Reciprocal and rational functions

IB Mathematics AA SL· Topic 2: Functions· 35 min read

1. Reciprocal Functions: Key Features★★☆☆☆⏱ 15 min

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📘 Definition

Reciprocal Function

$f (x) = \frac{k}{x - h} + v, k \neq = 0$

A transformed reciprocal function, formed by translating and stretching the basic reciprocal function $y = \frac{1}{x}$ . The denominator cannot equal zero, so there is a restriction on the domain.

Example:

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

For any reciprocal function in the standard form above: the vertical asymptote (where the function is undefined) occurs at $x = h$ , and the horizontal asymptote, approached as $x$ grows very large in magnitude, occurs at $y = v$ .

📐 Worked Example

For $f (x) = \frac{4}{x + 2} - 1$ , find: (i) domain, (ii) asymptote equations, (iii) range.

1
Find the excluded domain value by setting the denominator to zero:
2
$x+2 = 0 \implies x = -2$
3
Domain is all real numbers except $x = - 2$ , written as: $x \in R, x \neq = - 2$
4
Match to standard form $\frac{k}{x - h} + v$ : here $h = - 2$ , $v = - 1$ . So:
5
Vertical asymptote: $x = - 2$ , Horizontal asymptote: $y = - 1$
6
The function can never equal the horizontal asymptote value, so range is: $y \in R, y \neq = - 1$

Exam tip:

Always write asymptotes as full equations, not just numbers. Examiners will not award marks for just writing $- 2$ instead of $x = - 2$ .

2. Linear-over-Linear Rational Functions★★★☆☆⏱ 20 min

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📘 Definition

Simple Rational Function

$f (x) = \frac{a x + b}{c x + d}, c \neq = 0$

A rational function where both numerator and denominator are linear polynomials, the most common form tested in IB AA SL exams.

To find asymptotes for this form:

Vertical asymptote: Set denominator equal to zero, solve for $x$ .
Horizontal asymptote: Divide numerator and denominator by $x$ , the limit as $x \to \pm \infty$ is $\frac{a}{c}$ , so the asymptote is $y = \frac{a}{c}$ .

📐 Worked Example

For $f (x) = \frac{3 x + 6}{x - 2}$ , find all asymptotes and the x and y intercepts.

1
Find vertical asymptote: set denominator to zero:
2
$x - 2 = 0 \implies x = 2$
3
Vertical asymptote: $x = 2$
4
Find horizontal asymptote: leading coefficient of numerator is 3, leading coefficient of denominator is 1:
5
$y = \frac{3}{1} = 3$
6
Horizontal asymptote: $y = 3$
7
Find y-intercept: substitute $x = 0$ :
8
$f(0) = \frac{0 + 6}{0 - 2} = -3 \implies \text{y-intercept } (0, -3)$
9
Find x-intercept: set numerator equal to zero:
10
$3x + 6 = 0 \implies x = -2 \implies \text{x-intercept } (-2, 0)$

3. Solving Rational Equations★★★☆☆⏱ 15 min

To solve a rational equation, first eliminate denominators by multiplying both sides by the lowest common denominator, then solve the resulting linear or quadratic equation. Always check your solutions against the original domain restriction.

📐 Worked Example

Solve the equation $\frac{2 x}{x - 1} = 5 + \frac{2}{x - 1}$

1
First note the domain restriction: denominator $x - 1 \neq = 0 ⟹ x \neq = 1$
2
Multiply every term by $(x - 1)$ to eliminate denominators:
3
$2x = 5(x-1) + 2$
4
Expand and simplify:
5
$2x = 5x - 5 + 2 \implies -3x = -3 \implies x = 1$
6
Check against domain restriction: $x = 1$ is not allowed, so this is extraneous
7
Conclusion: the equation has no real solutions

Exam tip:

If a question asks for the number of solutions, always remember to count only valid non-extraneous solutions.

4. Common Pitfalls

Wrong move:

Writing only the number for an asymptote, e.g. writing $2$ instead of $x = 2$

Why:

Examiners require the full equation of the asymptote line, not just the intercept value

Correct move:

Always write vertical asymptotes as $x = a$ and horizontal asymptotes as $y = b$

Wrong move:

Forgetting to check for extraneous solutions after solving a rational equation

Why:

Multiplying by the denominator can introduce invalid solutions that do not satisfy the original equation

Correct move:

Always check any solution against the original domain, discard any that make a denominator zero

Wrong move:

Confusing vertical and horizontal asymptote formulas for $\frac{a x + b}{c x + d}$ , writing $y = - \frac{d}{c}$

Why:

Mixing up the calculation for vertical and horizontal asymptotes

Correct move:

Vertical comes from denominator: $x = - \frac{d}{c}$ , horizontal from leading coefficients: $y = \frac{a}{c}$

Wrong move:

Claiming a graph can cross a vertical asymptote

Why:

Misunderstanding the definition of an asymptote

Correct move:

Vertical asymptotes are lines where the function is undefined, the graph never crosses them

5. Quick Reference Cheatsheet

Feature	Reciprocal $\frac{k}{x - h} + v$	Linear-over-linear $\frac{a x + b}{c x + d}$
Vertical Asymptote	$x = h$	$x = - \frac{d}{c}$
Horizontal Asymptote	$y = v$	$y = \frac{a}{c}$
Domain	$x \neq = h$	$x \neq = - \frac{d}{c}$
Range	$y \neq = v$	$y \neq = \frac{a}{c}$
x-intercept	Solve $\frac{k}{x - h} + v = 0$	$x = - \frac{b}{a}$
y-intercept	Substitute $x = 0$	$y = \frac{b}{d}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find asymptotes of a rational function
2022 · 2
Solve rational equation, sketch graph

Going deeper

syllabusOfficial IB AA SL Topic 2 SyllabusFull learning outcomes reference

What's Next

Reciprocal and rational functions are foundational for more advanced topics in the IB AA SL course. You will apply your knowledge of asymptotes and domain restrictions when you study transformations of functions and solve rational inequalities. These functions also appear regularly in calculus topics, including differentiation and integration of non-polynomial functions, and understanding their key features will help you solve application problems in exams. Mastering this sub-topic also prepares you for working with other non-linear functions like logarithms and trigonometric functions.