Expert answer:Pre Calc Graphing Poly Functions and Rational Func

Solved by verified expert:There is a total of 20 question, some I answered and some that need to be answered. For the ones I answered, correct them if they’re wrong in answer or explanation. For all questions, you must show your work. Word doc and pdf included.
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Name:
Date:
Graded Assignment
Checkup: 3.04 and 3.05
Answer the following questions using what you’ve learned from these lessons. Write your
responses in the space provided, and turn the assignment in to your instructor.
For problems 1 and 2, state the x- and y-intercepts for each function.
1. f  x   x 4  5x 2  4
y = 4, and x = 2, -2, 1, -1
To solve for y, set x equal to zero.
(0)4 − 5(0)2 + 4 = 4 So, y = 4.
For x, use the root rational theorem, since f(x) = 0.
{±1,±2,±4}
= {(−2,0)(−1,0)(1,0)(2,0)} So, x = -2, -1, 1, and 2.
{±1}
You can graph the equation to check your answer.
2. f  x   x 5  x
y = 0, and x = 0, -1, 1
To solve, use the same processes as the last question.
(0)5 − (0) = 0 So, y = 0.
{±1}
{±1}
= {(−1,0)(1,0)} Since the y intercept is 0, that means another x intercept would also be
0. So, x = -1, 0, and 1.
You can check your answer by graphing the function.
For problems 3 and 4, describe the end behavior for each of the functions by writing a limit expression.
1
3
3. f  x   x 3  2
4.
f  x   2x 6  5x  4
For problems 5 and 6, use the graphs to state the zeros for each polynomial function. State the
multiplicity of any roots if the multiplicity is 2 or higher.
5. f  x   x 3  2x 2
6. f  x   3x  12x 3
For problems 7 and 8, identify the extrema for each function. Classify each as a relative (local) or
absolute (global) maximum or minimum value.
7. f  x   4x 3  12x 2  15
All of the information can be found by using the largest degree of the polynomial, n. In this
case, n is 3. The graph will have at the most n – 1 maximum or minimum values.
3 – 1 = 2. There will be a total of 2 turns in the graph.
Input the extreme value of x = 0 into the equation to get y = 15. (4(0)3 − 12(0)2 + 15 = 15)
Absolute Maximum: (0, 15)
Plug in the value of x = 2 into the equation to get y = -1. (4(2)3 − 12(2)2 + 15 = 4(8) −
12(4) + 15 = 32 − 48 + 15 = −1)
Absolute Minimum: (2, -1)
8. f  x   4x 3  15x 2  8x  3
Use the same method as the last question.
3 – 1 = 2. There will be a total of 2 turns in the graph.
Input x = 0 into the equation to get y = -3. (−4(0)3 + 15(0)2 − 8(0) − 3 = −3)
Absolute Maximum: (0, -3)
For problems 9 and 10, sketch a graph of each function.
9.
f  x   4x 3  15x 2  8x  3
10. f  x   x 4  5x 2  4
Identify the vertical and/or horizontal asymptotes for each function.
11. y 
5x
x 1
= 1. To find the answer, you start by solving x – 1 from the
denominator. (x – 1 = 0, x = 1). This means that the domain includes all real numbers except
x = 1 or { : ≠ 1}.
Vertical asymptote(s): x
Horizontal asymptote: y = 5. To find the answer, you would use the 5 in 5x from the
numerator. This means that the range is all real numbers except for 5 or { : ≠ 5}.
You can check both of these by graphing the equation.
12. y 
3x
2x 2  1
Vertical asymptote(s): There is
no vertical asymptote because there are no undefined points.
Horizontal asymptote: y = 0. Since the denominators degree (2) is greater than the
numerators degree (1), the horizontal asymptote is the x axis.
Identify the oblique asymptote for each function.
13. y 
2x 2
x 1
State the domain and range for each function.
14. y 
Domain:
Range:
2x 2
x2  1
15. y 
x
x 1
2
Domain:
Range:
Match each equation with its graph.
16. y 
x 2
x 6
17. y  1 
5
x2  9
A.
B.
C.
D.
E.
F.
Sketch the graph of each function.
18. y 
5
x 9
19. y 
x 2  2x
x2  4
2
Write a function with the following characteristics:
20. A vertical asymptote at x = 3
A horizontal asymptote at y = 2
An x-intercept at x  5
Name:
Date:
Graded Assignment
Checkup: 3.04 and 3.05
Answer the following questions using what you’ve learned from these lessons. Write your
responses in the space provided, and turn the assignment in to your instructor.
