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Math Help - Parabola Graph

  1. #1
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    Parabola Graph

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

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

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

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

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

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

     (y+1)^2=-2x-24

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

     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
     y^2 +2y +12x+25=0

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

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

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

     (y+1)^2=-2x-24

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

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

    What kind of parabola is this?

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

    I finally figured it out. The equation can be factored (y+1)^2=-12(x-2) This is just a transformation of the parabola y^2=-12x which can be written as 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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