# How do you find the vertex of y = 8x + 5x^2 – 3?

Vertex $\left(h , k\right) = \left(- \frac{4}{5} , - \frac{31}{5}\right)$

#### Explanation:

By completing the square method

$y = 8 x + 5 {x}^{2} - 3$

rearrange first

$y = 5 {x}^{2} + 8 x - 3$

Do factor out the 5 to make the coefficient of ${x}^{2}$ be one.

$y = 5 \left({x}^{2} + \frac{8}{5} x\right) - 3$

$y = 5 \left({x}^{2} + \frac{8}{5} x + \frac{16}{25} - \frac{16}{25}\right) - 3$

$y = 5 {\left(x + \frac{4}{5}\right)}^{2} - \frac{16}{5} - 3$

$y = 5 {\left(x + \frac{4}{5}\right)}^{2} - \frac{31}{5}$

$y + \frac{31}{5} = 5 {\left(x + \frac{4}{5}\right)}^{2}$

$y - - \frac{31}{5} = 5 {\left(x - - \frac{4}{5}\right)}^{2}$

kindly see the graph.
graph{y=8x+5x^2-3[-20,20,-10,10]}

Have a nice day from the Philippines