Question

How do you find the equation of the tangent line to the graph of `f(x)=x^3` at given point (2,8)?

Answer 1

In the point slope form it would be y-8=12(x-2)

Explanation:

To find the equation of the tangent line, its slope is needed first. This can be had by determining `(df)/dx` to be = `3x^2`. At (2,8) this would be = `3(2)^2`= 12.

Now equation can be written in point slope form as y-8=12(x-2)