# What is the equation of the line tangent to # f(x)=x(x-1)^2-x+3 # at # x=1 #?

When y equals 2, x equals 1.

Thus, P (1, 2) is the tangent's point of contact.

At P, the tangent's slope is f'.

Thus, the tangent's equation is

graph{(x^3-2x^2+3-y)(x+y-2.95)((x-1)^2+(y-2)^2-.01)=0 [0, 10,0, 5]}

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The equation of the line tangent to f(x) = x(x-1)^2 - x + 3 at x = 1 is y = 4x - 1.

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