. . .y = -(x - 5)^2 + 11

I worked it by using FOIL:

. . .(x - 5)(x - 5)

. . .x^2 - 10x + 25

But because of the "minus" in front, I put:

. . .-x^2 + 10x - 25

Then I used the formula for the vertex, and put x = -b/(2a), which I did as:

. . .-(10) / (2(-1))

. . .-[(10) / (-2)]

. . .-(-5)

...or x = 5. Then I put x back into the expression:

. . .(-5)^2 + 10(5) - 25 + 11

This worked out would be y = 39, so the vertex is (5, 39).