The Parent Reciprocal Function $f(x) = \frac{1}{x}$
- Core Characteristics:
- Domain: $\{x \in \mathbb{R} \mid x \ne 0\}$ (or $(-\infty, 0) \cup (0, \infty)$).
- Range: $\{y \in \mathbb{R} \mid y \ne 0\}$ (or $(-\infty, 0) \cup (0, \infty)$).
- Vertical Asymptote: $x = 0$ ($y$-axis).
- Horizontal Asymptote: $y = 0$ ($x$-axis).
- Symmetry: Odd function ($f(-x) = -f(x)$) $\implies 180^\circ$ rotational symmetry about $(0, 0)$.
- Self-inverse: $f^{-1}(x) = \frac{1}{x} = f(x)$.
- Shifted Reciprocal Form: $g(x) = \frac{a}{x - h} + k$.
- Vertical asymptote: $x = h$.
- Horizontal asymptote: $y = k$.
- Center of hyperbolic symmetry: $(h, k)$.
- Domain: $x \ne h$; Range: $y \ne k$.
(a) State the equations of the asymptotes and coordinates of the center of symmetry.
(b) Find the coordinates of all axial intercepts.
Part (a): Asymptotes & Symmetry Center
Vertical asymptote: $x = 2$
Horizontal asymptote: $y = 4$
Center of symmetry: $(2, 4)$
Part (b): Axial Intercepts
$y$-intercept (set $x = 0$):
$$g(0) = \frac{3}{0 - 2} + 4 = -\frac{3}{2} + 4 = 2.5 \implies (0, 2.5)$$$x$-intercept (set $g(x) = 0$):
$$\frac{3}{x - 2} + 4 = 0 \implies \frac{3}{x - 2} = -4 \implies 3 = -4(x - 2) = -4x + 8$$ $$4x = 5 \implies x = \frac{5}{4} = 1.25 \implies (1.25, 0)$$The General Linear Rational Form $f(x) = \frac{ax + b}{cx + d}$
- Standard Asymptote Equations ($c \ne 0$):
- Vertical Asymptote: Set the denominator equal to zero: $$cx + d = 0 \implies x = -\frac{d}{c}$$
- Horizontal Asymptote: Take the limit as $x \to \pm\infty$ (ratio of leading coefficients): $$\lim_{x \to \pm\infty} \frac{ax + b}{cx + d} = \frac{a}{c} \implies y = \frac{a}{c}$$
- Domain and Range: $$\text{dom } f = \left\{x \in \mathbb{R} \;\middle|\; x \ne -\frac{d}{c}\right\}, \qquad \text{range } f = \left\{y \in \mathbb{R} \;\middle|\; y \ne \frac{a}{c}\right\}$$
- Axial Intercepts:
- $y$-intercept: set $x = 0 \implies \left(0, \frac{b}{d}\right)$ (provided $d \ne 0$).
- $x$-intercept: set numerator equal to zero $\implies ax + b = 0 \implies \left(-\frac{b}{a}, 0\right)$ (provided $a \ne 0$).
(a) State the equations of the vertical and horizontal asymptotes.
(b) State the domain and range of $f$.
(c) Find the coordinates of the $x$- and $y$-intercepts.
Part (a): Asymptotes
Vertical: $2x + 8 = 0 \implies 2x = -8 \implies x = -4$.
Horizontal: $y = \frac{a}{c} = \frac{4}{2} = 2 \implies y = 2$.
Part (b): Domain and Range
Domain: $\{x \in \mathbb{R} \mid x \ne -4\}$. Range: $\{y \in \mathbb{R} \mid y \ne 2\}$.
Part (c): Intercepts
$y$-intercept: $f(0) = \frac{-6}{8} = -\frac{3}{4} = -0.75 \implies (0, -0.75)$.
$x$-intercept: $4x - 6 = 0 \implies 4x = 6 \implies x = 1.5 \implies (1.5, 0)$.
Algebraic Division & Shifted Form Conversion
- Polynomial Long Division Technique: $$\frac{ax + b}{cx + d} = \frac{a}{c} + \frac{b - \frac{ad}{c}}{cx + d}$$ Writing the function in this form immediately makes the horizontal asymptote ($y = \frac{a}{c}$) and vertical stretch factor obvious.
Rewrite numerator:
$$3x + 11 = 3(x + 2) + 5$$ $$\frac{3x + 11}{x + 2} = \frac{3(x + 2) + 5}{x + 2} = \frac{3(x + 2)}{x + 2} + \frac{5}{x + 2} = 3 + \frac{5}{x + 2}$$Answer: $A = 3$ and $B = 5$, so $f(x) = 3 + \frac{5}{x + 2}$.
