A rational function is $R(x) = P(x)/Q(x)$ where $P, Q$ are polynomials, $Q \not\equiv 0$.
Domain: ${x \in \mathbb{R} : Q(x) \neq 0}$.
Range: The set of all $y$ for which $P(x)/Q(x) = y$ has at least one solution. Often found by solving $Q(x)y = P(x)$ for $x$ and finding values of $y$ with solutions.
Example: Find the range of $f(x) = \dfrac{x}{x^2+1}$.
Set $y = \dfrac{x}{x^2+1}$, so $yx^2 - x + y = 0$.
For real $x$ (and $y \neq 0$): discriminant $\geq 0$: \$1 - 4y^2 \geq 0 \Rightarrow -\dfrac{1}{2} \leq y \leq \dfrac{1}{2}$.
When $y=0$: $x=0$ is a solution. So range $= \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$.
Vertical asymptotes (VAs): at zeros of $Q(x)$ that are not also zeros of $P(x)$ (no cancellation).
Horizontal asymptote (HA): determined by degree comparison (see table in previous note).
Oblique asymptote: divide $P$ by $Q$ when $\deg P = \deg Q + 1$; the polynomial quotient is the asymptote.
No asymptote: when $\deg P \geq \deg Q + 2$, the function grows without bound (may be parabolic etc.).
If $(x - a)$ is a common factor of $P$ and $Q$, there is a hole (removable discontinuity) at $x = a$, not a vertical asymptote.
Example: $f(x) = \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$.
Graph is the line $y = x+1$ with a hole at $(1, 2)$. Domain: $\mathbb{R} \setminus {1}$.
Divide numerator and denominator by $x^m$ (highest power in denominator):
$$\frac{2x^3 - x + 3}{x^3 + 4x} = \frac{2 - x^{-2} + 3x^{-3}}{1 + 4x^{-2}} \to \frac{2}{1} = 2 \quad \text{as } x \to \pm\infty$$
So HA is $y = 2$. Determine whether $f$ approaches from above or below:
For large positive $x$: numerator $\approx 2x^3 - x$, denominator $\approx x^3 + 4x$.
Ratio $\approx 2 - 9x^{-2}$, so $f(x) < 2$ for large $x$ (approaches from below).
| $\deg P$ vs $\deg Q$ | As $x \to \pm\infty$ |
|---|---|
| $\deg P < \deg Q$ | $f(x) \to 0$ |
| $\deg P = \deg Q$ | $f(x) \to a_n/b_m$ |
| $\deg P = \deg Q + 1$ | $f(x) \to$ oblique asymptote |
| $\deg P > \deg Q + 1$ | $f(x) \to \pm\infty$ |
Functions like $f(x) = \dfrac{\sqrt{x}}{x+1}$ or $f(x) = \dfrac{\sin x}{x}$ are not rational but share similar analysis techniques.
Domain restrictions come from both the denominator and any additional restrictions (e.g., $x \geq 0$ for $\sqrt{x}$).
KEY TAKEAWAY: Domain excludes zeros of the denominator; range requires solving for $x$ and checking the discriminant. Asymptote type depends on the degree relationship.
EXAM TIP: Always check for common factors before declaring a vertical asymptote — it may be a hole instead.
APPLICATION: Rational functions model situations with competing rates: population per area, concentration after dilution, cost per unit as production scales.
