1. Section 2.2 Polynomial Functions of Higher Degree

2. What you should learn
How to use transformations to sketch graphs of polynomial functions
How to use the Leading Coefficient Test to determine the end behavior of graphs of polynomial functions
How to use zeros of polynomial functions as sketching aids
How to use the Intermediate Value theorem to help locate zeros of polynomial functions

3. A polynomial function is a function of the form
where n is a nonnegative integer and each ai (i = 0, , n) is a real number.
The polynomial function has a leading
coefficient an and degree n.
Examples: Find the leading coefficient and degree of each polynomial function.
Polynomial Function Leading Coefficient Degree
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Polynomial Function

4. Graphs of Polynomial Functions
Graphs of polynomial functions are continuous. That is, they have no breaks, holes, or gaps. x y
f (x) = x3 – 5x2 + 4x + 4 x y x y
continuous
not continuous
continuous
smooth
not smooth
polynomial
not polynomial
not polynomial
Polynomial functions are also smooth with rounded turns. Graphs with points or cusps are not graphs of polynomial functions.
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Graphs of Polynomial Functions

5. If n is even, their graphs resemble the graph of f (x) = x2.
Polynomial functions of the form f (x) = x n, n  1 are called power functions.
f (x) = x5 x y
f (x) = x4 x y
f (x) = x2
f (x) = x3
If n is even, their graphs resemble the graph of f (x) = x2.
If n is odd, their graphs resemble the graph of f (x) = x3.
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Power Functions

6. Example: Graph of f(x) = – (x + 2)4
Example: Sketch the graph of f (x) = – (x + 2)4 .
This is a shift of the graph of y = – x 4 two units to the left. This, in turn, is the reflection of the graph of y = x 4 in the x-axis. x y
y = x4
f (x) = – (x + 2)4
y = – x4
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Example: Graph of f(x) = – (x + 2)4

7. Leading Coefficient Test
As x grows positively or negatively without bound, the value f (x) of the polynomial function
f (x) = anxn + an – 1xn – 1 + … + a1x + a0 (an  0)
grows positively or negatively without bound depending upon the sign of the leading coefficient an and whether the degree n is odd or even. x y x y
an positive
an negative
n even
n odd
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Leading Coefficient Test

8. Example: Right-Hand and Left-Hand Behavior
Example: Describe the right-hand and left-hand behaviour for the graph of f(x) = –2x3 + 5x2 – x + 1.
Negative -2
Leading Coefficient
Odd 3
Degree
As , and as , x y
f (x) = –2x3 + 5x2 – x + 1
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Example: Right-Hand and Left-Hand Behavior

9. Real Zeros of Polynomial Functions
A real number a is a zero of a function y = f (x) if and only if f (a) = 0.
Real Zeros of Polynomial Functions
If y = f (x) is a polynomial function and a is a real number then the following statements are equivalent.
1. a is a zero of f.
2. a is a solution of the polynomial equation f (x) = 0.
3. x – a is a factor of the polynomial f (x).
4. (a, 0) is an x-intercept of the graph of y = f (x).
A turning point of a graph of a function is a point at which the graph changes from increasing to decreasing or vice versa.
A polynomial function of degree n has at most n – 1 turning points and at most n zeros.
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Zeros of a Function

10. Factor completely: f (x) = x 4 – x3 – 2x2 = x2(x + 1)(x – 2).
Example: Find all the real zeros and turning points of the graph of f (x) = x 4 – x3 – 2x2.
Factor completely: f (x) = x 4 – x3 – 2x2 = x2(x + 1)(x – 2).
The real zeros are x = –1, x = 0, and x = 2. y x
f (x) = x4 – x3 – 2x2
These correspond to the x-intercepts (–1, 0), (0, 0) and (2, 0).
Turning point
The graph shows that there are three turning points. Since the degree is four, this is the maximum number possible.
Turning point
Turning point
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Example: Real Zeros

11. 1. If k is odd the graph of f (x) crosses the x-axis at (a, 0).
Repeated Zeros
If k is the largest integer for which (x – a) k is a factor of f (x) and k > 1, then a is a repeated zero of multiplicity k.
1. If k is odd the graph of f (x) crosses the x-axis at (a, 0).
2. If k is even the graph of f (x) touches, but does not cross
through, the x-axis at (a, 0).
Example: Determine the multiplicity of the zeros of f (x) = (x – 2)3(x +1)4. x y
Zero Multiplicity Behavior 2 3
odd
crosses x-axis at (2, 0) –1 4
even
touches x-axis at (–1, 0)
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Repeated Zeros

12. Example: Graph of f(x) = 4x2 – x4
Example: Sketch the graph of f (x) = 4x2 – x4.
1. Write the polynomial function in standard form:
f (x) = –x4 + 4x2
The leading coefficient is negative and the degree is even.
as ,
2. Find the zeros of the polynomial by factoring.
f (x) = –x4 + 4x2 = –x2(x2 – 4) = – x2(x + 2)(x –2) x y
Zeros: x = –2, 2 multiplicity 1 x = 0 multiplicity 2
(–2, 0)
(2, 0)
(0, 0)
x-intercepts: (–2, 0), (2, 0)
crosses through
(0, 0)
touches only
Example continued
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Example: Graph of f(x) = 4x2 – x4

13. Example continued: Sketch the graph of f (x) = 4x2 – x4.
3. Since f (–x) = 4(–x)2 – (–x)4 = 4x2 – x4 = f (x), the graph is
symmetrical about the y-axis.
4. Plot additional points and their reflections in the y-axis:
(1.5, 3.9) and (–1.5, 3.9 ), ( 0.5, 0.94 ) and (–0.5, 0.94)
5. Draw the graph. x y
(1.5, 3.9)
(–1.5, 3.9 )
(– 0.5, 0.94 )
(0.5, 0.94)
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Example Continued

14. The Intermediate Value Theorem
Let a and b be real numbers such that a < b.
If f is a polynomial function such that f(a) ≠ f(b) then in the interval [a, b], f takes on every value between f(a) and f(b).

15. There is a zero Since f(1)= -6 f(3) = 4 We know that for
1< x < 3
There is an x such that
f(x)=0

16. Homework
1 - 8, Matching
13-21 odd,
odd

17. 1-8 A Matching f(x) = -2x + 3 f(x) = x2 – 4x f(x) = -2x2 – 5x
f(x) = 1/5 x5 -2x3 + 9/5 x A

18. 1-8 B Matching f(x) = -2x + 3 f(x) = x2 – 4x f(x) = -2x2 – 5x
f(x) = 1/5 x5 -2x3 + 9/5 x B

19. 1-8 C Matching C f(x) = -2x + 3 f(x) = x2 – 4x f(x) = -2x2 – 5xB

20. 1-8 D Matching C f(x) = x2 – 4x f(x) = -2x2 – 5x f(x) = 2x3 – 3x + 1 AB D

21. 1-8 E Matching C f(x) = x2 – 4x f(x) = -2x2 – 5x f(x) = 2x3 – 3x + 1 AD B E

22. 1-8 F Matching C f(x) = x2 – 4x f(x) = -2x2 – 5x f(x) = 2x3 – 3x + 1 AE D B F

23. 1-8 G Matching C
f(x) = x2 – 4x
f(x) = -2x2 – 5x F A E D B G

24. 1-8 H Matching C G
f(x) = -2x2 – 5x F A E D B H

25. #37 g(t) = t5 – 6t3 + 9t g(t) = t5 – 6t3 + 9t g(t) = t(t4 – 6t2 + 9)