SOLUTION: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola. (x+2) ^ 2

SOLUTION: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola. (x+2) ^ 2 - 16 (y-1) = 0 thank you!

Question 733804: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola.
(x+2) ^ 2 - 16 (y-1) = 0
thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find the axis of symmetry, the focus, directrix and vertex of the parabola.
(x+2) ^ 2 - 16 (y-1) = 0
--------------------------
Form of parabola opening up:
(x+2)^2 = 16(y-1)
------
Vertex: (-2,1)
Axis of symmetry: x = -2
---
Since 4p = 16, p = 4
---
Directrix: y = -3
Focus:: (-2,5)
==========================
16y-16 = (x+2)^2
16y = (x+2)^2+16
y = (1/16)(x+2)^2+1

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Cheers,
Stan H.
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