Final answer:
The functions f(x) = 1/x, f(x) = 1/(x+2), and f(x) = 1/(x-1) all have vertical asymptotes where their denominators are zero, and f(x) = x/(x+1) has both a vertical and a horizontal asymptote. Option C, f(x) = ⅛, has both a vertical asymptote at x = -1 and a horizontal asymptote at y = 1 since as x approaches infinity, the value of f(x) approaches 1.
Step-by-step explanation:
The functions f(x) = 1/x, f(x) = 1/(x+2), and f(x) = 1/(x-1) all have vertical asymptotes where their denominators are zero, and f(x) = x/(x+1) has both a vertical and a horizontal asymptote.
To create a function with asymptotes, you need to look for places where the function is undefined or where it approaches infinity. For example, the function f(x) = ⅛ would have a vertical asymptote at x = 0 because as x approaches 0, the function approaches infinity, which is one of the characteristics of an asymptote.
Options B, f(x) = ⅛, and D, f(x) = ⅛, also have vertical asymptotes at x = -2 and x = 1, respectively. The difference in the location of the asymptotes is due to the shift in the function's denominator. Option C, f(x) = ⅛, has both a vertical asymptote at x = -1 and a horizontal asymptote at y = 1 since as x approaches infinity, the value of f(x) approaches 1.