Solution manual for precalculus 1st edition by miller

Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.1 Chapter Functions and Relations Section 1.1 The Rectangular Coordinate System and Graphing Utilities x +x y +y  b M =  ,     −4 + ( −2) 11+  = ,   2    −6 18  =  ,  = −3,9  2 origin quadrants d = (x − x1 ) + ( y2 − y1 ) 2  x + x y + y2  M =  ,    solution 0; y ( 12 a d = (x ) − x1 ) + ( y2 − y1 ) 2 = 3− ( −1)  +  −7 − ( −3)  = ( 4) + ( −4) 2 = 16 + 16 = 32 = x +x y +y  b M =  ,     3+ ( −1) −7 + ( −3)  = ,      − 10   = ,  = 1,− 2  ( 10 13 a d = (x − x1 ) + ( y2 − y1 ) 2 ) = 2 − ( −7)  + 5 − ( −4)  = ( 9) + ( 9) 2 = 81+ 81 = 162 = 11 a d = (x − x1 ) + ( y2 − y1 ) 2 =  −4 − ( −2)  + (11− 7) = ( −2) + ( 4) 2 = + 16 = 20 = 155 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations  x + x y + y2  b M =  ,     31.1+ 37.1 −32.7 + ( −24.7)  = ,    2    x + x y + y2  b M =  ,     + ( −7) + ( −4)  = ,   2    −5    =  ,  = − ,   2  2 14 a d = = =  68.2 −57.4  = = 34.1,− 28.7 ,   ( (x − x ) + (y − y ) ( −4 − 3) + ( −1− 6) ( −7) + ( 7) 2 2 17 a d = 2 = = = 49 + 49 = 98 = x +x y +y  b M =  ,     −4 + −1+  = ,    −1    =  ,  = − ,   2  2 15 a d = = = 2 2 = = − x1 ) + ( y2 − y1 ) ( 31.1− 37.1) ( −6) + ( −8) 2 = ( ) ) (x − x1 ) + ( y2 − y1 ) ( − 7) +  ( ) + ( 5) 2 ) ( ) − −3   2 ( +  −32.7 − ( −24.7)  = + 80 = 87  x + x y + y2  b M =  ,      + + −3   , =   2    −2    = , = ,−          7.4 −8.8  = = 3.7,− 4.4 ,   (x ( ) ( ) + ( −6 ) 2 =  x + x y + y2  b M =  ,     5.2 + 2.2 −6.4 + ( −2.4)  = ,    2   16 a d = − +  −7 − −    18 a d = ( − x1 ) + ( y2 − y1 ) ( 2 = 45 + 72 = 117  x + x y + y2  b M =  ,      + −7 + −   , =   2    5 −8   5  = , = ,−         (x − x ) + (y − y ) ( 5.2 − 2.2) + −6.4 − ( −2.4) ( 3) + ( 4) = + 16 = 25 = (x ) 2 = 36 + 64 = 100 = 10 19 d1 = ( 3− 1) + (1− 3) ) = 4+ = 2 156 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.1 d2 = ( − 3) + ( −2 − 1) 2 22 d1 = = 9+ = d3 = (1− 0) 2 d2 = = 1+ 25 = 26 d12 + d22 = d32 = d3 = 26 2 ( 2) + ( 2) = ( 10) x2 + y = 4− 3= 1=  Yes x2 + y = b ( 4) + ( −17) = 2 16 − 17 = + 32 = 40 −1 = False 40 = 40  True No Yes 21 d1 =  − ( −2)  + ( − 4) = 49 + 16 = 65 2  1  3  2 +  4 =     + =1 4 1=  Yes 24 a x−3 − y = d3 =  −2 − ( −5)  + ( − 1) = 9+ = d12 + d32 = d22 2 ) x2 + y = c = 100 + = 101 ( 65) + (3 2) = ( (1) − − ( −2) = 2+ = 101 65 + 18 = 101 2 2 ( −2) + ( −3) = d3 = 1− ( −3)  + 2 − ( −2)  = 16 + 16 = d12 + d32 = d22 ( −5− 5) + (1− 0) + 2 − ( −2)  78 = 82 False No 23 a = 36 + = 10 d2 = 13+ 65 = 82 ( −3− 3) + ( −2 − 0) 2 2 = 4+ = 2 d2 = + 1− ( −2)  ( −6 − 1) Yes 2 ( 13) + ( 65) = ( 82) 26 = 26  True ( 3− 1) + ( − 2) ( 3− 1) = 49 + 16 = 65 d22 + d32 = d12 + 18 = 26 20 d1 = = + = 13 ( ) ( ) ( ) 2 = 81+ = 82 + 3 − ( −2)  2 + ( −6 − 3) + ( −1) 4=  Yes 83 = 101 False No 157 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations 