8/(x^2 + x +2) is my function and -((8 (1 + 2 x))/(2 + x + x^2)^2) is my calculated derivative according to mathematica. Now I need to find an equation which when solved will give the point(s) where the tangent line has y intercept of 5 (without mathematica but my next question will involve mathematica...).

Question

anonymous · Answer

A tangent line with a y-intercept of 5 will have the form
$$y=f'(x_0)x+5,\text{ where $x_0$ is the point of tangency.}$$

anonymous · Answer

So if
$$f(x) = \frac{8}{x^2+x+2}\text{ and}\
f'(x)=-\frac{8(2x + 1)}{(x^2+x+2)^2},$$
it looks like you're left with finding the points of intersection between f(x) and the tangent line y.

And a small correction on my first post: x_0 is **the x-coordinate of** the point of tangency.

anonymous · Answer

Alright so how should I go about this? plug in the derivative and y to solve for x?

anonymous · Answer

I'd say set f '(x) = y, and solve for x, yeah.

anonymous · Answer

The solution you get for x should be the x-coordinate of the intersection. If you get multiple solutions, make sure to check for tangency. 

For example, if you happen to get x = 1 and x = 4 as solutions, check which one is actually a point of tangency as opposed to a regular point of intersection.

anonymous · Answer

Alright, thank you Sith! The rest I can attempt on mathematica.