Download In Class Assignment 2: Mixed Solutions

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Solutions
In Class Assignment 2: Mixed Solutions
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 = 16
2. (3pts) Center: (−5, −2), Radius:
√
21
Solution.
(x + 5)2 + (y + 2)2 = 21
3. (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 = 25
4. (5pts) x2 + y2 + 12x − 14y + 84 = 0
Solution. Using the technique of completing the square we have
x2 + y2 + 12x − 14y + 84 = 0
⇒ x2 + 12x + 36 + y2 − 14y + 49 = 36 + 49 − 84
⇒ (x + 6)2 + (y − 7)2 = 1
5. (5pts) x2 + y2 − 10x + 4y − 20 = 0
Solution. Using the technique of completing the square we have
x2 + y2 − 10x + 4y − 20 = 0
⇒ x2 − 10x + 25 + y2 + 4y + 4 = 20 + 25 + 4
⇒ (x − 5)2 + (y + 2)2 = 49
6. (5pts) x2 + y2 + 22x − 4 = 0
1
Solution. Using the technique of completing the square we have
x2 + y2 + 22x − 4 = 0
⇒ x2 + 22x + 121 + y2 = 4 + 121
⇒ (x + 11)2 + (y − 0)2 = 125
7. (5pts) 2x2 + 2y2 − 32x + 12y + 90 = 0
Solution. Using the technique of completing the square we have
2x2 + 2y2 − 32x + 12y + 90 = 0
⇒ x2 + y2 − 16x + 6y + 45 = 0
⇒ (x − 8)2 + (y + 3)2 = −45 + 64 + 9
⇒ (x − 8)2 + (y + 3)2 = 28
8. (5pts) x2 + y2 − 10x − 22y + 155 = 0
Solution. Using the technique of completing the square we have
x2 + y2 − 10x − 22y + 155 = 0
⇒ (x − 5)2 + (y − 11)2 = −155 + 25 + 121
⇒ (x − 5)2 + (y − 11)2 = −9
⇒ A circle cannot have a negative radius so this is not a circle.
9. (5pts) 4x2 + 4y2 − 12x + 9 = 0
Solution. Using the technique of completing the square we have
4x2 + 4y2 − 12x + 9 = 0
9
⇒ x2 + y2 − 3x + = 0
4
(
)2
3
9 9
⇒ x−
+ (y − 0)2 = − +
2
4 4
(
)2
3
⇒ x−
+ (y − 0)2 = 0
2
Since this is a circle of radius 0 this is equation represents a single point.
5
5
2
10. (5pts) x2 + y2 − x − y − = 0
3
3
9
Solution. Using the technique of completing the square we have
5
5
2
x2 + y2 − x − y − = 0
3
3
9
(
)2 (
)2
1
5
5 1 25
x−
+ y−
= + +
3
6
9 9 36
)2 (
)2
(
1
5
49
⇒ x−
+ y−
=
3
6
36
2
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 13-20 determine if the graphs or equations define:
(a) y as a function of x. State which test it fails or passes to justify your response.
(b) x as a function of y. State which test it fails or passes to justify your response.
13. (4pts)
(a)
Solution. Not a function. Fails the vertical line test.
(b)
Solution. Not a function. Fails the horizontal line test.
14. (4pts)
3
(a)
Solution. Is a function. Passes the Vertical line test.
(b)
Solution. Is not a function. Fails the Horizontal line test.
15. (4pts)
(a)
Solution. Is not a function. Fails the Vertical line test.
(b)
Solution. Is not a function. Fails the Horizontal line test.
16. (4pts)
(a)
Solution. Is not a function. Fails the vertical line test.
(b)
Solution. Is a function. Passes the horizontal line test.
17. (4pts)
4
(a)
Solution. Is not a function. Fails the vertical line test.
(b)
Solution. Is a function. Passes the horizontal line test.
18. (4pts)
(a)
Solution. Is a function. Passes the vertical line test.
(b)
Solution. Is not a function. Fails the horizontal line test.
19. (4pts) (x + 3)2 + (y + 4)2 = 0
(a)
Solution. Is a function. The equation represents a single point and thus passes the vertical line test.
