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Graphing Quadratic Functions – Standard Form

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

Graphing Quadratic Functions – Concept.

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

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SUMMARY

• Axis of symmetry: -value of vertexxx

Graphs of Quadratic Functions

• The graph of a quadratic function in called a parabola.

• The maximum or minimum y-value of a quadratic occurs at the vertex.

0a Face Up 0a Face Down•

x-int: y = f (x) = 0 and solve for x.

y-int: x = 0 and solve for y.

•

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2( ) ( )f x a x h k

The vertex is given by V(h,k).

• Example 12( ) 3( 2) 4f x x

The vertex is given by: (2,4)V

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

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• Example 2

2( ) 2( 3) 5f x x

The vertex is given by: ( 3, 5)V

2( ) ( )f x a x h k

Put the function in the form of …

2( ) 2( ( 3)) ( 5)f x x

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2( ) 2( 3) 5f x x

The vertex is given by: ( 3, 5)V

Here is an easier way to work the last problem:

For the h value, take the opposite sign … For the k value, take

the same sign …

3h 5k

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• Example 3

22( ) ( 5) 7

3f x x

The vertex is given by: ( 5,7)V

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• Recall that the Axis of Symmetry has the equation

Since the vertex of the standard quadratic function given by

( , )V h k

-value of vertexxx

has an x-value of h, we can write the equation of the axis of symmetry as

x h• Put all of the tools learned so far together to sketch the graph of a quadratic function in standard form.

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2( ) ( )f x a x h k

Sketch the Graph of a Quadratic in Standard Form

0a Face Up

0a Face Down

Vertex ( , )V h k

Axis of symmetry

x h

x-int: f (x) = 0 and solve for x

y-int: x = 0 and solve for y

Draw the parabola

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• Example 4

2( ) 2 1f x x

Sketch the graph of the following function:

Face Up1 0a

Vertex: 2, 1V

Axis of Symmetry: 2x

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• Start the sketch of the graph with what we have so far.

2, 1V

4

2

5

4

2

5

4

2

5

Axis of Symmetry

2x

The parabola is face up

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2( ) 2 1f x x

x-intercepts

22 1 0x

22 1x

2 1x 2 1x

1,0 3,0

y-intercept

0x

2(0) 0 2 1f

4 1 3

0,3

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• Plot the intercepts

4

2

5

x-intercepts

1,0

3,0

4

2

5

y-intercept 0,3

4

2

5

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4

2

5

• Sketch the parabola, using the points and axis of symmetry.

4

2

5

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• Example 5

2( ) 2 3 4f x x

Sketch the graph of the following function:

Face Down2 0a

Vertex: 3,4V

Axis of Symmetry: 3x

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• Start the sketch of the graph with what we have so far.

3,4V

Axis of Symmetry

3x

The parabola is face down

4

2

-2

-5

4

2

-2

-5

4

2

-2

-5

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x-intercepts

22 3 4 0x

22 3 4x

23 2x

4.4,0 1.6,0

2( ) 2 3 4f x x

3 2x

3 2x

4.41, 1.59x

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4

2

-2

-5

• Plot the x-intercepts

x-intercepts

4.4,0

1.6,0

4

2

-2

-5

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y-intercept

0x

2(0) 2 0 3 4f

0, 14

2( ) 2 3 4f x x

(0) 2 9 4f (0) 14f

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• Skip plotting the y-intercept since it is off of the graph.

4

2

-2

-5

•Sketch the parabola, using the points and axis of symmetry.

4

2

-2

-5

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