Chapter Graphs and Functions 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 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions 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) d ( B, C ) = = (−3) + 12 = + = 10 22 d ( P1 , P2 ) = (1 − (−2) )2 + (3 − 5)2 = 32 + (−2) = + = 13 = (−2) + 12 = + = 21 d ( P1 , P2 ) = (−2 − 1) + (2 − 1) 2 ( −1 − 1)2 + (0 − 3)2 = (−2) + (−3)2 = + = 13 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 2 = + (− 2) = 81 + = 85 26 d ( P1 , P2 ) = ( − )2 + ( − (−3) )2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 2 = 22 + = + 49 = 53 26 ) 13 + 13 = 26 = 22 + 52 = + 25 = 29 27 d ( P1 , P2 ) = (6 − 4) + ( − (−3) ) ( 13 ) + ( 13 ) = ( 26 = 26 The area of a triangle is A = problem, Copyright © 2016 Pearson Education, Inc ⋅ bh In this Section 1.1: The Distance and Midpoint Formulas 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 ) = (12 − (−2) )2 + (3 − 5)2 problem, A = ⋅ [ d ( A, B ) ] ⋅ [ d ( B, C ) ] = ⋅10 ⋅10 2 = ⋅100 ⋅ = 100 square units 33 A = (− 5,3), B = (6, 0), C = (5,5) 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: Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: 2 [ d ( A, B)] + [ d ( B, C )] (10 ) + (10 ) 2 = [ d ( A, C ) ] = ( 20 ) 200 + 200 = 400 400 = 400 The area of a triangle is A = bh In this [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 ( 104 ) +( 26 ) =( 130 ) 104 + 26 = 130 130 = 130 The area of a triangle is A = bh In this Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions 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) 34 A = (−6, 3), B = (3, −5), C = (−1, 5) ( − (−6) )2 + (−5 − 3)2 d ( A, B) = = 92 + (−8) = 81 + 64 = 145 d ( B, C ) = ( −1 − 3) + (5 − (−5)) = (−4) + 102 = 16 + 100 = 116 = 29 d ( A, C ) = ( −1 − (− 6) ) 2 + (5 − 3) Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 ) + (2 29 ) =( 145 = (− 4)2 + 02 = 16 + = 16 =4 d ( B, C ) = ( − )2 + ( − (−3) )2 = 42 + 52 = 16 + 25 = 41 d ( A, C ) = (4 − 4) + ( − (−3) ) = 25 =5 = 29 29 = 02 + 52 = + 25 = + = 25 + ( d ( A, B ) = (0 − 4) + ( −3 − (−3) ) ) Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ 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 = Copyright © 2016 Pearson Education, Inc 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 = A= 36 A = (4, −3), B = (4, 1), C = (2, 1) d ( A, B ) = (4 − 4) + (1 − (−3) ) = 02 + 42 = + 16 = 16 =4 d ( B, C ) = ( − )2 + (1 − 1)2 37 The coordinates of the midpoint are: x +x y +y ( x, y ) = , 3+5 −4+ 4 = , 8 0 = , 2 2 = (4, 0) 38 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , = (−2) + 02 = + = =2 d ( A, C ) = (2 − 4) + (1 − (−3) ) ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅4⋅2 = square units bh In this problem, 2 = (−2) + 42 = + 16 = 20 =2 −2 + + , = 0 4 = , 2 2 = ( 0, ) 39 The coordinates of the midpoint are: x +x y +y ( x, y ) = , −3 + + = , 3 2 = , 2 2 3 = ,1 2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 42 + 22 = ( 16 + = 20 20 = 20 ) 40 The coordinates of the midpoint are: x +x y +y ( x, y ) = , + −3 + = , −1 = , 2 1 = 3, − 2 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions 41 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , + −3 + , = 10 − = , 2 = (5, −1) 52 + b = 132 25 + b = 169 b = 144 b = 12 Thus the coordinates will have an y value of −1 − 12 = −13 and −1 + 12 = 11 So the points are ( 3,11) and ( 3, −13) 42 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , − + −3 + , = − −1 , = 2 b Consider points of the form ( 3, y ) that are a distance of 13 units from the point ( −2, −1) d= ( x2 − x1 )2 + ( y2 − y1 )2 = ( − (−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 ) = , a+0 b+0 = , a b = , 2 y + y + 26 13 = 132 = ( y + y + 26 ) 169 = y + y + 26 = y + y − 143 = ( y − 11)( y + 13) 44 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , a+0 a+0 = , a a = , 2 2 45 The x coordinate would be + = and the y coordinate would be − = Thus the new point would be ( 5,3) 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 ) 47 a If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length One of the legs of the triangle will be 2+3=5 Thus the other leg will be: y − 11 = or y + 13 = y = 11 y = −13 Thus, the points ( 3,11) and ( 3, −13) are a distance of 13 units from the point ( −2, −1) 48 a If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length One of the legs of the triangle will be 2+6=8 Thus the other leg will be: 82 + b = 17 64 + b = 289 b = 225 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 ) Copyright © 2016 Pearson Education, Inc 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 distance of 13 units from the point (1, ) 49 Points on the x-axis have a y-coordinate of Thus, we consider points of the form ( x, ) that are a distance of units from the point ( 4, −3) ( x2 − x1 )2 + ( y2 − y1 )2 ( − x )2 + ( −3 − )2 = 16 − x + x + ( −3) = 16 − x + x + = x − x + 25 = x − x + 25 ( x − x + 25 ) 2 36 = x − x + 25 = x − x − 11 x= −(−8) ± (−8) − 4(1)(−11) 2(1) ± 64 + 44 ± 108 = 2 8±6 = = 4±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 62 = ( y + y + 25 ) 36 = y + y + 25 = y + y − 11 y= ( −6) ± (6)2 − 4(1)( −11) 2(1) −6 ± 36 + 44 −6 ± 80 = 2 −6 ± = = −3 ± y = −3 + or y = −3 − = ( ) ( Thus, the points 0, −3 + and 0, −3 − ) are on the y-axis and a distance of units from the point ( 4, −3) 51 a 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) Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions 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 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) y + y2 y= + y2 4= = + y2 = y2 = x1 and Thus, P1 = (3, −6) 56 Let P1 = (0, 0), P2 = (0, 4), P = ( x, y ) d ( P1 , P2 ) = (0 − 0) + (4 − 0) = 16 = d ( P1 , P ) = ( x − 0) + ( y − 0) = x2 + y = → x + y = 16 d ( P2 , P ) = ( x − 0) + ( y − 4) = x + ( y − 4) = 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 + ( − )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) = y + y2 y= y1 + (−2) −4 = −8 = y1 + (−2) → x + ( y − 4) = 16 Therefore, y2 = ( y − 4) y = y − y + 16 y = 16 y=2 which gives x + 22 = 16 x = 12 −6 = y1 0+6 0+0 , 55 The midpoint of AB is: D = = ( 3, ) x = ±2 Two triangles are possible The third vertex is (− ) ( ) 3, or 3, 0+4 0+4 The midpoint of AC is: E = , = ( 2, ) 6+4 0+4 , The midpoint of BC is: F = = ( 5, ) Copyright © 2016 Pearson Education, Inc Section 1.1: The Distance and Midpoint Formulas 57 Let P1 = ( 0, ) , P2 = ( 0, s ) , P3 = ( s, ) , and 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 58 Let P1 = ( 0, ) , P2 = ( a, ) , and 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 3a = + = = a2 = a 4a = a2 = a ( x2 − x1 )2 + ( y2 − y1 )2 a 3a = − + − 2 a a 3a + = 16 16 2 a 3a = − + 2 a a 3a + = 16 16 2 3a 3a a a d ( P5 , P6 ) = − + − 4 2 a = + 02 2 2 2 a a 3a d ( P4 , P6 ) = − + − 4 2 = a 3a = − a + − 2 2 3a a a d ( P4 , P5 ) = − + − 2 ( x2 − x1 )2 + ( y2 − y1 )2 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 a a2 = Since the sides are the same length, the triangle is equilateral 4a a 3a + = = a2 = a 4 Since all three distances have the same constant value, the triangle is an equilateral triangle Now find the midpoints: = Copyright © 2016 Pearson Education, Inc = Chapter 1: Graphs