Question

What is the equation of the line tangent to `f(x)=x^2 -x-9` at `x=-1`?

Answer 1

You have to use geometric derivative concept. See below

Explanation:

We know that the derivative `f´(x_0)` is the value of the slope of tangent line to `f(x)` in `x=x_o`

We have to do the derivative of `f(x)`

`f´(x)=2x-1` this function has the value in `x_0=-1`

`f´(-1)=2·(-1)-1=-3`

So the slope of tangent line to `f(x)` in `x_0=-1` is `-3`

Then the equation of tangent line has the form `y=-3x+b` where `b` is the intercept with `y` axis. Lets determine `b`

However `f(x)` and line are tangents in `x_0=-1` the value in both equations must be the same in `x_0=-1`.

`f(-1)=1+1-9=-7`
`y_(-1)=-3·(-1)+b=-7`. From here we can obtain `b` which is `b=-4`. Then the equation of tangent line to `f(x)` in `x_0=-1` is `y=-3x-4`