Question 761279
Find the vertex and directrix for the parabola 
y + 2 = 1/3 (x+2) ^ 2
rewrite equation in basic form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
{{{y+2=(1/3)(x+2)^2}}}
{{{3y+6)=(x+2)^2}}}
{{{(x+2)^2=3(y+2)}}}
This is an equation of a parabola that opens upward.
vertex: (-2,-2)
axis of symmetry: x=-2
4p=3
p=3/4
directrix: y=11/4 (p-units below vertex on the axis of symmetry.

see graph below as a visual check
y= (x^2+4x+4)/3-2

{{{ graph( 300, 300, -10, 10, -10, 10, (x^2+4x+4)/3-2) }}}