# What is the vertex form of #y=5x^2 + 4x - 3?

Nov 16, 2015

$y = 5 {\left(x + \frac{2}{5}\right)}^{2} - \frac{19}{5}$

#### Explanation:

A parabola has the form:
$y = a {x}^{2} + b x + c$
$y = 5 {x}^{2} + 4 x - 3$

The vertex form of a parabola is:
$y = a {\left(x - h\right)}^{2} + k$

To find the vertex $\left(h , k\right)$ of a parabola we use:

$h = - \frac{b}{2 a} = - \frac{4}{2 \cdot 5} = - \frac{4}{10} = - \frac{2}{5} = - 0.4$

And to find the y coordinate we just evaluate in the original function with the value we just found:

$k = f \left(- \frac{2}{5}\right) = 5 {\left(- \frac{2}{5}\right)}^{2} + 4 \left(- \frac{2}{5}\right) - 3 = - 3.8$

The vertex is $\left(- 0.4 , - 3.8\right)$. Rewriting the equation using the vertex form, we have:

$y = 5 {\left(x - \left(- 0.4\right)\right)}^{2} + \left(- 3.8\right)$
$y = 5 {\left(x + 0.4\right)}^{2} - 3.8$

If you want to have it in a nicer way:

$y = 5 {\left(x + \frac{2}{5}\right)}^{2} - \frac{19}{5}$

graph{5(x+2/5)^2-19/5 [-10, 10, -5, 5]}
Click on the graph to see its vertex