Question 980451
Best thing is use the derived equation for parabola translated from standard position,  {{{y^2=4px}}} becomes  {{{(y-k)^2=4p(x-h)}}} for which the vertex is (h,k).  The distance from vertex to either focus or directrix is absolute value of p.


{{{8x+8=-y^2}}}
{{{8(x+1)=-y^2}}}
{{{-8(x+1)=y^2}}}


This is concave to the left and has vertex (-1,0).
Axis of symmetry is {{{y=0}}}.


{{{-8=4p}}}
{{{p=-2}}}
Meaning the focus and directrix are both two units away from the point (-1,0).  Focus must be on the concave side, so  (-3,0) is the focus, and  (1,0) is a point on the directrix.  Directrix is {{{x=1}}}.