X squared asks x cubed if he is a religious variable I do believe in higher powers, if that’s what...

• X squared asks x cubed if he is a religious variable

• I do believe in higher powers, if that’s what you mean.

• student notes MADE for 2-2 and 2-3

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 1

Section 2.2 Polynomial Functions of

Higher Degree

Warm up - Graph this business


246 3613)( xxxxf

Psych!


A polynomial function is a function of the form1

1 1 0( ) n nn nf x a x a x a x a

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 Degree5 3( ) 2 3 5 1f x x x x

3 2( ) 6 7f x x x x

( ) 14f x

-2 5

1 3

14 0


Graphs of polynomial functions are continuous. That is, they have no breaks, holes, or gaps.

Polynomial functions are also smooth with rounded turns. Graphs with points or cusps are not graphs of polynomial functions.

x

y

x

y

continuous not continuous continuoussmooth not smooth

polynomial not polynomial not polynomial

x

y f (x) = x3 – 5x2 + 4x + 4


Polynomial functions of the form f (x) = x n, n 1 are called

power functions.

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.

x

y

x

y

f (x) = x2

f (x) = x5f (x) = x4

f (x) = x3

What do you think even functions look like? What do you think odd

functions look like?


Ex 1: Sketch the graph of f (x) = – (x + 2)4 .

Reflect the graph of y = x 4 in the x-axis.

Then shift the graph two units to the left.

x

y

y = x4

y = – x4f (x) = – (x + 2)4


Leading Coefficient TestAs x grows positively or negatively without bound, the value f (x) of the polynomial function

f (x) = anxn + an – 1x

n – 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

n odd n even

an positive

an negative


Ex 2: Describe the right-hand and left-hand behaviour for the graph of f(x) = –2x3 + 5x2 – x + 1.

As , and as , xx )(xf)(xf

Negative-2Leading Coefficient

Odd3Degree

x

y

f (x) = –2x3 + 5x2 – x + 1


A real number a is a zero of a function f (x)if and only if f (a) = 0.

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.

Real Zeros of Polynomial FunctionsIf a is a zero of a function f(x), then the following are true:

1. a is a solution of the polynomial equation f (x) = 0.2. x – a is a factor of the polynomial f (x).

3. (a, 0) is an x-intercept of the graph of y = f (x).


Repeated ZerosFor any factor (x – a)

k of f (x)a is a zero of the function and k is called the multiplicity. 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).


Ex 3: Determine end behavior and the multiplicity of the zeros, the graph: f (x) = (x – 2)3(x +1)4.

Zero Multiplicity Behavior

2

–1

3

4

odd

even

crosses x-axis at (2, 0) touches x-axis at (–1, 0)

Repeated ZerosFor any factor (x – a)

k of f (x)a is a zero of the function and k is called the multiplicity. 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).

x

yAs , and as , xx )(xf)(xf

Now graph this business


246 3613)( xxxxf


Ex 4: Find the end behavior, 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.

These correspond to the x-intercepts (–1, 0), (0, 0) and (2, 0).

The graph shows that there are three turning points. Since the degree is four, this is the maximum number possible.

y

x

f (x) = x4 – x3 – 2x2

Turning pointTurning point

Turning point


Ex 5: 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.

2. Find the zeros of the polynomial by factoring.

f (x) = –x4 + 4x2 = –x2(x2 – 4) = – x2(x + 2)(x –2)

Zeros: x = –2, 2 multiplicity 1 x = 0 multiplicity 2

x-intercepts: (–2, 0), (2, 0) crosses through (0, 0) touches only

Example continued

as , )(xfx

x

y

(2, 0)

(0, 0)

(–2, 0)


Ex 5 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)

g(t) = t5 – 6t3 + 9tg(t) = t5 – 6t3 + 9t

g(t) = t(t4 – 6t2 + 9)

g(t) = t(t2 – 3)2

22 )3()3()( ttttg

}3,3,0{ t

Cross

1ty Multiplici 0

Touch

2ty Multiplici 3

Touch

2ty Multiplici 3

Ex 6: Find the zeros and determine their multiplicity

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).

The Intermediate Value Theorem• Think of an elevator:

– If it reaches the 2nd floor and the 30th floor, it must reach every floor in between the 2nd and 30th.

The Intermediate Value Theorem

4

2

-2

-4

-6

-8

5

C: (3.00, 4.00)

A: (1.00, -5.99)

B

Why do we care?

Since

f(1)= -6

f(3) = 4

We know that for

1< x < 3

There is an x such that

f(x)=0

1-8 A Matching

1. f(x) = -2x + 3

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. f(x) = -¼ x4 + 3x2

6. f(x) = -1/3 x3 + x2 – 4/3

7. f(x) = x4 + 2x3

8. f(x) = 1/5 x5 -2x3 + 9/5 x

10

8

6

4

2

-2

-5 5

B

A

1-8 B Matching

1. f(x) = -2x + 3

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. A

6. f(x) = -1/3 x3 + x2 – 4/3

7. f(x) = x4 + 2x3

8. f(x) = 1/5 x5 -2x3 + 9/5 x

6

4

2

-2

-4

-6

-8

-5 5

B

1-8 C Matching

1. f(x) = -2x + 3

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. A

6. f(x) = -1/3 x3 + x2 – 4/3

7. f(x) = x4 + 2x3

8. B

C

6

4

2

-2

-4

-6

-8

-5 5

1-8 D Matching

1. C

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. A

6. f(x) = -1/3 x3 + x2 – 4/3

7. f(x) = x4 + 2x3

8. B

D

8

6

4

2

-2

-4

-6

-5 5

1-8 E Matching

1. C

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. A

6. f(x) = -1/3 x3 + x2 – 4/3

7. D

8. B

E

8

6

4

2

-2

-4

-6

-5 5

1-8 F Matching

1. C

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. f(x) = 2x3 – 3x + 1

5. A

6. E

7. D

8. B

F

8

6

4

2

-2

-4

-6

-5 5

1-8 G Matching

1. C

2. f(x) = x2 – 4x

3. f(x) = -2x2 – 5x

4. F

5. A

6. E

7. D

8. B

G

8

6

4

2

-2

-4

-6

5

1-8 H Matching

1. C

2. G

3. f(x) = -2x2 – 5x

4. F

5. A

6. E

7. D

8. B

H

6

4

2

-2

-4

-6

-8

-5 5

H Dub

• 2.2 Page 148 #1-8all, 13-21ODD, 27-41ODD (part c by hand – no calculators!!!), 65, 66, 92

X squared asks x cubed if he is a religious variable I do believe in higher powers, if that’s what...

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