where $P(x)$ and $Q(x)$ are polynomials, $Q(x) \neq 0$.

Domain

Exclude values where the denominator equals zero.

$f(x) = \frac{x+1}{x^2 - 4} = \frac{x+1}{(x-2)(x+2)}$

Domain: $x \neq 2$ and $x \neq -2$

Vertical Asymptotes

Occur where denominator $= 0$ (after canceling common factors).

$f(x) = \frac{x+1}{(x-2)(x+2)}$

Vertical asymptotes: $x = 2$ and $x = -2$

Horizontal Asymptotes

Compare degrees of numerator ($n$) and denominator ($m$):

Holes

A hole occurs when a factor cancels from both numerator and denominator.

$f(x) = \frac{(x-3)(x+1)}{(x-3)(x+2)}$

Hole at $x = 3$. To find the $y$-coordinate: $f(3) = \frac{3+1}{3+2} = \frac{4}{5}$

Hole at $\left(3, \frac{4}{5}\right)$

Oblique (Slant) Asymptotes

When degree of numerator = degree of denominator + 1, divide:

$f(x) = \frac{x^2 + 2x + 1}{x - 1}$

Long division gives: $y = x + 3$ (oblique asymptote)

Graphing Steps

1. Factor numerator and denominator
2. Find holes (cancel common factors)
3. Find x-intercepts (numerator = 0)
4. Find y-intercept ($f(0)$)
5. Find vertical asymptotes (remaining denominator = 0)
6. Find horizontal/oblique asymptote
7. Plot additional points as needed

Important: A graph can cross a horizontal asymptote in the middle but approaches it as $x \to \pm\infty$. It can NEVER cross a vertical asymptote.