Also we have a 3 and b 5 so the points are right most

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Unformatted text preview: Note that since the y coordinate of this point is zero it is also an xintercept. In fact it will be the only x-intercept for this graph. This makes sense if we consider the fact that the vertex, in this case, is the lowest point on the graph and so the graph simply can’t touch the x-axis anywhere else. The fact that this parabola has only one x-intercept can be verified by solving as we’ve done in the other examples to this point. 0 = x2 + 4x + 4 0 = ( x + 2) 2 Þ x = -2 Sure enough there is only one x-intercept. Note that this will mean that we’re going to have to use the axis of symmetry to get a second point from the y-intercept in this case. Speaking of which, the y-intercept in this case is ( 0, 4 ) . This means that the second point is ( -4, 4 ) . © 2007 Paul Dawkins 213 College Algebra Here is a sketch of the graph. [Return to Problems] As a final topic in this section we need to briefly talk about how to take a parabola in the general form and convert i...
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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.

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