(b)
Solution. Is a function. Same reasoning as before and thus passes the horizontal line test.
5
20. (4pts)
(a) y = |x|
Solution. This is a function. It passes the vertical line test.
(b) x = |y|
Solution. This is a function. It passes the horizontal line test.
• For problem 21-32 define the functions:
f(x) = x2 + 3
g(t) =
1
t
h(z) = 10
k(m) =
√
m−1
21. (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 x2 can take on we see that the smallest number x2 + 3 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
10.
24. (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)2 + 3 = 4 + 3 = 7
26. (1pts) Evaluate g(−1)
Solution. g(−1) =
1
= −1
−1
1
27. (1pts) Evaluate h( )
2
1
Solution. h( ) = 10
2
5
28. (1pts) Evaluate k( )
4
6
5
Solution. k( ) =
4
√
1
5
−1=
4
2
29. (1pts) Evaluate f(a)
Solution. f(a) = a2 + 3
30. (2pts) Evaluate g(x + h)
Solution. g(x + h) =
1
x+h
31. (2pts) Evaluate h(201h + 302t)
Solution. h(201h + 302t) = 10
32. (2pts) Evaluate k(x + h)
Solution. k(x + h) =
√
x+h−1
33. (2pts) Find and simplify f(x + h) if
f(x) = −2x2 + 6x − 3
Solution. f(x + h) = −2(x + h)2 + 6(x + h) − 3
34. (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) = x3 − 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 36-38
36. (1pts) f(−1)
Solution. f(−1) = 6
37. (1pts) For what value of x is f(x) = 7?
Solution. x = 9
7
38. (1pts) For what value of x is f(x) = 4?
Solution. x = 3
39. (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) = 0 + 3 = 3 thus y − int=(0,3) .
41. (4pts)
(a) Determine the displayed domain
Solution. Domain= {−6, −4, −2, 0, 2, 4, 6}
(b) Determine the displayed range
Solution. Range = {−4, −2, 0, 2, 4, 6, 8}
Assume the following are functions 42-58.
(a) Find their natural domains. (2pts)
(b) Find the x& y intercepts if they exist (4pts)
(c) Find there associated ranges. (2pts)
42. (6pts) k(x) =
x+6
x−2
(a)
Solution. Domain R \ {2}
(b)
Solution. x-int (−6, 0)
y-int (0, −3)
8
(c)
Solution. Range R \ {1}
43. (6pts) k(x) =
x+6
x2 + 2
(a)
Solution. Domain R
(b)
Solution. x-int (−6, 0)
y-int (0, 3)
(c)
Solution. Range (≈ −0.0411103, 3.04110324358)
44. (6pts) k(x) =
x+6
x2 − 2
(a)
√
Solution. Domain R \ { 2}
(b)
Solution. x-int (−6, 0) y-int (0, −3)
(c)
Solution. Range R \ {(−2.958, 0]}
45. (6pts) k(t) =
√
16 − t
(a)
Solution. Domain 16 − t ≥ 0 ⇒ t ≤ 16 ⇒ (−∞, 16]
(b)
Solution. x-int (16, 0)
y-int (0, 4)
(c)
Solution. Range [0, ∞)
46. (6pts) k(t) =
√
t − 16
(a)
Solution. Domain t − 16 ≥ 0 ⇒ t ≥ 16 ⇒ [16, ∞)
(b)
9
Solution. x-int (16, 0)
y-int does not exist as a real number
(c)
Solution. Range [0, ∞)
47. (6pts) k(t) = √
1
16 − t
(a)
Solution. Domain t ̸= 16 and 16 − t ≥ 0 ⇒ t ≤ 16. Thus (−∞, 16)
(b)
Solution.
x-int does not exist
(
)
1
y-int 0,
4
(c)
Solution. Range (0, ∞)
48. (6pts) k(x) =
√
5
3+x
(a)
Solution. Domain R
(b)
Solution.
√ x-int (−3, 0)
y-int (0, 5 3)
(c)
Solution. Range R
49. (6pts) k(x) =
√
5
x−3
(a)
Solution. Domain R
(b)
Solution.
