Rewrite 3y^2 + 6y + 108x - 969=0 in standard form, show work.

That'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