Graphs of Quadratic Graphs of Quadratic FunctionsFunctions

Graphs of Quadratic Graphs of Quadratic FunctionsFunctions

Quadratic Functions:

Definition of Quadratic Function:

2

Let , and be real numbers with 0. The function

( )

is called a quadratic function.

a b c a

f x ax bx c

The graph of a quadratic function is a special type of “U”-shaped curve that is called a parabola. (U shaped)

All parabolas are symmetric with respect to a line called theaxis of symmetry, or simply the axis of the parabola.

The point where the axis intersects the parabola is the vertex of theparabola.

Quadratic Equations

y = ax2 + bx + cf(x) = y = a(x - h)2 + k

The graph is “U-shaped” and is called a parabola.

Standard form Vertex form

(later)

The highest or lowest point on the parabola is called the vertex.

The axis of symmetry for

the parabola is the vertical

line through the vertex.

y = x2

y = -x2

Opens up

Opens down

Vertex is a minimum

Vertex is a maximum

Finding the vertex of a parabola:

2

Vertex of a Parabola:

The vertex of the graph of ( ) is

2

f x ax bx c

bx

a

This gives you the x-value and plug the x-value into the originalfunction to find the y-value

Therefore, the vertex is : ,2 2

b bf

a a

EXAMPLES:

Sketch the graph of the quadratic function without using a graphingutility. Identify the vertex and x-intercepts.

You know the graph will go up b/ca is a positive number

Find the vertex first!!

2

bx

a

So the x part of the vertex

is -2

Now, to find the y-value, plug in -2 wherever yousee an x.

Vertex: (-2,-7)

142 xx

14)( 2 xxxf

152 xx

22

4

)1(2

4

71)2(5)2( 2

Now make the table!

Remember the vertex goes in the middle!!!

x y (x, y)142 yx

-2

EXAMPLES:

Sketch the graph of the quadratic function without using a graphingutility. Identify the vertex and x-intercepts.

21. ( ) 5f x x You know the graph will go up b/ca is a positive number

Find the vertex first!!2( ) 5f x x

2

bx

a

0

2(1) 0

2 0 So the x part of the vertex

is 02( ) 5f x x Now, to find the y-value,

plug in zero wherever yousee an x.

Vertex: (0,-5)

2( ) (0) 5 5f x

Where is

B???

So, we know that the vertex is (0,-5) now all we have to do is a T-chart and pick 4 more points and then graph.

x y

0 -5

y = 2x2 + 1

a = 2 b = 0 c = 1

Name the vertex, axis of symmetry and whether it opens up or down

Axis of sym: x = 0

Opens up

Vertex: (0, 1)

AND GRAPH.

One More….

y = 2x2 + 1

Name the vertex, axis of symmetry and whether it opens up or down

Vertex: (0, 1)

AND GRAPH.

x y

0 1

-1 3

1 3

-2 9

2 9

Put vertex in the middle of the t-table

Example

Solution:Step 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.

Step 2 Find the vertex. We know the x-coordinate of the vertex is –b/2a.

We identify a, b, and c to substitute the values into the equation for the x-coordinate:

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

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

the parabola has its vertex at (3,7).

Graph the quadratic function f (x) x2 6x

2(3) 3 6(3) 2 9 18 2 7y f

ExampleGraph the quadratic function f (x) x2 6x

Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) x2 6x 2. 0 = x2 6x 2

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

Example

Graph the quadratic function f (x) x2 6x

Step 4 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). Step 5 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)).

Strategy for Solving Problems Involving Maximizing or

Minimizing Quadratic Functions1. Read the problem carefully and decide which

quantity is to be maximized or minimized.2. Use the conditions of the problem to express the

quantity as a function in one variable.3. Rewrite the function in the form f(x) = ax2 + bx +c.4. Calculate -b/(2a). If a > 0, then f has a minimum

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

5. Answer the question posed in the problem.