2.13 Rational Functions – Partial Fractions (HL)
1. Characteristics of Rational Functions
A rational function adheres to the structural form $f(x) = \frac{p(x)}{q(x)}$, where both $p(x)$ and $q(x)$ operate as independent polynomials.
The generalized asymptotic boundaries of a rational function behave according to strict polynomial rules:
- Vertical Asymptotes (VA): Exist precisely at points where the function is algebraically undefined, located by solving the denominator $q(x) = 0$.
Example: For $f(x) = \frac{x^2 + 4x - 5}{x^2 - 4x + 3}$, solving $x^2 - 4x + 3 = 0$ yields roots $x=1$ and $x=3$. Thus, the lines $x=1$ and $x=3$ constitute vertical asymptotes. - Horizontal Asymptotes (HA): Evaluated strictly by assessing the leading coefficients and degrees of $p(x)$ and $q(x)$.
- If $\deg p(x) = \deg q(x)$, then $y = \frac{\text{leading coefficient of } p(x)}{\text{leading coefficient of } q(x)}$.
- If $\deg p(x) < \deg q(x)$, then $y = 0$.
- If $\deg p(x) > \deg q(x)$, there is absolutely no horizontal asymptote.
EXAMPLE 1 (Determining Asymptotes)
| Function | Vertical Asymptote | Horizontal Asymptote |
|---|---|---|
| $f(x) = \frac{7x^2 + 5x - 3}{x^2 - 3x + 2}$ | $x=1, x=2$ | $y = 7$ |
| $f(x) = \frac{7x^2 + 5x - 2}{2x^2 - 6x + 4}$ | $x=1, x=2$ | $y = 7/2$ |
| $f(x) = \frac{5x - 2}{x^2 - 3x + 2}$ | $x=1, x=2$ | $y = 0$ |
| $f(x) = \frac{5x - 2}{x^2 - 3x + 2} + 8$ | $x=1, x=2$ | $y = 8$ |
| $f(x) = \frac{7x^3 + 5x - 3}{x^2 - 3x + 2}$ | $x=1, x=2$ | None |
EXAMPLE 2 & 3 (Intercepts and Range)
Factored denominator identifies VA at $x=1, x=3$. HA rests at $y=1$. The analytical range spans $y \in \mathbb{R}$, demonstrating that a function graph may safely cross its own horizontal asymptote within localized domains.
Factored components: $f(x) = \frac{(x-4)(x+1)}{(x-1)(x-3)}$.
Roots evaluate to $x=-1, x=4$. y-intercept maps to $y=-4/3$.
Graphing evaluation identifies localized turning points defining an absolute void in the central range. The functional range evaluates exactly to $y \in ]-\infty, 1.05] \cup [5.95, +\infty[$.
2. Oblique Asymptotes
When analyzing a rational function where the numerator polynomial degree is exactly one greater than the denominator ($\deg p(x) = \deg q(x) + 1$), an oblique (slanted) asymptote inherently emerges.
Polynomial long division splits the function into a linear quotient and a remainder fraction:
Because the remainder fraction trends asymptotically to zero as $x \rightarrow \pm\infty$, the graph aggressively approaches the foundational linear equation $y = Ax + B$, formulating the oblique asymptote.
EXAMPLE 4
Determine the complete asymptotic structure of $f(x) = \frac{4x^2 - 12x + 1}{2x - 6}$.
3. Partial Fractions (Linear Denominator Roots)
A complex rational function with a factorable quadratic denominator can be algebraically dismantled into isolated linear fractional components. If the denominator evaluates roots $r_1$ and $r_2$:
EXAMPLE 5
Decompose $f(x) = \frac{3x - 5}{x^2 - 4x + 3}$ entirely into its partial fractions.
Solution: Factoring the denominator yields $(x-1)(x-3)$. Construct the partial formulation:
Combine the fractions using a common denominator: $\frac{A(x-3) + B(x-1)}{(x-1)(x-3)} = \frac{Ax - 3A + Bx - B}{(x-1)(x-3)}$.
Group coefficients: $(A+B)x - (3A+B) = 3x - 5$.
Extract the linear system: $\begin{cases} A + B = 3 \\ 3A + B = 5 \end{cases}$.
Solving evaluates parameters $A=1$ and $B=2$.
Multiply uniformly by $(x-1)(x-3)$ to isolate the numerator:
$3x - 5 = A(x-3) + B(x-1)$.
Substitute $x=3$ intentionally to eliminate $A$: $3(3) - 5 = B(3-1) \Rightarrow 4 = 2B \Rightarrow \mathbf{B=2}$.
Substitute $x=1$ intentionally to eliminate $B$: $3(1) - 5 = A(1-3) \Rightarrow -2 = -2A \Rightarrow \mathbf{A=1}$.