Alg_Complete

# Here is the standard form of an ellipse x h a2 2

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Unformatted text preview: w that ( 2,5 ) must also be on the parabola. Here is a sketch of the parabola. [Return to Problems] (b) In this case a = -1 , b = 8 and c = 0 . From these we see that the parabola will open downward since a is negative. Here are the vertex evaluations. x=- 8 8 ==4 2 ( -1) -2 y = f ( 4 ) = - ( 4 ) + 8 ( 4 ) = 16 2 So, the vertex is ( 4,16 ) and we also can see that this time there will be x-intercepts. In fact, let’s go ahead and find them now. © 2007 Paul Dawkins 212 http://tutorial.math.lamar.edu/terms.aspx College Algebra 0 = - x2 + 8x 0 = x ( - x + 8) Þ x = 0, x = 8 So, the x-intercepts are ( 0, 0 ) and ( 8, 0 ) . Notice that ( 0, 0 ) is also the y-intercept. This will happen on occasion so don’t get excited about it when it does. At this point we’ve got all the information that we need in order to sketch the graph so here it is, [Return to Problems] (c) In this final part we have a = 1 , b = 4 and c = 4 . So, this parabola will open up. Here are the vertex evaluations. x=- 4 4 = - = -2 2 (1) 2 y = f ( -2 ) = ( -2 ) + 4 ( -2 ) + 4 = 0 2 So, the vertex is ( -2,0 ) ....
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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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