#### Answer

$(0,0)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find the vertex of the given quadratic function, $
f(x)=\dfrac{1}{2}x^2
,$ convert the function in the form $f(x)=a(x-h)^2+k.$ Once in this form, the vertex is located at $(h,k).$
$\bf{\text{Solution Details:}}$
In the form $f(x)=a(x-h)^2+k,$ the function above is equivalent to
\begin{array}{l}\require{cancel}
f(x)=\dfrac{1}{2}(x)^2
\\\\
f(x)=\dfrac{1}{2}(x-0)^2+0
.\end{array}
In the function above, $h=
0
$ and $k=
0
.$ Hence, the vertex is
\begin{array}{l}\require{cancel}
(0,0)
.\end{array}