Now `y = 3x − x^3`

`(dy)/(dx)=3-3x^2` and the value of this derivative at `x=2` is given by:

`(dy)/(dx)=3-3(2)^2=-9`

Since `y = 3x − x^3`,* *then when `x= 2`, `y=
-2.`

So we need the equation of the line passing through `(2,-2)` with slope `-9`.

Using the general equation of the line `y-y_1=m(x-x_1)`, we have:

`y + 2 = -9(x - 2)`

So the required equation is:

`y = -9x + 16`

Or, in general form: `9x + y - 16 = 0.`