Rational functions and vertical asymptotes

Near an input $a$ where the denominator heads to zero and the numerator does not, the fraction's value is (something near a nonzero number) divided by (something shrinking to 0), and that quotient grows without bound in magnitude. That is a vertical asymptote: the line $x = a$ that the graph hugs while the outputs run off to $+\infty$ or $-\infty$.

Two things are true at once and students often keep only one: the input $a$ is not in the domain (the graph never touches the line), and the outputs near $a$ are unbounded (the graph goes vertical trying). An asymptote is a claim about both.

When numerator and denominator share the zero $a$, cancellation competes with blow-up, and multiplicity referees. Suppose $(x-a)$ appears $m$ times in the numerator and $n$ times in the denominator:

1. n > m (denominator wins): after cancelling, at least one $(x-a)$ survives downstairs. Vertical asymptote at $x = a$.
2. m ≥ n (numerator holds or wins): every downstairs copy cancels. Hole at $x = a$, at the height of the reduced form.

So $\dfrac{x-1}{(x-1)^2}$ still has an asymptote at $x = 1$ (one copy survives below), while $\dfrac{(x-1)^2}{x-1}$ has a hole at $(1, 0)$. Same factors, opposite outcomes, decided entirely by the exponent count. "It cancels, so nothing happens there" is the trap; cancelling changes the kind of break, never erases the break.

The AP course writes unbounded behavior in one-sided limit notation. For $f(x) = \dfrac{1}{x-2}$:

1. As $x \to 2^{+}$ (from the right), $x - 2$ is a tiny positive number, so $f(x) \to +\infty$.
2. As $x \to 2^{-}$ (from the left), $x - 2$ is a tiny negative number, so $f(x) \to -\infty$.

Read $\lim_{x \to 2^{+}} f(x) = +\infty$ as a sentence about outputs: "as inputs approach 2 from the right, outputs increase without bound." It does not say $f(2)$ is infinity; $f(2)$ does not exist. And nothing forces the two sides to match: each side's sign is its own little sign analysis.

Given a factored rational function, take inventory in one pass: numerator-only zeros are zeros, denominator-only zeros are asymptotes, and shared zeros go to the multiplicity test. For example, $$f(x) = \dfrac{(x+4)(x-1)}{(x+4)^2 (x+7)}$$ has a zero at $x = 1$, an asymptote at $x = -7$, and at the shared $x = -4$ the denominator's power (2) beats the numerator's (1): asymptote there too.

The payoff for the discipline: you can sketch the skeleton of any rational graph, breaks and crossings in their right places, before plotting a single point.