# How do you find the vertex and axis of symmetry, and then graph the parabola given by: g(x)= 3x^2 + 12x + 15?

Oct 9, 2015

Vertex: $\left(- 2 , 3\right)$
Axis of symmetry: $x = - 2$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 {x}^{2} + 12 x + 15$

Part 1: The Vertex
Convert into vertex form ($g \left(x\right) = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$)

$\textcolor{w h i t e}{\text{XXX}}$Extract the $m$
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 \left({x}^{2} + 4 x\right) + 15$

$\textcolor{w h i t e}{\text{XXX}}$Complete the square
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 \left({x}^{2} + 4 x + {2}^{2}\right) + 15 - 3 \left({2}^{2}\right)$

$\textcolor{w h i t e}{\text{XXX}}$Re-write as a squared binomial of form ${\left(x - a\right)}^{2}$ and simply
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 {\left(x - \left(- 2\right)\right)}^{2} + 3$

This is in vertex form with the vertex at $\left(- 2 , 3\right)$

Part 2: The Axis of Symmetry
An parabolic equation in the form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
has a vertical axis (i.e. $x = c$ for some constant $c$) through the vertex.

Since the given $g \left(x\right)$ is of this form and the vertex is at $\left(x , y\right) = \left(- 2 , 3\right)$
the axis of symmetry is
$\textcolor{w h i t e}{\text{XXX}} x = \left(- 2\right)$