√ x-int (3, 0)
y-int (0, 5 −3)
(c)
Solution. Range R
50. (6pts) k(x) = √
5
1
x−3
(a)
10
Solution. I can take the 5th 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}.
(b)
Solution. x-int: DNE since I can never divide 1 by a real number and get 0.
−1
y-int: (0, √
5
3
(c)
Solution. Range R \ {0}
51. (6pts) k(x) = x2 − 4x − 12
(a)
Solution. Domain R
(b)
Solution. x-int: Setting 0 = x2 − 4x − 12 and solving I (x − 6)(x + 2) = 0 or x = 6, −2 giving intercepts
(6, 0), (−2, 0) .
y-int: k(0) = 0 − 0 − 12 = −12 giving an intercept of (0, −12)
(c)
Solution. Range [−16, ∞)
52. (6pts) k(x) =
x2 − 4x − 12
x+1
(a)
Solution. Domain: R \ {−1} since I cannot divide by zero and x + 1 = 0 when x = −1.
(b)
x2 − 4x − 12
which happens when the numerator is 0
x+1
or when x2 − 4x − 12 = 0. This we know gives us intercepts (6, 0), (−2, 0)
−12
= −12 thus the intercept is (0, −12)
y-int: k(0) =
1
Solution. x-int: Keeping in mind x ̸= −1 we solve 0 =
(c)
Solution. Range R
53. (6pts) k(x) =
x+1
x2 − 4x − 12
(a)
Solution. Domain: Cannot divide by zero which happens when 0 = x2 − 4x − 12 or when x = 6, −2. Thus our
domain is R \ {−2, 6}.
(b)
11
Solution. x-int: Keeping in mind x ̸= −2, 6 we solve 0 =
x+1
which happens when x + 1 = 0 or when
x2 − 4x − 12
x = −1. Thus our intercept is (−1, 0) .
y-int: k(0) =
1
−1
thus our intercept is (0,
).
−12
12
(c)
Solution. Range R
54. (6pts) k(x) = 8 − |x − 2|
(a)
Solution. Domain: R
(b)
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) .
(c)
Solution. Range (−∞, 8]
55. (6pts) k(x) =
5
8 − |x − 2|
(a)
Solution. Domain: I cannot divide by zero thus x ̸= 10, −6 giving a domain of R \ {−6, 10}
(b)
Solution. x-int: We cannot divide 5 by a real number and get 0 thus the intercept DNE.
5
5
5
y-int: k(0) =
= giving an intercept of (0, )
8 − | − 2|
6
6
(c)
Solution. Range R \ {[0, 0.625)}
56. (6pts) k(x) =
5
8 + |x − 2|
(a)
Solution. Domain R
(b)
Solution. x-int: DNE
5
1
y-int: k(0) =
thus (0, )
10
2
(c)
12
Solution. Range (0, 0.625]
57. (6pts) k(x) = 3x − 7 s.t. x < 0
(a)
Solution. Domain (−∞, 0)
(b)
Solution. x-int: Keeping in mind that x < 0 we solve 0 = 3x − 7 to get x =
7
but this is greater than 0 and
3
thus there is no x intercept DNE .
y-int: DNE since x < 0 and thus cannot be zero.
(c)
Solution. Range (−∞, −7)
58. (6pts) k(x) = 3x − 7 s.t. −2 < x < 2
(a)
Solution. Domain (−2, 2)
(b)
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. x-int: Since
(c)
Solution. Range (−13, −1)
13
59. (13pts) Define f(x) by the graph below.
(a) (1pts)Find f(−1)
Solution. −1
(b) (1pts)Find f(1)
Solution. 2
(c) (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}
√
e
2
. b.) passes through (e, π) with slope
60. (4pts total) Write equation of line a.) Passes through (5, 4) with slope
3
π
Solution. (a) Using point slope form of a line (y − y1 = m(x − x1 )) we find the equation of the line to be
√
2
(x − 5)
y−4=
3
(b) Using point slope form of a line (y − y1 = m(x − x1 )) we find the equation of the line to be
y−π=
14
e
(x − e)
π