Question

What is the equation of the line tangent to `f(x)=-x^2 -2x - 1` at `x=-1`?

Answer 1

The equation of the tangent line to `f` at the given point is `y=0` (the tangent line is the `x`-axis).

Explanation:

If `f(x)=-x^2-2x-1`, then `f'(x)=-2x-2` so that `f(-1)=-1+2-1=0` and `f'(-1)=2-2=0`.

In general, the equation of the tangent line to `f` at the point `(a,f(a))` is `y=f(a)+f'(a)(x-a)`. In the present case, this becomes

`y=f(-1)+f'(-1)(x+1)=0+0(x+1)=0`.