# What is the vertex form of y= 5x^2 + 9x − 4 ?

Feb 7, 2018

$y = 5 {\left(x + \frac{9}{10}\right)}^{2} - \frac{161}{20}$

#### Explanation:

Vertex form of equation for $y = a {x}^{2} + b x + c$ is $y = a {\left(x - h\right)}^{2} + k$ and vertex is $\left(h , k\right)$.

As $y = 5 {x}^{2} + 9 x - 4$, we have

$y = 5 \left({x}^{2} + \frac{9}{5} x\right) - 4$

= $5 \left({x}^{2} + 2 \times \frac{9}{10} x + {\left(\frac{9}{10}\right)}^{2} - {\left(\frac{9}{10}\right)}^{2}\right) - 4$

= 5((x+9/10)^2-5*(9/10)^2-4

= $5 {\left(x + \frac{9}{10}\right)}^{2} - \frac{81}{20} - 4$

= $5 {\left(x + \frac{9}{10}\right)}^{2} - \frac{161}{20}$

and as such vertex is $\left(- \frac{9}{10} , - \frac{161}{20}\right)$ or $\left(- \frac{9}{10} , - 8 \frac{1}{10}\right)$

graph{5x^2+9x-4 [-3.54, 1.46, -8.43, -5.93]}