# What is the vertex form of  f(x) = 6x^2+4x-3 ?

Feb 1, 2017

The answer is $= 6 {\left(x + \frac{1}{3}\right)}^{2} - \frac{11}{9}$

#### Explanation:

We look for the vertex form by completing the squares

$f \left(x\right) = 6 {x}^{2} + 4 x - 3$

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

$= 6 \left({x}^{2} + \frac{2}{3} x + \frac{1}{9}\right) - 3 - \frac{6}{9}$

$= 6 {\left(x + \frac{1}{3}\right)}^{2} - \frac{11}{9}$

This is the vertex form