Question

find the equation of the tangent line to f(x)=x^3 where x=2

Answer 1

ok, we'll need the derivate of the function

f'(x)=3x^2

now we can calculate the slope at x=2
f'(2)=3*2^2=12

now we need the point at x=2
f(x)=2^3=8

now a line looks like this: y = ax + b
we have a (the slope) =12
we have x = 2
we have y =8

now we substitue and find b

8 = 12*2 +b
b = -16

so the equation of the line is

y = 12x - 16