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

Oct 5, 2016

Vertex $\left(- \frac{3}{16} , - \frac{73}{32}\right)$

#### Explanation:

We will need to complete the square to solve this equation.

First move the constant to the other side of the equation by adding $2$ to both sides.

$8 {x}^{2} + 3 x = 2$

Factor out the coefficient, $8$, from the x^2 term.

$8 \left({x}^{2} + \frac{3}{8} x\right) = 2$

Take the coefficient of the $x$ term and divide it by 2 and then square it.

${\left(\frac{\frac{3}{8}}{2}\right)}^{2} = {\left(\frac{3}{8} \cdot \frac{1}{2}\right)}^{2} = {\left(\frac{3}{16}\right)}^{2} = \frac{9}{256}$

Add this value to the left hand side

$8 \left({x}^{2} + \frac{3}{8} x + \frac{9}{256}\right) = 2$

Add $8 \left(\frac{9}{256}\right)$ to the right hand side because of the factoring we did earlier.

$8 \left({x}^{2} + \frac{3}{8} x + \frac{9}{256}\right) = 2 + 8 \left(\frac{9}{256}\right)$

You now have a perfect square trinomial

$8 {\left(x + \frac{3}{16}\right)}^{2} = 2 + 8 \left(\frac{9}{256}\right)$

Simplify

$8 {\left(x + \frac{3}{16}\right)}^{2} = 2 + \cancel{8} \left(\frac{9}{\cancel{256} 32}\right)$

Convert $2$ to an improper fraction

$8 {\left(x + \frac{3}{16}\right)}^{2} = \frac{64}{32} + \left(\frac{9}{32}\right)$

Simplify

$8 {\left(x + \frac{3}{16}\right)}^{2} = \frac{73}{32}$

$y = 8 {\left(x + \frac{3}{16}\right)}^{2} - \frac{73}{32}$

Vertex form

$y = {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex

Vertex $\left(- \frac{3}{16} , - \frac{73}{32}\right)$

Click here to see more tutorials on how to complete the square.