Answer
The axis of symmetry is $x=\frac{1}{3}$.
The vertex is $(\frac{1}{3},-\frac{1}{3})$.
Work Step by Step
The given function is
$f(x)=3x^2-2x$
$f(x)=3x^2-2x+0$
It is in the form of $f(x)=ax^2+bx+c$
We have $a=3,b=-2$ and $c=0$
(a)
Equation for the axis of symmetry.
$\Rightarrow x=-\frac{b}{2a}$
Substitute $3$ for $a$ and $-2$ for $b$.
$\Rightarrow x=-\frac{-2}{2(3)}$
Simplify.
$\Rightarrow x=-\frac{-1}{3}$
$\Rightarrow x=\frac{1}{3}$
The axis of symmetry is $x=\frac{1}{3}$.
(b)
Substitute $\frac{1}{3}$ for $x$ in the given function.
$\Rightarrow f(\frac{1}{3})=3(\frac{1}{3})^2-2(\frac{1}{3})$
Simplify.
$\Rightarrow f(\frac{1}{3})=3(\frac{1}{9})-\frac{2}{3}$
$\Rightarrow f(\frac{1}{3})=\frac{3}{9}-\frac{2}{3}$
$\Rightarrow f(\frac{1}{3})=\frac{3}{9}-\frac{6}{9}$
$\Rightarrow f(\frac{1}{3})=\frac{3-6}{9}$
$\Rightarrow f(\frac{1}{3})=\frac{-3}{9}$
$\Rightarrow f(\frac{1}{3})=-\frac{1}{3}$
The vertex is $(\frac{1}{3},-\frac{1}{3})$.
