precalc
posted by 95 .
Parabola:
Rewrite 3y^2 + 6y + 108x  969=0 in standard form, show work.

3(y^2 + 2y + ...)  969 = 108x
3(y^2 + 2y + 1)  969  3 = 108x
3(y+1)^2  972 = 108x
(1/36) (y+1)^2 + 9 = x
I will assume you realize that this is a "horizontal" parabola and that you know how to read the vertex from that form. 
Just realized that Steve had already answered this question before you reposted it
http://www.jiskha.com/display.cgi?id=1324054249
Have patience and always check if your question has been answered before reposting it.
It saves unnecessary work on our part.