Chapter 2 Graphs
Section 2.1
1. 24yx=− () ()() ()() ()() 24, 222482,8 020440,4 222402,0 xyxxy y y y =− −=−−=−−−
=−= 2 5 5 10 5 y x (0, 4) (2, 0) (2, 8)
Based on the graph, the intercepts are () 2,0 and () 0,4
2. ()() 2 4120 620 xx xx −−= −+= 60 6 x x −= = or 20 2 x x += =−
The solution set is {} 2,6 .
3. intercepts
4. y-axis
5. 4
6. () 3,4
7. True
8. 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: 22 1 xy+=
9. a
10. c 11. 2 yx=+ x-intercept: y-intercept: 02 2 x x =+ −= 02 2 y y =+ = The intercepts are () 2,0 and () 0,2
12. 6 yx=− x-intercept: y-intercept: 06 6 x x =− = 06 6 y y =− =− The intercepts are () 6,0 and () 0,6 13. 28yx=+ x-intercept: y-intercept: 028 28 4 x x x =+ =− =− () 208 8 y y =+ =
14.
The intercepts are () 4,0 and () 0,8
yx
Section2.1: Intercepts; Symmetry; Graphing Key Equations
Chapter2: Graphs
The intercepts are () 1,0 , () 1,0 , and () 0,1
19. 236 xy+=
x-intercepts: y-intercept: () 2306 26 3 x x x += = = () 2036 36 2 y y y += = =
The intercepts are () 3,0 and () 0,2 .
20. 5210 xy+=
x-intercepts: y-intercept: () 52010 510 2 x x x += = = () 50210 210 5 y y y += = = The intercepts are () 2,0 and () 0,5
21. 2 9436 xy+=
x-intercepts: y-intercept: () 2 2 2 94036 936 4 2 x x x x += = = =± () 2 90436 436 9 y y y += = = The intercepts are () 2,0 , () 2,0 , and () 0,9
22. 2 44 xy+= x-intercepts: y-intercept: 2
Section2.1: Intercepts; Symmetry; Graphing Key Equations
33. a. Intercepts: () 1,0 and () 1,0
b. Symmetric with respect to the x-axis, y-axis, and the origin.
34. a. Intercepts: () 0,1
b. Not symmetric to the x-axis, the y-axis, nor the origin
Chapter2: Graphs
35. a. Intercepts: () 2 0 , π , () 0,1 , and () 2 ,0 π
b. Symmetric with respect to the y-axis.
36. a. Intercepts: () 2,0 , () 0,3 , and () 2,0
b. Symmetric with respect to the y-axis.
37. a. Intercepts: () 0,0
b. Symmetric with respect to the x-axis.
38. a. Intercepts: ()2,0, ()0,2, ()0,2, and () 2,0
b. Symmetric with respect to the x-axis, y-axis, and the origin.
39. a. Intercepts: () 2,0 , () 0,0 , and () 2,0
b. Symmetric with respect to the origin.
40. a. Intercepts: () 4,0 , () 0,0 , and () 4,0
b. Symmetric with respect to the origin.
41. a. x-intercept: [] 2,1 , y-intercept 0
b. Not symmetric to x-axis, y-axis, or origin.
42. a. x-intercept: [] 1,2 , y-intercept 0
b. Not symmetric to x-axis, y-axis, or origin.
43. a. Intercepts: none
b. Symmetric with respect to the origin.
44. a. Intercepts: none
b. Symmetric with respect to the x-axis.
45.
The intercepts are () 4,0 , () 0,2 and () 0,2
Test x-axis symmetry: Let yy =− () 2 2 4 4 same yx yx −=+ =+
Test y-axis symmetry: Let x x =− 2 4 yx=−+ different
Test origin symmetry: Let x x =− and yy =− () 2 2 4 4 different yx yx −=−+ =−+
Therefore, the graph will have x-axis symmetry.
50. 2 9 yx=+
x-intercepts: y-intercepts:
2 (0)9 09 9 x x x =−+ =−+ = 2 2 09 9 3 y y y =+ = =±
The intercepts are () 9,0 , () 0,3 and () 0,3 .
Test x-axis symmetry: Let yy =−
() 2 2 9 9 same yx yx −=+ =+
Test y-axis symmetry: Let x x =−
2 9 yx=−+ different
Test origin symmetry: Let x x =− and yy =−
() 2 2 9 9 different yx yx −=−+ =−+
Therefore, the graph will have x-axis symmetry.
51. 3 y x =
x-intercepts: y-intercepts:
3 0 0 x x = = 3 00 y ==
The only intercept is () 0,0
Test x-axis symmetry: Let yy =−
3 different yx−=
Test y-axis symmetry: Let x x =−
33 different yxx =−=−
Test origin symmetry: Let x x =− and yy =−
33 3 same yxx yx −=−=− =
Therefore, the graph will have origin symmetry.
52. 5 y x =
x-intercepts: y-intercepts:
3 0 0 x x = = 5 00 y ==
The only intercept is () 0,0
Test x-axis symmetry: Let yy =−
5 different yx−=
Test y-axis symmetry: Let x x =−
55 different yxx =−=−
Section2.1: Intercepts; Symmetry; Graphing Key Equations
Test origin symmetry: Let x x =− and yy =− 55 5 same y xx yx −=−=− =
Therefore, the graph will have origin symmetry.
53. 2 90 xy+−=
x-intercepts: y-intercepts: 2 2 90 9 3 x x x −= = =± 2 090 9 y y +−= =
The intercepts are () 3,0 , () 3,0 , and () 0,9 .
Test x-axis symmetry: Let yy =− 2 90 different xy−−=
Test y-axis symmetry: Let x x =−
() 2 2 90 90 same xy xy −+−= +−=
Test origin symmetry: Let x x =− and yy =− () 2 2 90 90 different xy xy −−−= −−=
Therefore, the graph will have y-axis symmetry.
54. 2 40 xy−−=
x-intercepts: y-intercept: 2 2 040 4 2 x x x −−= = =± 2 040 4 4 y y y −−= −= =−
The intercepts are () 2,0 , () 2,0 , and () 0,4 .
Test x-axis symmetry: Let yy =− () 2 2 40 40 different xy xy −−−= +−=
Test y-axis symmetry: Let x x =− () 2 2 40 40 same xy xy −−−= −−=
Test origin symmetry: Let x x =− and yy =−
()() 2 2 40 40 different xy xy −−−−= +−=
Therefore, the graph will have y-axis symmetry.
Chapter2: Graphs
55. 22 9436 xy+=
x-intercepts: y-intercepts: () 2 2 2 2 94036 936 4 2 x x x x += = = =± () 2 2 2 2 90436 436 9 3 y y y y += = = =±
The intercepts are () 2,0 , ()2,0, ()0,3, and () 0,3
Test x-axis symmetry: Let yy =− () 2 2 22 9436 9436 same xy xy +−= +=
Test y-axis symmetry: Let x x =− () 2 2 22 9436 9436 same xy xy −+= +=
Test origin symmetry: Let x x =− and yy =− ()() 22 22 9436 9436 same xy xy −+−= +=
Therefore, the graph will have x-axis, y-axis, and origin symmetry.
56. 22 44 xy+=
x-intercepts: y-intercepts: 22 2 2 404 44 1 1 x x x x += = = =± () 2 2 2 404 4 2 y y y += = =±
The intercepts are () 1,0 , () 1,0 , () 0,2 , and () 0,2
Test x-axis symmetry: Let yy =− () 2 2 22 44 44 same xy xy +−= +=
Test y-axis symmetry: Let x x =− () 2 2 22 44 44 same xy xy −+= +=
Test origin symmetry: Let x x =− and yy =−
()()22 22 44 44 same xy xy −+−= +=
Therefore, the graph will have x-axis, y-axis, and origin symmetry.
57. 3 27 yx=− x-intercepts: y-intercepts: 3 3 027 27 3 x x x =− = = 3 027 27 y y =− =−
The intercepts are () 3,0 and () 0,27
Test x-axis symmetry: Let yy =− 3 27 different yx −=−
Test y-axis symmetry: Let x x =− ()3 3 27 27 different yx yx =−− =−−
Test origin symmetry: Let x x =− and yy =− ()3 3 27 27 different yx yx −=−− =+
Therefore, the graph has none of the indicated symmetries.
58. 4 1 yx=−
x-intercepts: y-intercepts: 4 4 01 1 1 x x x =− = =± 4 01 1 y y =− =−
The intercepts are () 1,0 , () 1,0 , and () 0,1
Test x-axis symmetry: Let yy =− 4 1 different yx −=−
Test y-axis symmetry: Let x x =− () 4 4 1 1 same yx yx =−− =−
Test origin symmetry: Let x x =− and yy =− () 4 4 1 1 different yx yx −=−− −=−
Therefore, the graph will have y-axis symmetry.
59. 2 34yxx=−− x-intercepts: y-intercepts: ()() 2 034 041 4 or 1 xx xx x x =−− =−+ ==− () 2 0304 4 y y =−− =−
The intercepts are () 4,0 , () 1,0 , and () 0,4
Test x-axis symmetry: Let yy =−
2 34 different yxx −=−−
Test y-axis symmetry: Let x x =− ()() 2 2 34 34 different yxx yxx =−−−− =+−
Test origin symmetry: Let x x =− and yy =− ()() 2 2 34 34 different yxx yxx −=−−−− −=+−
Therefore, the graph has none of the indicated symmetries.
60. 2 4 yx=+
x-intercepts: y-intercepts: 2 2 04 4 no real solution x x =+ =− 2 04 4 y y =+ =
The only intercept is () 0,4 .
