posted by 95 .
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
Have patience and always check if your question has been answered before re-posting it.
It saves unnecessary work on our part.