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Transcript of Graphing Quadratic Functions – Standard Form It is assumed that you have already viewed the...
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