Question 968532
Skipping the fundamental derivation, but for directrix (-p,y) using p as a positive number, focus (p,0), for parabola in standard form a axis of symmetry being the x-axis,  equation is {{{y^2=4px}}}.  If vertex were some point (h,k), then the equation of this parabola is  {{{(y-k)^2=4p(x-h)}}}.


Your example problem equation is easily transformed into {{{(y+2)^2=4(x+2)}}}.  You can make the correspondances to find p=1  IF you had parabola in standard form and standard position, but your case is that vertex is pushed to the left by 2 units.  See, your vertex is (-2,-2);  your directrix is {{{x=-1-2}}}, or simply now {{{highlight(x=-3)}}}.


Remember, p was taken as nonnegative in the fundamental derivation for standard position, so directrix is p units to the left of the vertex point, and vertex would be on the origin.