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Pre-Cal

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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

  • Pre-Cal -

    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.

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