Y=9x^2+9x-1 is to be re-written in "vertex form," i. e., y-k = a(x-h)^2, where a is a constant coefficient and (h,k) is the vertex.
Y=9x^2+9x-1 can be re-written as y - 1 = 9(x^2 + x).
We can complete the square of x^2 + x:
x^2 + 2(1/2) + (1/2)^2 - (1/2)^2
and this can be simplified as follows: (x+1/2)^2 - 1/4
Go back to y - 1 = 9(x^2 + x) and subst. (x+1/2)^2 - 1/4 for (x^2 + x):
y - 1 = 9[ (x+1/2)^2 - 1/4]
Then the desired equation in vertex form is y-1 = 9[ (x+1/2)^2 -(-1/4) ]
or y - 1 = 9 [(x+1/2)^2 ] + 9/4, or y - 1 - (9/4) = 9 (x+1/2)^2, or,
finally, y - (13/4) = 9 (x+1/2)^2. The vertex is at (1/2, 13/4).
