QUADRATIC FUNCTIONS

By: Brooke Tellinghuisen

Kelli Peters

Austin Steinkamp

Chapter 2 Section 1

VOCABULARY

Term

Polynomials

Degree

Leading

coefficient

2x-2

2562 234 xxx

2463 24 xxx

225 3 xx

DEFINITION OF POLYNOMIAL FUNCTION

012

21

1 ... axaxaxaxaxf nn

nn

Example of Polynomial Functions

Polynomials are classified by degree.

Formula of Polynomial Function

baxxf cxf 2

xxf

DEFINITION OF QUADRATIC FUNCTION

2nd degree polynomials functions are called quadratic functions.

Example of Quadratic Functions

Formula of Quadratic Function

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

12

312

262

2

xxxf

xxf

xxxf

cbxaxxf 2

QUADRATIC FORMULA

a

acbbx

2

42

Used to find zeros (roots) in a quadratic function.

QuadraticsGeneral form y=ax2+bx+ c

Vertex Formy= a(x-h)2+k

Factored formy=(x-r1)(x-r2)

Vertex(h,k)

Vertex

a

bf

a

b

2,

2

Roots:Standard Form or factored form

Roots: Quadratic formula or factored form

Vertex: standard form or vertex form

Roots:x= r1,r2

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.

0,2 acbxaxxf

0,2 acbxaxxf

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.

PRACTICE PROBLEM F(x)=(x-2)^2 Tell what direction the graph moves and if it

opens up or down.

PRACTICE PROBLEM ANSWER

Since the 2 is connected with the x in the parentheses the graph moves the opposite way of what u think it would. Since it’s a subtraction problem it moved to the right.

PRACTICE PROBLEM 2 Find the vertex and x-intercepts of the equation

f(x)=x2-5

ANSWER TO PRACTICE PROBLEM 2 Take (x2-5) and set equal to zero x2-5=0

+5 =+5x2 =5

5x

To find your vertex

Those would be your x-intercepts

a

bf

a

b

2,

2

Use the formula to find your vertex

)1(2

0Plug 0 back into the equation and solve.

Your answer is (0, -5)

AREAEXAMPLE

Area ProblemA(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=200-2x2L/2=200-2x/2L=100-x

X-100=x(-x)-x2+100xFinding the Vertex-100/ 2(-1)(50,2500)

PROJECTILE MOTION FUNCTION

cbxaxxh 2 A function of height that depends on

time.

a = acceleration of gravity.b = initial velocity in which

object is thrown. c = initial height.

PROJECTILE MOTION PROBLEM 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?

ANSWER TO PROJECTILE MOTION PROBLEM

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.