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Question 74598: anybody good at parabolas?
on my "Y" axis from 1 through 5 the top of the parbola (known as V(-3,4)) is at 4 where it begins its downward peak.
on the "X" axis the right leg comes through -1 curves through the "Y" axis of 4 and the left leg goes back down through -5 on the "x" axis.
Question: Determine the standard equation of the parabola.
I know you people are whizzes and I would really be impressed. I have missed every question concering parabolas.
Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! on my "Y" axis from 1 through 5 the top of the parbola (known as V(-3,4)) is at 4 where it begins its downward peak.
on the "X" axis the right leg comes through -1 curves through the "Y" axis of 4 and the left leg goes back down through -5 on the "x" axis.
Question: Determine the standard equation of the parabola.
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The V point is the vertex.
The vertex form of this parabola that opens downward is (x-h)^2 = 4p(y-k)
where (h,k) is the vertex.
So, (x+3)^2=4p(y-4)
When x=-5, y=0 so substitute and solve for "p".
(-5+3)^2 = 4p(0-4)
4=4p(-4)
p=-1/4
EQUATION:
on my "Y" axis from 1 through 5 the top of the parbola (known as V(-3,4)) is at 4 where it begins its downward peak.
on the "X" axis the right leg comes through -1 curves through the "Y" axis of 4 and the left leg goes back down through -5 on the "x" axis.
Question: Determine the standard equation of the parabola.
(x+3)^2=4(-1/4)(y-4)
(x+3)^2 = -y+4
y=-(x+3)^2+4
In standard form:
y=-x^2-6x-5
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Cheers,
Stan H.
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