Test x-axis symmetry: Let yy =−
2 4 different yx −=+
Test y-axis symmetry: Let x x =− () 2 2 4 4 same yx yx =−+ =+
Test origin symmetry: Let x x =− and yy =− () 2 2 4 4 different yx yx −=−+ −=+
Therefore, the graph will have y-axis symmetry.
61. 2 3 9 x y x = + x-intercepts: y-intercepts: 2 3 0 9 30 0 x x x x = + = = () 2 30 0 0 9 09 y === +
The only intercept is () 0,0
Test x-axis symmetry: Let yy =−
2 3 different 9 x y x −= +
Section2.1: Intercepts; Symmetry; Graphing Key Equations
Test y-axis symmetry: Let x x =− () () 2 2 3 9 3 different 9 x y x x y x = −+ =− +
Test origin symmetry: Let x x =− and yy =− () () 2 2 2 3 9 3 9 3 same 9 x y x x y x x y x −= −+ −=− + = +
Therefore, the graph has origin symmetry.
62. 2 4 2 x y x = x-intercepts: y-intercepts: 2 2 2 4 0 2 40 4 2 x x x x x = −= = =± () 2 044 200 undefined y ==
The intercepts are () 2,0 and () 2,0 .
Test x-axis symmetry: Let yy =− 2 4 different 2 x y x −=
Test y-axis symmetry: Let x x =− () () 2 2 4 2 4 different 2 x y x x y x = =−
Test origin symmetry: Let x x =− and yy =− () () 2 2 2 4 2 4 2 4 same 2 x y x x y x x y x −= −= =
Therefore, the graph has origin symmetry.
Copyright © 2017 Pearson Education, Inc.
Chapter2: Graphs
63. 3 2 9 x y x =
Test origin symmetry: Let x x =− and yy =−
The only intercept is () 0,0
Test
Test origin symmetry: Let x x =− and yy =−
Therefore, the graph has origin symmetry. 64. 4 5 1 2 x y x + = x-intercepts: y-intercepts: 4 5 4 1 0 2 1 x x x + = =− () 4 5 011 0 20 undefined y +
There are no intercepts for the graph of this equation.
Test
Therefore, the graph has origin symmetry.
67. y x =
68. 1 y x =
69. If the point () 3, b is on the graph of 41 y x =+ , then we have () 43112113 b =+=+= Thus, 13 b =
70. If the point () 2, b is on the graph of 232 xy+= , then we have ()() 2232 432 36 2 b b b b −+= −+= = = Thus, 2 b = .
71. If the point () ,4 a is on the graph of 2 3 y xx=+ , then we have ()() 2 2 43 034 041 aa aa aa =+ =+− =+−
40 4 a a += =− or 10 1 a a −= = Thus, 4 a =− or 1 a = .
Section2.1: Intercepts; Symmetry; Graphing Key Equations
72. If the point () ,5 a is on the graph of 2 6 y xx=+ , then we have ()() 2 2 56 065 051 aa aa aa −=+ =++ =++
50 5 a a += =− or 10 1 a a += =−
Thus, 5 a =− or 1 a =− .
73. a. 2 2 05 5 5 x x x =− = =±
The x-intercepts are 5 x =− and 5 x = () 2 055 y =−=−
The y-intercept is 5 y =−
The intercepts are () 5,0 , () 5,0 , and () 0,5
b. x-axis (replace y by y ): 2 2 5 5 different yx yx −=− =−
y-axis (replace x by x ): () 2 2 55yxx=−−=− same origin (replace x by x and y by y ): () 2 2 5 5 different yx yx −=−− =−
The equation has y-axis symmetry.
c. 2 5 yx=−
Additional points:
Chapter2: Graphs
2 2 2 5, 11541,4 1from symmetry1,4
22512,1
() () () () ()
2from symmetry2,1 xyxxy y y =− =−=−−
74. a. 2 2 08 8 22 x x x =− = =±
The x-intercepts are 22 x =− and 22 x = . () 2 088 y =−=−
The y-intercept is 8 y =−
The intercepts are () 22,0 , () 22,0 , and () 0,8
b. x-axis (replace y by y ):
−=− =−
2 2 8 8 different yx yx
y-axis (replace x by x ):
() 2 2 88yxx=−−=− same
origin (replace x by x and y by y ):
−=−− =−
() 2 2 8 8 different yx yx
The equation has y-axis symmetry.
c. 2 8 yx=−
2 2 2 5, 11871,7 1from symmetry1,7
22842,4
2from symmetry2,4 xyxxy y y =− =−=−−
Additional points: () () () () ()
75. a. () 2 09 9 x x −=− =−
The x-intercept is 9 x =− . () 2 2 2 09 9 9 3 y y yy
−=− −=− =→=±
The y-intercepts are 3 y =− and 3 y =
The intercepts are () 9,0 , () 0,3 , and () 0,3 .
b. x-axis (replace y by y ):
() 2 2 9 9 same xy xy −−=− −=−
y-axis (replace x by x ): 2 2 9 9 different xy xy
−−=− +=
origin (replace x by x and y by y ):
−−−=− −−=− +=
() 2 2 2 9 9 9 different xy xy xy
The equation has x-axis symmetry.
Chapter2: Graphs
c. 22 9 xy+=
78. a. () 2 2 2 016 16 4 x x x += = =±
The x-intercepts are 4 x =− and 4 x = . () 2 2 2 016 16 4 y y y += = =±
The y-intercepts are 4 y =− and 4 y =
The intercepts are () 4,0 , () 4,0 , () 0,4 , and () 0,4 .
b. x-axis (replace y by y ):
() 2 2 22 16 16 same xy xy +−= +=
y-axis (replace x by x ):
() 2 2 22 16 16 same xy xy −+= +=
origin (replace x by x and y by y ):
()() 22 22 16 16 same xy xy −+−= +=
The equation has x-axis symmetry, y-axis symmetry, and origin symmetry.
c. 22 16 xy+=
79. a. () 3 2 04 04 x x xx =− =− 0 x = or 2 2 40 4 2 x x x −= = =±
The x-intercepts are 0 x = , 2 x =− , and 2 x = () 3 0400 y =−=
The y-intercept is 0 y =
The intercepts are () 0,0 , () 2,0 , and () 2,0
b. x-axis (replace y by y ): 3 3 4 4 different yxx yxx −=− =−
y-axis (replace x by x ): ()() 3 3 4 4 different yxx yxx =−−− =−+
origin (replace x by x and y by y ): ()() 3 3 3 4 4 4 same yxx yxx yxx −=−−− −=−+ =−
The equation has origin symmetry.
c. 3 4 y xx=−
Additional points: () ()() () 3 3 4, 114131,3 1from symmetry1,3 x yxxxy y =− =−=−−
80. a. () 3 2 0 01 x x xx =− =−
0 x = or 2 2 10 1 1 x x x −= = =±
The x-intercepts are 0 x = , 1 x =− , and 1 x = () 3 000 y =−=
The y-intercept is 0 y =
The intercepts are () 0,0 , () 1,0 , and () 1,0
b. x-axis (replace y by y ):
3 3 different yxx yxx −=− =− y-axis (replace x by x ): ()() 3 3 different yxx yxx =−−− =−+
Section2.1: Intercepts; Symmetry; Graphing Key Equations
81. For a graph with origin symmetry, if the point () , ab is on the graph, then so is the point () , ab . 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.
82. For a graph with y-axis symmetry, if the point () , ab is on the graph, then so is the point () , ab . 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 also be on the graph. Therefore, 6 is another xintercept.
83. For a graph with origin symmetry, if the point () , ab is on the graph, then so is the point () , ab . 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.
−=−+ =−
origin (replace x by x and y by y ): ()() 3 3 3 same yxx yxx yxx −=−−−
The equation has origin symmetry.
c. 3 y xx=−
Additional points: () ()() () 3 3 4, 22262,6 2from symmetry2,6 xyxxxy y =− =−=
84. For a graph with x-axis symmetry, if the point () , ab is on the graph, then so is the point () , ab . Since 2 is a y-intercept in this case, the point () 0,2 is on the graph of the equation. Due to the x-axis symmetry, the point () 0,2 must also be on the graph. Therefore, 2 is another yintercept.
85. a. () 2 2222 x yxxy+−=+ x-intercepts: () ()() () () 2 2222 2 22 4322 43 3 00 2 20 20 xxx xxx xxxx xx xx +−=+ −= −+= −= −= 3 0or20 02 xx xx =−= ==
Chapter2: Graphs
y-intercepts: () ()() () () 2 2222 2 22 42 42 22 000 0 10 yy yy yy yy yy +−=+ = = −= −= 22 2 0or10 0 1 1 yy y y y =−= == =±
The intercepts are ()0,0, ()2,0, ()0,1, and () 0,1
b. Test x-axis symmetry: Let yy =− () ()() () 2 2222 2 2222 same xyxxy xyxxy +−−=+− +−=+
Test y-axis symmetry: Let x x =− ()() ()() () 2 2222 2 2222 different xyxxy xyxxy −+−−=−+ ++=+
Test origin symmetry: Let x x =− and yy =− ()()() ()()() () 2 2222 2 2222 different xyxxy xyxxy −+−−−=−+− ++=+
Thus, the graph will have x-axis symmetry.
86. a. 2 16120225 yx=− x-intercepts: () 2 2 2 161200225 16225 225 16 no real solution y y y =− =− =−
y-intercepts: () 2 160120225 0120225 120225 22515 1208 x x x x =− =− −=− ==
The only intercept is 15 ,0 8
b. Test x-axis symmetry: Let yy =− () 2 2 16120225 16120225 same yx yx −=− =−
Test y-axis symmetry: Let x x =− () 2 2 16120225 16120225 different yx yx =−− =−−
Test origin symmetry: Let x x =− and yy =− ()() 2 2 16120225 16120225 different yx yx −=−− =−−
Thus, the graph will have x-axis symmetry.
87. a.
