# What is the vertex of y=-x^2+4x+12? (This replaces the same question that I accidentally deleted before my first cup of coffee).

Jun 29, 2015

The vertex of $- {x}^{2} + 4 x + 12$ is at $\left(2 , 16\right)$

#### Explanation:

By rewriting $y = - {x}^{2} + 4 x + 12$
into "vertex form": $y = m {\left(x - a\right)}^{2} + b$ (with vertex at $\left(a , b\right)$)
we can simply "read off" the vertex values.

$y = - {x}^{2} + 4 x + 12$
$\textcolor{w h i t e}{\text{XXXX}}$extract $m$
$y = \left(- 1\right) \left({x}^{2} - 4 x - 12\right)$
$\textcolor{w h i t e}{\text{XXXX}}$complete the square
$y = \left(- 1\right) \left(\textcolor{b l u e}{{x}^{2} - 4 x + 4} - 12 - 4\right)$
$\textcolor{w h i t e}{\text{XXXX}}$rewrite as a square plus an external term
$y = \left(- 1\right) {\left(x - 2\right)}^{2} + 16$

This is in vertex form with the vertex at $\left(2 , 16\right)$