Question 181825
Given the Parabola: x^2+2y+2=0
A. find the vertex
B. find the focus
C. Find the directrix
D. graph the function
:
{{{x^2+2y+2=0}}}
:
{{{x^2=-2(y+1)}}}
:
vertex is at (0,-1) (h,k)
:
4p=-2-->p=-1/2 
:
to get focus coordinates-subtract p from k( the vertex's y coordinate) to find the focus's y coordinate.....the x coordinate is h(the vertex's x coordinate)
:
directrix for  vertical parabola is y=k-p
:
focus(0,-1-1/2)-->(0,-3/2)
:
directrix(y=-1+1/2)--->y=-1/2
:
{{{graph(300,300,-10,10,-10,10,(-1/2)x^2-1)}}}