( ), slope = 4 ( ) ( ) ( ) ( ) ( ) 2 x 1 ) y = 4( x 1) PreCalculus Basics Homework Answer Key. 4-1 Free Response. 9. ( 3, 0) parallel to 4x 3y = 0

Transcription

1 PreCalculus Basics Homework Answer Ke 4-1 Free Response 1. ( 1, 1), slope = , 0 +1= 1 ( 2 x 1 ), slope = 4 0 = 4( x 1) = 4( x 1) 5. ( 1, 1) and ( 3, 5) m = = 2 1 = 2 x 1 or 5 = 2 x 3 7. ( 1, 4) and ( 4, 1) m = = 1 4 = 1 x 1 or 1 = 1 x 4 9. ( 3, 0) parallel to 4x 3 = 0 Solve for : = 4 3 x m = = 4 ( 3 x 3) = 4 3 x 4 3 = 4x 12 4x 3 = ( 7, 3) perpendicular to 2x 4 = 5 Solve for : = 1 2 x 5 4 m = 1 2 m = 2 3 = 2( x + 7) 2x + = = 4 3 x + 7 zero: 0 = 4 3 x x = 7 x = int: = 4 ( 3 0) + 7 = 7 intercepts: 21 4, 0 ; 0, 7 65

2 15. 3x + 5 = 7 zero: 3x = 7 x = 7 3 -int: = 7 = intercepts: 3, 0 ; 0, Multiple Choice 1. A f 2 = 1 means the point ( 2, 1) is on the line. Similarl, 4, 2 is on the line. m = 2 1 x 2 x 1 = 1 = 3 ( 2 x 2) = 3 2 x = A Horizontal lines have a constant -value. In this case, = Free Response 1. = x 2 4x 5 x-int: 0 = x 5 x = 5, 1 -int: = ( x +1) intercepts: 1, 0 5 = 5 and 5, 0 ( 0, 5) 3. = 4x 2 12x + 9 x-int: 0 = 4x 2 12x = 2x 3 x = 3 2 ( 2x 3) ; (repeated zero) -int: = 4( 0) 2 12( 0) + 9 = 9 intercepts: 3 2, 0 ; 0, 9 5. Zeros: 2, 1; -int: 4 = a x 2 ( x +1) ( 0 1) 4 = a 0 2 a = 2 = 2 x 2 x +1 = 2x 2 + 2x B f 0 = 10 means that our - intercept is m = 0 2 = 2 = 2x

3 7. 1 2, 0, 1 2, 0 = zeros ( 0, 1) = -int = a x 1 2 x = a = 1 4 a a = 4 = 4 x 1 2 x = 4 x = 4x = x 2 6x = x 2 6x ; v( 3, 2) = x = x 2 5x + 4 = x 2 5x = x or = x ; v 5 2, ; v 2.5, = 5x 2 30x i 9 = 5 x 2 + 6x = 5 x 2 + 6x + 9 = 5( x + 3) ; v( 3,96) 15. h = 16t t 20 h = 16( t 2 5t) 20 h = 16 t 2 5t h = 16 t max height = 80 ft 0 = 16t t 20 t = 80 ± 2 ( 20) t = sec, 4.736sec 17. = x 2 6x = x 3 v 3, 2 Parabola opens upwards Range: 2 or 2, ) 19. = 5x 2 30x = 5 x + 3 v 3, 96 Parabola opens downwards Range: 96 or (, 96 67

4 4-2 Multiple Choice 1. E = 2x 2 + x 6 0 = 2x + 4x 3x 6 3( x + 2) ( x + 2) 0 = 2x x = 2x 3 x = 3 2, Free Response , , x B = 3x = 3( x 0) vertex: ( 0, 4) x =0 = 4, line of smmetr: x = , , E Line of smmetr x = 3 means h = 3. Containing ( 3, 4) means k = 4. = a( x + 3) = a( 5 + 3) 2 4 a = 1 = ( x + 3) x , , 200 x

