Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 8.2 Parabolas and Modeling
Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 8.2 Parabolas and Modeling.
Mar 26, 2015
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Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Section 8.2
Parabolas and Modeling
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Objectives
• Vertical and Horizontal Translations
• Vertex Form
• Modeling with Quadratic Functions (Optional)
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).
All three graphs have the same shape.y = x2
y = x2 + 1 shifted upward 1 unity = x2 – 2 shifted downward 2 units
Such shifts are called translations because they do not change the shape of the graph only its position
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2
y = (x – 1)2 Horizontal shift to the right 1 unit
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2
y = (x + 2)2 Horizontal shift to the left 2 units
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated 3 units down.
The vertex is (0, 3).
2 3y x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated left 4 units.
The vertex is (4, 0).
2( 4)y x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated down 2 units and right 1 unit.
The vertex is (1, 2).
2( 1) 2y x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane.
SolutionThe graph is translated to the right 2 units and upward 3 units.The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0).The graph opens upward and is wider.
21( ) ( 2) 3
4f x x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c.SolutionThe vertex form of the parabola is where the vertex is (h, k).a = 3, h = 2 and k = 1
To write the equation in y = ax2 + bx + c, do the following:
2( ) ,y a x h k
2)3( 12y x
2)3( 12y x 2( 4 43 1)y x x
23 12 12 1y x x 23 12 13y x x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Write each equation in vertex form. Identify the vertex. a. b.Solutiona. Because , add and subtract 16 on the right.
2 8 13y x x 22 8 7y x x
2 28
162 2
b
2 8 13y x x
2 16 1 68 13y x x
24 3y x
The vertex is (4, 3).
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example (cont)
b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side.
22 8 7y x x
2 24
42 2
b
2
2
2 8 7
2( 4 ) 7
y x x
x x
2 42 4 74y x x
2 42 4 7 8y x x
22 2 1y x The vertex is ( 2, 1).
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