Section2.1: Intercepts; Symmetry; Graphing Key Equations
b. Since 2 x x = for all x , the graphs of 2 and y xyx== are the same.
c. For () 2 yx = , the domain of the variable x is 0 x ≥ ; for yx = , the domain of the variable x is all real numbers. Thus, () 2 only for 0. xxx=≥
d. For 2 yx = , the range of the variable y is 0 y ≥ ; for yx = , the range of the variable y is all real numbers. Also, 2 x x = only if 0 x ≥ . Otherwise, 2 x x =−
88. 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.
89. Answers will vary. One example: y x
90. Answers will vary
91. Answers will vary
92. Answers will vary.
Case 1: Graph has x-axis and y-axis symmetry, show origin symmetry.
()() , on graph, on graph (from -axis symmetry) xyxy x →−
()() () , on graph, on graph from -axis symmetry xyxy y −→−−
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.
()() () , on graph, on graph from -axis symmetry xyxy x →−
()() () , on graph, on graph from origin symmetry xyxy −→−
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.
()() () , on graph, on graph from -axis symmetry xyxy y →−
()() () , on graph, on graph from origin symmetry xyxy −→−
Since the point () , x y is also on the graph, the graph has x-axis symmetry.
93. 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 xaxis symmetry, then it will also have origin symmetry.
94. 6(2)41 6(2)82 +− ==
Chapter2: Graphs
95. 2 2 2 33075 3(1025) 3(5)(5)3(5) xx xx xxx −+= −+= −−=−
96. 196(1)(196)14i −=−=
97. () 2 2 2 2 840 84 816416 412 412 412 423 xx xx xx x x x −+= −=− −+=−+ −= −=± =± =±
Section 2.2
1. undefined; 0
2. 3; 2 x-intercept: 23(0)6 26 3 x x x += = = y-intercept: 2(0)36 36 2 y y y += = =
3. True
4. False; the slope is 3 2 235 35 22 yx yx =+ =+
5. True; ()() ? ? 2124 224 44 True += += =
6. 12mm = ; y-intercepts; 12 1 mm⋅=−
7. 2
8. 1 2
9. False; perpendicular lines have slopes that are opposite-reciprocals of each other.
10. d
11. c
12. b
13. a. 101 Slope 202 ==
b. If x increases by 2 units, y will increase by 1 unit.
14. a. 101 Slope 202 ==−
b. If x increases by 2 units, y will decrease by 1 unit.
15. a. 121 Slope 1(2)3 ==−
b. If x increases by 3 units, y will decrease by 1 unit.
16. a. 211 Slope 2(1)3 ==
b. If x increases by 3 units, y will increase by 1 unit.
17. 21 21 033 Slope 422 yy xx ===−
18. 21 21 422 Slope2 341 yy xx ====−
19. 21 21 1321 Slope 2(2)42 yy xx ====−
20. 21 21 312 Slope 2(1)3 yy xx ===
21. 21 21 1(1)0 Slope0 2(3)5 yy xx ====
22. 21 21 220 Slope0 549 yy xx ====
23. 21 21 224 Slope undefined. 1(1)0 yy xx ===
24. 21 21 202 Slope undefined. 220 yy xx ===
25. ()1,2;3Pm==
Chapter2: Graphs
26. ()2,1;4Pm==
27. () 3 2,4; 4 Pm==−
28. () 2 1,3; 5 Pm==−
29. ()1,3;0Pm=−=
30. ()2,4;0Pm=−=
31. ()0,3; slope undefined P =
(note: the line is the y-axis)
32. ()2,0; slope undefined P =−
33. 4 Slope4 1 == ; point: () 1,2
If x increases by 1 unit, then y increases by 4 units.
Answers will vary. Three possible points are: () () () 112 and 246 2,6 213 and 6410 3,10 314 and 10414 4,14 xy xy
34. 2 Slope2 1 == ; point: () 2,3
If x increases by 1 unit, then y increases by 2 units.
Answers will vary. Three possible points are: () () () 211 and 325 1,5 110 and 527 0,7 011 and 729 1,9 xy xy xy
35. 33 Slope 22 =−= ; point: () 2,4
If x increases by 2 units, then y decreases by 3 units.
Answers will vary. Three possible points are: () () () 224 and 437 4,7 426 and 7310 6,10 628 and 10313 8,13 xy xy xy =+==−−=− =+==−−=− =+==−−=−
36. 4 Slope 3 = ; point: () 3,2
If x increases by 3 units, then y increases by 4 units.
Answers will vary. Three possible points are: () () () 330 and 246 0,6 033 and 6410 3,10 336 and 10414 6,14 xy
37. 2 Slope2 1 =−= ; point: () 2,3
If x increases by 1 unit, then y decreases by 2 units.
Answers will vary. Three possible points are: () () () 211 and 325 1,5 110 and 527 0,7 011 and 729 1,9 xy xy xy =−+=−=−−=− =−+==−−=− =+==−−=−
38. 1 Slope1 1 =−= ; point: () 4,1
If x increases by 1 unit, then y decreases by 1 unit.
Answers will vary. Three possible points are: () () () 415 and 110 5,0 516 and 011 6,1 617 and 112 7,2 xy xy xy =+==−= =+==−=− =+==−−=−
39. (0, 0) and (2, 1) are points on the line. 101 Slope 202 -intercept is 0; using : yymxb == =+ 1 0 2 2 02 1 20 or 2 yx yx xy x yyx =+ = =− −==
40. (0, 0) and (–2, 1) are points on the line. 1011 Slope 2022 -intercept is 0; using : yymxb ===− =+ 1 0 2 2 20 1 20 or 2 yx yx xy x yyx =−+ =− += +==−
Chapter2: Graphs
41. (–1, 3) and (1, 1) are points on the line.
11 132 Slope1 1(1)2
Using ()yymxx ===− −=− 11(1) 11 2 2 or 2 yx yx yx xyyx −=−− −=−+ =−+ +==−+
42. (–1, 1) and (2, 2) are points on the line. 11 211 Slope 2(1)3
Using ()yymxx == −=− () 1 1(1) 3 1 1(1) 3 11 1 33 14 33 14 34 or 33 yx yx yx yx xyyx −=−− −=+ −=+ =+ −=−=+
43. 11(),2yymxxm −=−= 32(3) 326 23 23 or 23 yx yx yx xyyx −=− −=− =− −==−
44. 11(),1yymxxm −=−=− 21(1) 21 3 3 or 3 yx yx yx xyyx −=−− −=−+ =−+ +==−+
45. 11 1 (), 2 yymxxm −=−=− 1 2(1) 2 11 2 22 15 22 15 25 or 22 yx yx yx xyyx −=−− −=−+ =−+ +==−+
46. 11(),1yymxxm −=−= 11((1)) 11 2 2 or 2 yx yx yx xyyx −=−− −=+ =+ −=−=+
47. Slope = 3; containing (–2, 3) 11() 33((2)) 336 39 39 or 39 yymxx yx yx yx xyyx −=− −=−− −=+ =+ −=−=+
48. Slope = 2; containing the point (4, –3) 11() (3)2(4) 328 211 211 or 211 yymxx yx yx yx xyyx −=− −−=− +=− =− −==−
49. Slope = 2 3 ; containing (1, –1) 11() 2 (1)(1) 3 22 1 33 21 33 21 231 or 33 yymxx yx yx yx xyyx −=− −−=−− +=−+ =−− +=−=−−
50. Slope = 1 2 ; containing the point (3, 1) 11() 1 1(3) 2 13 1 22 11 22 11 21 or 22 yymxx yx yx yx xyyx −=− −=− −=− =− −==−
51. Slope = –3; y-intercept =3 33 33 or 33 ymxb yx xyyx =+ =−+ +==−+
52. Slope = –5; y-intercept = –7 5(7) 57 or 57 ymxb yx xyyx =+ =−+− +=−=−−
53. Containing (1, 3) and (–1, 2) 2311 1122 m === 11() 1 3(1) 2 11 3 22 15 22 15 25 or 22 yymxx yx yx yx xyyx
54. Containing the points (–3, 4) and (2, 5) 541 2(3)5 m == 11() 1 5(2) 5 12 5 55 123 55 123 523 or 55 yymxx yx yx yx xyyx −=−
55. x-intercept = 2; y-intercept = –1 Points are (2,0) and (0,–1) 1011 0222 m === 1 1 2 1 22 or 1 2 ymxb yx x yyx =+ =− −==−
56. x-intercept = –4; y-intercept = 4 Points are (–4, 0) and (0, 4) 404 1 0(4)4 m === 14 4 4 or 4 ymxb yx yx xyyx =+ =+ =+
57. Slope undefined; containing the point (2, 4)
This is a vertical line.
2 No slope-intercept form. x =
58. Slope undefined; containing the point (3, 8)
This is a vertical line.
