In fact lets go ahead and find them now 2007 paul

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Unformatted text preview: ans that there can’t possibly be x-intercepts 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 x-intercepts. Now, we do want points on either side of the vertex so we’ll use the y-intercept and the axis of symmetry to get a second point. The y-intercept 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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