## Algebra 2 (1st Edition)

Given $$(x-3)^2=16y$$ Rewrite as: $$(x-3)^2=16(y-0)$$ Thus, we have $h=3\\k=4\\p=4 \gt 0$ The parabola opens upward. The vertex is $(h,k)=(3,0)$ The focus is $(h,p+k)=(3,4)$