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

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Answer 1 · 2017-02-09T13:12:32

#x+y-3=0#. See the tangent-inclusive Socratic graph.

At x = 1, y = 2.

So, the point of contact of the tangent is P ( 1, 2 ).

Slope of tangent at P is f' at at P

#=3x^2-4x#, at x = 1

#=-1.#

So, the equation to the tangent is

#y-2=-(x-1)#, giving

#x+y-3=0#.

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