Solution manual for precalculus 10th edition by sullivan

Chapter Graphs Section 1.1 (f) Quadrant IV − ( −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 or abscissa; y-coordinate or ordinate 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 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 =  ,   13 b 14 a 15 (a) (b) (c) (d) (e) Quadrant II x-axis Quadrant III Quadrant I y-axis C P E I Chapter 1: Graphs 18 The points will be on a horizontal line that is three units above the x-axis 28 d ( P1 , P2 ) = ( − (− 4) )2 + ( − (−3) )2 = 102 + 52 = 100 + 25 = 125 = 5 29 d ( P1 , P2 ) = (0 − a) + (0 − b) = ( − a ) + ( −b ) = a + b 30 d ( P1 , P2 ) = (0 − a ) + (0 − a) = (−a)2 + (−a)2 = a + a = 2a = a 19 d ( P1 , P2 ) = (2 − 0) + (1 − 0) = 22 + 12 = + = 31 A = (−2,5), B = (1,3), C = (−1, 0) d ( A, B ) = 20 d ( P1 , P2 ) = (−2 − 0) + (1 − 0) (1 − (−2) )2 + (3 − 5)2 = 32 + (−2) = + = 13 = (−2) + 12 = + = d ( B, C ) = 21 d ( P1 , P2 ) = (−2 − 1) + (2 − 1) ( −1 − 1)2 + (0 − 3)2 = (−2) + (−3)2 = + = 13 = (−3) + 12 = + = 10 22 d ( P1 , P2 ) = d ( A, C ) = ( − (−1) )2 + (2 − 1)2 ( −1 − (−2) )2 + (0 − 5)2 = 12 + (−5) = + 25 = 26 = 32 + 12 = + = 10 23 d ( P1 , P2 ) = (5 − 3) + ( − ( −4 ) ) 2 = 22 + ( ) = + 64 = 68 = 17 24 d ( P1 , P2 ) = = 25 d ( P1 , P2 ) = ( − ( −1) ) + ( − )2 ( 3)2 + 42 = + 16 = 25 = ( − (−3) )2 + (0 − 2)2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: = + (− 2) = 81 + = 85 26 d ( P1 , P2 ) = [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 ( − )2 + ( − (−3) )2 26 = 26 The area of a triangle is A = = 22 + = + 49 = 53 problem, C 26 ) 13 + 13 = 26 = 22 + 52 = + 25 = 29 27 d ( P1 , P2 ) = (6 − 4) + ( − (−3) ) ( 13 ) + ( 13 ) = ( P E I ⋅ bh In this Section 1.1: The Distance and Midpoint Formulas problem, A = ⋅ [ d ( A, B ) ] ⋅ [ d ( B, C ) ] = ⋅10 ⋅10 2 = ⋅100 ⋅ = 100 square units A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅ 13 ⋅ 13 = ⋅13 2 13 = square units 32 A = (−2, 5), B = (12, 3), C = (10, − 11) d ( A, B ) = 33 A = (− 5,3), B = (6, 0), C = (5,5) (12 − (−2) )2 + (3 − 5)2 d ( A, B) = = 142 + (−2) ( − (− 5) )2 + (0 − 3)2 = 196 + = 200 = 112 + (− 3) = 121 + = 10 = 130 d ( B, C ) = d ( B, C ) = (10 − 12 )2 + (−11 − 3)2 ( − )2 + (5 − 0)2 = (−2) + (−14) = (−1) + 52 = + 25 = + 196 = 200 = 26 d ( A, C ) = = 10 d ( A, C ) = (10 − (−2) )2 + (−11 − 5)2 ( − (− 5) )2 + (5 − 3)2 = 102 + 22 = 100 + = 122 + (−16) = 104 = 144 + 256 = 400 = 26 = 20 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: 2 [ d ( A, B)] + [ d ( B, C )] (10 ) + (10 ) ( = [ d ( A, C ) ] = ( 20 ) 104 26 ) =( 130 ) 104 + 26 = 130 130 = 130 The area of a triangle is A = bh In this 2 200 + 200 = 400 400 = 400 The area of a triangle is A = bh In this C ) +( P E I Chapter 1: Graphs problem, A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] = ⋅ 29 ⋅ 29 = ⋅ ⋅ 29 = 29 square units problem, A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] = ⋅ 104 ⋅ 26 = ⋅ 26 ⋅ 26 = ⋅ ⋅ 26 = 26 square units 35 A = (4, −3), B = (0, −3), C = (4, 2) d ( A, B ) = (0 − 4) + ( −3 − (−3) ) 34 A = (−6, 3), B = (3, −5), C = (−1, 5) = (− 4)2 + 02 = 16 + ( − (−6) )2 + (−5 − 3)2 d ( A, B) = = 16 =4 = 92 + (−8) = 81 + 64 = 145 d ( B, C ) = ( −1 − 3) + (5 − (−5)) d ( B, C ) = = 41 = 116 = 29 d ( A, C ) = ( −1 − (− 6) ) + (5 − 3) d ( A, C ) = (4 − 4) + ( − (−3) ) = 25 =5 = 29 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: 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 = 02 + 52 = + 25 = + = 25 + ( ( − )2 + ( − (−3) )2 = 42 + 52 = 16 + 25 = (−4) + 102 = 16 + 100 2 ) [ d ( A, B)]2 + [ d ( A, C )]2 = [ d ( B, C )]2 + 52 = 29 + ⋅ 29 = 145 16 + 25 = 41 41 = 41 29 + 116 = 145 145 = 145 The area of a triangle is A = bh