3 No slope-intercept form. x =
59. Horizontal lines have slope 0 m = and take the form yb = . Therefore, the horizontal line passing through the point () 3,2 is 2 y =
60. Vertical lines have an undefined slope and take the form x a = . Therefore, the vertical line passing through the point () 4,5 is 4 x =
61. Parallel to 4 y x = ; Slope = 4 Containing (–1, 2) 11() 24((1)) 2446 46 or 46 yymxx yx yxyx xyyx −=−
−=−−
−=+→=+
−=−=+
62. Parallel to 3 y x =− ; Slope = –3; Containing the point (–1, 2) 11() 23((1)) 23331 31 or 31 yymxx yx yxyx xyyx −=−
−=−−−
−=−−→=−−
+=−=−−
Chapter2: Graphs
63. Parallel to 52 xy−=− ; Slope = 5 Containing the point (0, 0) 11() 05(0) 5 50 or 5 yymxx yx yx x yyx −=− −=− = −==
64. Parallel to 25xy−=− ; () 1 Slope; Containing the point 0,0 2 = 11() 11 0(0)22 1 20 or 2 yymxx yxyx xyyx −=− −=−→= −==
65. Parallel to 5 x = ; Containing (4,2)
This is a vertical line. 4 No slope-intercept form. x =
66. Parallel to 5 y = ; Containing the point (4, 2)
This is a horizontal line. Slope = 0 2 y =
67. Perpendicular to 1 4; 6 yx=+ Containing (1, –2)
Slope of perpendicular = –6 11() (2)6(1) 2626 60 or 6 yymxx yx yxyx xyyx −=−
68. Perpendicular to 83yx=− ; Containing the point (10, –2) 1
Slope of perpendicular 8 =− 11() 1 (2)(10) 8 1513 2 8484 13 86 or 84 yymxx yx yxyx xyyx
69. Perpendicular to 252 xy+= ; Containing the point (–3, –6 ) 5
Slope of perpendicular 2 = 11() 5515 (6)((3))6222 53 22 53 523 or 22 yymxx yxyx yx xyyx −=− −−=−−→+=+ =+ −=−=+
70. Perpendicular to 312xy−=− ; Containing the point (0, 4)
Slope of perpendicular = –3 34 34 or 34 ymxb yx xyyx =+ =−+ +==−+
71. Perpendicular to 8 x = ; Containing (3, 4)
Slope of perpendicular = 0 (horizontal line) 4 y =
72. Perpendicular to 8 y = ; Containing the point (3, 4)
Slope of perpendicular is undefined (vertical line). 3 x = No slope-intercept form.
73. 23yx=+ ; Slope = 2; y-intercept = 3
74. 34yx=−+ ; Slope = –3; y-intercept = 4
75. 1 1 4 yx=− ; 44yx=− Slope = 4; y-intercept = –4
76. 1 2 3 xy+= ; 1 2 3 yx=−+ 1 Slope 3 =− ; y-intercept = 2
77. 1 2 2 yx=+ ; 1 Slope 2 = ; y-intercept = 2
78. 1 2 2 yx=+ ; Slope = 2; 1 -intercept 2 y =
Section2.2: Lines
79. 44xy+= ; 1 441 4 yxyx =−+→=−+ 1 Slope 4 =− ; y-intercept = 1
80. 36xy −+= ; 1 362 3 yxyx =+→=+ 1 Slope 3 = ; y-intercept = 2
81. 236 xy−= ; 2 3262 3 yxyx −=−+→=− 2 Slope 3 = ; y-intercept = –2
Chapter2: Graphs
82. 326 xy+= ; 3 2363 2 yxyx =−+→=−+ 3 Slope 2 =− ; y-intercept = 3
83. 1 xy+= ; 1 y x =−+ Slope = –1; y-intercept = 1
84. 2 xy−= ; 2 yx=− Slope = 1; y-intercept = –2
85. 4 x =− ; Slope is undefined y-intercept - none
86. 1 y =− ; Slope = 0; y-intercept = –1
87. 5 y = ; Slope = 0; y-intercept = 5
88. 2 x = ; Slope is undefined y-intercept - none
89. 0 yx−= ; yx = Slope = 1; y-intercept = 0
90. 0 xy+= ; yx =− Slope = –1; y-intercept = 0
91. 230 yx−= ; 3 23 2 yxyx =→=
3 Slope 2 = ; y-intercept = 0
92. 320 xy+= ; 3 23 2 yxyx =−→=−
3 Slope 2 =− ; y-intercept = 0
93 a. x-intercept: () 2306
26 3 x x x += = =
The point () 3,0 is on the graph.
y-intercept: () 2036 36 2 y y y += = =
The point () 0,2 is on the graph.
b. y x 5 5 5 5 (3,0) (0,2)
94. a. x-intercept: () 3206
36 2 x x x −= = =
The point () 2,0 is on the graph.
y-intercept: () 3026 26 3 y y y −= −= =−
The point () 0,3 is on the graph.
b. (0,−3) y x 5 5 5 5 (2,0)
Chapter2: Graphs
95. a. x-intercept: () 45040
440 10 x x x −+= −= =−
The point () 10,0 is on the graph.
b.
y-intercept: () 40540
540 8 y y y −+= = =
The point () 0,8 is on the graph.
96. a. x-intercept: () 64024
b.
624 4 x x x −= = =
The point () 4,0 is on the graph.
y-intercept: () 60424
424 6 y y y −= −= =−
The point () 0,6 is on the graph.
97. a. x-intercept: () 72021
b.
721 3 x x x += = =
The point () 3,0 is on the graph.
y-intercept: () 70221
221 21 2 y y y += = =
The point 21 0, 2 is on the graph.
98. a. x-intercept: () 53018 518 18 5 x x x += = =
The point 18 ,0 5
is on the graph.
b.
y-intercept: () 50318
318 6 y y y += = =
The point () 0,6 is on the graph.
99.
x-intercept:
The point () 2,0 is on the graph. y-intercept: () 11 01
100.
The point () 0,3 is on the graph. b.
-intercept:
The point () 4,0 is on the graph.
101.
x-intercept:
The point () 5,0 is on the graph.
The point () 0,2 is on the graph.
102. a. x-intercept: ()
The point () 0,3 is on the graph.
103. The equation of the x-axis is 0 y = . (The slope is 0 and the y-intercept is 0.)
104. The equation of the y-axis is 0 x = . (The slope is undefined.)
105. The slopes are the same but the y-intercepts are different. Therefore, the two lines are parallel.
Chapter2: Graphs
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. 2 or 2 yxxy=+−=−
110. Intercepts: () 0,1 and () 1,0 . Thus, slope = –1. 1 or 1 yxxy =−++=
111. Intercepts: () 3,0 and () 0,1 . Thus, slope = 1 3 1 1 or 33 3 yxxy =−++=
112. Intercepts: () 0,1 and () 2,0 . Thus, slope = 1 2 1 1 or 22 2 yxxy =−−+=−
113. () 1 2,5 P =− , () 2 1,3 P = : 1 5322 2133 m ===− () 2 1,3 P = , () 3 1,0 P =− : () 2 303 112 m == Since 12 1 mm⋅=− , the line segments 12PP and 23PP are perpendicular. Thus, the points 1P , 2P , and 3P are vertices of a right triangle.
114. () 1 1,1 P =− , () 2 4,1 P = , () 3 2,2 P = , () 4 5,4 P = () 12 11 2 413 m == ; 24 41 3 54 m == ; 34 422 523 m == ; () 13 21 3 21 m ==
Each pair of opposite sides are parallel (same slope) and adjacent sides are not perpendicular. Therefore, the vertices are for a parallelogram.
115. () 1 1,0 P =− , () 2 2,3 P = , () 3 1,2 P =− , () 4 4,1 P = () 12 303 1 213 m === ; 24 13 1 42 m ==− ; () 34 12 3 1 413 m === ; () 13 20 1 11 m ==−
Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is 1 ). Therefore, the vertices are for a rectangle.
116. () 1 0,0 P = , () 2 1,3 P = , () 3 4,2 P = , () 4 3,1 P =− 12 30 3 10 m == ; 23 231 413 m ==− ; 34 12 3 34 m == ; 14 101 303 m ==− ()() 22 12 10301910 d =−+−=+=
()() 22 23 41239110 d =−+−=+=
()() 22 34 34121910 d =−+−−=+=
()() 22 14 30109110 d =−+−−=+=
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.
117. Let x = number of miles driven, and let C = cost in dollars.
Total cost = (cost per mile)(number of miles) + fixed cost 0.6039Cx=+
When x = 110, ()() 0.6011039$105.00 C =+=
When x = 230, ()() 0.6023039$177.00 C =+= .
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 8500Cx=+
When x = 400, ()()8400500$3700 C =+=
When x = 740, ()()8740500$6420 C =+=
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 0.161461Cx=+
120. Let x = profit in dollars, and let S = salary in dollars.
Weekly salary = (% share of profit)(profit) + weekly pay 0.05375Sx=+
121. a. 0.075715.14Cx=+ ; 0800 x ≤≤
b.
c. For 200 kWh, 0.0757(200)15.14$30.28 C =+=
d. For 500 kWh, 0.0757(500)15.14$52.99 C =+=
e. For each usage increase of 1 kWh, the monthly charge increases by $0.0757 (that is, 7.57 cents).
122. a. 0.09017.57Cx=+ ; 01000 x ≤≤
b.
c. For 200 kWh, () 0.09012007.57$25.59 C =+=
d. For 500 kWh, () 0.09015007.57$52.62 C =+=
e. For each usage increase of 1 kWh, the monthly charge increases by $0.0901 (that is, 9.01 cents).
Section2.2: Lines
123. (,)(0,32);(,)(100,212) CFCF °°=°°= 212321809 slope 10001005 9 32(0) 5 9 32() 5 5 (32) 9 FC FC CF === °−=°− °−=° °=°−
If 70 F °= , then 55(7032)(38) 99 21.1 C C °=−= °≈°
124. a. º273KC=+ b. 5 º(º32) 9 CF=− () 5 (32)273 9 5160 º273 99 52297 º 99 1 52297 9 KF KF KF KF =°−+ =−+ =+ =°+
125. a. The y-intercept is (0, 30), so b = 30. Since the ramp drops 2 inches for every 25 inches of run, the slope is 22 2525 m ==− . Thus, the equation is 2 30 25 yx=−+ .
b. Let y = 0. () 2 030 25 2 30 25 25225 30 2252 375 x x x x =−+ =
=
= 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.
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 12 inches of run. This means 1 12 m ≤
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 301 36012 = . That is, 1 12 m ≥ . Thus, the
only possible slope is 1 12 m = . The 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 requirements is 1 12 m =− . In words, for every 12 inches of run, the ramp must drop exactly 1 inch.
