If a rational function has a canceled factor producing a hole, how does this affect vertical asymptotes?

If a rational function has a canceled factor producing a hole, how does this affect vertical asymptotes?

• A. It creates a vertical asymptote where the hole was
• B. It has no impact on asymptotes
• C. It may remove a potential vertical asymptote by canceling a factor; vertical asymptotes are determined by the reduced denominator zeros.
• D. It shifts vertical asymptotes to the left

Answer: (C)

When a common factor cancels between the numerator and denominator, you create a hole (a removable discontinuity) at that x-value. Vertical asymptotes, on the other hand, come from where the reduced denominator equals zero (provided the numerator isn’t zero there). If the canceled factor was part of what would have forced a zero in the original denominator, that zero disappears in the reduced form, so the associated vertical asymptote can disappear as well. In the simplified expression, vertical asymptotes occur only at the zeros of the reduced denominator. For example, (x−1)/[(x−1)(x+2)] simplifies to 1/(x+2). There’s a hole at x=1, and the vertical asymptote is at x=−2.

When a common factor cancels between the numerator and denominator, you create a hole (a removable discontinuity) at that x-value. Vertical asymptotes, on the other hand, come from where the reduced denominator equals zero (provided the numerator isn’t zero there). If the canceled factor was part of what would have forced a zero in the original denominator, that zero disappears in the reduced form, so the associated vertical asymptote can disappear as well. In the simplified expression, vertical asymptotes occur only at the zeros of the reduced denominator.

For example, (x−1)/[(x−1)(x+2)] simplifies to 1/(x+2). There’s a hole at x=1, and the vertical asymptote is at x=−2.