Quick questions on The sign of a function and sign tables

Polynomials are the easy case because they have no discontinuities at all, so the test values only have to dodge around the zeroes. Take $y = (x - 1)(x - 3)(x - 5)$, with zeroes at $x = 1, 3, 5$. Testing $x = 0, 2, 4, 6$ gives

When the function has a denominator, the test values must dodge around the discontinuities (zeroes of the denominator) as well as the zeroes. The method is otherwise identical. Take $y = \dfrac{x - 1}{x - 4}$: a zero at $x = 1$ (numerator zero) and a discontinuity at $x = 4$ (denominator zero). Testing $x = 0, 2, 5$ gives $\tfrac{1}{4}, -\tfrac{1}{2}, 4$, so the sign row is $+, -, +$.

For an inequation with the unknown in the denominator you now have two complete methods, and the HSC examines both. The companion page develops the first; this page develops the second.