126. a. The year 2000 corresponds to x = 0, and the year 2013 corresponds to x = 13. Therefore, the points (0, 20.6) and (13, 8.5) are on the line. Thus, 20.68.512.1 0.93 01313 m ===−
The y-intercept is 20.6, so b = 20.6 and the equation is 0.9320.6yx=−+
b. x-intercept: 00.9320.6 22.15 x x =−+ = y-intercept: () 0.93020.620.6 y =−+=
The intercepts are (22.15, 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. () 0.932520.62.65 y =−+=−
This prediction is not reasonable since it is negative.
127. a. Let x = number of boxes to be sold, and A = money, in dollars, spent on advertising. We have the points 11 (,)(100,000,40,000); xA =
22 (,)(200,000,60,000) xA = () 60,00040,000 slope 200,000100,000 20,0001 100,0005 1 40,000100,000 5 1 40,00020,000 5 1 20,000 5 Ax Ax Ax = == −=− −=− =+
b. If x = 300,000, then () 1 300,00020,000$80,000 5 A =+=
c. Each additional box sold requires an additional $0.20 in advertising.
128. Find the slope of the line containing () , ab and () , ba : slope1 ab ba ==−
The slope of the line yx = is 1.
Since 111−⋅=− , the line containing the points (,) and (,)abba is perpendicular to the line yx =
The midpoint of (,) and (,)abba is , 22 abba M ++ = .
Since the coordinates are the same, the midpoint lies on the line yx =
Note: 22 abba ++ =
129. 2 x yC−=
Graph the lines: 24 20 22 xy xy xy −=− −= −=
All the lines have the same slope, 2. The lines
are parallel.
130. Refer to Figure 33. ()
2 1 2 2 12 length of ,1 length of ,1 length of , OAdOAm
Now consider the equation
22 2 22 111212 mmmm +++=−
If this equation is valid, then AOBΔ is a right triangle with right angle at vertex O
But we are assuming that 12 1 mm =− , so we have () 2222 1212 2222 1212 221 22 00 mmmm mmmm ++=−−+ ++=++ =
Therefore, by the converse of the Pythagorean Theorem, AOBΔ is a right triangle with right angle at vertex O. Thus Line 1 is perpendicular to Line 2.
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.
133. (c)
The equation 2 xy−=− has slope 1 and yintercept (0,2). The equation 1 xy−= has 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 22yx−= has slope 2 and yintercept (0,2). The equation 21xy+=− has slope 1 2 and y-intercept 1 0,. 2
The lines are perpendicular since 1 21 2
−=−
. One line has a positive y-intercept and the other with a negative y-intercept.
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.
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).
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.
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 1ymxb =+ and Line 2 has equation 2ymxb =+ ,
Line 1 has x-intercept 1b m and y-intercept 1b
Line 2 has x-intercept 2b m and y-intercept 2b .
Assume also that Line 1 and Line 2 have unequal x-intercepts.
If the lines have the same y-intercept, then 12bb =
Chapter2: Graphs
But 12 bb mm
Line 1 and Line 2 have the 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 0 x =
Assume Line 1 has equation 1 ymxb =+ and Line 2 has equation 2 ymxb =+ ,
Line 1 has x-intercept 1 b m and y-intercept b
Line 2 has x-intercept 2 b m and y-intercept b
Assume also that Line 1 and Line 2 have unequal slopes, that is 12mm ≠ . If the lines have the same x-intercept, then 12 bb
Since we are assuming that 12mm ≠ , the only way that the two lines can have the same x-intercept is if 0. b =
144. Answers will vary.
145.
It appears that the student incorrectly found the slope by switching the direction of one of the subtractions.
The solution set is: {} |14xx<< .
Interval notation: () 1,4
2. () 2 29 29 23 23 x x x x −= −=± −=± =± 5 or 1 xx==−
The solution set is {1,5}.
3. False. For example, 22 2280xyxy++++= is not a circle. It has no real solutions.
4. radius
5. True; 2 93rr=→=
6. False; the center of the circle ()() 22 3213xy++−= is () 3,2
7. d
8. a
9. Center = (2, 1)
Radiusdistance from (0,1) to (2,1) (20)(11)42 = =−+−==
22
Equation: 22 (2)(1)4 xy−+−=
10. Center = (1, 2)
Radiusdistance from (1,0) to (1,2) (11)(20)42 = =−+−==
22
Equation: 22 (1)(2)4 xy−+−=
11. Center = midpoint of (1, 2) and (4, 2) ()() 1422 5 222,,2 ++ == () 2 2 5 2
Radiusdistance from ,2 to (4,2) 4(22)593 242 = =−+−==
Equation: 2 2 59 (2) 24xy −+−=
13. ()()222 x hykr −+−= 222 22 (0)(0)2 4 xy xy −+−= +=
General form: 22 40 xy+−=
14. ()()222 x hykr −+−= 222 22 (0)(0)3 9 xy xy −+−= +=
General form: 22 90 xy+−=
15. ()()222 x hykr −+−= 222 22 (0)(2)2 (2)4 xy xy −+−= +−=
General form: 22 22 444 40 xyy xyy +−+= +−=
Radiusdistance from 1,2 to (2,3) 21(32)2 = =−+−=
12. Center = midpoint of (0, 1) and (2, 3) () 0213 ,1,2 22 ++ == () () 2 2
Equation: () 2 2 1(2)2xy−+−=
Chapter2: Graphs
16. ()()222 x hykr −+−= 222 22 (1)(0)3 (1)9 xy xy −+−= −+=
General form: 22 22 219 280 xxy xyx −++= +−−=
17. ()()222 x hykr −+−= 222 22 (4)((3))5 (4)(3)25 xy xy −+−−= −++=
General form: 22 22 8166925 860 xxyy xyxy −++++= +−+=
18. ()()222 x hykr −+−= 222 22 (2)((3))4 (2)(3)16 xy xy −+−−= −++=
General form: 22 22 446916 4630 xxyy xyxy −++++= +−+−=
19. ()()222 x hykr −+−= () 222 22 (2)(1)4 (2)(1)16 xy xy −−+−= ++−=
General form: 22 22 442116 42110 xxyy xyxy +++−+= ++−−=
20. ()()222 x hykr −+−= () 222 22 (5)((2))7 (5)(2)49 xy xy −−+−−= +++=
General form: 22 22 10254449 104200 xxyy xyxy +++++= +++−=
21. ()()222 x hykr −+−= 22 2 2 2 11 (0) 22 11 24 xy xy
−+−=
−+=
General form: 22 22 11 44 0 xxy xyx −++= +−=
Section2.3: Circles
22. ()()222 x hykr −+−=
23. 22 4 xy+= 222 2 xy+=
a. Center: (0,0); Radius2 = b.
c. x-intercepts: () 2 2 2 04 4 42 x x
-intercepts:
The intercepts are ()() 2,0,2,0, ()0,2, and ()0,2.
24. 22(1)1xy+−= 222 (1)1xy+−=
a. Center:(0,1); Radius1 = b.
c. x-intercepts: 22 2 2 (01)1 11 0 00 x x x x +−= += = =±=
-intercepts:
The intercepts are () 0,0 and ()0,2.
Chapter2: Graphs
25. () 2 2 2328 xy−+= () 2 2 34xy−+=
a. Center: (3, 0); Radius 2 = b.
c. x-intercepts: ()() () 22 2 304 34 34 32 32 x x x x x −+= −= −=± −=± =± 5 or 1 x x ==
y-intercepts: () () 2 2 2 2 2 2 034 34 94 5 y y y y −+= −+= += =− No real solution. The intercepts are () 1,0 and ()5,0.
26. ()() 22 31316 xy++−= ()() 22 112xy++−=
a. Center: (–1,1); Radius = 2 b.
c. x-intercepts: ()() ()() () () 22 22 2 2 1012 112 112 11 11 11 11 x x x x x x x ++−= ++−= ++= += +=± +=± =−± 0 or 2 xx==− y-intercepts: ()() ()() () () 22 22 2 2 0112 112 112 11 11 11 11 y y y y y y y ++−= +−= +−= −= −=± −=± =± 2 or 0 yy== The intercepts are ()() 2,0,0,0, and ()0,2.
27. 22 2440xyxy+−−−= 22 22 222 244 (21)(44)414 (1)(2)3 xxyy xxyy xy −+−= −++−+=++ −+−=
a. Center: (1, 2); Radius = 3 b.
c. x-intercepts: () () 222 222 2 2 (1)(02)3 (1)(2)3 149 15 15 15 x x x x x x −+−= −+−= −+= −= −=± =±
y-intercepts: () () 222 222 2 2 (01)(2)3 (1)(2)3 129 28 28 222 222 y y y y y y y −+−= −+−= +−= −= −=± −=± =±
The intercepts are ()() 15,0,15,0, −+ () 0,222, and () 0,222. +
28. 22 42200xyxy+++−= 22 22 222 4220 (44)(21)2041 (2)(1)5 xxyy xxyy xy +++= +++++=++ +++=
a. Center: (–2,–1); Radius = 5 b.
c. x-intercepts: 222 2 2 (2)(01)5 (2)125 (2)24 224 226 226 x x x x x x +++= ++= += +=± +=± =−± y-intercepts: 222 2 2 (02)(1)5 4(1)25 (1)21 121 121 y y y y y +++= ++= += +=± =−±
The intercepts are () 226,0, () 226,0, −+() 0,121, and () 0,121. −+
Section2.3: Circles
29. 22 22 22 222 4410 441 (44)(44)144 (2)(2)3 xyxy xxyy xxyy xy ++−−= ++−= +++−+=++ ++−=
a. Center: (–2, 2); Radius = 3 b. y x −55 5 −5 (−2,2)
c. x-intercepts: 222 2 2 (2)(02)3 (2)49 (2)5 25 25 x x x x x ++−= ++= += +=± =−± y-intercepts: 222 2 2 (02)(2)3 4(2)9 (2)5 25 25 y y y y y ++−= +−= −= −=± =±
The intercepts are () 25,0, () 25,0, −+() 0,25, and () 0,25. +
30. 22 22 22 222 6290 629 (69)(21)991 (3)(1)1 xyxy xxyy xxyy xy +−++= −++=− −++++=−++ −++=
a. Center: (3, –1); Radius = 1 b.
