# What is the vertex of y= 3x^2 - 4x - 12?

Aug 10, 2018

Vertex is $\left(\frac{2}{3} , \frac{96}{9}\right)$

#### Explanation:

To find vertex convert the equation $y = 3 {x}^{2} - 4 x - 12$ to the form

$y = a {\left(x - h\right)}^{2} + k$ and then vertex is $\left(h , k\right)$.

We have $y = 3 {x}^{2} - 4 x - 12$

= $3 \left({x}^{2} - 2 \times \frac{2}{3} x + {\left(\frac{2}{3}\right)}^{2} - {\left(\frac{2}{3}\right)}^{2}\right) - 12$

= $3 \left({x}^{2} - 2 \times \frac{2}{3} x + {\left(\frac{2}{3}\right)}^{2}\right) - 3 \cdot \frac{4}{9} + 12$

= $3 {\left(x - \frac{2}{3}\right)}^{2} + \frac{96}{9}$

Hence, vertex is $\left(\frac{2}{3} , \frac{96}{9}\right)$