Question

Which statement is true about the discontinuities of the function f(x)?

f (x) = StartFraction x squared minus 4 Over x cubed minus x squared minus 2 x EndFraction
There is a hole at x = 2.
There are asymptotes at x = 0 and x = –1.
There are asymptotes at x = 0 and x = –1 and a hole at (2, two-thirds).
There are holes at x = 0 and x = –1 and an asymptote at x = 2.

Answer 1

Factor: f(x) = (x^2 − 4)/(x^3 − x^2 − 2x) = (x−2)(x+2)/[x(x−2)(x+1)]. Cancel (x−2) (removable):

f(x) = (x+2)/[x(x+1)] for x ≠ 2.

Thus x = 2 is a removable discontinuity (hole). Evaluating the simplified form at x = 2 gives y = 4/(2·3) = 2/3, so hole at (2, 2/3). x = 0 and x = −1 remain zeros of the denominator with nonzero numerator, so they are vertical asymptotes.

So the true statement is: “There are asymptotes at x = 0 and x = −1 and a hole at (2, two-thirds).”