How do you find the equation of the line tangent to #f(x) = x^3# at x = 2?

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Answer 1 · 2016-02-17T18:24:31

#y=2x-16#

#f(2)=2^3=8# and hence #(2,8)# lies on the curve and the tangent.

The gradient (slope) of the tangent at any point is represented by the derivative at that point.

#therefore f'(x)=3x^2#

#therefore f'(2)=3xx2^2=12#.

The tangent is a straight line so has linear equation #y=mx+c#.

Substituting the point #(2,8)# in this equation, we get :

#8=(12)(2)+c#

#therefore c=-16#.

Hence the equation of the required tangent line to the curve is

#y=2x-16#