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find the vertex f(x)= x^2-2x-8
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first we derive f(x)
f´(x) = 2x - 2
then 2x -2 = 0 x = 1
we evaluate x = 1 at original equation
f(x) = y = 1^2 - 2 (1) - 8 = 1 - 2 - 8 = -9
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vertex i located at ( 1 ; -9)
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