Answer
$y^2=16x$
Work Step by Step
Vertex: $V(h,k)=V(0,0)$
$h=0$ and $k=0$
The focus is $4$ units to the rigth of the vertex. We have a parabola with horizontal axis that opens to the right: $(y-h)^2=4p(x-k)$
Focus: $F(p,0)=(4,0)$
$p=4$
$(y-0)^2=4(4)(x-0)$
$y^2=16x$
