Question

What is the equation of the line tangent to `f(x)= -1/(5x-x^2)` at `x=-3`?

Answer 1

Tangent Line
`y=11/576 x +19/192`

Explanation:

`f(x) = -1/(5x-x^2`

determine the point of tangency

`f(-3)=-1/(5(-3)-(-3)^2`

`f(-3)=1/24`

Determine the slope `m`

`m=f'(x)=(5-2x)/(5x-x^2)^2`

`m=(5-2(-3))/(5(-3)-(-3)^2)^2`

`m=11/576`

Use point-slope form
`y-y_1=m(x-x_1)`

`y-1/24=11/576(x-(-3))`

`576y-24=11x+33`

`y=11/576 x+57/576`

`y=11/576 x + 19/192`