26 x + ≠ x−3 − y = b ( −2) − − ( −3) = x ≠ −7 {x |x ≠ −7} 5+ = 27 x − 10 ≥ = False x ≥ 10 No {x |x ≥ 10} c x − − y = 28 x + 11 ≥ x ≥ −11  1  11   10  − −  − 10  =       11 10    − +  = 10  4 10    10  {x |x ≥ −11} 29 1.5 − x ≥ − x ≥ −1.5 x ≤ 1.5 1− 30 + 11 = 40 29 + 11 = 40 {x |x≤ 1.5} 40 = 40  Yes 25 x − ≠ 30 2.2 − x ≥ − x ≥ −2.2 x ≤ 2.2 x≠3 {x |x ≠ 3} {x |x ≤ 2.2} 31 y = x x y y=x −3 −3 y = −3 −2 −2 y = −2 −1 −1 y = −1 0 y=0 1 y =1 2 y=2 3 y=3 Ordered pair ( −3,− 3) ( −2,− 2) ( −1,− 1) ( 0,0) (1,1) ( 2,2) ( 3,3) 158 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.1 32 y = x2 x y y = x2 −3 y = ( −3) = y = ( −2) = −1 y = ( −1) = 0 y = ( 0) = 1 y = (1) = y = ( 2) = y = ( 3) = x y y= x Ordered pair 0 y= 0=0 1 y = 1=1 y= 4=2 ( 0,0) (1,1) ( 4,2) 2 −2 2 2 Ordered i −3,9 ( ) ( −2,4) ( −1,1) ( 0,0) (1,1) ( 2,4) (3,9) 33 y = x 159 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations y= 9=3 x y y= x −3 y = −3 = −2 y = −2 = −1 y = −1 = 0 y= =0 1 y = =1 2 y= =2 3 y= 3=3 y y = x3 ( 9,3) 34 y = x Ordered pair ( −3,3) ( −2,2) ( −1,1) ( 0,0) (1,1) ( 2,2) ( 3,3) 35 y = x3 x Ordered pair 160 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.1 −2 −8 y = ( −2) = −8 −1 −1 y = ( −1) = −1 0 y = ( 0) = 1 y = (1) = y = ( 2) = 36 y = 3 3 ( −2,− 8) ( −1,− 1) ( 0,0) (1,1) ( 2,8) x y −4 − −2 − −1 −1 − −2 2 1 x 1 y= =− ( −4) Ordered pair y= 1 =− ( −2) 1   −2,−    y= = −1 ( −1) ( −1,1) y= = −2  1  − 2      − ,− 2   y= =2  1  2   1   ,2   y= =1 (1) (1,1) y= x 1   −4,−    161 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations 2 y= 1 = ( 2) 4 y= 1 = ( 4)  1  2,2     1  4,4    37 y − x = y = x +2 x y y = x +2 −3 y = −3 + = −2 y = −2 + = −1 y = −1 + = y = + 2= y = 1+ 2= y = + 2= y = + 2= Ordered pair ( −3,5) ( −2,4) ( −1,3) ( 0,2) (1,3) ( 2,4) (3,5) 162 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.1 38 x + y = y = 3− x x y y = 3− x −3 y = 3− −3 = −2 y = 3− −2 = −1 y = 3− −1 = y = 3− = y = 3− = 2 y = 3− = y = 3− = Ordered pair ( −3,0) ( −2,1) ( −1,2) ( 0,3) (1,2) ( 2,1) (3,0) 39 y − x − = y2 = x + y = ± x+ x y y = ± x+2 −2 y=± ( −2) + = ( −2,0) −1 ±1 y=± ( −1) + = ±1 ( −1,1) ,( −1,− 1) ±2 y=± ( 2) + = ±2 ( 2,2) ,( 2,− 2) Ordered pairs 163 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations y=± ±3 ( 7) + = ±3 ( 7,3) ,( 7,− 3) 40 y − x + = y2 = x − y = ± x −1 x y y = ± x −1 y=± (1) − = ±1 y=± ( 2) − = ±1 ±2 y=± ( 5) − = ±2 10 ±3 y=± (10) − = ±3 Ordered pairs (1,0) ( 2,1) ,( 2,− 1) ( 5,2) ,( 5,− 2) (10,3) ,(10,− 3) 41 x = y + y = x −1 y = ± ( x − 1) x y y = ± ( x −1) y = ± (1) − 1 = Ordered pairs (1,0) 164 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2 b 20 a ( x − h) + ( y − k ) = r ( x − 6) +  y − ( −2) = ( 6) ( x − 6) + ( y + 2) = 36 2 2 2 2 b 18 a ( x − h) + ( y − k ) = r  x − ( −3)  + ( y − 2) = ( 4) ( x + 3) + ( y − 2) = 16 2 2 2 2 b 21 a ( x − h) + ( y − k)  x − ( −4)  +  y − ( −3)  ( x + 4) + ( y + 3) 2 2 = r2 = = 11 ( 11) b 19 a ( x − h) + ( y − k ) = r  x − ( −4)  + ( y − 1) = ( 3) ( x + 4) + ( y − 1) = 2 2 2 2 b 22 a ( x − h) + ( y − k )  x − ( −5)  +  y − ( −2)  ( x + 5) + ( y + 2) 2 2 = r2 = = 21 ( 21) b 175 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations ( x − h) + ( y − k ) ( x − 2) + ( y − 1) ( x − 2) + ( y − 1) 23 a ( x − h) + ( y − k ) = r 2 ( x − 0) + ( y − 0) = ( 2.6) 2 x2 + y2 = 6.76 b = r2 2 = ( 5) 2 = 25 b 24 a ( x − h) + ( y − k ) = r  x + x y + y2  26 a C =  ,     + + ( −1)  , =   2    12  =  ,  = 6,1  2 ( x − 0) + ( y − 0) = ( 4.2) 2 x2 + y2 = 17.64 b ( ) r= = (x − x ) + (y − y ) ( − 6) + ( 3− 1) 2 2 2 = 1+ = ( x − h) + ( y − k ) ( x − 6) + ( y − 1) ( x − 6) + ( y − 1)  x + x y + y2  25 a C =  ,     −2 + + ( −2)  , =   2    2 =  ,  = 2,1  2 = r2 2 = 2 =5 ( ) b ( ) r= = (x − x ) + (y − y ) ( −2 − 2) + ( − 1) 2 2 2 = 16 + = 25 = 176 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2 27 a r = = (x − x ) + (y − y ) ( −2 − 6) + ( −1− 5) 2 2 r= 2 = = 64 + 36 = 100 = 10 ( x − h) + ( y − k ) = r  x − ( −2)  +  y − ( −1)  = (10) ( x + 2) + ( y + 1) = 100 2 2 2 2 2 = 16 = ( x − h) + ( y − k ) ( x − 4) + ( y − 6) ( x − 4) + ( y − 6) 2 (x − x ) + (y − y ) ( − 0) + ( − 6) 2 b 2 = r2 2 = ( 4) 2 = 16 b 28 a r = = (x − x ) + (y − y ) ( 3− 6) + (1− 5) 2 2 30 a The circle must touch the x-axis at − 2,0 , at one side of a diameter ( r= = + 16 = 25 = ( x − h) + ( y − k ) ( x − 3) + ( y − 1) ( x − 3) + ( y − 1) 2 = r2 2 = ( 5) 2 = 25 ) (x − x1 ) + ( y2 − y1 ) 2 =  −2 − ( −2)  + ( −4 − 0) = 16 = 2 ( x − h) + ( y − k ) = r  x − ( −2)  +  y − ( −4)  = ( 4) ( x + 2) + ( y + 4) = 16 2 b 2 2 2 b 29 a The circle must touch the y-axis at 0,6 , at one side of a diameter ( ) 31 a The circle must touch the x-axis and the y-axis at a distance r 177 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations from the center Since the center is in quadrant IV, the center must be 5,− ( 34 ) ( x − h) + ( y − k) = r ( x − 5) +  y − ( −5) = ( 5) ( x − 5) + ( y + 5) = 25 2 2 2 35 ( x − h) + ( y − k) = r  x − ( −4)  + ( y − 16) = ( 9) ( x + 4) + ( y − 16) = 81 ( x − h) + ( y − k) = r 2 2 2 2 2 ( ( ) ( 2 ) ( ) ) {( −1,5)} ) = r2 37 ( x + 1) + ( y − 5) = The sum of two squares will equal zero only if each individual term is zero Therefore, x = −1 and y = 32 a The circle must touch the x-axis and the y-axis at a distance r from the center Since the center is in quadrant II, the center must be −3,3 ) 2    x − − 11    = 11  2    + y − − 11      2 x + 11 + y + 11 = 11 2 2 ( 2 ( x − h) + ( y − k ) 36 ( x − h) + ( y − k) = r  x − ( −3)  + ( y − 3) = ( 3) ( x + 3) + ( y − 3) = ( x − 7) + ( y − 7) = ( 7) ( x − 7) + ( y − 7) = b ( 38 ( x − 3) + ( y + 12) = The sum of two squares will equal zero only if each individual term is