How do you find the equation of the line tangent to #y=x^3# at the point (-1,-1)?

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Answer 1 · 2016-09-22T14:37:07

#y=3x+2#

#(x_0,y_0) = (-1,-1)#

#y=kx+n# where #k=f'(x_0)#

#f'(x) = 3x^2 => k=f'(x_0) = 3#

The tangent line goes through #(-1,-1)# so:

#-1 = 3*(-1) + n => n = 2#

Finally, the equation of the tangent line is:

#y=3x+2#