Note that the general vertex form is
color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)XXXy=m(x−a)2+b
for a parabola with vertex at (color(red)a,color(blue)b)(a,b)
Given
color(white)("XXX")y=-x^2+12x-11XXXy=−x2+12x−11
Extract the color(green)(m)m factor from the first two terms:
color(white)("XXX")y=color(green)(""(-1))(x^2-12x) -11XXXy=(−1)(x2−12x)−11
If (x^2-12x)(x2−12x) are the first two terms of a squared binomial the third term must be 6^262
[since (x-a)^2=(x^2-2ax+a^2)](x−a)2=(x2−2ax+a2)]
and anything we add to complete the square must also be subtracted.
Completing the square
color(white)("XXX")y=color(green)(""(-1))(x^2-12xcolor(magenta)(+6^2))-11 color(magenta)(-color(green)(""(-1)) * (6^2))XXXy=(−1)(x2−12x+62)−11−(−1)⋅(62)
Simplifying:
color(white)("XXX")y=color(green)(""(-1))(x-color(red)(6))^2+color(blue)(25)XXXy=(−1)(x−6)2+25
Here is a graph of the original equation to help verify this result:
graph{-x^2+12x-11 [-3.106, 12.697, 18.77, 26.67]}
