# What is the vertex form of y= 3x^2 + 9x-6?

Mar 2, 2018

$3 {\left(x - \frac{3}{2}\right)}^{2} - \frac{51}{4}$

#### Explanation:

The vertex form of the quadratic expression $a {x}^{2} + b x + c$ is $a {\left(x - h\right)}^{2} + k$.

We have $3 {x}^{2} + 9 x - 6$

Or:

$3 \left({x}^{2} + 3 x\right) - 6$

To complete the square, we take the coefficient of $3 x$, divide it by $2$, square it and add it:

$3 \left({x}^{2} + 3 x + \frac{9}{4}\right) - 6 - \frac{27}{4}$

$3 {\left(x - \frac{3}{2}\right)}^{2} - \frac{51}{4}$

So the vertex is, incidentally, $\left(\frac{3}{2} , - \frac{51}{4}\right)$