Final answer:
The function f(x) = 1/(x^2 - 2x - 8) has a domain of all real numbers except x = 4 and x = -2, and a range of all real numbers except y = 0. There are no x-intercepts and the y-intercept is (0, -1/8). The function has vertical asymptotes at x = 4 and x = -2 and no horizontal asymptote, and it decreases between the asymptotes while increasing outside them.
Step-by-step explanation:
Key Features of the Function f(x) = 1/(x^2 - 2x - 8)
The function f(x) = 1/(x^2 - 2x - 8) has several key features that we can analyze, such as its domain, range, intercepts, asymptotes, and intervals of increase or decrease.
Domain and Range
The domain of f(x) is all real numbers except for x-values that make the denominator zero. After factoring, we find x^2 - 2x - 8 = (x - 4)(x + 2); hence, the domain is all real numbers except x = 4 and x = -2. The range is all real numbers y such that y ≠ 0, since the function can never take the value of zero.
Intercepts
There are no x-intercepts because the function will never equal zero. The y-intercept is found by plugging in x = 0, which gives us the point (0, -1/8).
Equations of Asymptotes
The equations of the vertical asymptotes are x = 4 and x = -2. There is no horizontal asymptote, but there is an oblique asymptote since the degree of the denominator is two and the degree of the numerator is zero.
Intervals of Increase and Decrease
The function f(x) increases and decreases based on the sign of its derivative, but without the explicit derivative provided, we generally look for changes in the function's behavior on a graph. The function will decrease between the vertical asymptotes and increase outside them due to the nature of rational functions.