How do you write y=x2+12x11 in vertex form?

1 Answer
Oct 10, 2016

x=(1)(x6)2+25

Explanation:

Note that the general vertex form is
XXXy=m(xa)2+b
for a parabola with vertex at (a,b)

Given
XXXy=x2+12x11

Extract the m factor from the first two terms:
XXXy=(1)(x212x)11

If (x212x) are the first two terms of a squared binomial the third term must be 62
[since (xa)2=(x22ax+a2)]
and anything we add to complete the square must also be subtracted.

Completing the square
XXXy=(1)(x212x+62)11(1)(62)

Simplifying:
XXXy=(1)(x6)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]}