Question

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

f (x) = StartFraction x + 1 Over 6 x squared minus 7 x minus 3 EndFraction
There are asymptotes at x = three-halves and x = negative one-third.
There are holes at x = three-halves and x = negative one-third.
There are asymptotes at x = negative three-halves and x = one-third.
There are holes at x = negative three-halves and x = one-third.

Answer 1

Factor the denominator: 6x^2 − 7x − 3 = (2x − 3)(3x + 1), so the denominator is zero at x = 3/2 and x = −1/3. The numerator x + 1 is nonzero at those points, so there are no cancellations (no holes). Therefore the function has vertical asymptotes at x = 3/2 and x = −1/3.