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Math Help - parabolas - urgent

  1. #1
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    Post parabolas - urgent

    Hey guys, kinda stumped here on these 2. Especially the first one since it's a horizontal parabola, I don't really know how the form is supposed to be. Any help appreciated.

    I know that standard form for a vertical is y = a(x - h)^2 + k, if that helps.

    1. Write in standard form, showing work: -2y^2 + 4y + 6 = x

    2. Fint the equation(in standard form) of the parabola that satisfies the following conditions: the vertex is at the origin, the parabola passes through the point (3,1), and the focus lies on the x-axis.

    Again, any help is MUCH appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post

    I know that standard form for a vertical is y = a(x - h)^2 + k, if that helps.

    1. Write in standard form, showing work: -2y^2 + 4y + 6 = x
    let's complete the square:

    -2y^2 + 4y + 6 = x
    => -2(y^2 - 2y - 3) = x
    => -2(y^2 - 2y + (-1)^2 - (-1)^2 - 3) = x
    => -2((y - 1)^2 - 4) = x
    => -2(y - 1)^2 + 8 = x
    => x = -2(y - 1)^2 + 8
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    2. Fint the equation(in standard form) of the parabola that satisfies the following conditions: the vertex is at the origin, the parabola passes through the point (3,1), and the focus lies on the x-axis.
    the equation of a parabola with focus (p,0) and directrix x = -p is

    y^2 = 4px

    since (3,1) lies on the parabola,

    1^2 = 4p(3)
    => 1 = 12p
    => p = 1/12

    so y^2 = (1/3)x
    so x = 3y^2 is our parabola
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