Graphing Quadratic Functions – Standard Form

• It is assumed that you have already viewed the previous slide show titled

Graphing Quadratic Functions – Concept.

• A quadratic function in what we will call Standard Form is given by:

2( ) ( )f x a x h k

• The summary of the Concept slide show is given again on the next page.

SUMMARY

0a Face Up

0a Face Down

1a Narrow

0 1a Wide

2( ) ( )f x a x h k

Vertex ( , )V h k

Axis of symmetry

x h

• One more thing is needed before sketching the graph of a quadratic function. A point is plotted to know just how narrow or how wide the graph is.

• When the graph is narrow, choose an x-value that is only one unit from the vertex.

• In the graph on the right, a good choice would be

x = 1

4

2

Narrow

4

2

• If the value x = 2 were chosen, then the corresponding y-value would be off the graph.

• When the graph is wide, choose an x-value that is more than one unit from the vertex.

• In the graph on the right, a good choice would be

x = 2 or x = 3

• Note that x = 1 would not be very helpful in determining just how wide the graph would be.

4

2

-2

Wide

1a

4

2

Narrow

4

2

-2

Wide

0 1a SUMMARY

Choose a value for x 1 unit away

from the vertex.

Choose a value for x more than 1 unit

away from the vertex

• Example 1:

2( ) 3 2 1f x x

Sketch the graph of the following function:

Face Up3a

Narrow3 1a

Vertex: 2,1V

Axis of Symmetry: 2x

2( ) 3 2 1f x x

2,1V• Plot the vertex:

5

4

2

-2

• Draw the axis of symmetry: 2x

4

2

-2

55

4

2

-2

2( ) 3 2 1f x x

5

4

2

-2

• Since the graph is narrow, find a point that is only 1 unit from the vertex.

Try x = 3. 2

2

(3) 3 3 2 1

3(1) 1

4

f

3,4

5

4

2

-2

2( ) 3 2 1f x x

• Draw the right branch of the parabola using the vertex and the point (3,4).

5

4

2

-2

• Now use symmetry to draw the left branch.

5

4

2

-2

5

4

2

-2

• Label the axis and important points.

5

4

2

-2

(3,4)

(2,1)

x = 2

• Example 2:

21( ) 1 2

5f x x

Sketch the graph of the following function:

1

5a Face Down

11

5a Wide

Vertex: 1, 2V

Axis: 1x

1, 2V • Plot the vertex:

• Draw the axis of symmetry: 1x

21( ) 1 2

5f x x

-5

-2

-4

-6

-5

-2

-4

-6

-5

-2

-4

-6

• Since the graph is wide, find a point that is more than 1 unit from the vertex (-1,-2).

21( ) 1 2

5f x x

• This problem presents another challenge, which is to avoid fractions if possible.

21( ) 1 2

5f x x

• Therefore, we want to meet two goals:1. Select an x-value more than one unit to the right

of the vertex (-1,-2).2. Avoid fractions.

• To meet goal #2, all that is needed is for the quantity that is squared to be divisible by 5.

• An x-value of 4 meets this condition, and also satisfies goal #1.

2

2

1(4) 4 1 2

51

5 251

25 255 2

7

f

4, 7

-5

-2

-4

-6

-8

• Draw the right branch of the parabola using the vertex and the point (4,-7).

• Now use symmetry to draw the left branch. -5

-2

-4

-6

-8

-5

-2

-4

-6

-8

-5

-2

-4

-6

-8

• Label the axis and important points.

-5

-2

-4

-6

-8

(4,-7)

(-1,-2)

x = -1