How do you find the vertex and intercepts for y = 8x + 5x^2 – 3?

Jul 12, 2016

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

Explanation:

$y = 5 {x}^{2} + 8 x - 3$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{8}{10} = - \frac{4}{5}$
y-coordinate of vertex:
$y \left(- \frac{4}{5}\right) = 5 \left(\frac{16}{25}\right) - \frac{32}{5} - 3 = \frac{16}{5} - \frac{32}{5} - 3 = - \frac{31}{5}$
Vertex $\left(- \frac{4}{5} , - \frac{31}{5}\right)$
To find y-intercept, make x = 0 --> y = -3
To find 2 x-intercepts, make y = 0 and solve the quadratic equation $y = 5 {x}^{2} + 8 x - 3 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 64 + 60 = 124$ --> $d = \pm 2 \sqrt{31}$
There are 2 x-intercepts (2 real roots):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{8}{10} \pm \frac{2 \sqrt{31}}{10} = - \frac{4}{5} \pm \frac{\sqrt{31}}{5}$
$x = \frac{- 4 \pm \sqrt{31}}{5}$