and Functions 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 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , the triangle is a right triangle the triangle is also a right triangle Therefore, the triangle is an isosceles 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 2 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , the triangle is a right triangle 2 Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , 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 10 Copyright © 2016 Pearson Education, Inc Section 1.7: One-to-One Functions; Inverse Functions c T = 1815 + 0.15 ( g − 18,150) T − 1815 = 0.15 ( g − 18,150) T − 1815 = g − 18,150 0.15 T − 1815 + 18,150 = g 0.15 T − 1815 + 18,150 We would write g (T ) = 0.15 Domain: {T | 1815 ≤ T ≤ 10162.50} 98 a b The graph of H is symmetric about the y-axis Since t represents the number of seconds after the rock begins to fall, we know that t ≥ The graph is strictly decreasing over its domain, so it is one-to-one b H + 4.9t = 100 99 100 − H 4.9 (Note: we only need the principal square root since we know t ≥ ) Therefore, we would write t ( H ) = 100 − H H ( t ( H ) ) = 100 − 4.9 4.9 100 − H = 100 − 4.9 4.9 = 100 − 100 + H =H t ( H (t )) = 100 − 100 − 4.9t ( c 4.9t = t2 = t 4.9 l 32.2 8.05T π2 T = 8.05 , T > π 3 l ( 3) = 8.05 ≈ 7.34 π A pendulum whose period is seconds will be about 7.34 feet long ax + b cx + d ax + b y= cx + d ay + b x= Inverse cy + d x(cy + d ) = ay + b cxy + dx = ay + b cxy − ay = b − dx y (cx − a ) = b − dx b − dx y= cx − a −dx + b −1 f ( x) = cx − a f ( x) = Now, f = f −1 provided that ax + b −dx + b = cx + d cx − a This is only true if a = − d ) 4.9 = T = 2π l (T ) = H = 100 − 4.9t 4.9t = 100 − H 100 − H t2 = 4.9 100 − H t= 4.9 l 32.2 T l = 2π 32.2 T2 l = 32.2 4π 32.2T l= 4π Range: { g |18,150 ≤ g ≤ 73,800} 97 a T (l ) = 2π (since t ≥ 0) 100 − 80 t ( 80 ) = ≈ 2.02 4.9 It will take the rock about 2.02 seconds to fall 80 meters 100 Yes In order for a one-to-one function and its inverse to be equal, its graph must be symmetric about the line y = x One such example is the function f ( x ) = x 101 Answers will vary 111 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions 102 Answers will vary One example is 1 , if x < f ( x) = x x, if x ≥ P1 = (1, −1) and P2 = ( −2,3) a d ( P1 , P2 ) = ( −2 − 1)2 + ( − ( −1) ) = + 16 = 25 = b The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , + ( −2 ) −1 + = , −1 = , = − ,1 2 P1 = ( 4, −4 ) and P2 = ( 4,8 ) This function is one-to-one since the graph passes the Horizontal Line Test However, the function is neither increasing nor decreasing on its domain 103 No, not every odd function is one-to-one For example, f ( x) = x3 − x is an odd function, but it is not one-to-one 104 C −1 (800, 000) represents the number of cars manufactured for $800,000 105 If a horizontal line passes through two points on a graph of a function, then the y value associated with that horizontal line will be assigned to two different x values which violates the definition of one-to-one a d ( P1 , P2 ) = ( − )2 + (8 − ( −4 ) ) = + 144 = 144 = 12 b The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , + −4 + , = = , = ( 4, ) 2 2 y = x + y (2, 8) (−2, 8) (−1, 5) (1, 5) (0, 4) 106 Answers may vary −5 −1 x x-intercepts: −4, 0, ; y-intercepts: −2, 0, Intercepts: (−4, 0), (0, 0), (2, 0), (0, −2), (0, 2) 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 ) = , 0+4 0+2 2 , = = , = ( 2,1) 2 2 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) Test x-axis symmetry: Let y = − y x = 3(− y ) 2 x = y same 112 