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April 1, 2015

April 1, 2015

Posted by **Asher** on Tuesday, September 14, 2010 at 12:18am.

- Pre Calc -
**drwls**, Tuesday, September 14, 2010 at 2:53amIt will be a parabola. The vertical axis will be at x = -3. F is the focal point. The vertex of the parabola is equidistant from F and the y = -2 line, at x = -3, y = 1.

y = a (x+3)^2 + 1 is the equation.

Do some reading up on parabolas to figure out what a is. I think you will find it is 4 times the distance from the focus to the vertex, or 4*3 = 12

That would make the equation

y = 12(x+3)^2 + 1

Expand and rearrange that to the "general form".

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