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Question 555284: Hi i am wondering how to solve for the "p" value in parabolas. from what i understand you are supposed to do 1/4a but everytime i try this it is unsuccessful. if anyone could help it would be very appreciated. the problem i need the p value for is as follows. 1/4(x-3)^2-1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Hi i am wondering how to solve for the "p" value in parabolas. from what i understand you are supposed to do 1/4a but everytime i try this it is unsuccessful. if anyone could help it would be very appreciated. the problem i need the p value for is as follows. 1/4(x-3)^2-1
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Rewrite the formula with the y in it.
y=1/4(x-3)^2-1
y+1=1/4(x-3)^2
(x-3)^2=4(y+1)
This is the standard form of equation for parabolas showing the focus and directrix:
(x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex.
For given equation:
vertex: (3,-1)
axis of symmetry: x=3 (parabola opens upwards)
4p=4
p=1
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