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When you graph a parabola with a vertical directrix and the directrix is x=-1/12y^2 and the form they give you is x=a(y-k)^2+h how do write x=-1/12y^2 to that format
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you need the equation of parabola ?
yea like i know the format i just dont know what numbers to plug in my original equation is x=-1/12y^2
would it look like this x=3(y-0)+_?
it is in that format only, x= (-1/12) (y-0)^2 +0
so, here a = -1/12 k=0 h=0
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oh ok i understand now thanks
you're welcome ^_^
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