Copyright © 2016 Pearson Education, Inc Chapter Review Exercises Test y-axis symmetry: Let x = − x 2(− x) = y Test y-axis symmetry: Let x = − x −2 x = y different Test origin symmetry: Let x = − x and y = − y y = x4 + x2 + 2(− x) = 3(− y ) −2 x = y different Therefore, the graph will have x-axis symmetry 2 x +4 y =16 y-intercepts: x +4 ( ) =16 ( )2 +4 y =16 x = 16 x = ±4 y = 16 y2 = y = ±2 The intercepts are (−4, 0), (4, 0), (0, −2), and (0, 2) Test x-axis symmetry: Let y = − y x + ( − y ) =16 x + y =16 same Test y-axis symmetry: Let x = − x ( − x )2 + y =16 x + y =16 same + ( − y ) =16 2 x +4 y =16 same Therefore, the graph will have x-axis, y-axis, and origin symmetry y = x +2 x +1 = x +1 x +1 ( − y = (−x) + 2(−x) +1 − y = x4 + x2 + y = − x4 − x2 − different y = x3 − x x-intercepts: = x3 − x y-intercepts: y = (0)3 − =0 = x x2 − ( ) = x ( x + 1)( x − 1) x = 0, x = −1, x = The intercepts are (−1, 0), (0, 0), and (1, 0) Test x-axis symmetry: Let y = − y − y = x3 − x y = − x3 + x different Test y-axis symmetry: Let x = − x y = ( − x )3 − ( − 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 10 x + x + y + y = x-intercepts: = x +2 x +1 same Test origin symmetry: Let x = − x and y = − y y = − x3 + x Test origin symmetry: Let x = − x and y = − y (−x) Therefore, the graph will have y-axis symmetry x-intercepts: y = (−x) + 2(−x) +1 )( ) y-intercepts: y = (0) +2(0) +1 =1 x-intercepts: x + x + (0) + 2(0) = x2 + x = x( x + 1) = x = 0, x = −1 x +1 = 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 y-intercepts: (0) + + y + y = y2 + y = y ( y + 2) = y = 0, y = −2 The intercepts are (−1, 0), (0, 0), and (0, −2) 113 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions Test x-axis symmetry: Let y = − y ( x + x + (− y ) + 2(− y ) = x + x + y − y = different Test y-axis symmetry: Let x = − x (− x) + (− x) + y + y = )( The intercepts are − 3, , ) 3, , ( 0, −1) , and ( 0, 3) x2 + y − x + y − = 14 x2 − x + y + y = x − x + y + y = different Test origin symmetry: Let x = − x and y = − y (x − 2x + + y2 + y + = + + ) ( ) 2 + ( y + ) = 32 Center: (1, –2) Radius = ( x − 1) (− x) + (− x) + (− y ) + 2(− y ) = x − x + y − y = different The graph has none of the indicated symmetries ( x − h) + ( y − k ) = r 11 ( x − ( −2 ) ) + ( y − 3)2 = 42 ( x + )2 + ( y − 3)2 = 16 ( x − h) + ( y − k ) = r 12 ( x − ( −1) ) + ( y − ( −2 ) ) 2 = 12 ( x − 1)2 + = ( x − 1)2 = ( x + 1)2 + ( y + )2 = 13 x + ( y − 1) = 2 x-intercepts: ( x − 1) + ( + ) = 32 x −1 = ± x + ( y − 1) = Center: (0,1); Radius = x = 1± 2 y-intercepts: ( − 1) + ( y + ) = 32 + ( y + 2) = ( y + )2 = y+2= ± y + = ±2 y = −2 ± 2 ( ) ( ) The intercepts are − 5, , + 5, , x-intercepts: x + ( − 1) = ( 0, −2 − 2 ) , and ( 0, −2 + 2 ) x2 + = x2 = 3x + y − x + 12 y = 15 x=± x2 + y2 − x + y = x2 − 2x + y2 + y = y-intercepts: 02 + ( y − 1) = ( y − 1) = y − = ±2 y = 1± y = or y = −1 (x − 2x + + y2 + y + = + ) ( ) ( x − 1)2 + ( y + )2 = ( Center: (1, –2) Radius = 114 Copyright © 2016 Pearson Education, Inc 5 ) Chapter Review Exercises 19 This relation does not represent a function, since is paired with two different values 20 2 x-intercepts: ( x − 1) + ( + ) = ( 5) ( x − 1)2 + = ( x − 1)2 = x − = ±1 x = 1±1 x = or x = 2 y-intercepts: ( − 1) + ( y + ) = ( 5) f ( x) = 3x x2 − 3(2) 6 = = =2 (2) − − a f (2) = b f (−2) = 3(−2) −6 −6 = = = −2 − (−2) − c f (− x) = 3(− x) −3 x = (− x)2 − x − d 3x −3 x − f ( x) = − = x − x2 − e f ( x − 2) = + ( y + 2) = 3( x − 2) ( x − 2) − 3( x − 2) 3x − = x − 4x + −1 x − 4x + = ( y + )2 = f y + = ±2 y = −2 ± y = or y = −4 21 The intercepts are ( 0, ) , ( 