SOLUTION: (y+1)^2 = 12(x-1) Graph, including the foci

Question 617191:
(y+1)^2 = 12(x-1)
Graph, including the foci
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
(y + 1)� = 12(x - 1)
Graph, including the foci.
----------------------------
We
(y - k)� = 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

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)

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