y1 m( x x1 ) y (1) ( x 1) 2 y 1 x 3 y x 3 Perpendicular line Any line perpendicular to x y has slope The line contains (0, 3) : y y1 m( x x1 ) m ( x 0) y 3 x y x3 y 3 75 Copyright © 2017 Pearson Education, Inc Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; ... https://TestbankDirect.eu/ Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities; ... this Solution Manual for Precalculus Enhanced with Graphing Utilities 7th Edition by Sullivan Full file at https://TestbankDirect.eu/ Section 1.1: The Distance and Midpoint Formulas; Graphing Utilities;
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