## Intermediate Algebra (12th Edition)

$(0,0)$
$\bf{\text{Solution Outline:}}$ 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}