The tangent equation of curve f (x) = X2 (X-2) + 1 at point (0, f (0)) is

f(x)=x³-2x²+1
Then f '(x) = 3x & # 178; - 4x
So the tangent slope k = f '(0) = 0
f(0)=1
So tangent point (0,1)
So the tangent is Y-1 = 0

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