Chapter2: Graphs
b.
33.
34.
a. Center: (3,–2); Radius = 5
b.
c. x-intercepts:
Section2.3: Circles
are
a. Center: () 2,0 ; Radius: 2 r
Chapter2: Graphs
c. x-intercepts: ()() () 22 2 2 2 202 (2)4 24 22 22 x x x x x ++= += +=± +=± =−± 0 or 4 xx==−
y-intercepts: () 2 22 2 2 022 44 0 0 y y y y ++= += = =
The intercepts are () 4,0 and ()0,0.
36. () 22 22 22 2 2 33120 40 4404 24 xyy xyy xyy xy +−= +−= +−+=+ +−=
a. Center: () 0,2 ; Radius: 2 r =
b.
c. x-intercepts: () 2 2 2 2 024 44 0 0 x x x x +−= += = = y-intercepts: () () 2 2 2 024 24 24 22 22 y y y y y +−= −= −=± −=± =± 4 or 0 yy==
The intercepts are () 0,0 and ()0,4.
37. Center at (0, 0); containing point (–2, 3). ()() 22 20304913 r =−−+−=+=
Equation: () 2 22 22 (0)(0)13 13 xy xy −+−= +=
38. Center at (1, 0); containing point (–3, 2). ()() 22 31201642025 r =−−+−=+==
Equation: () 2 22 22 (1)(0)20 (1)20 xy xy −+−= −+=
39. Center at (2, 3); tangent to the x-axis. 3 r =
Equation: 222 22 (2)(3)3 (2)(3)9 xy xy −+−= −+−=
40. Center at (–3, 1); tangent to the y-axis. 3 r =
Equation: 222 22 (3)(1)3 (3)(1)9 xy xy ++−= ++−=
41. Endpoints of a diameter are (1, 4) and (–3, 2). The center is at the midpoint of that diameter:
Center: ()1(3)42,1,3 22 +−+ =−
Radius: 22 (1(1))(43)415 r =−−+−=+=
Equation: () 2 22 22 ((1))(3)5 (1)(3)5 xy xy −−+−= ++−=
42. Endpoints of a diameter are (4, 3) and (0, 1). The center is at the midpoint of that diameter:
Center: () 4031 22,2,2 ++ =
Radius: 22 (42)(32)415 r =−+−=+=
Equation: () 2 22 22 (2)(2)5 (2)(2)5 xy xy −+−= −+−=
43. Center at (–1, 3); tangent to the line y = 2. This means that the circle contains the point (–1, 2), so the radius is r = 1.
Equation: 222 22 (1)(3)(1) (1)(3)1 xy xy ++−= ++−=
44. Center at (4, –2); tangent to the line x = 1. This means that the circle contains the point (1, –2), so the radius is r = 3.
Equation: 222 22 (4)(2)(3) (4)(2)9 xy xy −++= −++=
45. (c); Center: () 1,2 ; Radius = 2
46. (d) ; Center: () 3,3 ; Radius = 3
47. (b) ; Center: () 1,2 ; Radius = 2
48. (a) ; Center: () 3,3 ; Radius = 3
49. Let the upper-right 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
The length of one side of the square is 2 x . Thus, the area is
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
The length of one side of the square is 2 x . Thus, the area of the square is () 2 23272 ⋅= square units. From the equation of the circle, we have 6 r = . The area of the circle is () 2 2 636 r πππ == square units. Therefore, the area of the shaded region is 3672 A π =− square units.
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 264125139 −= 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: ()() () 22 2 2 2 0139125 13915,625 xy xy −+−= +−=
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 550260290 −= 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: ()() () 22 2 2 2 0290260 29067,600 xy xy −+−= +−=
53. 22 2440910xyxy+++−= ()() 22 22 22 2440910 214440915 124096 xxyy xxyy xy +++−= +++++=+ +++=
The circle representing Earth has center () 1,2 and radius = 409664 = . So the radius of the satellite’s orbit is 640.664.6 += units. The equation of the orbit is ()()() 222 22 1264.6 244168.160 xy xyxy +++= +++−=
54. a. 222 22222 2222 () 2 (1)20 x mxbr x mxbmxbr mxbmxbr ++= +++= +++−=
There is one solution if and only if the discriminant is zero. 2222 22222222 2222 2222 222 (2)4(1)()0 444440 4440 0 (1) bmmbr bmbrbmmr brmr brmr rmb −+−= −+−+= −++=
b. Using the quadratic formula, the result from part (a), and knowing that the discriminant is zero, we get:
2222 (1)20 mxbmxbr +++−= 22 22 2 2 2 222222 2 2(1) bmbmbmrmr x b mb b r mr ymb b mrmrbr b bbb ====
c. The slope of the tangent line is m . The slope of the line joining the point of tangency and the center is: 2 2 2 2 0 1 0 r b rb bm
Therefore, the tangent line is perpendicular to the line containing the center of the circle and the point of tangency.
55. 22 9 xy+=
Center: (0, 0)
Slope from center to () 1,22 is 22022 22 101 ==
Slope of the tangent line is 12 4 22 =−
Equation of the tangent line is: () 2 221 4 22 22 44 48222 2492 24920 yx yx yx xy xy −=−−
56. 22 4640xyxy+−++= 22 22 (44)(69)449 (2)(3)9 xxyy xy −++++=−++ −++=
Center: (2, –3)
Slope from center to () 3,223 is 223(3)22 22 321 ==
Slope of the tangent line is: 12 4 22 =−
Equation of the tangent line: () 2 223(3) 4 232 223 44 48212232 24112120 yx yx yx xy −−=−−
57. Let (,) hk be the center of the circle. 240 24 1 2 2 xy yx yx −+= =+ =+
The slope of the tangent line is 1 2 . The slope from (,) hk to (0, 2) is –2. 2 2 0 22 k h kh =− −=
The other tangent line is 27yx=− , and it has slope 2.
Section2.3: Circles
The slope from (,) hk to (3, –1) is 1 2 . 11 32 223 21 12 k h kh kh hk =− +=− =− =−
Solve the two equations in and hk : 22(12) 224 30 0 kk kk k k −=− −=− = = 12(0)1 h =−=
The center of the circle is (1, 0).
58. Find the centers of the two circles: 22 22 22 4640 (44)(69)449 (2)(3)9 xyxy xxyy xy +−++= −++++=−++ −++=
Center: () 2,3 22 22 22 6490 (69)(44)994 (3)(2)4 xyxy xxyy xy ++++= +++++=−++ +++=
Center: () 3,2
Find the slope of the line containing the centers: 2(3)1 325 m ==−
Find the equation of the line containing the centers: 1 3(2) 5 5152 513 5130 yx yx xy xy +=−− +=−+ +=− ++=
59. Consider the following diagram: (2,2) Therefore, the path of the center of the circle has the equation 2 y =
60. 2 62 62 22 3 Cr r r r π ππ ππ ππ = = = =
The radius is 3 units long.
61. (b), (c), (e) and (g)
We need ,0hk > and () 0,0 on the graph.
62. (b), (e) and (g)
We need 0 h < , 0 k = , and hr >
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 ()() 22 2 x hykr −+−= . ()() () ()() 22 2 2 2 3216 324 xy xy ++−= −−+−=
Thus, ()() ,3,2hk =− and 4 r =
65. 2 2 2 (13) 169 cm 2 2(13) 26 cm Ar Cr π π π π π π = = = = = =
66. 2322 32 (32)(23)369246 38136 xxxxxxxx xxx −−+−+−+− = =−+−
67. () 2 2 2 22 2 2311 2311 23121 20 (2)(1)0 2 or 1 xxx xxx xxxx xx xx xx +−=+ +−=+ +−=++ +−= +−= =−=
We need to check each possible solution:
Chapter2: Graphs
Section 2.4
ykx =
False. If y varies directly with x, then ,ykx = where k is a constant.
b
c
11. () 22zkxy =+
22 22 534 5(25) 51 255 1 5 k k k zxy =+ = == =+
12. ()() 2 3 Tkxd =
13. 2 kd M
()322 zkxy =+
Section2.4: Variation
14. () 32zkxy =+
() 32 32 123 117 1 17 1 17 k k k zxy =+ = = =+
222 cab =+ 19. 1 2 A bh = 20. () 2 plw =+ 21. () 11 2 6.6710 mM F d
22. 2 32 Tl π = 23. ()6.491000 0.00649 pkB k k = = = Therefore we have the linear equation 0.00649 pB = . If 145000 B = , then () 0.00649145000$941.05 p == .
24. ()8.991000 0.00899 pkB k k = = = Therefore we have the linear equation 0.00899 pB = If 175000 B = , then () 0.00899175000$1573.25 p ==
25. () 2 2 161 16 s kt k k = = =
Therefore, we have equation 2 16. s t = If t = 3 seconds, then () 2 163144 s == feet.
If s = 64 feet, then 2 2 6416 4 2 t t t = = =±
Time must be positive, so we disregard 2. t =− It takes 2 seconds to fall 64 feet.
26. ()642 32 vkt k k = = = Therefore, we have the linear equation 32.vt = If t = 3 seconds, then () 32396 v == ft/sec.
27. ()320 3 20 EkW k k = = =
Therefore, we have the linear equation 3 20 EW = If W = 15, then () 3 152.25 20 E == .
