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

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Answer 1 · 2016-07-10T08:07:39

#f(x)=x^3# and #f(2)=8#

#f'(x)=3x^2# and #f'(2) = 12#

Knowing the point slope formula given by

#y-y_1=m(x-x_1)#

we can substitute our values for #y_1#, #m#, as well as #x_1#, giving us

#y-8=12(x-2)#

Simplifying yields

#y=12x-16# or #y=4(3x-4)#