Question

What is the equation of the line tangent to `f(x)=(-x^2-1)/(x+4)` at `x=5`?

Answer 1

`y+26/9=-64/81(x-5)`

Explanation:

the slope of the tangent line at `x=5` is the derivative of the function evaluated at `x=5`
use quotient rule or `d/dx(f(x)/g(x))=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)` to find the derivative:

`d/dx((-x^2-1)/(x+4))`
`=(-2x*(x+4)-1*(-x^2-1))/((x+4)^2)`

when `x=5`:

`=(-2(5)*(5+4)-1*(-5^2-1))/((5+4)^2)`

`=(-90+26)/(81)`
`=-64/81`

find the point `(5,f(5))` to complete the equation of the tangent line:
`f(5)=(-5^2-1)/(5+4)`
`f(5)=(-26)/(9)`

use point slope form to find the equation of the line:
`y-(-26/9)=-64/81(x-5)`
`y+26/9=-64/81(x-5)`