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By: Brooke Tellinghuisen

Kelli Peters

Austin Steinkamp

Chapter 2 Section 1

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TermPolynomials

Degree

Leading

coefficient

2x-2

2562234 xxx

246324

xxx

2253

xx

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012

2

1

1... aaaaaf

n

n

n

n!

Example of Polynomial Functions

Polynomials are classified by degree.

Formula of Polynomial Function

baxxf ! cxf !

2

xxf !

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2nd degree polynomials functions are called quadraticfunctions.

Example of Quadratic Functions

Formula of Quadratic Function

NOTE: a, b, and c are real numbers with a 0.

1

31

6

!

!

!

xxxf

xxf

xxxf

cbxaxxf ! 2

{

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a

acbb

2

42s!

Used to find zeros (roots) in a

quadratic function.

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The graph for a quadratic function is a U-shaped

graph, called a parabola.

If the leading coefficient is positive, the graph opens

upward.

If the leading coefficient is negative, the graph opens

downward.

0, "! acbaf

0, ! acbaf

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The point where the axis intersects the parabola is the

vertex.

If a > 0, the vertex is the point with the minimum y-

value on the graph.

If a < 0, the vertex is the point with the maximum y-

value on the graph.

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F(x)=(x-2)^2

Tell what direction the graph moves and if it

opens up or down.

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Since the 2 is connected with the x in the parentheses the graph

moves the opposite way of what u think it would. Since its a

subtraction problem it moved to the right.

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Find the vertex and x-intercepts of the equation

f(x)=x2-5

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Take (x2-5) and set equal to zero

x2-5=0

+5 =+5

x2 =5

5s!x

To find your vertex

Those would be your

x-intercepts

a

bf

a

b

2,

2

Use the formula to find

your vertex

)(2

Plug 0 back into

the equation and

solve.Your answer is (0, -5)

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Area Problem

A(x)=width x length

A farmer has 200 yards of fencing. Write the area

as a function of x, if the farmer encloses a

rectangular area letting the width equal to x.

What is my maximum area? What are my zeros?

Do the zeros match common sense?

L

XX

L

2x+2L=200

-2x -2x

2L=200-2x

2L/2=200-2x/2

L=100-x

X-100=x(-x)

-x2+100x

Finding the Vertex

-100/ 2(-1)

(50,2500)

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An object is launched at 19.6 meters per second

from a 58.8 meter tall platform.

The equation for the objects highest s at time t

seconds after launch is s(t)=-4.9t2+19.6t+58.8,where s is in meters.

When does the object strike the ground?

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0=-4.9t2+19.6t+58.8

0=t2-4t-12

0=(t-6)(t+2)

So T=6 and -2. The answer cant be negative so

the object hit the ground at 6 seconds after the

launch.