Question 932351
graph the parabola and identify the vertex, focus, directrix: x^2+16y=0
This is a parabola that opens downward.
Its basic equation: (x-h)^2=4p(y-k), (h,k)=coordinates of vertex
For given parabola:
x^2=-16y
vertex: (0,0)
axis of symmetry: x=0 or y-axis
4p=16
p=4
focus:(0,-4) (p-units below vertex on the axis of symmetry)
directrix:y=4 (p-units above vertex on the axis of symmetry)
see graph below:
{{{ graph( 300, 300, -10, 10, -10, 10, -x^2/16) }}}