Question 412001: find the diretrix, focus,vertex and then graph
y=1/12(x+1)^2-2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the diretrix, focus,vertex and then graph
y=1/12(x+1)^2-2
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Standard form of given parabola: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex.
y=1/12(x+1)^2-2
12y=(x+1)^2-24
12y+24=(x+1)^2
12(y+2)=(x+1)^2
(x+1)^2=12(y+2)
This is a parabola that opens upward with vertex at (-1,-2).
axis of symmetry: x=-1
4p=12
p=3
diretrix: y=-5
focus: (-1,1)
vertex:(-1,-2)
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