The graph of a rational function can have breaks of two kinds. Which kind appears depends on whether a factor of the denominator cancels. Far from the origin, its behavior follows the leading terms, just as for polynomials.
Key idea
Factor and cancel. A cancelled factor $(x - a)$ gives a hole at $x = a$. A factor $(x - a)$ left in the denominator gives a vertical asymptote $x = a$. For the horizontal asymptote, compare degrees: bottom bigger gives $y = 0$; equal degrees give $y = \frac{a}{b}$, the ratio of the leading coefficients; top bigger gives no horizontal asymptote.
Why it works
Near a vertical asymptote the denominator is tiny but the numerator isn't, so the quotient is huge. At a cancelled factor, top and bottom are both 0, but everywhere nearby the function equals the simplified form, which is a perfectly ordinary number: only that single point is missing. For large $|x|$, each polynomial behaves like its leading term, so $\frac{2x^2 + \dots}{x^2 + \dots}$ behaves like $\frac{2x^2}{x^2} = 2$.
Worked examples
Worked example: Holes and vertical asymptotes
Find the holes and vertical asymptotes of $f(x) = \frac{2x^2 + 2x - 4}{x^2 - 4x + 3}$.
- Factor: $\frac{2(x - 1)(x + 2)}{(x - 1)(x - 3)}$.
- $(x - 1)$ cancels: hole at $x = 1$. Its height comes from $\frac{2(x + 2)}{x - 3}$ at $x = 1$: $\frac{6}{-2} = -3$.
- $(x - 3)$ stays: vertical asymptote $x = 3$.
Answer: Hole $(1, -3)$; vertical asymptote $x = 3$
Worked example: Horizontal asymptotes
Find the horizontal asymptotes of $g(x) = \frac{3x^2 - 1}{x^2 + 5}$ and $h(x) = \frac{x + 4}{x^2 - 1}$.
- $g$: equal degrees, so $y = \frac{3}{1} = 3$.
- $h$: the denominator has the higher degree, so $y = 0$.
Answer: $y = 3$ and $y = 0$
Common mistake
Putting a vertical asymptote where a factor cancels. In the first example, $x = 1$ makes the original denominator 0, but the graph has only a missing point there, not an asymptote. Simplify first, then read asymptotes from what is left. Check: values of $f$ near $x = 1$ (like $f(1.01)$) are close to $-3$, not huge.