Monday
March 27, 2017

Post a New Question

Posted by on .

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.

  • calculus - ,

    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

  • calculus - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question