Question

For `f(x)=x-x^3-x^2` what is the equation of the tangent line at `x=12`?

Answer 1

`f(12)=12-1728-144=-1860`
So the point of tangency is `(x_0,y_0)=(12,-1860)`.

Since `f'(x)=1-3x^2-2x`, the slope of the tangent line at `x=12` is `m=f'(12)=1-432-24=-455`.

So the equation of the tangent line at the point `(12,-1860)` is

`y-y_0 = m(x-x_0)`

`y+1860=-455(x-12)`

`y+1860=-455x+5460`

`y=-455x+3600`