Unformatted text preview: ans that there can’t possibly be xintercepts since the x axis is above the vertex and the
parabola will always open down. This means that there is no reason, in general, to go through the
solving process to find what won’t exist.
However, let’s do it anyway. This will show us what to look for if we don’t catch right away that
they won’t exist from the vertex and direction the parabola opens. We’ll need to solve, © 2007 Paul Dawkins 209 http://tutorial.math.lamar.edu/terms.aspx College Algebra 0 =  ( x  2) 1
2 ( x  2) 2 = 1 x  2 = ±i
x = 2 ±i
So, we got complex solutions. Complex solutions will always indicate no xintercepts.
Now, we do want points on either side of the vertex so we’ll use the yintercept and the axis of
symmetry to get a second point. The yintercept is a distance of two to the left of the axis of
symmetry and is at y = 5 and so there must be a second point at the same y value only a
distance of 2 to the right of the axis of symmetry. The coordinates of this point must then be
( 4, 5 ) .
Here is the sketch of this parabola. [Return to Problems] (c) h ( x ) = x...
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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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