Question

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

Answer 1

The derivative does just that!
\[f(x) = 8 * (x ^{2}+ x +2)^{-1}\]
\[f'(x) = -8 * (x ^{2}+ x +2)^{-2} * 2x\]

Answer 2

Ok. im confused...i thought we would be using y=mx+b form. I dont know where i got this but i thought you would take f(x)=f'(x)+5, then move f(x) over to have it all equal to 0?? So confused!