# How do you find the axis of symmetry and vertex point of the function: f(x)= 3x^2+12x-6?

Oct 19, 2015

The vertex is $\left(1 , 9\right)$
The axis of symmetry is $x = 1$

#### Explanation:

To find the axis of symmetry of the function, use the coefficients of the equation and the formula $x = - \frac{b}{2} a$

For the function $f \left(x\right) = 3 {x}^{2} + 12 x - 6$
$a = 3$
$b = 12$
$c = - 6$

The x of the vertex and the axis of symmetry can be found by
$- \frac{b}{2} a$
$- \frac{12}{2 \left(- 6\right)}$
$x = 1$

This means the x of the vertex is x = 1 which is also the line of symmetry for the function.

To find the y of the vertex, plug x = 1 into the function and solve for y

$y = 3 {\left(1\right)}^{2} + 12 \left(1\right) - 6$
$y = 3 + 12 - 6$
$y = 9$

Therefore the vertex point is $\left(1 , 9\right)$