Question

What is the equation of the line tangent to `f(x)=-(x-3)^2-5x-3` at `x=5`?

Answer 1

Equation of tangent is `9x+y-13=0`

Explanation:

The line tangent to `f(x)` at `x=x_0` has the slope equal to `f'(x_0)` and as such the equation of tangent at `x=x_0` is `y-f(x_0)=f'(x_0))(x-x_0)`.

Here `f(x)=-(x-3)^2-5x-3` and hence we are seeking a tangent at `(5.f(5))` i.e. `(5,-32)`

As `f'(x)=-2(x-3)-5`, `f'(5)=-2xx2-5=-9`

and equation of tangent is `y-(-32)=-9(x-5)` i.e.

`9x+y-13=0`

graph{(9x+y-13)(y+(x-3)^2+5x+3)=0 [-10, 10.8, -60, 20]}