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
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.
2 answers
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
0=(5x^4+10x^3+x^2+4x+4)/(x^2+x+2)^2
2 answers