For problems 1 and 2, state the x- and y-intercepts for each function.
1. f  x   x 4  5x 2  4
y = 4, and x = 2, -2, 1, -1
To solve for y, set x equal to zero.
ሺ0ሻ4 − 5ሺ0ሻ2 + 4 = 4 So, y = 4.
For x, use the root rational theorem, since f(x) = 0.
ሼ±1,±2,±4ሽ
= ሼሺ−2,0ሻሺ−1,0ሻሺ1,0ሻሺ2,0ሻሽ So, x = -2, -1, 1, and 2.
ሼ±1ሽ
You can graph the equation to check your answer.
2. f  x   x 5  x
y = 0, and x = 0, -1, 1
To solve, use the same processes as the last question.
ሺ0ሻ5 − ሺ0ሻ = 0 So, y = 0.
ሼ±1ሽ
ሼ±1ሽ
= ሼሺ−1,0ሻሺ1,0ሻሽ Since the y intercept is 0, that means another x intercept would also be
0. So, x = -1, 0, and 1.
You can check your answer by graphing the function.
For problems 3 and 4, describe the end behavior for each of the functions by writing a limit expression.
1
3
3. f  x   x 3  2
4.
f  x   2x 6  5x  4
For problems 5 and 6, use the graphs to state the zeros for each polynomial function. State the
multiplicity of any roots if the multiplicity is 2 or higher.
5. f  x    x 3  2x 2
6. f  x   3x  12x 3
For problems 7 and 8, identify the extrema for each function. Classify each as a relative (local) or
absolute (global) maximum or minimum value.
7. f  x   4x 3  12x 2  15
All of the information can be found by using the largest degree of the polynomial, n. In this
case, n is 3. The graph will have at the most n – 1 maximum or minimum values.
3 – 1 = 2. There will be a total of 2 turns in the graph.
Input the extreme value of x = 0 into the equation to get y = 15. ሺ4ሺ0ሻ3 − 12ሺ0ሻ2 + 15 = 15ሻ
Absolute Maximum: (0, 15)
Plug in the value of x = 2 into the equation to get y = -1. ሺ4ሺ2ሻ3 − 12ሺ2ሻ2 + 15 = 4ሺ8ሻ −
12ሺ4ሻ + 15 = 32 − 48 + 15 = −1ሻ
Absolute Minimum: (2, -1)
8. f  x   4x 3  15x 2  8x  3
Use the same method as the last question.
3 – 1 = 2. There will be a total of 2 turns in the graph.
Input x = 0 into the equation to get y = -3. ሺ−4ሺ0ሻ3 + 15ሺ0ሻ2 − 8ሺ0ሻ − 3 = −3ሻ
Absolute Maximum: (0, -3)
For problems 9 and 10, sketch a graph of each function.
9.
f  x   4x 3  15x 2  8x  3
10. f  x   x 4  5x 2  4
Identify the vertical and/or horizontal asymptotes for each function.
11. y 
5x
x 1
= 1. To find the answer, you start by solving x – 1 from the
denominator. (x – 1 = 0, x = 1). This means that the domain includes all real numbers except
x = 1 or ሼ : ≠ 1ሽ.
Vertical asymptote(s): x
Horizontal asymptote: y = 5. To find the answer, you would use the 5 in 5x from the
numerator. This means that the range is all real numbers except for 5 or ሼ : ≠ 5ሽ.
You can check both of these by graphing the equation.
12. y 
3x
2x 2  1
Vertical asymptote(s): There
is no vertical asymptote because there are no undefined points.
Horizontal asymptote: y = 0. Since the denominators degree (2) is greater than the
numerators degree (1), the horizontal asymptote is the x axis.
Identify the oblique asymptote for each function.
13. y 
2x 2
x 1
State the domain and range for each function.
14. y 
Domain:
Range:
2x 2
x2  1
15. y 
x
x 1
2
Domain:
Range:
Match each equation with its graph.
16. y 
x 2
x 6
17. y  1 
5
x 9
2
A.
B.
C.
D.
E.
F.
Sketch the graph of each function.
18. y 
5
x 9
19. y 
x 2  2x
x2  4
2
Write a function with the following characteristics:
20. A vertical asymptote at x = 3
A horizontal asymptote at y = 2
An x-intercept at x  5
…
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