# What is the vertex of y= -6x^2 + 4x + 3?

Dec 21, 2015

Vertex: $\left(\frac{1}{3} , 3 \frac{2}{3}\right)$

#### Explanation:

Probably the easiest way to do this is to convert the equation into "vertex form": $y = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$

Given:
$\textcolor{w h i t e}{\text{XXX}} y = - 6 {x}^{2} + 4 x + 3$

Extract the $m$ factor
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 6\right) \left({x}^{2} - \frac{2}{3} x\right) + 3$

Complete the square
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 6\right) \left({x}^{2} - \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2}\right) + 3 - \left(- 6\right) \cdot {\left(\frac{1}{3}\right)}^{2}$

Rewrite with a squared binomial and simplified constant
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 6\right) {\left(x - \frac{1}{3}\right)}^{2} + 3 \frac{2}{3}$

which is in vertex form with vertex at $\left(\frac{1}{3} , 3 \frac{2}{3}\right)$