# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of g(x) = 3x^2 + 12x + 15 ?

##### 1 Answer
Sep 29, 2015

Vertex: $\left(- 2 , 3\right)$
Axis of Symmetry: $x = - 2$
Domain: $\mathbb{R}$
Range: $\mathbb{R} \ge 3$

#### Explanation:

The "vertex form" for a parabola is
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = m {\left(x - a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXXXX}}$with the vertex at $\left(a , b\right)$

To convert $g \left(x\right) = 3 {x}^{2} + 12 x + 15$ into "vertex form"

Extract the $m$
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 \left({x}^{2} + 4 x\right) + 15$
Complete the square
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 \left({x}^{2} + 4 x + {2}^{2}\right) + 15 - 3 \left({2}^{2}\right)$
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 {\left(x + 2\right)}^{2} + 3$
In complete vertex form:
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 3 {\left(x - \left(- 2\right)\right)}^{2} + 3$
with the vertex at $\left(- 2 , 3\right)$

The equation is of the form for a parabola with a vertical axis of symmetry opening upward and since the axis of symmetry passes through the vertex,
the axis of symmetry is $x = - 2$

$g \left(x\right) = 3 {x}^{2} + 12 x + 15$ is defined for all Real values of $x$,
therefore the Domain is $x \in \mathbb{R}$.

$g \left(x\right)$ has values for all points at or above the vertex,
therefore the Range is $g \left(x\right) \in \mathbb{R} \ge 3$
graph{3x^2+12x+15 [-7.85, 6.196, -0.84, 6.186]}