In this The area of a triangle is A = C ( P E I 41 ) bh In this Section 1.1: The Distance and Midpoint Formulas problem, A = ⋅ [ d ( A, B) ] ⋅ [ d ( A, C ) ] = ⋅4⋅5 = 10 square units The area of a triangle is A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅4⋅2 = square units A= 36 A = (4, −3), B = (4, 1), C = (2, 1) d ( A, B ) = (4 − 4) + (1 − (−3) ) 37 The coordinates of the midpoint are: x +x y +y  ( x, y ) =  ,     3+5 −4+ 4 = ,   8 0 = ,  2 2 = (4, 0) = 02 + 42 = + 16 = 16 =4 d ( B, C ) = ( − )2 + (1 − 1)2 38 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   = (−2) + 02 = + = =2 d ( A, C ) = (2 − 4) + (1 − (−3) ) bh In this problem,  −2 + +  , =   0 4 = ,  2 2 = ( 0, ) = (−2) + 42 = + 16 = 20 =2 39 The coordinates of the midpoint are: x +x y +y  ( x, y ) =  ,     −3 + +  = ,   3 2 = ,  2 2 3  =  ,1 2  40 The coordinates of the midpoint are: x +x y +y  ( x, y ) =  ,     + −3 +  = ,    −1  = ,  2  1  =  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 42 + 22 = ( ) 16 + = 20 20 = 20 C P E I Chapter 1: Graphs 41 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   52 + b = 132 25 + b = 169 b = 144 b = 12  + −3 +  , =    10 −  = ,   2  = (5, −1) 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) b Consider points of the form ( 3, y ) that are a 42 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   distance of 13 units from the point ( −2, −1)  − + −3 +  , =    − −1  ,  =  2  d= ( x2 − x1 )2 + ( y2 − y1 )2 = ( − (−2) )2 + ( −1 − y )2 = ( 5)2 + ( −1 − y )2 = 25 + + y + y 1  =  −1, −  2  y + y + 26 = 43 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   y + y + 26 13 = 132 = a+0 b+0 = ,   a b = ,   2 ( y + y + 26 ) 169 = y + y + 26 = y + y − 143 = ( y − 11)( y + 13) y − 11 = or y + 13 = y = 11 y = −13 44 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   Thus, the points ( 3,11) and ( 3, −13) are a distance of 13 units from the point ( −2, −1) a+0 a+0 = ,   a a = ,  2 2 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: 45 The x coordinate would be + = and the y coordinate would be − = Thus the new point would be ( 5,3) 82 + b = 17 64 + b = 289 b = 225 46 The new x coordinate would be −1 − = −3 and the new y coordinate would be + = 10 Thus the new point would be ( −3,10 ) b = 15 Thus the coordinates will have an x value of − 15 = −14 and + 15 = 16 So the points are ( −14, −6 ) and (16, −6 ) 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: C P E I Section 1.1: The Distance and Midpoint Formulas b Consider points of the form ( x, −6 ) that are x = + 3 or x = − 3 a distance of 17 units from the point (1, ) Thus, the points + 3, and − 3, are ( (1 − x )2 + ( − ( −6 ) ) = = x − x + + (8) ( ) on the x-axis and a distance of units from the point ( 4, −3) ( x2 − x1 )2 + ( y2 − y1 )2 d= ) 50 Points on the y-axis have an x-coordinate of Thus, we consider points of the form ( 0, y ) that = x − x + + 64 are a distance of units from the point ( 4, −3) = x − x + 65 d= 17 = x − x + 65 ( 17 = x − x + 65 ) = 42 + + y + y = 16 + + y + y = x − x − 224 62 = y= distance of units from the point ( 4, −3) ( −6) ± (6)2 − 4(1)( −11) 2(1) ( ) ( Thus, the points 0, −3 + and 0, −3 − = 16 − x + x + ) are on the y-axis and a distance of units from the point ( 4, −3) = x − x + 25 = x − x + 25 51 a 2 36 = x − x + 25 = x − x − 11 x= −6 ± 36 + 44 −6 ± 80 = 2 −6 ± = = −3 ± y = −3 + or y = −3 − ( − x )2 + ( −3 − )2 ) ) = ( x2 − x1 )2 + ( y2 − y1 )2 ( y + y + 25 = y + y − 11 49 Points on the x-axis have a y-coordinate of Thus, we consider points of the form ( x, ) that are a x − x + 25 ( 36 = y + y + 25 distance of 13 units from the point (1, ) = 16 − x + x + ( −3) y + y + 25 6= x = −14 