PreCal
posted by Lala .
I'm trying to put this parabola in standard form which is: (yk)^2=4p(xh)
The equation is x=(1/12)(y2)^2 +6
Here is what i have so far:
x=(1/12)(y2)^2 +6
6 6
x6=(1/12)(y2)^2
*12 *12
Would it be
12x6 OR 12x72?
thanks

from
x6=(1/12)(y2)^2
multiply both sides by 12
12(x6) = (y2)^2 , which of course is
(y2)^2 = 12(x6) , leave it like that, or you lose the form requested.