Question

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

Answer 1

`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)`