Polynomial Graphs Exercises: Intercepts, Zeros, and End Behavior, Assignments of Mathematics

This resource offers a comprehensive set of exercises on polynomial graphs, focusing on intercepts, zeros, multiplicity, and end behavior. Problems include finding zeros and their multiplicities, graphing polynomial functions, determining polynomial degrees, and constructing equations from graphs and descriptions. Designed for high school and early university students, it enhances understanding of polynomial functions and their graphical representations. Detailed answers are provided for odd-numbered exercises to aid self-assessment and learning. Its clear organization and comprehensive coverage make it ideal for classroom instruction and independent study, ensuring a solid foundation in polynomial functions.

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2022/2023

Uploaded on 08/14/2025

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3.4e.1
Exercises - Polynomial Graphs
A: Concepts
1) What is the difference between an -intercept and a zero of a polynomial function ?
2) If a polynomial function of degree has distinct zeros, what do you know about the graph of the function?
3) What is the relationship between the degree of a polynomial function and the maximum number of turning points in its
graph? .
4) Explain how the factored form of the polynomial helps us in graphing it.
5) If the graph of a polynomial just touches the -axis and then changes direction, what can we conclude about the factored
form of the polynomial?
Answers to odd exercises:
1. The -intercept is where the graph of the function crosses the -axis, and the zero of the function is the input value for
which .
3. The maximum number of turning points is one less than the degree of the polynomial.
5. There will be a factor raised to an even power.
B: Multiplicity from an Equation
Find the zeros and give the multiplicity of each.
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
Answers to odd exercises:
7. and with multiplicity , with multiplicity
9. with multiplicity , with multiplicity
11. with multiplicity , with multiplicity
13. with multiplicity , and with multiplicity
15. with multiplicity , with multiplicity
17. with multiplicity , with multiplicity
C: Multiplicity from a Graph
Use the graph to identify zeros and multiplicity.
Exercise
3.4
e
.
A
x f
n n
x
x x
f
(
x
) = 0
Exercise
3.4
e
.
B
f
(
x
) = (
x
+ 2 (
x
3)3)2
f
(
x
) = (2
x
+ 3 (
x
4
x
2)5)2
f
(
x
) = (
x
1 (
x
+ 2)
x
3)3
f
(
x
) = ( +4
x
+ 4)
x
2
x
2
f
(
x
) = (2
x
+ 1 (9 6
x
+ 1))3
x
2
f
(
x
) = (3
x
+ 2 ( 10
x
+ 25))5
x
2
f
(
x
) =
x
(4 12
x
+ 9)( + 8
x
+ 16)
x
2
x
2
f
(
x
) = 2
x
6
x
5
x
4
f
(
x
) = 3 + 6 + 3
x
4
x
3
x
2
f
(
x
) = 4 12 + 9
x
5
x
4
x
3
f
(
x
) = 2 ( 4 + 4
x
)
x
4
x
3
x
2
f
(
x
) = 4 (9 12 + 4 )
x
4
x
4
x
3
x
2
0 4 2 3
25
0 2 2 2
2
35 5 2
0 4 2 1 1
3
22 0 3
0 6 2
32
Exercise
3.4
e
.
C
pf3
pf4
pf5

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Exercises - Polynomial Graphs

A: Concepts

1) What is the difference between an -intercept and a zero of a polynomial function?

2) If a polynomial function of degree has distinct zeros, what do you know about the graph of the function?

3) What is the relationship between the degree of a polynomial function and the maximum number of turning points in its

graph?.

4) Explain how the factored form of the polynomial helps us in graphing it.

5) If the graph of a polynomial just touches the -axis and then changes direction, what can we conclude about the factored

form of the polynomial?

Answers to odd exercises:

1. The -intercept is where the graph of the function crosses the -axis, and the zero of the function is the input value for

which.

3. The maximum number of turning points is one less than the degree of the polynomial.

5. There will be a factor raised to an even power.

B: Multiplicity from an Equation

Find the zeros and give the multiplicity of each.

Answers to odd exercises:

  1. and with multiplicity , with multiplicity
  2. with multiplicity , with multiplicity
  3. with multiplicity , with multiplicity
    1. with multiplicity , and with multiplicity
    2. with multiplicity , with multiplicity
    3. with multiplicity , with multiplicity

C: Multiplicity from a Graph

Use the graph to identify zeros and multiplicity.

Exercise 3.4 e. A

x f

n n

x

x x

f ( x ) = 0

Exercise 3.4 e. B

f ( x ) = ( x + 2 )^3 (^ x − 3)^2 f ( x ) = x^2 (2^ x + 3 )^5 (^ x − 4)^2 f ( x ) = x^3 (^ x − 1 )^3 ( x + 2) f ( x ) = x^2 (^ x 2 + 4 x + 4) f ( x ) = (2 x + 1 )^3 (9 x^2 − 6 x + 1) f ( x ) = (3 x + 2 )^5 ( x^2 − 10 x + 25)

f ( x ) = x (4 x^2 − 12 x + 9)( x^2 + 8 x + 16) f ( x ) = x^6 − x^5 − 2 x^4 f ( x ) = 3 x^4 + 6 x^3 + 3 x^2 f ( x ) = 4 x^5 − 12 x^4 + 9 x^3 f ( x ) = 2 x^4 ( x^3 − 4 x^2 + 4 x ) f ( x ) = 4 x^4 (9 x^4 − 12 x^3 + 4 x^2 )

Exercise 3.4 e. C

    1. (^) 21) 22)

Answers to odd exercises:

  1. with multiplicity 21. each with multiplicity

D: Graph polynomials

Graph the polynomial functions. State the - and - intercepts, multiplicity, and end behavior.