x = 16 Thus, the points ( −14, −6 ) and (16, −6 ) are a 62 = y + y + 25 = = ( x + 14 )( x − 16 ) x + 14 = or x − 16 = = ( − ) + ( −3 − y ) = 289 = x − x + 65 d= ( x2 − x1 )2 + ( y2 − y1 )2 To shift units left and units down, we subtract from the x-coordinate and subtract from the y-coordinate (2 − 3,5 − 4) = ( −1,1) b To shift left units and up units, we subtract from the x-coordinate and add to the y-coordinate ( − 2,5 + 8) = ( 0,13) −(−8) ± (−8) − 4(1)(−11) 2(1) ± 64 + 44 ± 108 = 2 8±6 = = 4±3 = C P E I Chapter 1: Graphs 52 Let the coordinates of point B be ( x, y ) Using the midpoint formula, we can write  −1 + x + y  ( 2,3) =  ,    This leads to two equations we can solve −1 + x 8+ y =2 =3 2 −1 + x = 8+ y = x=5 y = −2 = (− 4) + (− 1) = 16 + = 17 d ( B, E ) = = 20 = d ( A, F ) = (2 − 0) + (5 − 0) = 22 + 52 = + 25 = 29 56 Let P1 = (0, 0), P2 = (0, 4), P = ( x, y )  x + x y + y2  53 M = ( x, y ) =  ,   P1 = ( x1 , y1 ) = (−3, 6) and ( x, y ) = (−1, 4) , so and = x2 Thus, P2 = (1, 2) d ( P1 , P2 ) = (0 − 0) + (4 − 0) = 16 = y + y2 y= + y2 4= = + y2 d ( P1 , P ) = ( x − 0) + ( y − 0) = x2 + y = → x + y = 16 d ( P2 , P ) = ( x − 0) + ( y − 4) 2 = y2 = x + ( y − 4) = → x + ( y − 4) = 16 Therefore,  x + x y + y2  54 M = ( x, y ) =  ,   P2 = ( x2 , y2 ) = (7, −2) and ( x, y ) = (5, −4) , so x +x x= 2 x1 + 5= 10 = x1 + = x1 and Thus, P1 = (3, −6) ( − )2 + (2 − 0)2 = (− 4) + 22 = 16 + Point B has coordinates ( 5, −2 ) x +x x= 2 −3 + x2 −1 = −2 = −3 + x2 ( − )2 + (3 − 4)2 d (C , D) = y2 = ( y − 4) y = y − y + 16 y = 16 y=2 which gives x + 22 = 16 y + y2 y= y1 + (−2) −4 = −8 = y1 + (−2) x = 12 −6 = y1 x = ±2 Two triangles are possible The third vertex is 0+6 0+0 , 55 The midpoint of AB is: D =    = ( 3, ) (− ) ( 0+4 0+4 The midpoint of AC is: E =  ,   = ( 2, ) 6+4 0+4 , The midpoint of BC is: F =    = ( 5, ) C P E ) 3, or 3, I Section 1.1: The Distance and Midpoint Formulas 57 Let P1 = ( 0, ) , P2 = ( 0, s ) , P3 = ( s, ) , and 0+a 0+0  a  , P4 = M P1P2 =   =  , 0  2     a   3a a  a  P5 = M P2 P3 =  a + +  =  ,  ,          3a  a 0+ 0+  2,  =a, 3a P6 = M P1P3 =       4      P4 = ( s, s ) 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  0+s s+0  s s  M 2,3 =  , = ,   2 2  The midpoints of the diagonals are the same Therefore, the diagonals of a square intersect at their midpoints   3a a   a d ( P4 , P5 ) =  −  +  −   2   a  3a =   +       = a 3a P3 =  ,  To show that these vertices 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 ) = = d ( P2 , P3 ) = ( x2 − x1 ) + ( y2 − y1 ) ( a − )2 + ( − )2 = d ( P1 , P3 ) = = a 3a + = 4 a a 3a + = 16 16 2 a a2 = Since the sides are the same length, the triangle is equilateral = 4a = a2 = a ( x2 − x1 )2 + ( y2 − y1 )2 a 3a 4a + = = a2 = a 4 Since all three distances have the same constant value, the triangle is an equilateral triangle Now find the midpoints: = C a =   + 02 2 2  a   3a =  −  +  −  2     2 3a  3a a   a d ( P5 , P6 ) =  −  +  −    4  = a2 = a ( x2 − x1 )2 + ( y2 − y1 )2 a a 3a + = 16 16  a  3a =  −  +       2  a   3a =  − a  +  −  2    2  a a  3a d ( P4 , P6 ) =  −  +  −  4 2   58 Let P1 = ( 0, ) , P2 = ( a, ) , and P E I Chapter 1: Graphs 59 d ( P1 , P2 ) = (− − 2) + (1 − 1) 61 d ( P1 , P2 ) = ( − (− 2) )2 + ( − (−1) )2 = (− 6) + 02 = 22 + 82 = + 64 = 68 = 36 =6 = 17 d ( P2 , P3 ) = ( − − (− 4) ) = + (− 4) + (−3 − 1) = 32 + (− 5) = + 25 = 34 = 16 =4 ( − (−2) )2 + ( − (−1) )2 d ( P1 , P3 ) = = 52 + 32 = 25 + d ( P1 , P3 ) = (− − 2) + (−3 − 1) = 34 Since d ( P2 , P3 ) = d ( P1 , P3 ) , the triangle is isosceles = (− 6) + (− 4) = 36 + 16 = 52 2 2 the triangle is also a right triangle Therefore, the triangle is an isosceles right