
Solving Higher Order Polynomials
Presentation
•
Mathematics
•
10th - 11th Grade
•
Hard
Joseph Anderson
FREE Resource
11 Slides • 28 Questions
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Think Quadratics
Concave up vs. Concave down
Think Linear
Positive slope vs. Negative slope
Summary of End Behaviors
12
Multiple Choice
Does the graph show an EVEN or ODD degree polynomial?
EVEN degree
ODD degree
13
Multiple Choice
Does the graph have a POSITIVE or NEGATIVE leading coefficient?
POSITIVE
NEGATIVE
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Multiple Choice
Does the graph show an EVEN or ODD degree polynomial?
EVEN degree
ODD degree
15
Multiple Choice
Does the graph have a POSITIVE or NEGATIVE leading coefficient?
POSITIVE
NEGATIVE
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Multiple Choice
Does the graph show an EVEN or ODD degree polynomial?
EVEN degree
ODD degree
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Multiple Choice
Does the graph have a POSITIVE or NEGATIVE leading coefficient?
POSITIVE
NEGATIVE
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Multiple Select
Which functions' graph would have a decreasing left end behavior and increasing right end behavior?
NOTE: There may be more than one correct answer.
f(x)=2x3+4x2+6x+3
g(x)=−8x3+2x2+4x
h(x)=−4x2+6x3+x+2
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Multiple Choice
Which of the function's graph would have a INCREASING left end behavior and INCREASING right end behavior?
f(x)=−4x2+6x+3
g(x)=8x3+2x2+4x
h(x)=2x4+6x3+x+2
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Multiple Choice
Which graphs would have a NEGATIVE leading coefficient?
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Open Ended
What do you notice/wonder about the cubic functions and their graphs? How are they similar? How are they different?
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Multiplicity
A zero has MULTIPLICITY.
Multiplicity is the number of times the zero's associated factor appears in the polynomials.
You can determine if the graph will bounce or cross the x-axis from the multiplicity
If the multiplicity is odd, the graph will cross at that zero
If the multiplicity is even, the graph will bounce at that zero
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Theorem: Turning Points
If f is a polynomial function of degree n, then the graph of f has at most n-1 turning points.
If the graph of a polynomial function f has n-1 turning points, the the degree of f is a least n.
The graph on the right has 3 turning points so that means the degree of the function is at least 4.
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Multiple Select
Select all the zeros (with multiplicities) of
f(x)=(x+2)(x−1)2
x= -2 multiplicity 2
x= 2 multiplicity 1
x= -2 multiplicity 1
x= -1 multiplicity 2
x= 1 multiplicity 2
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Multiple Select
f(x)=(x+2)(x−1)2
From the last question, we know that f(x) has zeros at
x= -2 multiplicity 1 and x= 1 multiplicity 2.
Select whether the graph of f(x) will cross or bounce at each of the zeros.
cross at x= -2
bounce at x= -2
cross at x=1
bounce at x=1
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Multiple Choice
Which of the following is the graph of
f(x)=(x+2)(x−1)2 ?
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Multiple Select
Select all the zeros (with multiplicities) of
g(x)=−2x2(x+4)2
x= -4 multiplicity 2
x= 4 multiplicity 2
x= 0 multiplicity 2
x= 2 multiplicity 2
x= -2 multiplicity 2
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Multiple Select
g(x)=−2x2(x+4)2
From the last question, we know that g(x) has zeros at
x= 0 multiplicity 2 and x= -4 multiplicity 2.
Select whether the graph of g(x) will cross or bounce at each of the zeros.
cross at x= 0
bounce at x= 0
cross at x= -4
bounce at x= -4
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Multiple Choice
Which of the following is the graph of
g(x)=−2x2(x+4)2 ?
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Multiple Select
x=0 multiplicity of 1 is a zero of h(x).
Select all other zeros (with multiplicities) of
h(x)=x(x−7)2(x+5)2
x= -7 multiplicity 2
x= 7 multiplicity 2
x= -5 multiplicity 2
x= 5 multiplicity 2
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Multiple Choice
h(x)=x(x−7)2(x+5)2
From the last question, we know that h(x) has zeros at
x= 0 multiplicity 1, x= 7 multiplicity 2, and x= -5 multiplicity 2
Select the graph of h(x).
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Multiple Select
Select all other zeros (with multiplicities) of
k(x)=−(x+1)2(x+4)
x= 0 multiplicity 1
x= -1 multiplicity 2
x= 1 multiplicity 2
x= -4 multiplicity 1
x=4 multiplicity 1
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Multiple Choice
k(x)=−(x+1)2(x+4)
From the last question, we know that k(x) has zeros at
x= -1 multiplicity 2, and x= -4 multiplicity 1
Select the graph of h(x).
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Multiple Select
Select ALL the possible factors in the equation of the graph.
(x−2)
(x+1)
(x−1)2
(x+1)2
(x+2)
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Writing Polynomial Functions
Use smallest degree possible
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Open Ended
Which of these doesn't belong? Explain your choice.
Show answer
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