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Unformatted text preview: ll not all that
difficult.
Note as well that we will get the yintercept for free from this form. The yintercept is, f ( 0) = a (0) + b ( 0) + c = c
2 Þ ( 0, c ) so we won’t need to do any computations for this one.
Let’s graph some parabolas. Example 2 Sketch the graph of each of the following parabolas.
© 2007 Paul Dawkins 211 http://tutorial.math.lamar.edu/terms.aspx College Algebra (a) g ( x ) = 3 x 2  6 x + 5 [Solution]
(b) f ( x ) =  x 2 + 8 x [Solution]
(c) f ( x ) = x 2 + 4 x + 4 [Solution]
Solution
(a) For this parabola we’ve got a = 3 , b = 6 and c = 5 . Make sure that you’re careful with
signs when identifying these values. So we know that this parabola will open up since a is
positive.
Here are the evaluations for the vertex. x= 6
6
==1
2 ( 3)
6 y = g (1) = 3 (1)  6 (1) + 5 = 3  6 + 5 = 2
2 The vertex is then (1, 2 ) . Now at this point we also know that there won’t be any xintercepts for
this parabola since the vertex is above the xaxis and it opens upward.
The yintercept is ( 0,5 ) and using the axis of symmetry we kno...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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