Rational functions and zeros

Write a rational function as $r(x) = p(x)/q(x)$, a polynomial over a polynomial. A fraction equals zero exactly when its numerator equals zero and its denominator does not. So the real zeros of r are the real zeros of the numerator $p$ that are in the domain of r.

The denominator's zeros are a different species entirely: they are inputs where the function is undefined. Depending on multiplicities they produce vertical asymptotes or holes (Topics 1.9 and 1.10), but they never, under any circumstances, produce zeros. Setting the denominator to zero finds where the graph breaks, not where it crosses.

There is one subtlety, and the AP exam loves it. Suppose $r(x) = \dfrac{(x-3)(x+3)}{x-3}$. The numerator is zero at $x = 3$ and $x = -3$. But at $x = 3$ the denominator is also zero, so $x = 3$ is not in the domain: the function has no value there, and something with no value cannot equal zero. The only zero is $x = -3$.

So finding zeros is a two-step discipline:

1. Solve numerator = 0 for candidates.
2. Check each candidate against the denominator. If the denominator is also zero there, strike it: that input is a hole or an asymptote, not a zero.

A struck candidate can still matter to the graph. In the example, the reduced form is $x + 3$ with a hole at $(3, 6)$: the graph sails through height 6 with a puncture, and never touches the axis there.

Solving $r(x) \le 0$ or $r(x) \ge 0$ uses both kinds of special input as fence posts, but they behave differently at the fence:

1. Numerator zeros (in the domain) can be included when the inequality allows equality: the function really is 0 there.
2. Denominator zeros are never included: the function has no value there, so no inequality about its value can hold.

Example: $\dfrac{x-1}{x+3} \le 0$. The fence posts are $x = 1$ (numerator) and $x = -3$ (denominator). Testing the three intervals gives negative values only between them, so the solution is $-3 < x \le 1$: square bracket at the numerator's post, round bracket at the denominator's, always.

From a formula in factored form, the zeros are on display: $r(x) = \dfrac{(x-3)(x+2)}{x-5}$ has zeros at $x = 3$ and $x = -2$, and a domain exclusion at $x = 5$. From a graph, the zeros are the x-intercepts, and the breaks (asymptotes, holes) are not intercepts no matter how dramatic they look.

Multiplicity still works the way it did for polynomials: a numerator factor with even multiplicity makes the graph touch the axis and turn; odd multiplicity crosses. The denominator has no vote in that behavior, only in where the function exists.