Algebra 2 (Math)

What is the vertex form of the equation? y = -x^2 + 12x - 4. I have no clue how to do this.

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  1. If you have no clue, then clearly you have not studied your material, since it must have been covered.

    If a parabola has vertex at (h,k), then the equation is

    y = a(x-h)^2+k

    You have
    y = -x^2+12x-4
    = -(x^2-12x) -4

    Now recall that (x+n)^2 = x^2+2nx+n^2
    That means that you have 2n = -12, or n=-6.

    So, add (-6)^2=36 inside the parentheses and you have a perfect square. But, having done that, you have changed the expression, so you have to subtract it as well, so the value is unchanged:

    y = -(x^2-12x+36) - 4 + 36
    y = -(x-6)^2 + 32

    To check you answer, expand things out; it should give you what you started with:

    -(x-6)^2+32
    = -(x^2-12x+36)+32
    = -x^2+12x-36+32
    = -x^2+12x-4

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