Homework 6. (x 3) 2 + (y 1) 2 = 25. (x 5) 2 + (y + 2) 2 = 49

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1 Name: Solutions Due Date: Monday May 16th. Homework 6 Directions: Show all work to receive full credit. Solutions always include the work and problems with no work and only answers will receive NO CREDIT for that problem. For problems 1-10, write the equation for the circle in standard form. 1. (3pts) Center: ( 3, 2), Radius: 4 Solution. (x + 3) 2 + (y 2) 2 = (3pts) Center: ( 5, 2), Radius: 21 Solution. (x + 5) 2 + (y + 2) 2 = (5pts) The center is (3, 1) and another point on the circle is (6, 5) Solution. The distance from the center of a circle to any point on the circle is the radius of the circle. In this case r = (6 3) 2 + (5 1) 2 = 25 = 5 thus the equation for the circle is (x 3) 2 + (y 1) 2 = (5pts) x 2 + y x 14y + 84 = 0 Solution. Using the technique of completing the square we have x 2 + y x 14y + 84 = 0 x x y 2 14y + 49 = (x + 6) 2 + (y 7) 2 = 1 5. (5pts) x 2 + y 2 10x + 4y 20 = 0 Solution. Using the technique of completing the square we have x 2 + y 2 10x + 4y 20 = 0 x 2 10x y 2 + 4y + 4 = (x 5) 2 + (y + 2) 2 = 49 1

2 6. (5pts) x 2 + y x 4 = 0 Solution. Using the technique of completing the square we have x 2 + y x 4 = 0 x x y 2 = (x + 11) 2 + (y 0) 2 = (5pts) 2x 2 + 2y 2 32x + 12y + 90 = 0 Solution. Using the technique of completing the square we have 2x 2 + 2y 2 32x + 12y + 90 = 0 x 2 + y 2 16x + 6y + 45 = 0 (x 8) 2 + (y + 3) 2 = (x 8) 2 + (y + 3) 2 = (5pts) x 2 + y 2 10x 22y = 0 Solution. Using the technique of completing the square we have x 2 + y 2 10x 22y = 0 (x 5) 2 + (y 11) 2 = (x 5) 2 + (y 11) 2 = 9 A circle cannot have a negative radius so this is not a circle. 9. (5pts) 4x 2 + 4y 2 12x + 9 = 0 Solution. Using the technique of completing the square we have 4x 2 + 4y 2 12x + 9 = 0 x 2 + y 2 3x = 0 ( x 3 ) 2 + (y 0) 2 = ( x (y 0) 2) 2 = 0 Since this is a circle of radius 0 this is equation represents a single point. 10. (5pts) x 2 + y x 5 3 y 5 9 = 0 Solution. Using the technique of completing the square we have x 2 + y x 5 3 y 5 9 = 0 ( x 3) 1 2 ( + y 5 ) 2 = ( x 1 2 ( + y 3) 5 ) 2 =

3 11. (2pts) Determine the solution set for (x 3) 2 + (y + 12) 2 = 0 Solution. The only way that the sum of two non-negative numbers is 0 is if both numbers are 0. Thus x 3 = 0 and y+12 = 0 giving the only solution of {(3, 12)} 12. (2pts) Determine the solution set for (x + 15) 2 + (y 3) 2 = 25 Solution. The sum of two non-negative numbers can never be negative thus there is no solution or the solution set is the empty set {}. For questions determine if the graphs or equations define: y as a function of x. State which test it fails or passes to justify your response. x as a function of y. State which test it fails or passes to justify your response. 13. (4pts) Solution. Not a function. Fails the vertical line test. 14. (4pts) Solution. Not a function. Fails the horizontal line test. 3

4 Solution. Is a function. Passes the Vertical line test. 15. (4pts) Solution. Is not a function. Fails the Horizontal line test. Solution. Is not a function. Fails the Vertical line test. 16. (4pts) Solution. Is not a function. Fails the Horizontal line test. Solution. Is not a function. Fails the vertical line test. 17. (4pts) Solution. Is a function. Passes the horizontal line test. 4

5 Solution. Is not a function. Fails the vertical line test. 18. (4pts) Solution. Is a function. Passes the horizontal line test. Solution. Is a function. Passes the vertical line test. Solution. Is not a function. Fails the horizontal line test. 19. (4pts) (x + 3) 2 + (y + 4) 2 = 0 Solution. Is a function. The equation represents a single point and thus passes the vertical line test. Solution. Is a function. Same reasoning as before and thus passes the horizontal line test. 5

