How do you find the equation of the tangent line to the graph of #y=x^3-3x^2+x# at the point (2,-2)?

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How do you find the equation of the tangent line to the graph of #y=x^3-3x^2+x# at the point (2,-2)?

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Answer 1 · 2015-02-03T22:44:01

I would find the derivative of your function, #y'(x)#, evaluate the derivative at the point of interest (#x=2#) and use it as the slope of the line through #(2,-2)#.
Step 1
Derive: #y'(x)=3x^2-6x+1#
Step2:
Evaluate: #y'(2)=12-12+1=1#
Step3:
Tangent: line through #(2,-2)# and slope #1#;
#y-(-2)=1*(x-2)#
#y+2=x-2#
#y=x-4# which is your tangent.

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How do you find the equation of the tangent line to the graph of #y=x^3-3x^2+x# at the point (2,-2)?

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