28. 256 48 12,288 k R l k k = = =
Therefore, we have the equation 12,288 . R l = If 576, R = then 12,288 576 57612,288 12,28864 inches 5763 l l l = = == 29. ()47.4012 3.95 Rkg k k = = = Therefore, we have the linear equation 3.95 Rg = If 10.5 g = , then ()() 3.9510.5$41.48 R =≈ .
30. ()23.755 4.75 CkA k k = = = Therefore, we have the linear equation 4.75. CA = If 3.5 A = , then ()() 4.753.5$16.63 C ==
31. k D p = a. 156 D = , 2.75 p = ; 156 2.75 429 k k = = So, 429 D p = . b. 429 143 bags of candy 3 D == 32. k t s = a. 40 t = , 30 s = ; 40 30 1200 k k = = So, we have the equation 1200 t s = . b. 1200 30 minutes 40 t ==
33. k V P =
600,150VP== ;
600 150 90,000 k k = =
So, we have the equation 90,000 V P =
If 200 P = , then 3 90,000 450 cm. 200 V ==
34. k i R =
If 30,8iR== , then 30 and 240 8 k k ==
So, we have the equation 240 i R =
If 10, R = then 240 24 amperes 10 i == .
35. 2 k W d =
If 125,3960Wd== then 2 125 and 1,960,200,000 3960 k k ==
So, we have the equation 1,960,200,000 W d =
At the top of Mt. McKinley, we have 39603.83963.8 d =+= , so () 2 1,960,200,000 124.76 pounds. 3963.8 W =≈
36. 2 2 55 3960 862,488,000 k W d k k = = =
So, we have the equation 2 862,488,000 W d =
If =3965, d then 2 862,488,000 54.86 pounds. 3965 W =≈
37. 2 Vrh =π
38. 2 3 Vrh π =
39. 2 k I d =
If 0.075,2Id== , then 2 0.075 and 0.3 2 k k ==
So, we have the equation 2 0.3 I d =
If 5, d = then 2 0.3 0.012 foot-candles. 5 I ==
40. 2 2 11(20)(22) 119860 111 9680880 FkAv k k k = = = ==
So, we have the equation 2 1 . 880 F Av = If 47.125 A = and 36.5 v = , then ()() 2 1 47.12536.571.34 pounds. 880 F =≈
41. 3 36(75)(2)3 36600 0.06 hksd k k k = = = =
So, we have the equation 3 0.06.hsd = If 45 h = and 125, s = then 3 3 3 3 45(0.06)(125) 457.5 6 61.82 inches d d d d = = = =≈
42. (300) 100 15 10020 5 kT V P k k k = = = =
So, we have the equation 5T V P =
Chapter2: Graphs
If 80 V = and 310, T = then 5(310) 80 801550 1550 19.375 atmospheres 80 P P P = = ==
43. 2 2 1250(25)(10) 12502500 0.5 Kkmv k k k = = = =
So, we have the equation 2 0.5. K mv = If 25 m = and 15, v = then ()() 2 0.525152812.5 K == Joules
44. () 2 2 432 1.24 (4) 1.2427 1.24 27 kl R d k k k = = = =
So, we have the equation 2 1.24 . 27 l R d = If R = 1.44 and d = 3, then 2 1.24 1.44 27(3) 1.24 1.44 243 349.921.24 349.92 282.2 feet 1.24 l l l l = = = =≈
45. (25)(5) 100 0.75 75125 0.6 kpd S t k k k = = = =
So, we have the equation 0.6 pd S t = If 40, 8, pd== and 0.50, t = then 0.6(40)(8) 384 psi. 0.50 S == 46. 2 (4)(2)2 750 8 7502 375 kwt S l k k k = = = =
So, we have the equation 2 375 . wt S l = If 10, 6, lw== and 2, t = then 2 375(6)(2) 900 pounds. 10 S ==
47 – 50. Answers will vary. 51. 32 32 2 2 32512100 (325)(12100) (325)4(325) (4)(325) (2)(2)(325) xxx xxx xxx xx xxx +−− =+−+ =+−+ =−+ =−++
2 2 (3)(4) 5252 33(3)(4) 712 5(4) (3)(4) 5(4)(2) (3)(4) 5202 (3)(4) 618 (3)(4) 6(3)6 (3)(4)(4) x xx xx xxxx xx x xx xx xx xx xx x xx x xxx ++ +=+ ++++ ++ + =+ ++ ++− = ++ ++− = ++ + = ++ + == +++
3 31 22 3 44 2525 28 5125 = =
54. The term needed to rationalize the denominator is 72 +
Chapter 2 Review Exercises
1. ()()12 0,0 and 4,2 PP==
a. 2021 slope 4042 y x Δ− ==== Δ−
b. For each run of 2, there is a rise of 1.
2. ()()12 1,1 and 2,3 PP=−=−
a. ()31 44 slope 2133 y x Δ ====− Δ−−−
b. For each run of 3, there is a rise of 4.
3. ()()12 4,4 and 4,8 PP=−=
a. ()84 12 slope,undefined 440 y x Δ === Δ−
b. An undefined slope means the points lie on a vertical line. There is no change in x
4. ()()12 2,1 and 3,1 PP=−−=−
a. ()11 0 slope0 3(2)5 y x Δ ==== Δ−−
b. A slope of 0 means the points lie on a horizontal line. There is no change in y.
5. 2 23 x y =
x-intercepts: y-intercepts: 2 23(0) 20 0 x x x = = = 2 2 2(0)3 0 0 y y y = = =
The only intercept is (0,0).
Test x-axis symmetry: Let yy =−
2 2 23() 23 same xy xy =− =
Test y-axis symmetry: Let x x =− 2 2 2()3 23 different xy xy −= −=
Test origin symmetry: Let x x =− and yy =− 2 2 2()3() 23 different xy xy −=− −=
Therefore, the graph will have x-axis symmetry.
6. 22+4=16xy
x-intercepts: y-intercepts: () 2 2 2 +40=16 16 4 x x x = =± () 2 2 2
The intercepts are (4,0),(4,0),(0,2), and (0,2).
Test x-axis symmetry: Let yy =− () 2 2 22 4=16 4=16 same xy xy +− +
Test y-axis symmetry: Let x x =− () 2 2 22 4=16 4=16 same xy xy −+ +
Test origin symmetry: Let x x =− and yy =− . ()() 22 22 4=16 +4=16 same xy xy −+−
Therefore, the graph will have x-axis, y-axis, and origin symmetry.
7. 4234yxx=−−
x-intercepts: y-intercepts: ()() 42 22 2 2 034 041 40 4 2 xx xx x x x =−− =−+ −= = =± 42 (0)3(0)4 4 y =−− =−
The intercepts are (2,0),(2,0),(0,4), and (0,2).
Test x-axis symmetry: Let yy =− 42 42 34 34 different yxx yxx −=−− =−++
Test y-axis symmetry: Let x x =− ()() 42 42 34 34 same yxx yxx =−−−− =−−
Test origin symmetry: Let x x =− and yy =− . ()() 42 42 42 34 34 34 different yxx yxx yxx −=−−−− −=−− =−++
Therefore, the graph will have y-axis symmetry.
Chapter2: Graphs
8. 3 y xx=−
x-intercepts: y-intercepts: () ()() 3 2 0 01 011 xx xx xxx =− =− =+− 3 (0)0 0 y =− =
0,1,1xxx==−=
The intercepts are (1,0), (0,0), and (1,0).
Test x-axis symmetry: Let yy =− 3 3 different yxx yxx −=− =−+
Test y-axis symmetry: Let x x =− 3 3 ()() different yxx yxx =−−− =−+
Test origin symmetry: Let x x =− and yy =− . 3 3 3 ()() same yxx yxx yxx −=−−− −=−+ =−
Therefore, the graph will have origin symmetry.
9. 22 20xxyy+++=
x-intercepts: 22 2 (0)2(0)0 0 (1)0 xx xx xx +++= += += 0,1xx==−
y-intercepts: 22 2 (0)020 20 (2)0 yy yy yy +++= += += 0,2yy==−
The intercepts are (1,0), (0,0), and (0,2).
Test x-axis symmetry: Let yy =−
22 22 ()2()0 20 different xxyy xxyy ++−+−= ++−=
Test y-axis symmetry: Let x x =−
22 22 ()()20 20 different xxyy xxyy −+−++= −++=
Test origin symmetry: Let x x =− and yy =− 22 22 ()()()2()0 20 different xxyy xxyy −+−+−+−= −+−=
The graph has none of the indicated symmetries.
10. () ()() ()() 222 2 2 2 22 ()() 234 2316 x hykr xy xy −+−= −−+−= ++−=
11. () ()() () ()() 222 22 2 22 ()() 121 121 x hykr xy xy −+−= −−+−−= +++=
12. () () 2 2 2 22 14 12 xy xy +−= +−=
Center: (0,1); Radius = 2
x-intercepts: () 2 2 2 2 014 14 3 3 x x x x +−= += = =± y-intercepts: () 2 2 2 014 (1)4 12 12 y y y y +−= −= −=± =± 3 or 1 yy==−
The intercepts are ()3,0, ()3,0, ()0,1, and ()0,3.
13. ()() ()() 22 22 22 22 2 2440 244 2144414 123 xyxy xxyy xxyy xy +−+−= −++= −++++=++ −++=
Center: (1, –2) Radius = 3
x-intercepts:
-intercepts:
The intercepts are () 15,0, () 15,0, + () 0,222, and () 0,222. −+
Center: (1, –2) Radius = 5
x-intercepts: ()()()
-intercepts: ()()()
The intercepts are ()0,0, ()2,0, and ()0,4.
