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Wednesday, April 05, 2006

PARABOLA

Hey guys
I have nothing to say so im just going stray 2 explaining what happen 2day. But the thing is we did not do much today, it was just 2 question we had to work it out 2gether and play with this circle Mr. K gave it 2 us. Here are the 2 questions:

1) A parabola has vertex (2,-3).
And contains the point (9,-10).
Find its equation and sketch the graph if it is oriented:

a) Vertically

(x-h)^2=4p(y-k)
(x-2)^2=4p(y+3)
(9-2)^2=4p(-10+3)
(7)^2=4p(-7)
49=-28p
-7/4=p
Therefore the equation is (x-2)^2=-7(y+3)
And the graph looks like this:

The point inside the parabola is the focus point which is (2,-19/4)
The line on top of the parabola is axis of symmetry which is Y=-5/4

b) Horizontally

(y-k)^2=4p(x-h)
(-10+3)^2=4p(9-2)
49=28p
49/28=p
7/4=p
Therefore the equation is (y+3)^2=7(x-2)
And the graph looks like the first graph with the same vertex but the graph opens right, the focus is (15/4,-3), and the axis of symmetry is X=1/4.

Anyways i think this was it for today's class, so the next scribe for tomorrow is a-b33-l



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