5 7. Find on calculator using 2 nd +Trace+Zero ( 2, 0), ( 0.382, 0), 2.618, 0 9. Find on calculator using 2 nd +Trace+Zero ( 0.435, 0), ( 3.704, 0) 11. Find on calculator using 2 nd +Trace+Minimum 2.644, Use shape of graph to determine range, from absolute minimum to or , ) 13. Find on calculator using 2 nd +Trace+Maximum/Minimum ( 1.2, ), ( 6, 2587) Range: All Reals or because the degree, of the polnomial is odd. 4-3 Multiple Choice 1. D Test each option and see which shows the zeroes and vertex of the parabola. 3. B Max and min values means the -coordinates of the max/min points. Find on calculator. 5. E Since it s a negative quartic, the ends should both be down. 4-4 Free Response 1. Let w = width, l = length P = 2w + l = 3000 l = w A = lw A = ( w)w A = 3000w 2w 2 Window: [ 3000, 10,000] 1000, 1,200,000 Max: ( 750, 1,125,000) w = 750 l = [ ] l = 1500 Dimensions: 750 d x 1500 d Area: A = 1,125,000 d 3 3. Girth in this case is circumference, because the cross section is a circle. Girth plus length is 2πr + l 2πr + l = 108 l = 108 2πr Volume is V = πr 2 l V = πr πr V = 108πr 2 2π 2 r 3 Window: [ 20, 20] [ 150, 15,000] Max: ( , 14, ) Dimensions: in x 36 in Max Volume: 14,851 in 3 69

6 5. A = lw = 16 l = 16 w P = 2l + 2w P = 2 16 w + 2w P = 32 w + 2w Window: [ 20, 20] 20, 20 Min: ( 4, 16) Min perimeter: 16 cm 7. V = πr 2 h = 32π h = 32π r 2 π = 32 r 2 SA = 2πr 2 + 2πrh SA = 2πr 2 + 2πr 32 r 2 SA = 2πr π r Window: [ 10, 10] 100, 200 Min: 2.520, [ ] r = Min SA: in 2 9. Via Pthagorean theorem: x 2 + r 2 = 10 2 x 2 + r 2 = 100 r 2 = 100 x 2 V = πr 2 h 2x V = π 100 x 2 V = 200π x 2π x 3 [ ] Window: [ 20, 20] [ 3000, 3000] Max: ( 5.774, ) Max Volume: in Multiple Choice 1. E Label the top side. Perimeter: 2x + = 160 = 160 2x Area: A = x A = x 160 2x A = 160x 2x 2 3. B Let x = length; = width x = + 7 = x 7 A = x A = x x 7 5. B height = x width = 15 2x length = 24 2x A = lwh A = ( 24 2x) ( 15 2x)x = 4x 3 78x x 400 4x 3 78x x Graph left hand side and make its sign pattern to determine positive region(s). 70

7 4-5 Free Response 1. = x 3 3x 2 4x = = = x 4 3x = = = x 3 x 2 5x + 2 Via rational roots theorem, possible rational roots are ±1, ± 2. Guess and check or use calculator to see: x 2 3x +1 = x + 2 Use quadratic formula to solve: 3± 5 ( 2, 0),, = 2x 3 + 5x 2 + x 2 Do snthetic division twice: Factor: = x + 2 ( x +1) ( 2x 1) Zeros: ( 1, 0), ( 2, 0), 9. = x 4 2x 3 3x 2 + 4x ( x 2) x 2 x 2 2 ( x 2) 2 = x +1 Factor: = x +1 Zeros: ( 1, 0), ( 2, 0) 11. = 6x 3 +13x 2 14x = x , 0 6x 2 5x +1 ( 3x 1) ( 2x 1) Factor: = x + 3 Zeros: ( 3, 0), 1 2, 0, 1 3, 0 71

