Question

Find the vertical asymptote(s) of f(x) = (2x^2 + 3x + 6) / (x^2 - 1).

A) No vertical asymptote

B) Vertical asymptote at x = 1

C) Vertical asymptote at x = -1

D) Vertical asymptote at x = 2

Answer 1

The function f(x) = (2x^2 + 3x + 6) / (x^2 - 1) has vertical asymptotes at x = 1 and x = -1, where the denominator equals zero.

Step-by-step explanation:

To find the vertical asymptote(s) of the function f(x) = (2x^2 + 3x + 6) / (x^2 - 1), we need to determine where the denominator is equal to zero, as this is typically where vertical asymptotes occur. Factoring the denominator, we get x^2 - 1 = (x - 1)(x + 1). Thus, the denominator is zero when x = 1 and when x = -1, which implies that the function has vertical asymptotes at these points.

Therefore, the correct answer is that the function has vertical asymptotes at x = 1 and x = -1.