triangle Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , the triangle is a right triangle 62 d ( P1 , P2 ) = ( − (−1) )2 + (2 − 4)2 ( − − )2 + ( − )2 = + (− 2) = (−11) + (− 2) = 49 + = 121 + = 125 =5 = 53 d ( P2 , P3 ) = ( − )2 + (−5 − 2)2 d ( P2 , P3 ) = = 100 = 10 = + 49 = 53 d ( P1 , P3 ) = ( − (−1) ) + (−5 − 4)2 = + (− 9) ( − )2 + ( − )2 = (−3) + 42 = + 16 = 25 =5 = 25 + 81 = 106 ( − (− 4) )2 + (6 − 0)2 = 82 + 62 = 64 + 36 = (− 2)2 + (− 7) d ( P1 , P3 ) = Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , = 13 60 d ( P1 , P2 ) = ( − )2 + (2 − 7)2 d ( P2 , P3 ) = 2 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , 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 C 2 Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , 10 P E I Section 1.4: Circles y-intercepts: (0 − 1) + ( y − 2) = 32 29 (−1) + ( y − 2) = 32 + ( y − 2) = ( y − )2 = x2 + y2 + x − y − = x2 + x + y − y = ( x + x + 4) + ( y − y + 4) = + + ( x + 2) + ( y − 2) = 32 a Center: (–2, 2); Radius = y b y−2 = ± y − = ±2 y = 2±2 ( )( ) , and ( 0, + 2 ) (−2, 2) ) The intercepts are − 5, , + 5, , ( 0, − x + y + x + y − 20 = 28 x −5 −5 x + x + y + y = 20 c ( x + x + 4) + ( y + y + 1) = 20 + + x-intercepts: ( x + 2) + (0 − 2) = 32 ( x + 2) + = ( x + 2) + ( y + 1) = 52 ( x + 2) = a Center: (–2,–1); Radius = b x+2= ± x = −2 ± y-intercepts: (0 + 2) + ( y − 2) = 32 + ( y − 2)2 = ( y − 2)2 = y−2 = ± y = 2± c ( ( −2 + ( x + 2) + = 25 ( x + 2) = 24 30 x + = ± 24 x + = ±2 x = −2 ± y-intercepts: (0 + 2) + ( y + 1) = 52 + ( y + 1) = 25 b y + = ± 21 y = −1 ± 21 ) The intercepts are −2 − 6, , ) 21 , and 45 C P E ) Center: (3, –1); Radius = ( y + 1) = 21 ( −2 + 6, 0) , ( 0, − − ( 0, − + 21 ) )( ( I ) 5, , 0, − , and 0, + x2 + y2 − x + y + = x − x + y + y = −9 ( x − x + 9) + ( y + y + 1) = −9 + + ( x − 3) + ( y + 1) = 12 a ( ) The intercepts are −2 − 5, , x-intercepts: ( x + 2) + (0 + 1) = 52 Chapter 1: Graphs c x-intercepts: ( x − 3) + (0 + 1) = 12 ( x − 3) + = 32 ( x − 3)2 = x−3 = x=3 y-intercepts: (0 − 3) + ( y + 1) = 12 =0 x2 + x + y + y = 1  1 1  x + x+ + y + y+  = + + 4  4 4  x2 + y2 + x + y − + ( y + 1) = ( y + 1)2 = −8 a No real solution The intercept only intercept is ( 3, ) 31 1  1   x +  + y +  =1 2     b  1 Center:  − , −  ; Radius =  2 x + y − x + y +1 = x − x + y + y = −1 1  2  x − x +  + ( y + y + 1) = −1 + + 4   2 1  1  x −  + ( y + 1) =   2  2 a 1  Center:  , −1 ; Radius = 2  c b 2 1  1  x-intercepts:  x +  +  +  = 12 2  2  1  x +  + =1 2  c 1  1 x-intercepts:  x −  + (0 + 1) =   2  2 1   x −  +1 =   1  x+  = 2  x+ =± 2 −1 ± x= 2 2 1  x−  = − 2  No real solutions 1  1 y-intercepts:  −  + ( y + 1) =   2  2 1 + ( y + 1) = 4 ( y + 1)2 = y +1 = y = −1 1  y+  = 2  y+ =± 2 −1 ± y=  −1 −   −1 +  ,  ,  ,  , The intercepts are       −1 −   −1 +   0,  , and  0,  2     The only intercept is ( 0, −1) C 1  1  y-intercepts:  +  +  y +  = 12 2  2   1 + y +  =1  2 46 P E I Section 1.4: Circles b x + y − 12 x + y − 24 = 33 x + y − x + y = 12 x − x + y + y = 12 ( x − x + 9) + ( y + y + 4) = 12 + + ( x − 3)2 + ( y + 2) = 52 a Center: (3,–2); Radius = b c 2 = x-intercepts: ( x + 2) + ( ) = ( x + )2 2 x+2= ± x+2= ± c x-intercepts: ( x − 3) + (0 + 2) = 52 x = −2 ± ( x − 3) + = 25 ( x − 3)2 = 21 2 4+ y = y-intercepts: (0 + 2) + y = x − = ± 21 x = ± 21 y-intercepts: (0 − 3) + ( y + 2) = 52 No real solutions   ,  and The intercepts are  −2 −     ,   −2 +   y2 = − 2 + ( y + ) = 25 ( y + )2 = 16 y + = ±4 y = −2 ± y = or y = −6 ( ) (3 + The intercepts are − 21, , ( 0, −6 ) , 34 a ) 21, , 35 and ( 0, ) 2x2 + y2 + 8x + = x + x + y = −7 x2 + x + y2 = − 2 ( x + x + 4) + y = − + 2 ( x + 2) + y = 2  2 ( x + 2) + y =     Center: (–2, 0); Radius = x2 + 8x + y2 = x2 + x + y2 = x2 + x + + y = + a ( x + )2 + y = 22 Center: ( −2, ) ; b 2 47 C 2 P E I Radius: r = Chapter 1: Graphs 37 Center at (0, 0); containing point (–2, 3) x-intercepts: ( x + ) + ( ) = 22 c ( x + 2)2 = ( x + )2 = ± r= ( −2 − )2 + ( − )2 Equation: ( x − 0)2 + ( y − 0) = x + = ±2 y-intercepts: ( + ) + y = r= 4+ y = ( −3 − 1)2 + ( − )2 y =0 ( 20 ) ( x − 1) + y = 20 y=0 The intercepts are ( −4, ) and ( 0, ) 39 Center at (2, 3); tangent to the x-axis r =3 Equation: ( x − 2) + ( y − 3) = 32 36 x + y − 12 y = x2 + y − y = ( x − 2) + ( y − 3) = x + y − 4y + = + 40 Center at (–3, 1); tangent to the y-axis r =3 Equation: ( x + 3) + ( y − 1) = 32 x2 + ( y − 2) = Center: = 16 + = 20 = Equation: ( x − 1) + ( y − 0) = a 38 Center at (1, 0); containing point (–3, 2) 2 ( 13 ) x + y = 13 x = −2 ± x = or x = −4 = + = 13 ( 0, ) ; Radius: r=2 ( x + 3) + ( y − 1) = b 41 Endpoints of a diameter are (1, 4) and (–3, 2) The center is at the midpoint of that diameter:  + (−3) +  Center:  ,  = ( −1,3)   Radius: r = (1 − (−1)) + (4 − 3) = + = Equation: ( x − (−1)) + ( y − 3) = c x-intercepts: x + ( − ) = 42 Endpoints of a diameter are (4, 3) and (0, 1) The center is at the midpoint of that diameter:  + +1 Center:  ,  = ( 2, )   x2 + = x2 = x=0 Radius: r = (4 − 2) + (3 − 2) = + = y-intercepts: + ( y − ) = ( y − )2 = Equation: ( x − 2) + ( y − 2) = y−2 = ± ( 5) ( x − 2) + ( y − 2) = y − = ±2 y = 2±2 y = or y = The intercepts are ( 0, ) and ( 0, ) C ( x + 1) + ( y − 3) = 2 ( 5) 48 P E I Section 1.4: Circles 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 + y = 36 x + x = 36 x = 36 x = 18 ( x + 1) + ( y − 3) = x=3 The length of one side of the square is 2x Thus, 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 ( the area of the square is ⋅ ( x − 4) + ( y + 2) = 47 (b) ; Center: ( −1, ) ; Radius = 48 (a) ; Center: ( −3,3) ; Radius = Therefore, the area of the shaded region is A = 36π − 72 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 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: 49 Let the upper-right corner of the square be the point ( x, y ) The circle and the square are both ( x − )2 + ( y − 139 )2 = 1252 x + ( y − 139 ) = 15, 625 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 = 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: x2 + x2 = 2x2 = 9 x2 = ( x − 0)2 + ( y − 290)2 = 2602 x + ( y − 290) = 67, 600 = 2 The length of one side of the square is 2x Thus, the area is x=  2 A = s =  ⋅  = 2   ( ) 53 x + y + x + y − 4091 = x + x + y + y − 4091 = = 18 square units x + x + + y + y + = 4091 + ( x + 1)2 + ( y + )2 = 4096 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 representing Earth has center ( −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 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 ( x + 1)2 + ( y + )2 = ( 64.6 )2 x + y + x + y − 4168.16 = 49 C = 72 square π r = π ( ) = 36π square units 45 (c); Center: (1, −2 ) ; Radius = ( −3,3) ; Radius = units From the equation of the circle, we have r = The area of the circle is 46 (d) ; Center: ) P E I Chapter 1: Graphs (1 + m ) x + 2bmx + b − r = There is one solution if and only if the discriminant is zero (2bm) − 4(1 + m )(b − r ) = Equation of the tangent line is: y−2 = − ( x − 1) 2 y−2 = − x+ 4 4y −8 = − x + 4b m − 4b + 4r − 4b m + 4m r = x + 4y = x + (mx + b) = r 54 a x + m x + 2bmx + b = r 2 x + 4y − = − 4b + 4r + 4m r = − b2 + r + m2 r = 56 x + y − x + y + = r (1 + m ) = b b ( x − x + 4) + ( y + y + 9) = −4 + + Using the quadratic formula, the result from part (a), and knowing that the discriminant is zero, we get: (1 + m ) x + 2bmx + b − r = − 2bm −bm −bmr −mr = = = 2 b 2(1 + m )  b  b  2 r   −mr  y = m +b  b  ( x − 