zero Therefore, x = and y = −12 2 2 {(3,− 12)} b 39 ( x − 17) + ( y + 1) = −9 2 The sum of two squares cannot be negative, so there is no solution {} 40 ( x + 15) + ( y − 3) = −25 The sum of two squares cannot be negative, so there is no solution 33 ( x − h) + ( y − k) = r ( x − 8) +  y − ( −11) = ( 5) ( x − 8) + ( y + 11) = 25 2 2 2 {} 2 41 x2 + y2 + 6x − 2y + = ( x + 6x ) + ( y 2 − 2y ) = −6 178 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2 ( ( 1   ( −2)  =   )  x2 + 6x +    = −6 + +  + y2 − 2y +  2 ( x + 3) + ( y − 1) = ( ( ) 42 2 45 ( ( ) 43 2 1   ( −22)  = 121   ( + 6y 46 44 (x 2 1   ( 6)  =   ( x −10x ) + ( y 1   ( −10)  = 25   + 4y =4 ) + 22x + 121 + y2 = + 121 ( x + 11) ( + y2 = 125 ) Center: −11,0 ; Radius: 125 = 5 47 10x2 + 10y2 − 80x + 200y + 920 = 1=1 x2 + y2 − 8x + 20y + 92 = (x 2 2 x + y − 10x + 4y − 20 = ( x + 22x ) + y 1   ( 22)  = 121   ) = −129 ) x2 + y2 + 22x − = ) ) ( ) 104 = 26 1=1 Center: 11,− ; Radius: x2 + ( y − 10) = 104 (  x2 − 22x + 121    = −129 + 121+  + y2 + 6y +  2 ( x − 11) + ( y + 3) = ( ) Center: 0,10 ; Radius: x2 + y2 − 22x + 6y + 129 = ( x − 22x ) + ( y )=4 ) Center: − 6,7 ; Radius: ( x2 + y2 − 20y + 100 = + 100  x2 + 12x + 36    = −84 + 36 + 49  + y2 − 14y + 49  2 ( x + 6) + ( y − 7) = ( x2 + y2 − 20y − = 1   ( −14)  = 49   ) 49 = 1   ( −20)  = 100   ) = −84 − 14y ) x2 + y2 − 20y 4=2 1   (12)  = 36   ( ) Center: 5,− ; Radius: x2 + y2 + 12x − 14y + 84 = ( x + 12x ) + ( y ( ( ) Center: −3,1 ; Radius: )  x2 − 10x + 25    = 20 + 25 +  + y + 4y +  2 ( x − 5) + ( y + 2) = 49 1   ( 6)  =   − 8x ) + (y 1   ( −8)  = 16   ) = 20 ( 1   ( 4)  =   ) ) = −92 + 20y 1   ( 20)  = 100    x2 − 8x + 16    = −92 + 16 + 100  + y2 + 20y + 100  2 ( x − 4) + ( y + 10) = 24 ( ) 179 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations ( ) 51 4x2 + 4y2 − 20y + 25 = 25 =0 x2 + y2 − 5y + Center: 4,− 10 ; Radius: 24 = 2x2 + 2y2 − 32x + 12y + 90 = 48 (x ( x2 + y2 − 5y x2 + y2 − 16x + 6y + 45 = ) + (y − 16x + 6y 2 ) 1  25  ( −5)  =   1   ( −16)  = 64   ( ) = −45 1   ( 6)  =   25  25 25  x2 +  y2 − 5y +  = − + 4 4  5  x2 +  y −  = 2  Degenerate case (single point):     0,       x2 − 16x + 64    = −45 + 64 +  + y2 + 6y +  2 ( x − 8) + ( y + 3) = 28 ( ( ) ) Center: 8,− ; Radius: 49 28 = x2 + y2 − 4x − 18y + 89 = ( x − 4x ) + ( y 2 1   ( −4)  =   ( − 18y 52 4x2 + 4y − 12x + = x2 + y − 3x + = ) = −89 1   ( −18)  = 81   ) ( x − 3x )  x2 − 4x +    = −89 + + 81  + y − 18y + 81  2 ( x − 2) + ( y − 9) = −4 ( 50 2 2 ) = −155 1  1   ( −10)  = 25  ( −22)  = 121     ( )  x2 − 10x + 25    = −155 + 25 + 121  + y2 − 22y + 121  2 ( x − 5) + ( y − 11) = −9 ( Degenerate case: 9 9   x − 3x +  + y = − +   3   x− 2 + y =   Degenerate case (single point):     ,0     3 53 x2 + y2 − x − y − =   +  y2 − y x2 − x =   {} − 22y + y2 = − x2 + y2 − 10x − 22y + 155 = ( x −10x ) + ( y 