Jeff Bivin -- LZHS Graphs of Polynomial Functions By: Jeffrey Bivin Lake Zurich High School...

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Jeff Bivin -- LZHS Graphs of Polynomial Functions By: Jeffrey Bivin Lake Zurich High School [email protected] Last Updated: October 6, 2009

Transcript of Jeff Bivin -- LZHS Graphs of Polynomial Functions By: Jeffrey Bivin Lake Zurich High School...

Jeff Bivin -- LZHS

Graphs of Polynomial Functions

By: Jeffrey Bivin

Lake Zurich High School

[email protected]

Last Updated: October 6, 2009

Jeff Bivin -- LZHS

Continuous FunctionThe graph of a polynomial function

f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0 is continuous if it has no breaks, holes,

vertical asymptotes or sharp turns.

Jeff Bivin -- LZHS

Are these graphs continuous?

Jeff Bivin -- LZHS

Leading Coefficient Test& End Behavior

f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0

n is even n is odd

If an is positive If an is positive

If an is negative If an is negative

Jeff Bivin -- LZHS

Leading Coefficient Test& End Behavior

left (x ) right (x )

f(x) = 5x4 + 7x2 + 3 ↑ ↑

f(x) = -7x8 + x + 3 ↓ ↓

f(x) = 5x5 + 7x2 + 3x ↓ ↑

f(x) = -3x5 + 6x3 + 2 ↑ ↓

y y

y

y

y y

y

y

Jeff Bivin -- LZHS

Real Zeros & Relative Extrema

1. The function has at most n real zeros

2. The graph has at most n-1 relative extrema (relative minima or relative maxima)

f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0

Jeff Bivin -- LZHS

Finding Real Zeros

F(x) is a polynomial function….

1. x = a is a zero of the function

2. x = a is a solution of f(x) = 0

3. x – a is a factor of f(x)

4. (a, 0) is an x-intercept of the graph of f(x)

Jeff Bivin -- LZHS

Find all real zerosof the polynomial functions

1. F(x) = x2 – 121

2. F(x) = 2x2 – 14x + 24

3. F(x) =

4. F(x) = x4 – x3 – 20x2

5. F(x) = (x+2)4(x-3)5(x+4)3(x-5)2

25 163 3x + x + 4

Jeff Bivin -- LZHS

1. F(x) = x2 – 121 = 0

(x – 11)(x + 11) = 0

x – 11 = 0 or x + 11 = 0

x = 11 or x = – 11

Jeff Bivin -- LZHS

2. F(x) = 2x2 – 14x + 24 = 0

2(x – 4)(x – 3) = 0

2 = 0 or x – 4 = 0 or x – 3 = 0

x = 4 or x = 3

2(x2 – 7x + 12) = 0

Jeff Bivin -- LZHS

3. 25 163 3 4 0( )F x x x

213 5 16 12 0x x

13 5 6 2 0x x

13 0 5 6 0 2 0or x or x

65 2x or x

Jeff Bivin -- LZHS

4. F(x) = x4 – x3 – 20x2 = 0

x2(x – 5)(x + 4) = 0

x2 = 0 or x – 5 = 0 or x + 4 = 0

x = 0 or x = 5 or x = – 4

x2(x2 – x – 20) = 0

Jeff Bivin -- LZHS

5. F(x) = (x+2)4(x-3)5(x+4)3(x-5)2 = 0

x + 2 = 0 or x – 3 = 0 or x + 4 = 0 or x – 5 = 0

x = -2 or x = 3 or x = – 4 or x = 5

Jeff Bivin -- LZHS

Graph

F(x) = (x-2)2

F(x) = (x-2)3

F(x) = (x-2)4

F(x) = (x-2)5

Jeff Bivin -- LZHS

Multiplicity – repeated zeros

• If the multiplicity is odd graph crosses x axis

• If the multiplicity is even graph touches x axis

Jeff Bivin -- LZHS

F(x) = (x-5)2(x-3)5(x-1)4(x+4)2(x+7)5

Degree = 2+5+4+2+5 = 18

(left side): as x - ∞, y + ∞(right side): as x + ∞, y + ∞

x = 5 multiplicity 2 touchx = 3 multiplicity 5 cross

x = 1 multiplicity 4 touchx = -4 multiplicity 2 touchx = -7 multiplicity 5 cross

TC T C T

Jeff Bivin -- LZHS

Find zeros of the polynomial function f(x) = x3 – 4x2 – 25x + 100

x3 – 4x2 – 25x + 100 = 0

x2(x – 4) – 25(x – 4) = 0

( x – 4)(x2 – 25) = 0

( x – 4)(x – 5)(x+5) = 0x – 4 = 0 x – 5 = 0 x + 5 = 0

x = 4 x = 5 x = –5

Jeff Bivin -- LZHS

Find a polynomial function that has the given zeros

Zeros: 5, 3, -2

x = 5 x = 3 x = –2x – 5 = 0 x – 3 = 0 x + 2 = 0

( x – 5)(x – 3)(x + 2) = 0

(x – 5)(x2 – x – 6) = 0

x3 – x2 – 6x – 5x2 + 5x + 30 = 0

x3 – 6x2 – x + 30 = 0

f(x) = x3 – 6x2 – x + 30

Jeff Bivin -- LZHS

Find a polynomial function that has the given zeros

Zeros:

f(x) = x3 – 4x2 + 2x + 4 2 1 3 1 3, , 2 1 3 1 3 0x x x

2 1 3 1 3 0x x x

2 1 3 1 3 0x x x

222 1 1 3 1 3 3 0x x x x

22 2 1 3 0x x x

22 2 2 0x x x 3 2 22 2 2 4 4 0x x x x x

3 24 2 4 0x x x

Jeff Bivin -- LZHS

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

a b

f(a)f(b)

Jeff Bivin -- LZHS

How is this especially helpful?

f(5) = 7

f(4) = -3

We must have a real zero

between the x values of 4 and

5

We may have more than one real zero in the interval [4, 5]

Jeff Bivin -- LZHS

Another Look…….x y1

-2 9

-1 -2

0 -5

1 3

2 12

3 7

4 -6

At least one real zero between

-2 & -1

At least one real zero between

0 & 1

At least one real zero between

3 & 4

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