Answers to odd exercises:

  1. -intercepts, with multiplicity , with multiplicity , - intercept . As , , as .
  2. -intercepts with multiplicity , with multiplicity , - intercept As , as .
  3. -intercepts with multiplicity , -intercept (0, 0). As , as .
  4. multiplicity , multiplicity 1, y-intercept , end behaviour:
  5. multiplicity , multiplicity , y-intercept , end behaviour:
  6. , , , all multiplicity , y-intercept , end behaviour:

Exercise 3.4 e. D

★ x y

f ( x ) = ( x + 3 )^2 ( x − 2) g ( x ) = ( x + 4)( x − 1)^2 h ( x ) = ( x − 1 )^3 ( x + 3)^2 k ( x ) = ( x − 3 )^3 ( x − 2)^2 m ( x ) = −2 x ( x − 1)( x + 3)

n ( x ) = −3 x ( x + 2)( x − 4) a ( x ) = x ( x + 2)^2 g ( x ) = x ( x + 2)^3 f ( x ) = −2( x − 2 )^2 ( x + 1) g ( x ) = (2 x + 1 )^2 ( x − 3) f ( x ) = x^3 ( x + 2)^2

P ( x ) = ( x − 1)( x − 2)( x − 3)( x − 4) q ( x ) = ( x + 5 )^2 ( x − 3)^4 h ( x ) = x^2 ( x − 2 )^2 ( x + 2)^2 h ( t ) = (3 − t )( t^2 + 1) Z ( b ) = b (42 − b^2 )

x (1,0) 2 (– 4,0) 1 y (0,4) x → −∞ f ( x ) → −∞ x → ∞, f ( x ) → ∞

x (3,0) 3 (2,0) 2 y (0,– 108). x → −∞, f ( x ) → −∞ x → ∞, f ( x ) → ∞

x (0,0),(– 2,0),(4,0) 1 y x → −∞, f ( x ) → ∞ x → ∞, f ( x ) → −∞

E: Polynomial Degree from a Graph

Determine the least possible degree of the polynomial function shown.

Answers to odd exercises:

F: Construct an Equation from a graph

Use the graphs to write the formula for the polynomial function of least degree.

Answers to odd exercises:

69. or

71. or

73. or

Exercise 3.4 e. E

Exercise 3.4 e. F

f ( x ) = −( x + 3)( x + 1)( x − 3) f ( x ) = − 29 ( x + 3)( x + 1)( x − 3)

f ( x ) = ( x + 2 )^2 ( x − 3) f ( x ) = 14 ( x + 2 )^2 ( x − 3)

f ( x ) = −( x + 3)( x + 2)( x − 2)( x − 4) f ( x ) = − 241 ( x + 3)( x + 2)( x − 2)( x − 4)

Use the graphs to write a formula for the polynomial function of least degree.

  1. (^) 78)

  2. (^) 81)

Answers to odd exercises:

Use the graphs to write a formula for the polynomial function of least degree.

  1. 84(a).^ 84(b).

f ( x ) = ( x − 500 )^2 ( x + 200) f ( x ) = ( x + 300 )^2 ( x − 100)^3

f ( x ) = ( x + 3)( x − 3)( x^2 + 10) f ( x ) = 4 x ( x − 5)( x − 7)

110. Degree 4. Root of multiplicity 2 at = 4. Roots of multiplicity 1 at = 1 and = -2.

Vertical intercept at (0, -3)

111. Degree 5. Double zero at = 1. Triple zero at = 3. Passes through the point (2, 15)

112. Degree 5. Single zero at = -2 and = 3. Triple zero at = 1. Passes through the point (2, 4)

Answers to odd exercises:

  1. (f(x)=−2(x+3)(x+2)(x−1))

H: Turning Points

Determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the

number of turning points and the least possible degree for the function.

  1. (^) 120) 121)

Answers to odd exercises:

115. Yes. Number of turning points is. Least possible degree is.

117. Yes. Number of turning points is. Least possible degree is.

119. Yes. Number of turning points is. Least possible degree is.

212. Yes. Number of turning points is. Least possible degree is.

x x x

x x

x x x

f ( x ) = x^2 − 4 f ( x ) = x^3 − 4 x^2 + 4 x f ( x ) = x^4 + 1 f ( x ) = − 2 ( x + 2)( x − 1)( x − 3) 3 f ( x ) = 1 ( x − 3 ( x − 1 ( x + 3) 3

)^2 )^2

f ( x ) = −15( x − 1 )^2 ( x − 3)^3

f ( x ) = − 3 (2 x − 1 ( x − 6)( x + 2) 2

)^2

y = − ( x + 2)( x − 1)( x − 3)

y = ( x − 1 ( x − 3 ( x + 3)

)^2 )^2

y = −15( x − 1 )^2 ( x − 3)^3

Exercise 3.4 e. H