# What is the vertex form of y= 6x^2+16x-12 ?

Vertex form
${\left(x + \frac{4}{3}\right)}^{2} = \frac{1}{6} \left(y + \frac{68}{3}\right) \text{ }$with Vertex at $\left(- \frac{4}{3} , - \frac{68}{3}\right)$

#### Explanation:

Let us start from the given equation

$y = 6 {x}^{2} + 16 x - 12$

$y = 6 \left({x}^{2} + \frac{16}{6} x\right) - 12$

$y = 6 \left({x}^{2} + \frac{8}{3} x + \frac{16}{9} - \frac{16}{9}\right) - 12$

$y = 6 \left({x}^{2} + \frac{8}{3} x + \frac{16}{9}\right) - \left(\frac{6 \cdot 16}{9}\right) - 12$

$y = 6 {\left(x + \frac{4}{3}\right)}^{2} - \frac{68}{3}$

$y + \frac{68}{3} = 6 {\left(x + \frac{4}{3}\right)}^{2}$

$\frac{1}{6} \left(y + \frac{68}{3}\right) = {\left(x + \frac{4}{3}\right)}^{2}$

${\left(x + \frac{4}{3}\right)}^{2} = \frac{1}{6} \left(y + \frac{68}{3}\right)$

Kindly see the graph of ${\left(x + \frac{4}{3}\right)}^{2} = \frac{1}{6} \left(y + \frac{68}{3}\right) \text{ }$with Vertex at $\left(- \frac{4}{3} , - \frac{68}{3}\right)$

graph{y=6x^2+16x-12[-60,60,-30,30]}

God bless....I hope the explanation is useful.