At which values of x does the graph of f of x is equal to the fraction with numerator x squared minus x minus 2 and denominator open paren x squared minus 1 close paren times open paren x squared minus 16 close paren have a vertical asymptote? Choose three correct answers.

Factor the numerator and denominator: x² − x − 2 = (x − 2)(x + 1), and (x² − 1)(x² − 16) = (x − 1)(x + 1)(x − 4)(x + 4). The factor (x + 1) cancels, so x = −1 is a hole, not a vertical asymptote. The remaining denominator is zero at x = 1, 4, and −4, giving choices B, D, and C.

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