Answer
Please see the graph.
Work Step by Step
$f(x)=3x^2$
$a=3$, $b=0$, $c=0$
The vertex is found at $x=-b/2a$, on the line of symmetry.
$x=-0/2*3$
$x=0/6$
$x=0$
$f(x)=3x^2$
$f(0)=3*0^2$
$f(0)=3*0$
$f(0)=0$
$(0,0)$ is the vertex.