SOLUTION: I'm trying to graph a parabola and find its vertex and two other points. The equation given is y^2-4x=8y. I'm not sure which steps to take. Please help! Thanks!

Question 943079: I'm trying to graph a parabola and find its vertex and two other points. The equation given is y^2-4x=8y. I'm not sure which steps to take. Please help! Thanks!
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I'm trying to graph a parabola and find its vertex and two other points. The equation given is y^2-4x=8y.
y^2-4x=8y
complete the square:
y^2-8y=4x
(y^2-8x+16)=4x+16
(y-4)^2=4(x+4)
This is an equation of a parabola that opens right
Its basic form of equation: (y-h)^2=4p(x-h), (h,k)=coordinates of the vertex
For given parabola:
vertex:(-4,4)
axis of symmetry: y=4
4p=4
p=1
focus:(-3,4)(p-distance right of vertex on the axis of symmetry)
directrix:x=-5(p-distance left of vertex on the axis of symmetry)
see graph below:
y=�(4x+16)^.5+4

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