Inverses and Self-Inverse Rational Functions
- The Inverse Formula: For $f(x) = \frac{ax + b}{cx + d}$: $$f^{-1}(x) = \frac{-dx + b}{cx - a}$$ Notice the shortcut: swap the top-left ($a$) and bottom-right ($d$) coefficients and negate both of them.
- Self-Inverse Condition: $$f(x) \equiv f^{-1}(x) \iff a = -d \iff a + d = 0$$ Whenever the sum of the leading numerator coefficient and the denominator constant is zero, the rational curve is self-inverse!
Step 1: Identify coefficients $a$ and $d$.
Here $a = 5$ and $d = -5$.
Notice that $a + d = 5 + (-5) = 0$.
Step 2: Find the inverse $f^{-1}(x)$ explicitly.
$$x = \frac{5y - 7}{3y - 5} \implies x(3y - 5) = 5y - 7 \implies 3xy - 5x = 5y - 7$$ $$3xy - 5y = 5x - 7 \implies y(3x - 5) = 5x - 7 \implies y = \frac{5x - 7}{3x - 5}$$Since $f^{-1}(x) = \frac{5x - 7}{3x - 5} = f(x)$, $f$ is self-inverse. $\blacksquare$
Self-Inverse Properties & Composite Rational Functions
(a) Write down the equations of the vertical and horizontal asymptotes of $f$. [2 marks]
(b) Find the inverse function $f^{-1}(x)$. [3 marks]
(c) Hence, evaluate $(f \circ f \circ f \circ f)(7)$. [2 marks]
Part (a): Asymptotes of $f$:
Vertical asymptote: $x - 3 = 0 \implies x = 3$ [A1].
Horizontal asymptote: $y = \frac{3}{1} \implies y = 3$ [A1].
Part (b): Find $f^{-1}(x)$ by setting $x = \frac{3y + 1}{y - 3}$:
$$x(y - 3) = 3y + 1 \implies xy - 3x = 3y + 1 \quad [M1]$$ $$xy - 3y = 3x + 1 \implies y(x - 3) = 3x + 1 \quad [M1]$$ $$y = \frac{3x + 1}{x - 3} \implies f^{-1}(x) = \frac{3x + 1}{x - 3} \quad [A1]$$Notice that $f^{-1}(x) = f(x)$, so $f$ is a self-inverse function.
Part (c): Evaluate quadruple composite $(f \circ f \circ f \circ f)(7)$:
Since $f(f(x)) = f(f^{-1}(x)) = x$ is the identity function [R1]:
$$(f \circ f)(x) = x \implies (f \circ f \circ f \circ f)(x) = (f \circ f)((f \circ f)(x)) = x$$ $$\text{Therefore, } (f \circ f \circ f \circ f)(7) = 7 \quad [A1]$$Answers: (a) $x = 3, y = 3$; (b) $f^{-1}(x) = \frac{3x + 1}{x - 3}$; (c) $7$
10 Progressive Practice Exercises
Tier 1 tests asymptote formulas and intercept finding; Tier 2 introduces shifted conversions and self-inverse parameters; Tier 3 challenges hyperbola intersections and real-world average cost modeling.
Tier 1 Core Fluency (Problems 1–3)
Find the equations of the vertical and horizontal asymptotes and the coordinates of all axial intercepts for $f(x) = \frac{2x + 8}{x - 3}$.
View Full Worked Solution
Vertical asymptote: Denominator $= 0 \implies x - 3 = 0 \implies x = 3$.
Horizontal asymptote: $\lim_{x \to \pm\infty} \frac{2x + 8}{x - 3} = \frac{2}{1} \implies y = 2$.
$y$-intercept: $f(0) = \frac{8}{-3} = -\frac{8}{3} \implies \left(0, -\frac{8}{3}\right)$.
$x$-intercept: $2x + 8 = 0 \implies 2x = -8 \implies x = -4 \implies (-4, 0)$.
For the function $g(x) = \frac{-4}{x + 5} + 2$, state its domain, range, and the coordinates of its center of symmetry.
View Full Worked Solution
Asymptotes: Vertical $x = -5$, Horizontal $y = 2$.
Domain: $\{x \in \mathbb{R} \mid x \ne -5\}$.
Range: $\{y \in \mathbb{R} \mid y \ne 2\}$.
