Integrated Math 3 / Unit 2: Rational Expressions and Functions
2.4 Asymptotes and Holes
Notes
For $\displaystyle f(x) = \frac{p(x)}{q(x)}$ in lowest terms:
- Vertical asymptotes are at zeros of the denominator that remain after canceling.
- A factor that cancels leaves a hole (removable discontinuity) at that $x$-value.
- Horizontal asymptote: if degree of $p$ < degree of $q$, it is $y = 0$. If the degrees are equal, it is $\displaystyle y = \frac{\text{leading coeff. of } p}{\text{leading coeff. of } q}$. If degree of $p$ is bigger, there is no horizontal asymptote (one bigger gives a slant asymptote).
- Example: $\displaystyle f(x) = \frac{x^2 - 4}{x - 2} = x + 2$ for $x \neq 2$, so it is a line with a hole at $(2, 4)$.
Practice
What are the asymptotes of $\displaystyle f(x) = \frac{x + 1}{x - 3}$?
The denominator is 0 at $x = 3$. Equal degrees give $\displaystyle y = \frac{1}{1} = 1$.
What is the horizontal asymptote of $\displaystyle f(x) = \frac{2x^2}{x^2 + 1}$?
The degrees are equal, so $\displaystyle y = \frac{2}{1} = 2$.
Where is the hole in the graph of $\displaystyle f(x) = \frac{x - 1}{x^2 - 1}$?
$\displaystyle f(x) = \frac{1}{x + 1}$ for $x \neq 1$. The cancelled factor $x - 1$ leaves a hole at $x = 1$, where $\displaystyle \frac{1}{1 + 1} = \frac{1}{2}$. The asymptote is at $x = -1$.
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