SOLUTION: what is the vertex of the parabola x^2-4y+8=0

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Question 319959: what is the vertex of the parabola x^2-4y+8=0
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
what is the vertex of the parabola x^2-4y+8=0
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Solve for "y":
4y = x^2+8
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y = (1/4)x^2 +2
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Vertex occurs where x = -b/2a = 0/(1/2) = 0
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If x = 0, y = 2
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Vertex: (0,2)
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Cheers,
Stan H.
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Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
what is the vertex of the parabola x^2-4y+8=0
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x^2-4y+8=0
y = (1/4)x^2 + 2
The vertex is on the axis of symmetry, which is x = -b/2a
x = 0
@ x = 0, y = 2
Vertex at (0,2)