2, ) , and ( 0, −4 ) f (2 x) = 3(2 x ) 6x = (2 x) − x − f ( x) = x − a f (2) = 22 − = − = = b f (−2) = d B ,C = (− − 1) + (3 − 1) = + = 13 c f (− x) = (− x) − = x − d A,C = (− − 3) + (3 − 4) = 25 + = 26 d − f ( x) = − x − e f ( x − 2) = ( x − 2) − 16 Find the distance between each pair of points d A, B = (1 − 3) + (1 − 4) = + = 13 Since AB = BC, triangle ABC is isosceles 17 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) = x2 − x f f (2 x) = (2 x) − = x − = x2 − = x2 − ( = 32 = Equation: 4−4 = = = x2 − x + − = 16 + 16 ( −2 )2 − = ( x − 1) + ( y + 2) = ( x − 1) 2 ( + ( y + ) = 32 ) 55 f ( x) = a ) x2 − x2 f (2) = 18 This relation represents a function Domain = {–1, 2, 4}; Range = {0, 3} 115 Copyright © 2016 Pearson Education, Inc 22 − 4 − = = =0 4 22 Chapter 1: Graphs and Functions b f (−2) = ( −2 )2 − 4 − = = =0 4 ( −2 )2 c f (− x) = (− x) − x − = (− x) x2 d x2 − − x2 x2 − − f ( x) = − = = − x2 x2 x e f ( x − 2) = = f f (2 x) = = x2 − x x ( x − 4) = ( x − 2) ( x − 2) ( 4x )=x x −1 x ≠ −3 or {x 28 ( x + 3)( x − 3) ≠ 29 x x+8 The radicand must be non-negative and not zero: x+8> x > −8 Domain: { x x > −8} f ( x) = f ( x) = −2 x + x + f ( x + h) − f ( x) h = = x ≠ −3, x ≠ 3} x ≠ −2 or Also, the radicand must be non-negative: x +1 ≥ x ≥ −1 Domain: [ −1, 2) ∪ ( 2, ∞ ) (2 x) − 4 x − = (2 x) x2 x2 − x +1 x −4 The denominator cannot be zero: x2 − ≠ f ( x) = ( x + 2)( x − 2) ≠ ( x − 2) − x − x + − = ( x − 2) ( x − 2) x 23 f ( x) = x −9 The denominator cannot be zero: x2 − ≠ Domain: 27 −2 ( x + h ) + ( x + h ) + − −2 x + x + ( h −2 x + xh + h ( f ( x) = − x The radicand must be non-negative: 2− x ≥ x≤2 Domain: {x x ≤ 2} or ( −∞, 2] x 25 g ( x) = x The denominator cannot be zero: x≠0 Domain: { x x ≠ 0} 26 x x + 2x − The denominator cannot be zero: x2 + x − ≠ f ( x) = 2 ) + x + h + + 2x − x −1 h −2 x − xh − 2h + x + h + + x − x − = h −4 xh − 2h + h h ( −4 x − 2h + 1) = = h h = −4 x − 2h + 24 ) 30 a Domain: { x − ≤ x ≤ } ; [ −4, 3] Range: {y − ≤ y ≤ } ; [ −3, 3] b Intercept: ( 0, ) c f ( −2 ) = −1 d f ( x ) = −3 when x = –4 e f ( x) > when < x ≤ { x | < x ≤ 3} ( x + 3)( x − 1) ≠ x ≠ −3 or Domain:{ x x ≠ −3, x ≠ 1} 116 Copyright © 2016 Pearson Education, Inc Chapter Review Exercises f To graph y = f ( x − 3) , shift the graph of f f The function is neither horizontally units to the right g x-intercepts: −3, 0,3 ; y-intercept: f ( x) = x3 − x 32 f (− x) = (− x)3 − 4(− x) = − x3 + x = − x3 − x = − f ( x) ( ) f is odd + x2 + x4 + (− x) + x g (− x) = = = g ( x) + (− x) + x g is even 33 g ( x) = g 1 To graph y = f x , stretch the graph of 2 f horizontally by a factor of 34 G ( x) = − x + x3 G ( − x ) = − ( − x ) + ( − x )3 = + x − x3 ≠ −G ( x) or G ( x) G is neither even nor odd h x + x2 −x −x f (− x) = = = − f ( x) + (− x) + x2 f is odd 35 f ( x) = 36 f ( x ) = x3 − x + on the interval ( −3,3) To graph y = − f ( x ) , reflect the graph of f vertically about the y-axis Use MAXIMUM and MINIMUM on the graph of y1 = x3 − x + 20 −3 −3 31 a −20 Domain: ( −∞, 4] Range: ( −∞, 3] b Increasing: ( −∞, −2 ) and ( 2, 4) ; Decreasing: ( −2, ) c −20 local maximum: 4.04 when x ≈ −0.91 local minimum: −2.04 when x = 0.91 f is increasing on: (−3, −0.91) and (0.91,3) ; f is decreasing on: (−0.91, 0.91) Local minimum is −1 at x = ; Local maximum is at x = −2 d No absolute minimum; Absolute maximum is at x = e 20 The graph has no symmetry 117 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions f ( x ) = x − x3 + x + on the interval ( −2,3) 37 40 Use MAXIMUM and MINIMUM on the graph of y1 = x − x3 + x + 20 −10 20 −10 41 The graph does