For the following function, find the intervals on which the function is increasing or decreasing. \( f(x)=\frac{x+9}{x+1} \)

To determine the intervals where the function \( f(x) = \frac{x+9}{x+1} \) is increasing or decreasing, we first find the derivative \( f'(x) \). The quotient rule gives us: \[ f'(x) = \frac{(x+1)(1) - (x+9)(1)}{(x+1)^2} = \frac{x + 1 - x - 9}{(x+1)^2} = \frac{-8}{(x+1)^2} \] Now, since the derivative \( f'(x) = \frac{-8}{(x+1)^2} \) is negative for all \( x \neq -1 \) (the denominator is always positive except at \( x = -1 \)), we conclude that the function is decreasing on the intervals \( (-\infty, -1) \) and \( (-1, \infty) \), with a vertical asymptote at \( x = -1 \). This means no matter where you are (except at the asymptote), the function is always trending downwards! Also, here's a fun fact: functions like these that have asymptotes often have "behavior" that depends strongly on the way they approach these points. So, while \( f(x) \) is decreasing everywhere it's defined, it does strange things at the asymptote, offering a peek into calculus concepts like limits and continuity!