Question

f(x)=8/(x^2+x+2)
Set up an equation, which when solved will give the points where the tangent line has y-intercept 5.

Answer 1

f' = a = -8(2x+1)/(x^2+x+2)^2


y = a x + 5

8/(x^2+x+2) = -8(2x^2+x)/ (x^2+x+2)^2 + 5
etc

Answer 2

Damon, so what you did was apply the quotient rule to the ax part? Wouldn't the -8 be -8x then? And if I'm doing it right, would my final equation be this?
0=(5x^4+10x^3+x^2+4x+4)/(x^2+x+2)^2