not pass the Vertical Line Test and is therefore not a function 42 The graph passes the Vertical Line Test and is therefore a function −2 2 2 f (3) − f (2) 3 ( 3) − ( 3) − 3 ( ) − ( ) = 3− 3− ( − 36 ) − ( − 16 ) = = −27 + 10 = −17 20 −2 −2 f ( x) = 3x − x 43 f ( x) = x 44 f ( x) = x −10 local maximum: 1.53 when x = 0.41 local minima: 0.54 when x = −0.34 , −3.56 when x = 1.80 f is increasing on: (−0.34, 0.41) and (1.80, 3) ; f is decreasing on: (−2, −0.34) and (0.41, 1.80) 38 f ( x) = x − x a b c 39 f (2) − f (1) 8(2) − − [8(1) − 1] = −1 = 32 − − (7) = 23 2 f (1) − f (0) 8(1) − − [8(0) − 0] = 1− = − − ( 0) = f (4) − f (2) 8(4) − − [8(2) − 2] = 4−2 128 − − (30) 94 = = = 47 2 f ( x) = − x f (3) − f (2) − ( 3) − − ( ) = 3− 3−2 ( − 15 ) − ( − 10 ) = = −13 − ( −8 ) = −5 118 Copyright © 2016 Pearson Education, Inc Chapter Review Exercises 45 F ( x) = x − Using the graph of y = x , vertically shift the graph downward units Intercepts: (–4,0), (4,0), (0,–4) Domain: { x x is any real number} Range: {y y ≥ − 4} or [ −4, ∞ ) 48 f ( x) = − x = −( x − 1) Reflect the graph of y = x about the y-axis and horizontally shift the graph to the right unit Intercepts: (1, 0), (0, 1) Domain: { x x ≤ 1} or ( −∞, 1] Range: {y y ≥ 0} or [ 0, ∞ ) 46 g ( x) = − x Reflect the graph of y = x about the x-axis and vertically stretch the graph by a factor of Intercepts: (0, 0) Domain: { x x is any real number} Range: { y y ≤ 0} or ( −∞, 0] 47 h( x) = x − Using the graph of y = x , horizontally shift the graph to the right unit 49 h( x ) = ( x − 1) + Using the graph of y = x , horizontally shift the graph to the right unit and vertically shift the graph up units Intercepts: (0, 3) Domain: { x x is any real number} Range: {y y ≥ 2} or [ 2, ∞ ) 50 g ( x) = − 2( x + 2)3 − Using the graph of y = x3 , horizontally shift the graph to the left units, vertically stretch the graph by a factor of 2, reflect about the x-axis, and vertically shift the graph down units Intercept: (1, 0) Domain: { x x ≥ 1} or [1, ∞ ) Range: {y y ≥ 0} or [ 0, ∞ ) 119 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions b Intercept: (0, 1) y −9 (− − 4, c x ) Graph: −5 (−3, −6) (−2, −8) (−1, −10) Intercepts: (0,–24), − − 4, ≈ ( −3.6, ) ( ) Domain: { x x is any real number} d Range: Range: { y y is any real number} 51 3x f ( x) = x +1 a Domain: b Intercept: c e if − < x ≤ if x > {x x > −2 } or ( −2, ∞ ) ( 0, ) Graph: 53 {y y ≥ − 4, y ≠ 0} There is a jump at x = Therefore, the function is not continuous Ax + and f (1) = 6x − A(1) + =4 6(1) − A+5 =4 A + = 16 A = 11 f ( x) = 54 a The function is one-to-one because there are no two distinct inputs that correspond to the same output b The inverse is d Range: e 52 { y | y > −6 } 55 The function f is one-to-one because every horizontal line intersects the graph at exactly one point There is a jump in the graph at x = Therefore, the function is not continuous x f ( x) = 1 3x a or ( −6, ∞ ) Domain: if − ≤ x < if x = if x > {x {( 2,1) , ( 5,3) , (8,5) , (10, )} x ≥ − 4} or [ −4, ∞ ) 120 Copyright © 2016 Pearson Education, Inc Chapter Test 56 2x + 5x − 2x + y= 5x − 2y + x= 5y − x(5 y − 2) = y + xy − x = y + xy − y = x + y (5 x − 2) = x + 2x + y= 5x − 2x + −1 f ( x) = 5x − Domain of f = Range of f −1 = { x | x ≥ 2} or [ 2, ∞ ) f ( x) = Range of f = Domain of f −1 = { x | x ≥ 0} or [ 0, ∞ ) 59 f ( x) = x1/ + y = x1/ + Inverse x = y1/ + 1/ y = x −1 y = ( x − 1)3 f −1 ( x) = ( x − 1)3 Domain of f = Range of f −1 = All real numbers or ( −∞, ∞ ) Range of f = Domain of f −1 = All real numbers or ( −∞, ∞ ) Domain of f = Range of f −1 = All real numbers except Range of f = Domain of f −1 = All real numbers except 57 x −1 y= x −1 Inverse x= y −1 x( y − 1) = xy − x = xy = x + x +1 y= x x +1 −1 f ( x) = x Domain of f = Range of f −1 = All real numbers except Range of f = Domain of f −1 = All real numbers except f ( x) = 58 Chapter Test d ( P1 , P2 ) = y−2 x = y−2 = 52 = 13 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , −1 + + (−1) , = 4 2 = , 2 2 = ( 2, 1) y = x − f x≥0 x≥0 x≥0 y = x +2 −1 Inverse ( x) = x + 2 = 36 + 16 y = x−2 ( − (−1) )2 + ( −1 − 3)2 = 62 + ( −4 ) f ( x) = x − x= Inverse 121 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions y y = x y 5 (1, 1) (4, 2) (9, 3) (−2, 1) y =x x −5 10 x (0, 0) (1,−1) (4,−2) −5 −5 (9,−3) a x + y = x-intercepts: x2 + = {( 2,5) , ( 4, ) , ( 6, ) , (8,8)} This relation is a function because there are no ordered pairs that have the same first element and different second elements Domain: {2, 4, 6,8} y-intercept: (0) + y = y=9 x2 = x = ±3 The intercepts are ( −3, ) , ( 3, ) , and ( 0,9 ) Range: {5, 6, 7,8} b Test x-axis symmetry: Let y = − y {(1,3) , ( 4, −2 ) , ( −3,5) , (1, )} This relation is not a function because there are two ordered pairs that have the same first element but different second elements x + (− y) = x − y = different c Test y-axis symmetry: Let x = − x ( − x )2 + y = This relation is not a function because the graph fails the vertical line test d This relation is a function because it passes the vertical line test Domain: { x x is any real number} x + y = same Test origin symmetry: Let x = − x and y = − y ( − x )2 + ( − y ) = Range: { y | y ≥ 2} or [ 2, ∞ ) x − y = different Therefore, the graph will have y-axis symmetry ( x − h) + ( y − k ) = r ( x − )2 + ( y − (−3) )2 = 52 ( x − )2 + ( y + 3)2 = 25 General form: ( x − )2 + ( y + 3)2 = 25 x − x + 16 + y + y + = 25 x2 + y − 8x + y = x2 + y2 + x − y − = x2 + x + y − y = ( x + x + 4) + ( y − y + 1) = + + ( x + 2) + ( y − 1) = 32 f ( x ) = − 5x The function tells us to take the square root of − 5x Only nonnegative numbers have real square roots so we need − x ≥ − 5x ≥ − 5x − ≥ − −5 x ≥ −4 −5 x −4 ≤ −5 −5 x≤ 4 4 Domain: x x ≤ or −∞, 5 f ( −1) = − ( −1) = + = = Center: (–2, 1); Radius = 122 Copyright © 2016 Pearson Education, Inc Chapter Test 10 g ( x ) = x+2 x+2 e values such that the graph is below the xaxis The graph is below the x-axis for values in the domain that are less than −2 and greater than Therefore, the solution set is { x | −5 ≤ x < −2 or < x ≤ 5} In The function tells us to divide x + by x + Division by is undefined, so the denominator can never equal This means that x ≠ −2 Domain: { x | x ≠ −2} g ( −1) = ( −1) + ( −1) + = =1 interval notation we would write the solution set as [ −5, −2 ) ∪ ( 2,5] 13 x−4 x + x − 36 The function tells us to divide x − by 11 h ( x ) = To solve f ( x ) < , we want to find x- f ( x ) = − x + x3 + x − We set Xmin = −5 and Xmax = The standard Ymin and Ymax will not be good enough to see the whole picture so some adjustment must be made x + x − 36 Since division by is not defined, we need to exclude any values which make the denominator x + x − 36 = ( x + )( x − ) = x = −9 or x = Domain: { x | x ≠ −9, x ≠ 4} (note: there is a common factor of x − but we must determine the domain prior to simplifying) h ( −1) = 12 a ( −1) − −5 = = − ( −1) + ( −1) − 36 40 To find the domain, note that all the points on the graph will have an x-coordinate between −5 and 5, inclusive To find the range, note that all the points on the graph will have a y-coordinate between −3 and 3, inclusive Domain: { x | −5 ≤ x ≤ 5} or [ −5, 5] Range: { y | −3 ≤ y ≤ 3} or [ −3, 3] b The intercepts are ( 0, ) , ( −2, ) , and ( 2, ) x-intercepts: −2, y-intercept: c We see that the graph has a local maximum of −0.86 (rounded to two places) when x = −0.85 and another local maximum of 15.55 when x = 2.35 There is a local minimum of −2 when x = Thus, we have Local maxima: f ( −0.85 ) ≈ −0.86 f ( 2.35 ) ≈ 15.55 Local minima: f ( ) = −2 The function is increasing on the intervals ( −5, −0.85) and ( 0, 2.35 ) and decreasing on the f (1) is the value of the function when intervals ( −0.85, ) and ( 2.35,5 ) x = According to the graph, f (1) = d Since ( −5, −3) and ( 3, −3) are the only points on the graph for which y = f ( x ) = −3 , we have f ( x ) = −3 when x = −5 and x = 14 a x < −1 2 x + f ( x) = x ≥ −1 x−4 To graph the function, we graph each “piece” First we graph the line y = x + but only keep the part for which x < −1 Then we plot the line y = x − but only 123 Copyright © 2016 Pearson Education, Inc Chapter 1: Graphs and Functions keep the part for which x ≥ −1 y = ( x + 1) y 10 (0, 1) (−2, −1) −2 x −10 Next we reflect this graph about the x-axis b To find the intercepts, notice that the only piece that hits either axis is y = x − y = x−4 y = x−4 y = 0−4 = x−4 y = −4 4=x to obtain the graph of y = − ( x + 1) y 10 (−2, 1) (0, −1) x −2 The intercepts are ( 0, −4 ) and ( 4, ) c −10 To find g ( −5 ) we first note that x = −5 so we must use the first “piece” because −5 < −1 g ( −5 ) = ( −5 ) + = −10 + = −9 d To find g ( ) we first note that x = so we y = − ( x + 1) Next we stretch this graph vertically by a factor of to obtain the graph of y = −2 ( x + 1) must use the second “piece” because ≥ −1 g ( ) = − = −2 15 The average rate of change from to is given by Δy f ( ) − f ( 3) = 4−3 Δx (3( 4) = = 16 a ) ( − ( ) + − ( 3) − ( 3) + y 10 (−2, 2) −2 ) −10 y = −2 (x + 1) 44 − 25 19 = = 19 4−3 The last step is to shift this graph up units to obtain the graph of y = −2 ( x + 1) + The basic function is y = x3 so we start with the graph of this function y 10 (−2, 5) y = x3 10 (−1, −1) (0, 1) −2 (1, 1) −2 x 4−3 y (0, −2) x x −10 y = −2 (x + 1) + −10 Next we shift this graph unit to the left to obtain the graph of y = ( x + 1) b The basic function is y = x so we start with the graph of this function 124 Copyright © 2016 Pearson Education, Inc Chapter Project y 18 If the point (3, −5) is on the graph of f, then the y= x point (−5, 3) must be on the graph of f −1 (2, 2) (−2, 2) x Next we shift this graph units to the left to obtain the graph of y = x + Chapter Project Project I – Internet Based Project y = x+4 y (−6, 2) x (−2, 2) Next we shift this graph up units to obtain the graph of y = x + + y y = x+4 +2 (−2, 4) (−6, 4) x 17 3x − y= 3x − Inverse x= 3y − x(3 y − 5) = xy − x = 3xy = x + 5x + y= 3x x +2 f −1 ( x) = 3x Domain of f = Range of f −1 f ( x) = = x | x ≠ 53 { } Range of f = Domain of f −1 = { x | x ≠ 0} 125 Copyright © 2016 Pearson Education, Inc ... 21 f has local maxima at x = − and x = The local maxima are and 10, respectively 22 f has local minima at x = − 8, x = and x = The local minima are –4, 0, and 0, respectively 23 f has absolute... feet indicates that there are at most 60 gigabytes in a month e 118 a The graph is flat at first and then rises in a straight line h ( x) = 2x h ( a + b ) = ( a + b ) = 2a + 2b = h ( a ) + h (b... G ( a ) + G (b) a+ b a b does not have the property x 119 Answers will vary From a graph, the domain can be found by visually locating the x-values for which the graph is defined The range can