# How do you find the vertex and the intercepts for  f(x)= 3x^2 + 12x + 1?

May 24, 2017

Vertex $\left(- 2 , - 11\right)$

#### Explanation:

$f \left(x\right) = 3 {x}^{2} + 12 x + 1$

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{12}{6} = - 2$

y-coordinate of vertex: use $x = - 2$
$f \left(- 2\right) = 12 - 24 + 1 = - 11$

Vertex $\left(- 2 , - 11\right)$.

To find $x$-intercepts, make $f \left(x\right) = 0 ,$ and solve the quadratic equation:

$3 {x}^{2} + 12 x + 1 = 0$

$D = {d}^{2} = {b}^{2} - 4 a c = 144 - 12 = 132 \rightarrow$

$d = \pm 2 \sqrt{33}$

There are 2 real roots (two $x$-intercepts):

$x = - \frac{12}{6} \pm \frac{2 \sqrt{33}}{6} = - 2 \pm \frac{\sqrt{33}}{3}$