# What is the vertex of  y= 3x^2-2x-(3x+2)^2?

Dec 17, 2015

The vertex is at $\left(x , y\right) = \left(- \frac{7}{6} , \frac{25}{6}\right)$

#### Explanation:

Probably the easiest way to do this is to convert the given equation into "vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given:
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} - 2 x - {\left(3 x + 2\right)}^{2}$

Expand and simplify the expression on the right side:
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} - 2 x - \left(9 {x}^{2} + 12 x + 4\right)$

$\textcolor{w h i t e}{\text{XXX}} y = - 6 {x}^{2} - 14 x - 4$

Extract the $m$ factor
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{\left(- 6\right)} \left({x}^{2} + \frac{14}{6} x\right) - 4$

Complete the square
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{\left(- 6\right)} \left({x}^{2} + \frac{14}{6} x + {14}^{2} / {12}^{2}\right) - 4 + 6 \cdot \left({14}^{2} / \left({12}^{2}\right)\right)$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{\left(- 6\right)} {\left(x + \textcolor{red}{\frac{14}{12}}\right)}^{2} - 4 + \frac{196}{24}$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{\left(- 6\right)} \left(x - \textcolor{red}{\left(- \frac{7}{6}\right)}\right) + \textcolor{b l u e}{\frac{25}{6}}$
graph{3x^2-2x-(3x+2)^2 [-3.342, 2.815, 2.025, 5.102]}