Center of symmetry: $(-5, 2)$.
Find an expression for $f^{-1}(x)$ for $f(x) = \frac{x + 4}{2x - 1}$, where $x \ne 0.5$.
View Full Worked Solution
Swap variables:
$$x = \frac{y + 4}{2y - 1} \implies x(2y - 1) = y + 4 \implies 2xy - x = y + 4$$ $$2xy - y = x + 4 \implies y(2x - 1) = x + 4 \implies y = \frac{x + 4}{2x - 1}$$Answer: $f^{-1}(x) = \frac{x + 4}{2x - 1}$ for $x \ne 0.5$. (Notice that $f$ is self-inverse!).
Tier 2 Exam Modeling (Problems 4–7)
Express $f(x) = \frac{5x - 3}{x - 1}$ in the form $A + \frac{B}{x - 1}$ and state its horizontal asymptote.
View Full Worked Solution
Rewrite numerator:
$$5x - 3 = 5(x - 1) + 2$$ $$\frac{5x - 3}{x - 1} = \frac{5(x - 1) + 2}{x - 1} = 5 + \frac{2}{x - 1}$$As $x \to \pm\infty$, $\frac{2}{x - 1} \to 0$, so $y \to 5$.
Answer: $f(x) = 5 + \frac{2}{x - 1}$; horizontal asymptote is $y = 5$.
Find the value of $k$ such that the function $f(x) = \frac{kx + 9}{2x - 7}$ is self-inverse.
View Full Worked Solution
Self-inverse rule: For $f(x) = \frac{ax + b}{cx + d}$, $f$ is self-inverse if and only if $a + d = 0$.
Here $a = k$ and $d = -7$:
$$k + (-7) = 0 \implies k = 7$$Answer: $k = 7$.
A rational function $f(x) = \frac{ax + 6}{cx - 3}$ has vertical asymptote $x = 1.5$ and horizontal asymptote $y = -2$. Find the values of the constants $a$ and $c$.
View Full Worked Solution
Step 1: Vertical asymptote at $x = 1.5$:
$$c(1.5) - 3 = 0 \implies 1.5c = 3 \implies c = 2$$Step 2: Horizontal asymptote at $y = -2$:
$$\frac{a}{c} = -2 \implies \frac{a}{2} = -2 \implies a = -4$$Answer: $a = -4$ and $c = 2$.
Solve the equation $\frac{3x - 1}{x + 2} = x$ analytically and check for extraneous roots.
View Full Worked Solution
Domain restriction: $x \ne -2$.
$$3x - 1 = x(x + 2) \implies 3x - 1 = x^2 + 2x$$ $$x^2 - x + 1 = 0$$Check discriminant:
$$\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$$Since $\Delta < 0$, there are no real solutions.
Answer: No real solutions.
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Find the exact coordinates of the points of intersection between the hyperbola $f(x) = \frac{4}{x}$ and the line $y = -x + 5$.
View Full Worked Solution
Set equations equal ($x \ne 0$):
$$\frac{4}{x} = -x + 5 \implies 4 = -x^2 + 5x \implies x^2 - 5x + 4 = 0$$ $$(x - 1)(x - 4) = 0 \implies x = 1 \quad \text{or} \quad x = 4$$Find $y$-coordinates:
$$\text{At } x = 1: \quad y = -1 + 5 = 4 \implies (1, 4)$$ $$\text{At } x = 4: \quad y = -4 + 5 = 1 \implies (4, 1)$$Answer: $(1, 4)$ and $(4, 1)$.
Find the equations of the two lines with gradient $m = -1$ that are tangent to the curve $y = \frac{2x - 1}{x - 1}$.
View Full Worked Solution
Let line equation be $y = -x + c$:
$$\frac{2x - 1}{x - 1} = -x + c \implies 2x - 1 = (-x + c)(x - 1)$$ $$2x - 1 = -x^2 + x + cx - c \implies x^2 + (1 - c)x + (c - 1) = 0$$Set discriminant $\Delta = 0$ for tangency:
$$\Delta = (1 - c)^2 - 4(1)(c - 1) = 0$$ $$(c - 1)^2 - 4(c - 1) = 0 \implies (c - 1)[(c - 1) - 4] = 0$$ $$(c - 1)(c - 5) = 0 \implies c = 1 \quad \text{or} \quad c = 5$$Answer: $y = -x + 1$ and $y = -x + 5$.
