# How do you find the vertex and intercepts for y = = 5x^2 + 4x - 3?

Jan 18, 2016

Vertex (-2/5, -19/5)

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{4}{10} = - \frac{2}{5}$.
y-coordinate of vertex:
$y \left(- \frac{2}{5}\right) = 5 \left(\frac{4}{25}\right) - 4 \left(\frac{2}{5}\right) - 3 = \frac{4}{5} - \frac{8}{5} - 3 = - \frac{19}{5}$
To find y-intercept, make x = 0 --> y = -3
To find x-intercepts, make y = 0, and solve the quadratic equation:
5x^2 + 4x - 3 = 0
D = d^2 = b^2 - 4ac = 16 + 60 = 76 --> $d = \pm \sqrt{76}$
$x = - \frac{2}{5} \pm \frac{\sqrt{76}}{5}$