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Thread: Parabola Graph

  1. #1
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    Parabola Graph

    I'm having trouble graphing the parabola $\displaystyle y^2 +2y +12x+25=0$.

    I rearrange and factor: $\displaystyle (y+1)^2=12x-26$. This isn't really the form $\displaystyle y^2=4Px$ because of the extra constant. Is the parabola just a transformation of $\displaystyle y=12x$? I'm not nsure how to graph this.

    Edit:$\displaystyle (y+1)^2=-12x-24$
    Last edited by adkinsjr; Sep 11th 2009 at 01:37 PM.
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  2. #2
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    $\displaystyle y^2 +2y +12x+25=0$

    $\displaystyle (y^2 +2y +1)-1+12x+25=0$

    $\displaystyle (y^2 +2y +1)+12x+24=0$

    $\displaystyle (y+1)^2+12x+24=0$

    $\displaystyle (y+1)^2=-2x-24$

    $\displaystyle y+1=\pm\sqrt{-2x-24}$

    $\displaystyle y=\pm\sqrt{-2x-24}-1$

    What kind of parabola is this?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    $\displaystyle y^2 +2y +12x+25=0$

    $\displaystyle (y^2 +2y +1)-1+12x+25=0$

    $\displaystyle (y^2 +2y +1)+12x+24=0$

    $\displaystyle (y+1)^2+12x+24=0$

    $\displaystyle (y+1)^2=-2x-24$

    $\displaystyle y+1=\pm\sqrt{-2x-24}$

    $\displaystyle y=\pm\sqrt{-2x-24}-1$

    What kind of parabola is this?

    You made an error: , it should be $\displaystyle -12x-24$

    I finally figured it out. The equation can be factored $\displaystyle (y+1)^2=-12(x-2)$ This is just a transformation of the parabola $\displaystyle y^2=-12x$ which can be written as $\displaystyle x=-\frac{1}{12}y^2$. So it is a parabola where the x-axis is the axis of symmetry. In other words, it's sideways.
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