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Rewrite 3y^2 + 6y + 108x - 969=0 in standard form, show work.

  • precalc -

    3(y^2 + 2y + ...) - 969 = -108x
    3(y^2 + 2y + 1) - 969 - 3 = -108x
    3(y+1)^2 - 972 = -108x
    (-1/36) (y+1)^2 + 9 = x

    I will assume you realize that this is a "horizontal" parabola and that you know how to read the vertex from that form.

  • precalc -

    Just realized that Steve had already answered this question before you reposted it

    Have patience and always check if your question has been answered before re-posting it.
    It saves unnecessary work on our part.

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