Question 743933
Identify the vertex, focus, and directrix of the graph of 
y=1/12(x+5)^2-3
rewrite equation to basic form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
1/12(x+5)^2=y+3
(x+5)^2=12(y+3)
vertex:(-5,-3)
axis of symmetry: x=-5
4p=12
p=3
focus: (-5,0)
directrix:(-5,-6) 

{{{ graph( 300, 300, -10,10, -10, 10,(1/12)(x+5)^2-3) }}}