which function has a vertex at the origin

[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] ~\dotfill\\\\ f(x)=-x^2\implies y = -(x-\stackrel{h}{0})^2+\stackrel{k}{0}\qquad \qquad \stackrel{\textit{vertex}}{(0,0)}[/tex]
Answer:
The answer is the second choice.
Step-by-step explanation: