When the numerator's degree is exactly one more than the denominator's, a rational function has no horizontal asymptote; instead its ends follow a slanted line. Division finds that line, and it extends what you learned in Graphing Rational Functions.
Idea clave
If $\deg(\text{top}) = \deg(\text{bottom}) + 1$, divide: $f(x) = (mx + b) + \frac{r(x)}{d(x)}$ with $\deg r < \deg d$. The slant asymptote is $y = mx + b$.
Por qué funciona
The leftover fraction $\frac{r(x)}{d(x)}$ has a smaller degree on top, so it goes to 0 as $|x|$ grows, just like a fraction with a horizontal asymptote $y = 0$. What remains is the line $mx + b$, so the graph hugs that line far from the origin.
Worked examples
Ejemplo resuelto: Dividing by a linear factor
Find the slant asymptote of $f(x) = \frac{x^2 + 5x + 5}{x + 3}$.
- Synthetic division by $-3$: coefficients $1, 5, 5$ give $1, 2$ with remainder $-1$.
- $f(x) = x + 2 - \frac{1}{x + 3}$.
Respuesta: $y = x + 2$
Ejemplo resuelto: A quadratic denominator
Find the slant asymptote of $g(x) = \frac{2x^3 - x^2 + 1}{x^2 + 1}$.
- Long division: $2x^3 \div x^2 = 2x$; subtract $2x^3 + 2x$ to get $-x^2 - 2x + 1$.
- $-x^2 \div x^2 = -1$; subtract $-x^2 - 1$ to get $-2x + 2$.
- $g(x) = 2x - 1 + \frac{-2x + 2}{x^2 + 1}$, and the fraction goes to 0.
Respuesta: $y = 2x - 1$
Error común
Keeping the remainder as part of the asymptote, or using the ratio of leading coefficients (which is for equal degrees). Only the quotient $mx + b$ is the asymptote; the remainder fraction is exactly the part that fades away. Check with a big $x$: $f(100) = \frac{10505}{103} \approx 101.99$, very close to $100 + 2$.