Question 617191
(y + 1)² = 12(x - 1)

Graph, including the foci.
----------------------------
We
(y - k)² = 4p(x - h)

has vertex (h,k).  If p is positive it opens to the right.
If p is negative it opens to the left. its fccus is inside
the parabola on its axis on symmetry and |p| units from the
vertex.And the latus rectum or focal chord is |4p| units long
bisected by the focus.

Comparing your equation to that, we get

h=1, k= -1, 4p=12, so p=3

So the graph opens to the right because p is positive

{{{drawing(400,800,-2,7,-10,9,
red(arc(7,-1,-12,12,90,270)),locate(4,-1,"FOCUS_(4,-1)"),
graph(400,800,-2,7,-10,9,sqrt(12(x-1))-1), circle(4,-1,.1),

graph(400,800,-2,7,-10,9,-sqrt(12(x-1))-1)  )}}}



An the focus is a point inside the parabola |p| or 3 units
to the right of the vertex.  The vertex is (1,-1), so the 
focus is (4,-1)