Maths: Why Rational Functions Hide Holes and Asymptotes

Rational functions can hide surprising behaviour at the values their denominators forbid. This post explores discontinuities in R(x) = (x² + 4x − 5)/(x³ − 6x² + 11x − 6), where factorising both numerator and denominator reveals what happens at each excluded point. The domain is set by the denominator's roots, found by factoring x³ − 6x² + 11x − 6 into (x − 1)(x − 2)(x − 3), so x cannot equal 1, 2, or 3. The numerator factors as (x + 5)(x − 1), sharing the factor (x − 1) with the denominator. Cancelling it simplifies R(x) to (x + 5)/((x − 2)(x − 3)), though the restriction x ≠ 1 must remain. This distinction matters: a cancelled factor leaves a removable discontinuity, or hole, while factors surviving in the denominator drive vertical asymptotes, since the denominator approaches zero while the numerator stays nonzero. Evaluating R at a convenient point confirms the simplified form behaves consistently away from the excluded values.