# How do you find the axis of symmetry and vertex point of the function: y=3x^2+3 ?

Oct 8, 2015

Axis of symmetry: $x = 0$
Vertex: $\left(0 , 3\right)$

#### Explanation:

Note that any parabolic equation of the general form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
has a vertical axis of symmetry.

$y = 3 {x}^{2} + 3 \textcolor{w h i t e}{\text{XXX")iffcolor(white)("XXX}} y = 3 {x}^{2} + 0 x + 3$
is in this general form.

Further, it can be re-written as
$\textcolor{w h i t e}{\text{XXX}} y = 3 {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{3}$
which is the vertex form with a vertex at $\left(\textcolor{red}{0} , \textcolor{b l u e}{3}\right)$

Its (vertical) axis of symmetry (i.e. $x = c$ for some constant $c$) must pass through the vertex
therefore, the axis of symetry is
$\textcolor{w h i t e}{\text{XXX}} x = \textcolor{red}{0}$