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Math Help - Standard form....

  1. #1
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    Standard form....

    use 3y^2-x-12y+14=0 to do the following

    a. write the equation in standard form:
    3y^2-x-12y+14=0
    -x+14=-3y^2-12y
    -x+14=-3(y^2+4y)
    -x+14= -3[(y^2+4y+4)-4]
    -x+14= -3[(y+2)^2-4]
    -x+14=-3(y-2)^2+4
    -x+10=-3(y-2)^2
    -(x-10)= -3(y-2)^2/-1
    x-10=3(x-2)^2

    standard form:
    x-10=3(x-2)^2

    Is this right??
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  2. #2
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    Quote Originally Posted by george93 View Post
    use 3y^2-x-12y+14=0 to do the following

    a. write the equation in standard form:
    3y^2-x-12y+14=0
    -x+14=-3y^2-12y
    -x+14=-3(y^2+4y)
    -x+14= -3[(y^2+4y+4)-4]
    -x+14= -3[(y+2)^2-4]
    -x+14=-3(y-2)^2+4
    -x+10=-3(y-2)^2
    -(x-10)= -3(y-2)^2/-1
    x-10=3(x-2)^2

    standard form:
    x-10=3(x-2)^2

    Is this right??
    No exactly.
    You have x instead of y.
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