2) + ( y + 3) = Center: (2, –3) ( 2 − − (−3) 2 = =2 3− x= = c Slope of the tangent line is: ( 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  −1 2 =− C ) 2−k = −2 0−h − k = 2h The other tangent line is y = x − , and it has slope is 2 −0 2 = =2 1− Slope of the tangent line is 57 Let (h, k ) be the center of the circle x − 2y + = 2y = x + y = x+2 The slope of the tangent line is The slope from (h, k ) to (0, 2) is –2 55 x + y = Center: (0, 0) ) =− x + y − 11 + 12 = Therefore, the tangent line is perpendicular to the line containing the center of the circle and the point of tangency ( −1 2 Equation of the tangent line: ( x − 3) y − 2 −3 = − y−2 +3 = − x+ 4 y − + 12 = − x + −m2 r −m2 r + b2 r +b = = b b b Slope from center to 1, 2 ) Slope from center to 3, 2 − is 2 50 P E I Chapter Review Exercises Therefore, the path of the center of the circle has the equation y = The slope from (h, k ) to (3, –1) is − −1 − k =− 3−h 2 + 2k = − h 2k = − h 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) 60 C = 2π r 6π = 2π r 6π 2π r = 2π 2π 3=r The radius is units long 61 (b), (c), (e) and (g) We need h, k > and ( 0, ) on the graph 62 (b), (e) and (g) We need h < , k = , and h > r 58 Find the centers of the two circles: x2 + y2 − 4x + y + = 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 ( x − x + 4) + ( y + y + 9) = − + + ( x − 2) + ( y + 3) = 2 standard form ( x − h ) + ( y − k ) = r Center: ( 2, −3) ( x + 3) + ( y − ) x2 + y + x + y + = ( x + x + 9) + ( y + y + 4) = − + + = 16 ( x − ( −3)) + ( y − 2) ( x + 3) + ( y + 2)2 = = 42 Thus, ( h, k ) = ( −3, ) and r = 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 = Chapter Review Exercises P1 = ( 0, ) and P2 = ( 4, ) a d ( P1 , P2 ) = ( − )2 + ( − )2 = 16 + = 20 = b The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   59 Consider the following diagram: 0+4 0+2  2 = ,  =  ,  = ( 2,1)  2 2  (2,2) c slope = Δy − = = = Δx − d For each run of 2, there is a rise of 51 C P E I Chapter 1: Graphs P1 = (1, −1) and P2 = ( −2,3) a d ( P1 , P2 ) = ( −2 − 1)2 + ( − ( −1) ) x = y x-intercepts: y-intercepts: x = 3(0) 2(0) = y 2x = 0 = y2 y=0 x=0 The only intercept is (0, 0) = + 16 = 25 = b The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   Test x-axis symmetry: Let y = − y x = 3(− y ) 2 x = y same Test y-axis symmetry: Let x = − x 2(− x) = y  + ( −2 ) −1 +  = ,     −1    =  ,  =  − ,1  2   c −2 x = y different Test origin symmetry: Let x = − x and y = − y Δy − ( −1) slope = = = =− Δx −2 − −3 2(− x) = 3(− y ) −2 x = y different Therefore, the graph will have x-axis symmetry d For each run of 3, there is a rise of −4 P1 = ( 4, −4 ) and P2 = ( 4,8 ) a d ( P1 , P2 ) = ( − )2 + (8 − ( −4 ) ) x +4 y =16 = + 144 = 144 = 12 x-intercepts: b The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   slope = Δy − ( −4 ) 12 = = , undefined 4−4 Δx y = 16 Test x-axis symmetry: Let y = − y x + ( − y ) =16 x + y =16 same y = x + Test y-axis symmetry: Let x = − x y ( − x )2 + y =16 x + y =16 same (2, 8) Test origin symmetry: Let x = − x and y = − y (1, 5) (0, 4) ( − x )2 + ( − y )2 =16 x +4 y =16 −5 −1 y2 = (0, 2) (−1, 5) x = 16 y = ±2 The intercepts are (−4, 0), (4, 0), (0, −2), and d An undefined slope means the points lie on a vertical line There is no change in x (−2, 8) ( )2 +4 y =16 x = ±4  + −4 +    = ,  =  ,  = ( 4, )  2 2  c y-intercepts: x +4 ( ) =16 x-intercepts: −4, 0, ; y-intercepts: −2, 0, Intercepts: (−4, 0), (0, 0), (2, 0), (0, −2), (0, 2) C same Therefore, the graph will have x-axis, y-axis, and origin symmetry x 52 P E I Chapter Review Exercises 10 x + x + y + y = y = x +2 x +1 x-intercepts: = x +2 x +1 = x2 + x2 + ( )( x-intercepts: x + x + (0) + 