# What is the vertex form of y=3x^2+2x-8 ?

Oct 7, 2017

$y = 3 {\left(x + 0. \overline{3}\right)}^{2} - 8. \overline{3}$

#### Explanation:

Vertex form is written:

$y = a {\left(x - h\right)}^{2} + k$

Where $\left(h , k\right)$ is the vertex.

Currently the equation is in standard form, or:

$y = a {x}^{2} + b x + c$

Where $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$ is the vertex.

Let’s find the vertex of your equation:

$a = 3 \mathmr{and} b = 2$

So,

$- \frac{b}{2 a} = - \frac{2}{2 \cdot 3} = - \frac{2}{6} = - \frac{1}{3}$

Thus $h = - \frac{1}{3} = - 0. \overline{3}$

$f \left(- \frac{1}{3}\right) = 3 {\left(- \frac{1}{3}\right)}^{2} + 2 \left(- \frac{1}{3}\right) - 8$
$f \left(- \frac{1}{3}\right) = 3 \left(\frac{1}{9}\right) - \frac{2}{3} - 8$
$f \left(- \frac{1}{3}\right) = \frac{1}{3} - \frac{2}{3} - 8 = - 8. \overline{3}$

Thus $k = - 8. \overline{3}$

We already know that $a = 3$, so our equation in vertex form is:

$y = 3 {\left(x - \left(- 0. \overline{3}\right)\right)}^{2} + \left(- 8. \overline{3}\right)$

$y = 3 {\left(x + 0. \overline{3}\right)}^{2} - 8. \overline{3}$