Question 733804
find the axis of symmetry, the focus, directrix and vertex of the parabola. 
(x+2) ^ 2 - 16 (y-1) = 0 
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Form of parabola opening up:
(x+2)^2 = 16(y-1)
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Vertex: (-2,1)
Axis of symmetry: x = -2
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Since 4p = 16, p = 4
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Directrix: y = -3
Focus::   (-2,5)
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16y-16 = (x+2)^2
16y = (x+2)^2+16
y = (1/16)(x+2)^2+1
{{{graph(400,400,-10,10,-10,10,(1/16)(x+2)^2+1)}}}
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Cheers,
Stan H.
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