15. Slope = –2; containing (3,–1) () () 11 (1)23 126 25 or 25 yymxx yx yx yxxy
16. Slope = 0; containing the point (–5, 4) () () 11 40(5) 40 4 yymxx yx y y −=− −=−− −= =
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 3. x =− The equation does not have a slopeintercept form.
18. x-intercept = 2; containing the point (4, –5) Points are (2, 0) and (4, –5). 505 422 m ==− () () 11 5 02 2 5 5 or 5210 2 yymxx yx yxxy −=− −=−− =−++=
Chapter2: Graphs
19. y-intercept = –2; containing (5,–3) Points are (5,–3) and (0,–2) 2(3)11 0555 m ===− 1 2 or 510 5 ymxb yxxy =+ =−−+=−
20. Containing the points (3,–4) and (2, 1) 1(4)5 5 231 m ===− () () 11 (4)53 4515 511 or 511 yymxx yx yx yxxy
21. Parallel to 234 xy−=− 234 324 324 33 24 33 xy yx yx yx −=−
= =+ 2 Slope; containing (–5,3) 3 = () () () 11 2 3(5) 3 2 35 3 210 3 33 219 or 2319 33 yymxx yx yx yx yxxy
22. Perpendicular to 34 xy−=− 34 34 xy yx −=− =+ The slope of this line is 3, so the slope of a line perpendicular to it is 1 3 1 Slope 3 =− ; containing (–2, 4) () 11() 1 4(2) 3 12 4 33 110 or 310 33 yymxx yx yx yx xy
23. 4520 5420 4 4 5 xy yx yx
slope = 4 5 ; y-intercept = 4
x-intercept: Let y = 0. 45(0)20 420 5 x x x −=−
24. 111 236 111 326 31 22 xy yx yx −=− −=−− =+ slope = 3 2 ; 1 -intercept 2 y = x-intercept: Let y = 0. 111 (0) 236 11 26 1 3 x x x
25. 2312 xy−=
x-intercept: y-intercept: 23(0)12
The intercepts are () 6,0 and ()
26. 11 2 23xy+=
x-intercept: y-intercept:
The
27. 3 y x =
28. slope = 2 3 , containing the point (1,2)
29. Given the points (2,0),(4,4),AB=−=− and (8,5). C =
a. Find the distance between each pair of points. () () () 22 22 22 ,(4(2))(40) 416
Chapter2: Graphs
The Pythagorean Theorem is satisfied, so this is a right triangle.
b. Find the slopes:
Since 1 21 2 mmABAC⋅=−⋅=− , the sides AB and AC are perpendicular and the triangle is a right triangle.
30. Endpoints of the diameter are (–3, 2) and (5,–6). The center is at the midpoint of the diameter:
Center: ()()26 35 ,1,2 22 +−
Radius: 22 (1(3))(22) 1616 3242 r =−−+−−
Equation: ()()() ()() 2 22 22 1242 1232 xy xy −++= −++=
31. 15 slope of 1 62 15 slope of 1 82 11 slope of 1 86 AB AC BC ==− ==− ==−
Therefore, the points lie on a line.
32. () 854130,000 854427 130,00065,000 pkB k k = = ==
Therefore, we have the equation 427 65,000 pB = If 165,000 B = , then () 427 165,000$1083.92 65,000 p == .
33. 2 k w d = ()() 2 2 200 3960 20039603,136,320,000 k k = ==
Therefore, we have the equation 2 3,136,320,000 w d = . If 396013961 d =+= miles, then 2 3,136,320,000 199.9 3961 w =≈ pounds.
34. 135(7.5)(40) 135300 0.45 Hksd k k k = = = =
So, we have the equation 0.45. H sd = If 12 s = and 35, d = then ()() 0.451235189 H == BTU
Chapter 2 Test
1. a. 21 21 1342 5(1)63 yy m xx ====−
b. If x increases by 3 units, y will decrease by 2 units.
2. 2 9 yx=−
3. 2 y x = y x 10 5 −5 (1,1) (0,0) (4,(9,3) 2) 2 y x = (1,−1)(4,−2)(9,−3)
4. 2 9 xy+= x-intercepts: y-intercept: 2 2 09 9 3 x x x += = =± 2 (0)9 9 y y += =
The intercepts are ()3,0, ()3,0, and ()0,9.
Test x-axis symmetry: Let yy =−
() 2 2 9 9 different xy xy +−= −=
Test y-axis symmetry: Let x x =−
() 2 2 9 9 same xy xy −+= +=
Test origin symmetry: Let x x =− and yy =−
()() 2 2 9 9 different xy xy −+−= −=
Therefore, the graph will have y-axis symmetry.
5. Slope = 2 ; containing (3,4) 11() (4)2(3) 426 22 yymxx yx yx yx −=− −−=−− +=−+ =−+ 6. 239 329 2 3 3 xy yx yx += =−+ =−+ slope = 2 3 ; y-intercept = 3 x-intercept: Let y = 0. 23(0)9 39 3 x x x += = =
7. 3424 xy−= x-intercepts: y-intercept: 34(0)24 324 8 x x x −= = = 3(0)424 6 y y −= =−
The intercepts are () 8,0 and ()0,6.
8. ()()222 x hykr −+−= ()() ()() 22 2 22 4(3)5 4325 xy xy −+−−= −++=
General form: ()() 22 22 22 4325 8166925 860 xy xxyy xyxy −++= −++++= +−+=
Chapter2: Graphs
9. 22 22 22 222 4240 424 (44)(21)441 (2)(1)3 xyxy xxyy xxyy xy ++−−= ++−= +++−+=++ ++−=
Center: (–2, 1); Radius = 3 y x −55 5 −5 (−2,1)
10. 236 326 2 2 3 xy yx yx += =−+ =−+
Parallel line
Any line parallel to 236 xy+= has slope 2 3 m =− . The line contains (1,1) : 11() 2 (1)(1) 3 22 1 33 21 33 yymxx yx yx yx −=− −−=−− +=−+ =−−
Perpendicular line
Any line perpendicular to 236 xy+= has slope 3 2 m = . The line contains (0,3) : 11() 3 3(0) 2 3 3 2 3 3 2 yymxx yx yx yx −=−
11. Let R = the resistance, l = length, and r = radius. Then 2 l Rk r =⋅ . Now, R = 10 ohms, when l = 50 feet and 3 610 r =× inch, so () () 2 3 2 3 6 50 10 610 610 107.210 50 k k =⋅ × × =⋅=×
Therefore, we have the equation () 6 2 7.210 l R r =× . If 100 l = feet and 3 710 r =× inch, then ()() 6 2 3 100 7.21014.69 710 R =×≈ × ohms.
Chapter 2 Cumulative Review 1. 350 35 5 3 x x x −= = = The solution set is 5 3
2. ()() 2 120 430 xx xx −−= −+= 4or 3 xx==−
The solution set is {} 3,4 . 3. ()() 2 2530 2130 xx xx −−= +−= 1 or 3 2 xx=−=
The solution set is 1 ,3 2
.
4. 2 220xx−−= ()()()() () 2 22412 21 248 2 212 2 223 2 13 x −−±−−− = ±+ = ± = ± = =±
The solution set is {} 13,13 −+ .
5. 2 250xx++= ()() () 2 22415 21 2420 2 216 2 x −±− = −±− = −±− = No real solutions
6. () 2 2 213 213 219 28 4 x x x x x += += += = =
Check: 2(4)13? 93? 33 True += = = The solution set is {} 4
7. 21 x −= 21 or 21 3 1 xx xx −=−=− ==
The solution set is {} 1,3
8. () 2 2 22 2 2 42 42 44 440 xx xx xx xx += += += +−= 2 444(1)(4) 41616 2(1)2 432442 222 22 x −±−−−±+ == −±−± ===−±
Check 222 x =−+ : ()() 2 22242222?
48288822? 42 True −++−+= −+−+= =
Check 222 x =−− : ()() 2 22242222? 48288822? 42 True −−+−−= ++−−= =
The solution set is {} 222,222 −−−+ . 9. 2 9 9 3 x x x i =− =±− =± The solution set is {} 3,3ii .
10. 2 250xx−+= ()()()() () 2 22415 2420 212 21624 12 22 x i i −−±−−±− == ±−± ===±
The solution set is {} 12,12 ii−+
11. 237 210 5 x x x −≤ ≤ ≤ {}(] 5 or ,5 xx ≤−∞
Chapter2: Graphs
12. 145 51 x x −<+< −<< {}() 51 or 5,1 xx−<<−
13. 21 121 13 x x x −≤ −≤−≤ ≤≤ {}[] 13 or 1,3 xx≤≤
14. 23 x +> 23 or 23 5 or 1 xx xx +<−+> <−> {} 5 or 1 xxx<−> or ()() ,51, −∞−∪∞
15. ()()() () ()() 2 2 22 ,1432 55 2525 5052 dPQ =−−+−− =−+ =+ == ()32 1431 Midpoint,, 2222 +− −+ ==
16. 3 31 y xx =−+
a. () 2,1 : ()()() 3 23218611 −−−+=−++=− () 2,1 is on the graph.
b. () 2,3 : ()()() 3 23218613 −+=−+= () 2,3 is on the graph.
c. () 3,1 : ()()() 3 33312791191 −+=−+=≠ () 3,1 is not on the graph.
17. 3 y x =
18. The points (–1,4) and (2,–2) are on the line. 246 Slope2 2(1)3 ===− () () () 11() 421 421 224 22 yymxx yx yx yx yx −=− −=−−−
−=−+ =−−+ =−+
19. Perpendicular to 21 y x =+ ; Contains () 3,5 1 Slope of perpendicular = 2 11() 1 5(3) 2 13 5 22 113 22 yymxx yx yx yx −=− −=−− −=−+ =−+
Center: (2,–4); Radius = 5
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