# How do you find the VERTEX of a parabola y=-3x^2+12x-9?

Jul 22, 2015

Re-write the given equation in "vertex form" to get:
$\textcolor{w h i t e}{\text{XXXX}}$vertex at $\left(2 , 3\right)$

#### Explanation:

The general "vertex form" for a parabola is
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with a vertex at $\left(a , b\right)$

Given $y = - 3 {x}^{2} + 12 x - 9$

Extract $m$
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(- 3\right) \left({x}^{2} - 4 x\right) - 9$

Complete the square:
$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(- 3\right) \left({x}^{2} - 4 x + 4\right) - 9 + 3 \left(4\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$y = \left(- 3\right) {\left(x - 2\right)}^{2} + 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$which is the vertex form
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with vertex at $\left(2 , 3\right)$