2(0) = x2 + x = x( x + 1) = x = 0, x = −1 y-intercepts: y = (0) +2(0) +1 =1 ) x +1 = y-intercepts: (0) + + y + y = y2 + y = y ( y + 2) = y = 0, y = −2 The intercepts are (−1, 0), (0, 0), and (0, −2) x = −1 no real solutions The only intercept is (0, 1) Test x-axis symmetry: Let y = − y − y = x4 + x2 + y = − x − x − different Test x-axis symmetry: Let y = − y x + x + (− y ) + 2(− y ) = Test y-axis symmetry: Let x = − x x + x + y − y = different Test y-axis symmetry: Let x = − x (− x) + (− x) + y + y = x − x + y + y = different Test origin symmetry: Let x = − x and y = − y y = (−x) + 2(−x) +1 y = x4 + x2 + same Test origin symmetry: Let x = − x and y = − y − y = (−x) + 2(−x) +1 − y = x4 + x2 + y = − x4 − x2 − (− x) + (− x) + (− y ) + 2(− y ) = different x − x + y − y = different The graph has none of the indicated symmetries Therefore, the graph will have y-axis symmetry y = x3 − x 11 x-intercepts: = x3 − x ( ( x − ( −2 ) ) + ( y − 3)2 = 42 y-intercepts: y = (0)3 − =0 = x x2 − ) ( x − h) + ( y − k ) = r ( x + )2 + ( y − 3)2 = 16 = x ( x + 1)( x − 1) ( x − h) + ( y − k ) = r 12 x = 0, x = −1, x = ( x − ( −1) ) + ( y − ( −2 ) ) The intercepts are (−1, 0), (0, 0), and (1, 0) = 12 ( x + 1)2 + ( y + )2 = Test x-axis symmetry: Let y = − y − y = x3 − x y = − x + x different 13 x + ( y − 1) = x + ( y − 1) = 22 Center: (0,1); Radius = Test y-axis symmetry: Let x = − x y = ( − x )3 − ( − x ) y = − x3 + x different Test origin symmetry: Let x = − x and y = − y − y = ( − x )3 − ( − x ) − y = − x3 + x y = x3 − x same Therefore, the graph will have origin symmetry 53 C P E I Chapter 1: Graphs 3x + y − x + 12 y = 15 x-intercepts: x + ( − 1) = x2 + = x2 + y2 − x + y = x2 = x2 − 2x + y2 + y = (x x=± − 2x + + y2 + y + = + ) ( 2 y-intercepts: + ( y − 1) = ) ( x − 1)2 + ( y + )2 = ( ( y − 1) = Center: (1, –2) Radius = y − = ±2 ) y = 1± y = or y = −1 ( )( The intercepts are − 3, , ) 3, , ( 0, −1) , and ( 0, 3) x2 + y − x + y − = 14 x2 − x + y + y = (x − 2x + + y2 + y + = + + ) ( ) x-intercepts: ( x − 1) + ( + ) = ( x − 1)2 + ( y + )2 = 32 ( 5) ( x − 1)2 + = ( x − 1)2 = Center: (1, –2) Radius = x − = ±1 x = 1±1 x = or x = 2 y-intercepts: ( − 1) + ( y + ) = ( 5) 2 + ( y + 2) = ( y + )2 = y + = ±2 x-intercepts: ( x − 1) + ( + ) = 32 y = −2 ± y = or y = −4 ( x − 1) + = ( x − 1)2 = The intercepts are ( 0, ) , ( 2, ) , and ( 0, −4 ) x −1 = ± 16 Slope = –2; containing (3,–1) y − y1 = m ( x − x1 ) x = 1± 2 y-intercepts: ( − 1) + ( y + ) = y − (−1) = −2 ( x − 3) + ( y + 2) = y + = −2 x + ( y + )2 = y = −2 x + or x + y = y+2= ± 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 y + = ±2 y = −2 ± 2 ( ) ( ) The intercepts are − 5, , + 5, , ( 0, −2 − 2 ) , and ( 0, −2 + 2 ) C 54 P E I Chapter Review Exercises 18 y-intercept = –2; containing (5,–3) Points are (5,–3) and (0,–2) 1 − − (−3) m= = =− 0−5 −5 22 x − y = − 20 −5 y = −4 x − 20 y = x+4 slope = ; y-intercept = y = mx + b y = − x − or x + y = −10 x-intercept: Let y = x − 5(0) = − 20 x = − 20 x = −5 19 Containing the points (3,–4) and (2, 1) − (−4) m= = = −5 2−3 −1 y − y1 = m ( x − x1 ) y − (− 4) = −5 ( x − 3) y + = −5 x + 15 y = −5 x + 11 or x + y = 11 20 Parallel to x − y = −4 2x − 3y = − −3 y = −2 x − −3 y −2 x − = −3 −3 y = x+ 3 Slope = ; containing (–5,3) 23 y − y1 = m ( x − x1 ) ( x − (−5) ) y − = ( x + 5) 10 y −3 = x+ 3 19 y = x+ or x − y = −19 3 y −3 = 1 x− y = − 1 − y =− x− y = x+ 2 slope = ; y -intercept = 2 x-intercept: Let y = 1 x − (0) = − 1 x=− x=− 21 Perpendicular to x + y = x+ y = y = −x + The slope of this line is −1 , so the slope of a line perpendicular to it is Slope = 1; containing (4,–3) y − y1 = m( x − x1 ) y − (−3) = 1( x − 4) y+3= x−4 y = x − or x − y = 55 C P E I Chapter 1: Graphs 24 x − y = 12 x-intercept: x − 3(0) = 12 x = 12 x=6 26 y = x3 y-intercept: 2(0) − y = 12 −3 y = 12 y = −4 The intercepts are ( 6, ) and ( 0, −4 ) 1 x+ y = 25 x-intercept: 1 x + (0) = 2 x=2 x=4 27 y = x y-intercept: 1 (0) + y = 2 y=2 y=6 The intercepts are ( 4, ) and ( 0, ) 28 slope = , containing the point (1,2) 29 Find the distance between each pair of points d A, B = (1 − 3) + (1 − 4) = + = 13 d B,C = (− − 1) + (3 − 1) = + = 13 d A,C = (− − 3) + (3 − 4) = 25 + = 26 Since AB = BC, triangle ABC is isosceles 30 Given the points A = (− 2, 0), B = (− 4, 4), and C = (8, 5) C 56 P E I Chapter Test a Find the distance between each pair of points 1− = −1 6−2 −1 − slope of AC = = −1 8−2 Therefore, the points lie on a line 32 slope of AB = d ( A, B ) = (− − (− 2)) + (4 − 0) = + 16 = 20 = d ( B, C ) = (8 − (− 4)) + (5 − 4) = 144 + Chapter Test = 145 d ( A, C ) = (8 − (− 2)) + (5 − 0) d ( P1 , P2 ) = = 100 + 25 = 62 + ( −4 ) = 125 = 5 2  d ( A, B )  +  d ( A, C )  =  d ( B, C )  ( 20 ) +( 125 ) =( 145 ) = 52 = 13 2 The coordinates of the midpoint are:  x + x y + y2  ( x, y ) =  ,   145 = 145 The Pythagorean Theorem is satisfied, so this is a right triangle  −1 + + (−1)  , =   4 2 = ,  2 2 = ( 2, 1) b Find the slopes: mAB = 4−0 = = −2 − − (− 2) − mBC = 5−4 = − ( − ) 12 mAC = 5−0 = = − ( − ) 10 2 = 36 + 16 20 + 125 = 145 a m= y2 − y1 −1 − −4 = = =− x2 − x1 − (−1) b If x increases by units, y will decrease by units = −1 , the sides AB and AC are perpendicular and the triangle is a right triangle Since mAB ⋅ mAC = − ⋅ y = x − 31 Endpoints of the diameter are (–3, 2) and (5,–6) The center is at the midpoint of the diameter:  −3 + + ( − )  , Center:   = (1, − )   Radius: r = (1 − (−3)) + (− − 2) = 16 + 16 = 32 = Equation: ( − (−1) )2 + ( −1 − 3)2 ( x − 1)2 + ( y + )2 = ( ) ( x − 1)2 + ( y + )2 = 32 57 C P E I Chapter 1: Graphs y = x ( x − h) + ( y − k ) = r y ( x − )2 + ( y − (−3) )2 = 52 ( x − )2 + ( y + 3)2 = 25 General form: ( x − )2 + ( y + 3)2 = 25 (1, 1) (4, 2) (9, 3) y2 = x x − x + 16 + y + y + = 25 10 x (0, 0) (1,−1) (4,−2) −5 x2 + y − 8x + y = (9,−3) x + y = x-intercepts: x2 + = x2 = x2 + y2 + x − y − = x2 + x + y − y = ( x + x + 4) + ( y − y + 1) = + + ( x + 2) + ( y − 1) = 32 Center: (–2, 1); Radius = y y-intercept: (0) + y = y=9 x = ±3 The intercepts are ( −3, ) , ( 3, ) , and ( 0, ) (−2, 1) x −5 Test x-axis symmetry: Let y = − y x2 + ( − y ) = x − y = different −5 Test y-axis symmetry: Let x = − x 10 x + y = y = −2 x + y = − x+2 ( − x )2 + y = x + y = same Test origin symmetry: Let x = − x and y = − y ( − x )2 + ( − y ) = Parallel line Any line parallel to x + y = has slope x − y = different Therefore, the graph will have y-axis symmetry m = − The line contains (1, −1) : y − y1 = m( x − x1 ) Slope = −2 ; containing (3, −4) y − y1 = m( x − x1 ) y − (−1) = − ( x − 1) 2 y +1 = − x + 3 y = − x− 3 y − (−4) = −2( x − 3) y + = −2 x + y = −2 x + Perpendicular line Any line perpendicular to x + y = has slope m= C 58 P E The line contains (0, 3) : I Chapter Project y − y1 = m( x − x1 ) ( x − 0) y −3 = x y = x+3 y −3 = Chapter Project Internet Based Project 59 C P E I ... four sides is the same Therefore, the vertices are for a square 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 kWh,... c 12 b 13 a Slope = 1− = 2−0 b If x increases by units, y will increase by unit 14 a Slope = 1− =− −2−0 b If x increases by units, y will decrease by unit 25 C P E I Chapter 1: Graphs 20 Slope... increases by units, y will decrease by unit 3; x-intercept: x + 3(0) = 2x = x=3 y-intercept: 2(0) + y = 3y = y=2 16 a Slope = −1 = − (−1) b If x increases by units, y will increase by unit 17