# How do you find the vertex and the intercepts for f(x) = -3x^2 - 6x - 2?

Jul 19, 2016

Vertex (1, -11)
$x = - 1 \pm \frac{\sqrt{3}}{3}$

#### Explanation:

x-coordinate of the vertex:
$x = - \frac{b}{2 a} = \frac{6}{-} 6 = 1$
y-coordinate of vertex:
$y \left(1\right) = - 3 - 6 - 2 = - 11$
Vertex (1, -11)
Make x = 0, y-intercept = - 2
To find x-intercepts, make y = 0 and solve the quadratic equation:
$f \left(x\right) = - 3 {x}^{2} - 6 x - 2 = 0.$
$D = {d}^{2} = {b}^{2} - 4 a c = 36 - 24 = 12$ --> $d = \pm 2 \sqrt{3}$
There are 2 x-intercepts (real roots):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{6}{-} 6 \pm 2 \frac{\sqrt{3}}{6} = - 1 \pm \frac{\sqrt{3}}{3}$