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Find the vertex of f(x)=4x^2+8x-2

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Answer:


f(x) =4x^2 + 8x -2 \\f(x) = 4(x^2 + 2x -(1)/(2) )\\\\f(x) = 4(x^2 + 2x +1 - 1 -(1)/(2) )\\f(x) = 4((x+1)^2 - 1 -(1)/(2) ))\\f(x) = 4((x+1)^2 -(3)/(2) ))\\f(x) = 4(x+1)^2 -6\\

the vertices (-1, -6)

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