8 13. = 2x 3 28x x Factor: = ( x 2) 2x 2 24x + 33 Use quadratic formula to solve: x =2, 24 ± 242 4( 2) ( 33) 2( 2) Zeros: ( 1.584, 0), ( 2, 0), , = x5 5x 4 +10x 3 10x 2 + 5x x 2 2x +1 = x 1 5 Factor: = x 1 Zero: 1, Zeros: 3, 3 2, 4 = a( x + 3) ( 2x 3) ( x 4) Point: ( 1, 3) 3 = a( 1+ 3) ( 2( 1) 3) ( 1 4) 3 = a( 2) ( 5) ( 5) 3 = 50a 3 50 = a = 3 ( 50 x + 3 )( 3x 2) ( x 4) Or, if ou multipl the factors: = 3 ( 50 2x3 5x 2 21x + 36) 19. Zeros: ± 2, 3 (twice) ( x 2 ) x 3 = a x + 2 x 3 = a x = a ( 1) Point: 1, 3 3 = a 1 3 = 16a 1 3 ( 16) ( x 3) 3 16 = a = 3 16 ( x2 2) ( x 3) 2 Or, if ou multipl the factors: = 3 16 x4 6x 3 + 7x 2 +12x 18 72

9 4-5 Multiple Choice 1. B Divisible b x + k means that the remainder when divided b k is 0. k 1 3 k k 3k + k k 2k + k 2 2k + k 2 = 0 k = 0, 2 3. D Remainder is E Option 1: Write out all 32 columns of the snthetic division. Option 2: Interpret the question. The remainder when the function is divided b x +1 is the same thing as evaluating the function at 1. ( 1) = = Free Response 1. = x 3 + 3x 2 18x 40 Graph: x = 5, 2, 4 x < 5 : ( )( )( ) = 5 < x < 2 : 2 < x < 4 : ( )( )(+) = ( )(+)(+) = x > 4 : (+)(+)(+) = x = x 3 + 5x 2 9x x 45 9( x + 5) ( x + 5) ( x + 3) ( x + 5) = x 3 + 5x 2 = x 2 x + 5 = x 2 9 = x 3 x = 5, 3, x x 1 4 Based on the zeros, the factors are ( x +1) and ( x 4). Negative so that the end behavior matches the sign pattern: = x +1 ( x 4) = x 2 + 3x + 4 x 1, 4 Since the solution set matches the strictl positive regions, we use > 0. Therefore, x 2 + 3x + 4 > 0 73

10 x Based on the zeros, the factors are ( x + 2 ), ( x 0), and ( x 2 ). Negative so that the end behavior matches the sign pattern: ( x 2 ) = x x + 2 = x x 2 2 = x 3 + 2x x (, 2 ) ( 0, 2 ) Because the solution set calls for the strictl positive regions, we use > 0. Therefore, x 3 + 2x > x Based on the zeros, the factors are ( x + 5), ( x 0), and ( x 5). Double zero at 0 because the sign doesn t change there: = ( x + 5)x 2 ( x 5) = x 4 25x 2 0 x (, 5] [ 5, ) or { 0} Solution set is regions that are positive or 0, so we use 0. Therefore, x 4 25x x Based on the zeros, the factors are ( x + 3), ( x 4) and ( x 5). The signs alternate, so all the factors are of odd degree. The sign on the right of the pattern is negative, so the sign in front of the factors is negative. x ( 3, 4) ( 5, ) The intervals indicate the negatives, so the inequalit is a less than and parentheses indicate no equals. Therefore, x + 3 ( x 4) ( x 5) < x Based on the zeros, the factors are ( 3x +1), ( 3x 2) and 3x 4. The signs do not alternate around 2 3, so ( 3x 2) is squared. The sign on the right of the pattern is negative, so the sign in front of the factors is negative. x ( 1 3, 2 3) ( 2 3, 4 3) The intervals indicate the positives, so the inequalit is a greater than and parentheses indicate no equals. Therefore, ( 3x 2) 2 ( 3x 4) > 0 3x +1 74

