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March 29, 2015

March 29, 2015

Posted by **95** on Friday, December 16, 2011 at 11:50am.

- precalc -
**Steve**, Friday, December 16, 2011 at 1:27pmThat's pretty much in standard form.

ax^2 + bxy + cy^2 + dx + ey + f = 0

3y^2 + 108x + 6y - 969 = 0

Now if you want a "standard" form for a parabola x-h = a(y-k)^2, try this:

108x = -3y^2 - 6y + 969

108x = -3(y^2 + 2y + 1) + 972

x = -1/36 (y+1)^2 + 9

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