1   ( −3)  =   ) Degenerate case: ) = − 254 ) ( {} ) 1   ( −1)  =        −  =    16 180 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2   1  x − x +   4  = + +   4 16    +  y − y + 16      The approximate location of the earthquake is 8,7 ( ) 58 1  3 25   x −  +  y −  = 16     25  3 = Center:  ,  ; Radius: 16  4 5 54 x2 + y2 − x − y − = 3  2     x − 3x  +  y − 3y =     The fire is located at approximately coordinate 2,1 ( )      −  =        25   −  =     36 59 A circle is the set of all points in a plane that are equidistant from a fixed point called the center 60 The center and radius can easily be identified from an equation of a circle written in standard form  2  1  x − x +   25 9  = + + 25   9 36    +  y − y + 36      2 1  5 49   x −  +  y −  = 36      5 Center:  ,  ; Radius:  6 55 ( x − 4) + ( y − 6) = (1.5) 2 61 2 2 = 10 72 − 12y + y2 = 10 72 − 12y + y2 = 100 y2 − 12y − 28 = 2 36 + 36 − 12y + y2 = 10 49 = 36 ( x − 4) + ( y − 6) = 2.25 56  x − ( −32)  + ( y − 40) = ( 20) ( x + 32) + ( y − 40) = 400 ( − − 4) + ( − y ) ( y + 2)( y − 14) = y = −2 or y = 14 y = −2 and y = 14 57 62 ( − x) + 2 − ( −1)  = 16 − 8x + x2 + = 25 − 8x + x2 = 25− 8x + x2 = 25 x2 − 8x = x ( x − 8) = x = or x = x = and x = 181 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations ( − x) + ( − x) 63 =6 − 4x + x2 + 16 − 8x + x2 = 20 − 12x + 2x2 = 20 − 12x + 2x2 = 36 2x2 − 12x − 16 = x2 − 6x − = x= = ( 3+ ( 3− − ( −6) ± ( −6) − 4(1)( −8) 2(1) b ± 68 ± 17 = = 3± 17 2 ) 17 ) 17,3+ 17 and 17,3 − ( −4 − x) 64 + 6 − ( −x)  = ( −4 − x ) + ( + x ) 2 c =4 16 + 8x + x2 + 36 + 12x + x2 = 52 + 20x + 2x2 = 52 + 20x + 2x2 = 16 2x2 + 20x + 36 = x2 + 10x + 18 = x= − (10) ± (10) − 4(1)(18) 2(1) d −10 ± 28 −10 ± = 2 = −5± = ( −5 + ( −5 − ) 7) 7,5 − and 7,5 + 65 a 66 a 182 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2 b b c c d d 67 a 68 a 183 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations 2 1   ( −6)  =   ( 1   ( −12)  = 36   )  x2 − 6x +    = −41+ + 36  + y2 − 12y + 36  2 ( x − 3) + ( y − 6) = ( ) ( ) Center: 3,6 ; Radius: b 4=2 The distance between the origin (0, 0) and the center (3, 6) is 32 + 62 = Therefore, the distance between the origin and the point on the circle is − 70 x2 + y2 + 4x − 12y + 31 = ( x + 4x ) + ( y 2 ( ) = −31 − 12y 1   ( 4)  =   c 1   ( −12)  = 36   )  x2 + 4x +    = −31+ + 36  + y2 − 12y + 36  2 ( x + 2) + ( y − 6) = ( ( ) ) Center: −2,6 ; Radius: 9=3 The distance between the origin (0, 0) and the center (3, 6) is d 22 + 62 = 10 Therefore, the distance between the origin and the point on the circle is 10 − 71  −15.1,15.1,1 by  −10,10,1 69 x2 + y2 − 6x − 12y + 41 = ( x − 6x ) + ( y 2 − 12y ) = −41 184 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.2 72  −15.1,15.1,1 by  −10,10,1 74  −0.28,0.18,0.02 by  −0.1,0.2,0.02 73  −14,39,5 by  −30,5,5 Section 