Critical points/values/numbers occur for a function whenever the derivative = 0 or is undefined. Finding intervals on which the function is increasing or decreasing likewise involves finding the derivative and determining when it is positive or negative.
Because this is an elementary rational function, its graph is easy to visualize, and our analysis should lead us to conclude this function decreases everywhere (everywhere it is defined, that is), and has no critical points:
We differentiate using quotient rule: f'(x) = [1(x+2) - 1(x+8)]/(x+2)2 = -6/(x+2)2
Because a fraction only = 0 when its numerator = 0, f' has no zeros. It is undefined when x = -2, since that makes the denom = 0, but since f(-2) is also undefined, we do not consider that a critical point of f (it is not a point on f, it is where f has a vertical asymptote).
Since the expression (x+2)2 is positive when x≠-2, f' is negative everywhere, which means f decreases on (-∞ , -2)∪(-2,∞).