Asked by Lala
                I'm trying to put this parabola in standard form which is: (y-k)^2=4p(x-h)
The equation is x=(1/12)(y-2)^2 +6
Here is what i have so far:
x=(1/12)(y-2)^2 +6
-6 -6
x-6=(1/12)(y-2)^2
*12 *12
Would it be
12x-6 OR 12x-72?
thanks
            
        The equation is x=(1/12)(y-2)^2 +6
Here is what i have so far:
x=(1/12)(y-2)^2 +6
-6 -6
x-6=(1/12)(y-2)^2
*12 *12
Would it be
12x-6 OR 12x-72?
thanks
Answers
                    Answered by
            Reiny
            
    from 
x-6=(1/12)(y-2)^2
multiply both sides by 12
12(x-6) = (y-2)^2 , which of course is
(y-2)^2 = 12(x-6) , leave it like that, or you lose the form requested.
    
x-6=(1/12)(y-2)^2
multiply both sides by 12
12(x-6) = (y-2)^2 , which of course is
(y-2)^2 = 12(x-6) , leave it like that, or you lose the form requested.
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