1.3 Functions and Relations relation; domain; y 2,4 That is, the height is unique at any given time No There are numerous times during the year when the temperature in Fort Collins, is 70 o Therefore, given an input value of 70o, there is more than one output value (time) a {(Tom Hanks, 5), (Jack Nicholson, 12), (Sean Penn, 5), (Dustin Hoffman, 7)} ( ) y f( x) −5 Yes For a given time after tree is planted, there cannot be two or more different heights 185 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations one point This relation is not a function b {Tom Hanks, Jack Nicholson, Sean Penn, Dustin Hoffman} c {5, 12, 7} d Yes 10 a {(Albany, 285), (Denver, 5883), (Miami, 11), (San Francisco, 11)} b {Albany, Denver, Miami, San Francisco} c {285, 5883, 11} d Yes  −4,3 , −2,− , 1,4 ,   11 a    3,− , 3,1  b −4,− 2,1,3 ( ( )( )( ) )( ) { } c {3,− 3,4,− 2,1} d No  −2,2 , −1,0 , 0,−  12 a   1,0 , 1,2 b −2,− 1,0,1 ( )( )( ( )( ) { } c {2,0,− 2} ) ,   d No 13 False 14 True 15 No vertical line intersects the graph in more than one point This relation is a function 16 No vertical line intersects the graph in more than one point This relation is a function 17 There is at least one vertical line that intersects the graph in more than one point This relation is not a function 18 No vertical line intersects the graph in more than one point This relation is a function 19 No vertical line intersects the graph in more than one point This relation is a function 20 There is at least one vertical line that intersects the graph in more than 186 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.3 No two ordered pairs have the same x value but different y values This relation is a function 21 No vertical line intersects the graph in more than one point This relation is a function 22 There is at least one vertical line that intersects the graph in more than one point This relation is not a function 23 This mapping defines the set of ordered pairs: 27 ( x + 1) + ( y + 5) = 25 This equation represents the graph of a circle with center −1,− and ( 28 ( x + 3) + ( y + 4) = This equation represents the graph of a circle with center −3,− and All four ordered pairs have the same x value but different y values This relation is not a function 24 This mapping defines the set of ordered pairs: ) radius This relation is not a function because it fails the vertical line test 29 y = x + Two ordered pairs have the same x value but different y values This relation is not a function 25 This mapping defines the set of {(1,4) ,( 2,5) ,(3,5)} No two ordered pairs have the same x value but different y values This relation is a function 26 This mapping defines the set of ordered pairs: )( ( {( −3,11) ,( −6,13) ,( −6,9)} {( ) radius This relation is not a function because it fails the vertical line test {( −8,3) ,( −8,4) ,( −8,5) ,( −8,6)} ordered pairs: )( )( No vertical line intersects the graph in more than one point This relation is a function )} 5,− , 6,− , 7,− , 8,− 30 y = x − No vertical line intersects the graph in more than one point This relation is a function 31 a y = x2 187 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Functions and Relations No vertical line intersects the graph in more than one point This relation is a function b x = y2 y2 = x y=± x x y y=± x 0 y=± 0=0 ± y = ± = ±1 y = ± = ±2 ± ± y = ± = ±3 Ordered pairs 0,0 ( ) (1,± 1) ( 4,± 2) ( 9,± 3) Two or more ordered pairs have the same x value but different y values This relation is not a function 32 a y = x 0,0 0 y = ±0 y = ±x 1,± 1 ±1 y = ±1 ±2 y = ±2 ±3 ( ) ( ) ( 2,± 2) ( 3,± 3) y = ±3 Two or more ordered pairs have the same x value but different y values This relation is not a function 33 4,1 ( ) 34 ( 7,− 5) No vertical line intersects the graph in more than one point This relation is a function b x = y 35 f ( x) = x2 + 3x y =x a f ( −2) = ( −2) + 3( −2) = − = −2 y = ±x b f ( −1) = ( −1) + 3( −1) = 1− = −2 x y 2 y = ±x Ordered i c f ( 0) = ( 0) + 3( 0) = + = 188 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller Section 1.3 d f (1) = (1) + 3(1) = 1+ =  1 =3 41 g   =  3    3   42 h( 7) = e f ( 2) = ( 2) + 3( 2) = + = 10 36 g ( x ) = x a g ( −2) = 43 k ( −5) = 1 =− ( −2) b g ( −1) = 37 h( x) = 44 f ( 5) = ( 5) + 3( 5) = 25 + 15 = 40 = −1 ( −1) 45 k ( 8) = (8) + = = 46 f ( −5) = ( −5) + 3( −5) = 25 − 15 = 10 47 g (t) = 49 k ( x + h) = = a2 + 8a + 16 + 3a + 12 = a2 + 11a + 28 52 f(t− 3) = (t− 3) + 3(t− 3) = t2 − 6t+ + 3t− = t2 − 3t 53 g ( 0) = Undefined 54 k ( −10) = ( −10) + = −9 d h(1) = e h( 2) = 38 k ( x) = x + ( −2) + = Undefined −1 55 f ( x + h) = ( x + h) + 3( x + h) Undefined d k (1) = e k ( 3) = x + h+1 51 f ( a + 4) = ( a + 4) + 3( a + 4) c h( 0) = c k ( 0) = ( x + h) + = 50 h( x + h) = 1 = ( 2) b h( −1) = b k ( −1) = 1 = (t) t 48 f ( a) = ( a) + 3( a) = a2 + 3a a h( −2) = a k ( −2) = −4 Undefined  1 = −2 c g  −  =  2  − 1  2    1 =2 d g   =  2  1  2   e g ( 2) = ( −5) + = ( −1) + = = ( 0) + = = (1) + = ( 3) + = = = x2 + 2xh + h2 + 3x + 3h 56 g ( x + h) = x+ h 57 f( x + h) = −4( x + h) − 5( x + h) + 2 ( ) = −4 x2 + 2xh + h2 − 5x − 5h + = −4x2 − 8xh − 4h2 − 5x − 5h + 1 = 39 g ( 3) = ( 3) 58 f( x + h) = −2( x + h) + 6( x + h) − 40 h( −7) = 189 Full file at https://TestbankDirect.eu/Solution-Manual-for-Precalculus-1st-Edition-By-Miller ... https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller. .. https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller. .. https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller Solution Manual for Precalculus 1st Edition By Miller Full file at https://TestbankDirect.eu /Solution- Manual- for- Precalculus- 1st- Edition- By- Miller