11 x Based on the zeros, the factors are ( x +1), ( x 1), ( x 3), and ( x 5). The signs alternate, so all the factors are of odd degree. The sign on the right of the pattern is positive, so the sign in front of the factors is positive. x (, 1) ( 1, 3) ( 5, ) The intervals indicate the positives, so the inequalit is a greater than and parentheses indicate no equals. Therefore, x 1 ( x +1) ( x 5) ( x 3) > x 3 x 2 4x + 4 > 0 x 2 ( x 1) 4( x 1) ( x 2 4) ( x 1) x = 2, 2, x x ( 2, 1) ( 2, ) 21. 2x 3 + 4x 2 17x 39 0 Graph: x = x 2 +10x +13 = x 3 The quadratic has imaginar roots. 0 + x 3 x (, x 4 x 3 11x 2 + 9x +18 < 0 Graph: x = 3, 1, 2, x x ( 3, 1) ( 2, 3) 25. x 3 + 5x 2 7x +15 < x x 4.184, 19. x 4 + x 3 7x 2 x + 6 < 0 Graph: x = 3, 1, 1, x x ( 3, 1) ( 1, 2) 75

12 4-6 Multiple Choice 1. A = 2x 4 2x 3 4x 2 = 2x 2 x 2 x 2 = 2x 2 x 2 ( x +1) x = 0 (repeated), 2, 1 End behavior: positive 4 th degree means both ends go up. 0 as a repeated root means the signs should not switch around it, but the should switch around 2 and D Solve x 3 + 4x 2 12x 0 x x 2 + 4x D 0 ( x 2) 0 x x + 6 x = 0, 6, x means we are looking for zero or positive regions. 0 at x 6, 0 2, ) x + 5 Solve x 2 + 3x 2 0 x = 0.562, 3.562, x at x 5, , ) 76

13 PreCalculus Basics Practice Test Answer Ke Part 1: CALCULATOR REQUIRED Multiple Choice 1. E = 3x 2 5 = 0 x 2 = 5 3 x = ± 5 3 Free Response 1. = ( x 3) ( x + 2) ( 2x +1) ( x +1) ( 3, 0), ( 2, 0), 1 2, 0, 1, 0 Graph on calc and use snthetic division to check answers. Sum: = C Find zeros on calculator using 2 nd +Trace+Zero x = is between 0 and 1 3. C k k 3 0 k 3k 12 3k 12 = 0 k = 4 Other factor is 3x 2 + 0x + k so, 3x D V = l i wi h 8 2x = 10 2x x Graph using the window [ 0, 4] [ 0, 80] 2 nd +Trace+Maximum to find the critical value of the maximum: x = cm Window: [ 4.7, 4.7] 20, 5 Zeros: Graph on calc using 2 nd +Trace+Zero ( 1.750, 0), ( 1.318, 0) [ ] Extreme points: Graph on calc using 2 nd +Trace+ Minimum/Maximum Abs. min: 1.117, Rel. max: (0.250, 6.527) Rel. min: (0.716, 7.069) Range: Use -value from absolute minimum , ) x 77

14 3. f ( x) = x 3 + x 2 + 0x f 2 3 = Part 2: NO CALCULATOR ALLOWED Multiple Choice 5. D 3x = 0 5 = 3x 8 = m = 3 5 m = B = 2x 2 + 4x = 2 x 2 + 2x = 2 x 2 + 2x = 2 x +1 Vertex: 1, 7 7. D E x x 3 x + 2 > 0 Zeros: x = 0, 3, x > 0 means we are looking for positive regions. Therefore, 2 < x < 0 or x > 3 Free Response x Based on the zeros, the factors are ( x + 6), ( x 2) and ( x 4). Double zero at 0 because the sign doesn t change. The sign on the right of the pattern is negative, so the sign in front of the factors is negative. ( x 2) ( x 4) 2 = x

15 5a. = x (, 6) ( 2, 4) ( 4, ) The intervals indicate negatives, so the inequalit is less than and parentheses means not equal. Therefore, ( x 2) ( x 4) 2 < 0 x + 6 4x 2 1 6x 2 + x 2 i 27x x x 1 ( 2x +1 ) ( ( 3x + 2) ( 2x 1) i 3x + 2) 9x 2 6x + 4 ( 2x +1) 4x 2 2x +1 = 9x2 6x + 4 4x 2 2x +1 5b. 8x 4x 2 x 2 + 3x 6 3x = = = 8x 4x 2 (x 2) + 3(x 2) i + 3 3x + 6 4x (x 2) ( + 3) (x 2) i + 3 3(x + 2) 4x 3 x