Answer
The axis of symmetry is $x=-\frac{1}{3}$.
The vertex is $(-\frac{1}{3},-\frac{1}{3})$.
Work Step by Step
Comparing $f(x)=3x^{2}+2x$ with $f(x)=ax^{2}+bx+c$, we see that $a=3$ and $b=2$.
Axis of symmetry is given by
$x=-\frac{b}{2a}=-\frac{2}{2(3)}=-\frac{1}{3}$
The axis of symmetry is $x=-\frac{1}{3}$.
As the axis of symmetry is $x=-\frac{1}{3}$, the x-coordinate of the vertex is $-\frac{1}{3}$.
We can use the function to find the y-coordinate.
$f(x)=3x^{2}+2x$
$f(1)=3(-\frac{1}{3})^{2}+2(-\frac{1}{3})=-\frac{1}{3}$
The vertex is $(-\frac{1}{3},-\frac{1}{3})$.