6 20. (4pts) y = x Solution. This is a function. It passes the vertical line test. x = y Solution. This is a function. It passes the horizontal line test. For problem define the functions: f(x) = x g(t) = 1 t h(z) = 10 k(m) = m (4pts) Find the natural domain of f(x) within the real numbers and it s associated range Solution. We can square and then add 3 to any real number and still get a real number back thus the natural domain is R. Since 0 is the smallest number x 2 can take on we see that the smallest number x can take on is 3 and as we increase x, f(x) continues to increase without bounds. Thus the range is [3, ) 22. (4pts) Find the natural domain of g(t) within the real numbers and it s associated range Solution. We can divide 1 by any number except for 0 thus the domain is R \ {0}. Similarly given any real number y we can solve y = g(t) for t explicitly except for y = 0 thus the range is R \ {0}. 23. (4pts) Find the natural domain of h(z) within the real numbers and it s associated range Solution. Given any real number z h(z) = 10 thus the domain is R. There is only 1 number in the range which is (4pts) Find the natural domain of k(m) within the real numbers and it s associated range Solution. You can take the square root of any non-negative real number and get back a real number. Thus m 1 0, which implies m 1. The domain is then [1, ). As m increases so does k(m) without bound. The smallest k(m) can be is 0 with m = 1. Thus the range is [0, ). 25. (1pts) Evaluate f( 2) Solution. f( 2) = ( 2) = = (1pts) Evaluate g( 1) Solution. g( 1) = 1 1 = (1pts) Evaluate h( 1 2 ) Solution. h( 1 2 ) = (1pts) Evaluate k( 5 4 ) 6

7 Solution. k( 5 4 ) = = (1pts) Evaluate f Solution. f = a (2pts) Evaluate g(x + h) Solution. g(x + h) = 1 x + h 31. (2pts) Evaluate h(201h + 302t) Solution. h(201h + 302t) = (2pts) Evaluate k(x + h) Solution. k(x + h) = x + h (2pts) Find and simplify f(x + h) if f(x) = 2x 2 + 6x 3 Solution. f(x + h) = 2(x + h) 2 + 6(x + h) (2pts) Find and simplify f(x + h) if f(x) = 11 5x Solution. f(x + h) = 11 5(x + h) 35. (2pts) Find and simplify f(x + h) if f(x) = x 3 4x + 2 Solution. f(x + h) = (x + h) 3 4(x + h) + 2 Define f = {(2, 3), (9, 7), (3, 4), ( 1, 6)} for problems (1pts) f( 1) Solution. f( 1) = (1pts) For what value of x is f(x) = 7? Solution. x = 9 7

8 38. (1pts) For what value of x is f(x) = 4? Solution. x = (4pts) Find the x and y intercepts for the function f(x) = x + 3 Solution. x-intercept occurs when y = 0 so setting f(x) = 0 and solving gives x = 3 and thus we have x intercept=(-3,0). y-intercept occurs when x = 0 so evaluating f(0) we get f(0) = 0+3 = 3 thus y intercept=(0,3). 40. (4pts) Find the x and y intercepts for the function h(x) = x + 3 Solution. Setting h(x) = 0 and solving gives 0 = x + 3 or x = 3 which yields x = 9 and thus we have x int=(9,0). Evaluating h(0) we have h(0) = = 3 thus y int=(0,3). 41. (4pts) Determine the displayed domain Solution. Domain= { 6, 4, 2, 0, 2, 4, 6} Determine the displayed range Solution. Range = { 4, 2, 0, 2, 4, 6, 8} Assume the following are functions Find their natural domains. (2pts) Find the x& y intercepts if they exist (4pts) FOR EXTRA CREDIT find there associated ranges. (2pts) 42. (6pts) k(x) = x + 6 x 2 Solution. Domain R \ {2} Solution. x-int ( 6, 0) y-int (0, 3) 8

9 Solution. Range R \ {1} 43. (6pts) k(x) = x + 6 x Solution. Domain R Solution. x-int ( 6, 0) y-int (0, 3) Solution. Range (0, ) 44. (6pts) k(x) = x + 6 x 2 2 Solution. Domain R \ { 2} Solution. x-int ( 6, 0) y-int (0, 3) Solution. Range R \ {( 2.958, 0]} 45. (6pts) k(t) = 16 t Solution. Domain 16 t 0 t 16 (, 16] Solution. x-int (16, 0) y-int (0, 4) Solution. Range [0, ) 46. (6pts) k(t) = t 16 Solution. Domain t 16 0 t 16 [16, ) 9

