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Question 822255: I need to find the equation of a parabola with a vertex at (-4, 1) and a focus at (-5, 1), but some how I am not doing it right and I do not know how to arrange the equation so that it comes out looking like it does in the answers.
The way it looks in the answers is (for example): 
Found 2 solutions by TimothyLamb, stanbon:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
for quadratic: f(x) = ax^2 + bx + c = 0
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vertexX = -b/2a
vertexY = -D/4a
where D = bb - 4ac
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focusX = -b/2a
focusY = (1-D)/4a
where D = bb - 4ac
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if the symmetry axis is vertical, the above apply unchanged
if the symmetry axis is horizontal, flip x-y in all of the above
(your case is the second, so flip x-y)
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Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! I need to find the equation of a parabola with a vertex at (-4, 1) and a focus at (-5, 1), but some how I am not doing it right and I do not know how to arrange the equation so that it comes out looking like it does in the answers:
x - 4 = -(y + 1)^2 .
Note: That parabola has its vertex at (4,-1). Hmmmm??????
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Plot those two points.
Note that the parabola must be opening to the left from (-4,1)
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Form: (y-k)^2 = 4p(x-h)
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h = -4; k = 1; p = -1
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(y-1)^2 = -4(x+4)
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Cheers,
Stan H.
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