Question

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

Answer 1

`y+1=0`

Explanation:

As slope can be calculated using derivative and we can also identify coordinates of point using function, we can use point slope form to get equation of tangent.

At `x=-1`, as `f(x)=-x^2-2x-1`, `f(-1)=-(-1)^2-2(-1)-1=-2+2-1=-1`, hence we need a tangent at point `(-1,-1)`.

Further as `f(x)=-x^2-2x-1`, `f'(x)=-2x-2`, slope of tangent at `x=-1` will be `f'(-1)=-2(-1)-2=0`, which means the tangent is parallel to `x`-axis and is of type `y=k`. As tangent is desired at `(-1,-1)`, the tangent is `y=-1` or `y+1=0`.