Question

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

Answer 1

The derivative gives us the general (that is, in terms of `x`) slope of the line
Given `f(x) =x^3`

`(df(x))/(dx) = 3x^2`

At `x=2`
- the slope is `m=3(2^2) = 12`
- and `f(2) = 2^3 = 8 rarr (x,y) = (2,8)`

We can use the slope-point form of the line to write:
`y-8 = 12(x-2)`
or, in standard form:
`12x -y = 16`