10 Solution. x-int (16, 0) y-int does not exist as a real number 47. (6pts) k(t) = Solution. Range [0, ) 1 16 t Solution. Domain t 16 and 16 t 0 t 16. Thus (, 16) Solution. ( x-int does not exist y-int 0, 1 ) 4 Solution. Range (0, ) 48. (6pts) k(x) = x Solution. Domain R Solution. x-int ( 3, 0) y-int (0, 5 3) Solution. Range R 49. (6pts) k(x) = 5 x 3 Solution. Domain R Solution. x-int (3, 0) y-int (0, 5 3) Solution. Range R 50. (6pts) k(x) = 1 5 x 3 10

11 Solution. I can take the 5 th root of any real number and still get a real number back however I can never divide by 0 thus x 3 0 x 3. My domain is then R \ {3}. Solution. x-int: DNE since I can never divide 1 by a real number and get 0. y-int: (0, Solution. Range R \ {0} 51. (6pts) k(x) = x 2 4x 12 Solution. Domain R Solution. x-int: Setting 0 = x 2 4x 12 and solving I (x 6)(x + 2) = 0 or x = 6, 2 giving intercepts (6, 0), ( 2, 0). y-int: k(0) = = 12 giving an intercept of (0, 12) Solution. Range [ 16, ) 52. (6pts) k(x) = x2 4x 12 x + 1 Solution. Domain: R \ { 1} since I cannot divide by zero and x + 1 = 0 when x = 1. Solution. x-int: Keeping in mind x 1 we solve 0 = x2 4x 12 which happens when the numerator is 0 x + 1 or when x 2 4x 12 = 0. This we know gives us intercepts (6, 0), ( 2, 0) y-int: k(0) = 12 = 12 thus the intercept is (0, 12) (6pts) k(x) = Solution. Range R x + 1 x 2 4x 12 Solution. Domain: Cannot divide by zero which happens when 0 = x 2 4x 12 or when x = 6, 2. Thus our domain is R \ { 2, 6}. 11

12 x + 1 Solution. x-int: Keeping in mind x 2, 6 we solve 0 = x 2 which happens when x + 1 = 0 or when 4x 12 x = 1. Thus our intercept is ( 1, 0). y-int: k(0) = 1 12 thus our intercept is (0, 1 12 ). Solution. Range R 54. (6pts) k(x) = 8 x 2 Solution. Domain: R Solution. x-int: Solving 0 = 8 x 2 yields x 2 = 8 and x 2 = 8 or x = 10, 6. Our intercepts are then ( 6, 0), (10, 0). y-int: k(0) = 8 2 = 6 thus our intercept is (0, 6). 55. (6pts) k(x) = Solution. Range (, 8] 5 8 x 2 Solution. Domain: I cannot divide by zero thus x 10, 6 giving a domain of R \ { 6, 10} Solution. x-int: We cannot divide 5 by a real number and get 0 thus the intercept DNE. 5 y-int: k(0) = 8 2 = 5 6 giving an intercept of (0, 5 6 ) 56. (6pts) k(x) = Solution. Range R \ {[0, 0.625)} x 2 Solution. Domain R Solution. x-int: DNE y-int: k(0) = 5 10 thus (0, 1 2 ) 12

13 Solution. Range (0, 0.625] 57. (6pts) k(x) = 3x 7 s.t. x < 0 Solution. Domain (, 0) Solution. x-int: Keeping in mind that x < 0 we solve 0 = 3x 7 to get x = 7 3 thus there is no x intercept DNE. y-int: DNE since x < 0 and thus cannot be zero. but this is greater than 0 and Solution. Range (, 7) 58. (6pts) k(x) = 3x 7 s.t. 2 < x < 2 Solution. Domain ( 2, 2) Solution. x-int: Since 7 > 2 it is outside the domain and hence there is again no x-intercept. 3 y-int: 0 is in the domain and since k(0) = 7 we have an intercept of (0, 7) Solution. Range ( 13, 1) 13

14 59. (13pts) Define f(x) by the graph below. (1pts)Find f( 1) Solution. 1 (1pts)Find f(1) Solution. 2 (1pts)Find f(2) Solution. 1 (d) (2pts)Find the y-intercept Solution. (0, 0) (e) (2pts)Find the x-intercepts Solution. ( 3, 0), ( 2, 0), (0, 0), ( 9, 0), ( 2.5, 0), ( 3.5, 0), ( 4.5, 0) 5 (f) (2pts)Find the Domain as shown Solution. [ 3, 5] (g) (2pts)Find the range as shown Solution. [ 1, 2] (h) (2pts)Find all x s.t. f(x) = 1 Approximate as near as possible if not clear. Solution. { 1.5, 1.5, 2, 4} 14

Test 2 Review Math 1111 College Algebra

Test 2 Review Math 1111 College Algebra 1. Begin by graphing the standard quadratic function f(x) = x 2. Then use transformations of this graph to graph the given function. g(x) = x 2 + 2 *a. b. c. d.