Question 1119533
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parabola, given that F(3,2), d: y + 4 = 0
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Vertex is at  (3,-1), and this is 3 units away from directrix or focus.

Parabola has vertex as a maximum point.

{{{(x-3)^2=-4*3(y+1)}}}

{{{highlight_green((x-3)^2=-12(y+1))}}}-------no real zeros; but you can calculate y-intercept if you want.



{{{graph(300,300,-3,9,-9,3,-(1/2)(x-3)^2-1)}}}