Quadratic Functions

A Parabola.

Graphs of Quadratic Functions

The Quadratic FunctionA quadratic function can be expressed in three formats:

• f(x) = ax2 bx c is called the standard form

• is called the factored form

• f (x) a(x h)2 k, is called the vertex form

21 rxrxaxf

The Standard Form• f(x)=ax2 bx c

– If a > 0 then parabola opens up– If a < 0 then parabola opens down– (0,c) is the y-intercept–

– Axis of symmetry is

a

bf

a

bV

2,

2

ha

bx

2

The Vertex Form f (x) a(x h)2 k, a 0

***y – k = a(x-h)2

• The graph of f is a parabola whose vertex is the point (h, k)

• The parabola is symmetric to the line x h

• If a 0, the parabola opens upward

• if a 0, the parabola opens downward

Graphing Parabolas-Vertex Form

To graph f (x) a(x h)2 k:1. Determine whether the parabola opens up or down.

– If a 0, it opens up.

– If a 0, it opens down.

2. Determine the vertex of the parabola--- V(h, k).

3. Find any x-intercepts.– Replace f (x) with 0.

– Solve the resulting quadratic equation for x.

4. Find the y-intercept by replacing x with zero.

5. Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

The Factored FormFactored Form of a quadratic

r represents roots or x intercepts

Axis of symmetry:

21 rxrxaxf

221 rr

h

hfhV ,

Vertex form f (x) a(x h)2 k  

a-2 h k 8

Given equation f (x) 2(x 3)2 8

Example• Graph the quadratic function f (x) 2(x 3)2 8.

This is a parabola that opens down with vertex V(3,8)

2(x 3)2 8 Find x-intercepts, setting f (x) equal to zero.

Find the x-intercepts. Replace f (x) with 0: f (x) 2(x 3)2 8.

Find the vertex. The vertex of the parabola is at (h, k). Because h 3 and k 8, the parabola has its vertex at (3, 8).

(x 3)2 4 Divide both sides by 2.

(x 3) 2 Apply the square root method.

x 3 2 or x 3 2 Express as two separate equations.

x 1 or x 5 Add 3 to both sides in each equation.

The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).

2(x 3)2 8

Example cont.

Graph the parabola. •With a vertex at (3, 8) •x-intercepts at (1,0) and (5,0) •and a y-intercept at (0,–10)•the axis of symmetry is the vertical line x 3.

f 2(0 3)2 8 2(3)2 8 2(9) 8 10

Find the y-intercept. Replace x with 0 in f (x) 2(x 3)2 8.

The y-intercept is –10. The parabola passes through (0, 10).

Example cont.

The Vertex of a Parabola Whose Equation Is f (x) ax 2 bx c

• Consider the parabola defined by the quadratic function

• f (x) ax 2 bx c. The parabola's vertex is at b

2a, f

b

2a

ExampleGraph the quadratic function f (x) x2 6x

1. Determine how the parabola opens. – Note that a, the coefficient of x 2, is -1. Thus, a 0;– This negative value tells us that the parabola opens downward.

2. Find the vertex. – x = -b/(2a).– Identify a, b, and c to substitute the values into the equation for the x-

coordinate: – x = -b/(2a) = -6/2(-1)=3.

3. The x-coordinate of the vertex is 3. 4. We substitute 3 for x in the equation of the function to find the y-

coordinate: – y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, – the parabola has its vertex at (3,7).

ExampleGraph the quadratic function f (x) x2 6x

• Find the x-intercepts. Replace f (x) with 0 in f (x) x2 6x 2. • 0 = x2 6x 2 (If you cannot factor)

a 1,b 6,c 2

x b b2 4ac

2a

6 62 4( 1)( 2)2( 1)

6 36 8

2

6 28

2 6 2 7

2

3 7

ExampleGraph the quadratic function f (x) x2 6x

Find the y-intercept. Replace x with 0 in f (x) x2 6x 2.

• f 02 6 • 0 2 The y-intercept is –2. The parabola passes through (0, 2).

Graph the parabola.

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

Minimum and Maximum: Quadratic Functions

• Consider f(x) = ax2 + bx +c.1. If a > 0, then f has a minimum that occurs at

x = -b/(2a). This minimum value is f(-b/(2a)).

2. If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).

No x-intercepts

No real solution; two complex imaginary solutions

b2 – 4ac < 0

One x-intercept

One real solution (a repeated solution)

b2 – 4ac = 0

Two x-intercepts

Two unequal real solutionsb2 – 4ac > 0

Graph of y = ax2 + bx + c

Kinds of solutions to ax2 + bx + c = 0

Discriminantb2 – 4ac

The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0

Quadratic Functions