Rational functions and holes

Take $f(x) = \dfrac{x^2 - 4}{x - 2}$. Factor: $\dfrac{(x-2)(x+2)}{x-2}$. At $x = 2$ both floors are zero, and the multiplicity count is 1 upstairs, 1 downstairs: the numerator's supply matches, every downstairs copy cancels, and near $x = 2$ the function behaves exactly like the reduced form $x + 2$.

So the graph of $f$ is the line $y = x + 2$, except at the one input $x = 2$, where the original formula demands $0/0$ and refuses to answer. That missing point is the hole. The rule: a shared zero $a$ produces a hole when its multiplicity in the numerator is greater than or equal to its multiplicity in the denominator; if the denominator has more copies, the survivor downstairs makes a vertical asymptote instead (Topic 1.9).

A hole is a point, and points have two coordinates. The x-coordinate is the cancelled input; the y-coordinate is the reduced form evaluated there. For $f(x) = \dfrac{x^2-4}{x-2}$: reduced form $x + 2$, evaluated at 2, gives 4. The hole is at $(2, 4)$.

This is the course's first taste of a limit: the outputs approach 4 as inputs approach 2, even though no input actually produces 4 there. In notation, $\lim_{x \to 2} f(x) = 4$ while $f(2)$ does not exist. Try inputs marching in: $f(1.9) = 3.9$, $f(1.99) = 3.99$, $f(2.01) = 4.01$. The trend is unmistakable; the value is absent.

1. Factor numerator and denominator completely.
2. Cancel shared factors; note each fully cancelled input $a$.
3. Evaluate the reduced form at $a$: the hole is at $(a, \text{that value})$.

Both live at inputs excluded from the domain, and that is where the resemblance ends. Near a hole, outputs stay bounded and settle toward the hole's height: the graph could be repaired with a single dot. Near a vertical asymptote, outputs grow without bound: no dot can fix it.

Watch the two side by side: $\dfrac{(x-3)(x+5)}{(x-3)(x-1)}$ has a shared zero at 3 and a denominator-only zero at 1. The shared factor cancels fully, so $x = 3$ is a hole, at height $\dfrac{3+5}{3-1} = 4$; the unshared $x = 1$ is an asymptote. One function, both breaks, and the factor bookkeeping tells them apart before any graphing.

Cancelling a shared factor produces a simpler formula for the same function on the same domain. It does not enlarge the domain: $\dfrac{x^2-4}{x-2}$ and the bare expression $x + 2$ are different functions, because one is undefined at 2 and the other is not. When you reduce, carry the restriction: $f(x) = x + 2$, $x \ne 2$.

And cancelling only works on factors: whole multiplied pieces of top and bottom. Slashing an $x^2$ against an $x$, or a 9 against a 3, term by term, invents a new function unrelated to the old one. If the top and bottom share no factor, there is no hole to find, however tempting the digits look.