Question 319959
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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