Question 213455
what is the vertex of parabola y=8x+1-4x^2? I need all the steps so I can understand the solution
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a = -4, b = 8, c = 1
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Vertex occurs at x = -b/2a = -8/(2*-4) = 1
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f(1) = 8 + 1 - 4 = 5
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Vertex: (1,5)
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If you work it by completing the square you get:
-4x^2 + 8x + ? = y - 1 + ?
-4(x^2 - 2x + (-1)^2) = y - 1 + -4*1
-4(x - 1)^2 = y-5
From this you can see that h=1 and k = 5
So vertex is at (1,5)
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Cheers,
Stan H.