# Parabola Graph

• September 11th 2009, 12:45 PM
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: $(y+1)^2=-12x-24$
• September 11th 2009, 01:47 PM
pickslides
$y^2 +2y +12x+25=0$

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

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

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

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

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

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

What kind of parabola is this?
• September 12th 2009, 06:27 PM
Quote:

Originally Posted by pickslides
$y^2 +2y +12x+25=0$

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

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

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

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

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

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

What kind of parabola is this?

You made an error: http://www.mathhelpforum.com/math-he...864b9f68-1.gif , 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.