Question

What is the equation of the line tangent to `f(x)=x^3-6x` at `x=2`?

Answer 1

The equation of the tangent at `f(2)` is `y=6x-16`

Explanation:

`f(x) = x^3-6x`
`f'(x) = 3x^2 - 6`

The equation of any tangent to `f(x)` in slope and intercept form is:
`y = mx+c`
Where: `m` = slope of `f(x)` and `c` = the intercept on the y axis

Slope of the tangent to `f(x)` at `x=2` is:
`f'(2)= 3*2^2-6`
`=6 -> m=6`

Since the tangent touches `f(x)` at `f(2)`
`f(2) = m*2 +c`

Replacing for `f(2)` and `m=6`:
`2^3-6*2 = 6*2 +c`

`c = 8-12-12`
`c=-16`

Hence the equation of the tangent at `x=2` is:

`y = 6x-16`