The total cost $C(x)$ (in dollars) of manufacturing $x$ units of a product is given by $C(x) = 4800 + 35x$. The average cost per unit is modeled by:
$$\bar{C}(x) = \frac{C(x)}{x} = \frac{4800 + 35x}{x}, \quad x > 0$$
(a) Find the average cost per unit when producing $100$ units.
(b) State the horizontal asymptote of $\bar{C}(x)$ as $x \to \infty$ and explain its economic meaning.
(c) Find the minimum number of units that must be produced for the average cost per unit to be less than $\$45$.
View Full Worked Solution
Part (a): Production of 100 units
$$\bar{C}(100) = \frac{4800 + 35(100)}{100} = \frac{4800 + 3500}{100} = \frac{8300}{100} = \$83.00$$Part (b): Horizontal Asymptote
$$\lim_{x \to \infty} \frac{4800 + 35x}{x} = \lim_{x \to \infty} \left(\frac{4800}{x} + 35\right) = 0 + 35 = 35 \implies y = 35$$Economic meaning: As production scale becomes extremely large, the fixed costs are distributed over so many units that the fixed cost per unit approaches zero, and the average cost per unit approaches the marginal production cost of $\$35$.
Part (c): Average cost below $\$45$
$$\frac{4800 + 35x}{x} < 45 \implies 4800 + 35x < 45x \quad (\text{since } x > 0)$$ $$10x > 4800 \implies x > 480$$Since $x$ must be an integer, the minimum number of units is $481$.
Answer: (a) $\$83.00$; (b) $y = 35$ (limiting cost per unit); (c) $481\text{ units}$.
Syllabus 2.8 Formula Notes
- Reciprocal Function: $$f(x) = \frac{1}{x}, \quad x \ne 0$$ Note: Specific rational formulas are not printed in the formula booklet; learn the standard forms below.
Rational Functions Key Rules
- Linear Rational Form: $$f(x) = \frac{ax + b}{cx + d} \quad (c \ne 0)$$
- Vertical Asymptote: $$x = -\frac{d}{c}$$
- Horizontal Asymptote: $$y = \frac{a}{c}$$
- Inverse Function: $$f^{-1}(x) = \frac{-dx + b}{cx - a}$$
- Self-Inverse Condition: $$a + d = 0 \iff a = -d$$
Critical IB Examination Pitfalls
- Swapping Vertical and Horizontal Asymptotes: Confusing the vertical line $x = -\frac{d}{c}$ with the horizontal line $y = \frac{a}{c}$. Always recall: vertical asymptotes come from the denominator; horizontal asymptotes come from large $x$ limits.
- Sign Error in the Self-Inverse Test: A rational function is self-inverse when $a = -d$ (meaning $a + d = 0$). Students often mistakenly write $a = d$.
- Not Writing Asymptotes as Full Equations: Writing "vertical asymptote is 3" receives zero marks. The response must state the full equation of the line: $x = 3$.
- Multiplying Inequalities by Variable Denominators: In an inequality like $\frac{2x - 1}{x + 3} < 1$, multiplying by $(x + 3)$ without considering its sign reverses the inequality when $x < -3$. Always bring everything to one side or sketch the graph.
IB AA SL Topic Check: Reciprocal & Rational Functions
Section A has diagnostic questions with feedback. Section B has Paper 1 and Paper 2 exam-style problems with markschemes.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: What is the equation of the vertical asymptote of $f(x) = \frac{3x + 1}{2x - 8}$?
Question 2: What is the equation of the horizontal asymptote of $f(x) = \frac{5x - 2}{x + 4}$?
Question 3: What value of $a$ makes the function $f(x) = \frac{ax + 7}{2x - 6}$ self-inverse?
Question 4: What are the coordinates of the $x$-intercept of $f(x) = \frac{4x - 12}{x + 5}$?
Section B: Exam-Style Free Response Questions
Question 5: Rational Function Properties
Consider the function $f(x) = \frac{2x + 6}{x - 1}$, for $x \ne 1$.
(a) Write down the equations of the vertical and horizontal asymptotes. [2 marks]
(b) Find the coordinates of the $x$-intercept and the $y$-intercept. [2 marks]
(c) State the range of $f$. [2 marks]
View Official Markscheme & Model Solution
Part (a) [2 Marks]:
Vertical asymptote: $x = 1$ (A1)
Horizontal asymptote: $y = 2$ (A1)
Part (b) [2 Marks]:
$x$-intercept: $2x + 6 = 0 \implies x = -3 \implies (-3, 0)$ (A1)
$y$-intercept: $f(0) = \frac{6}{-1} = -6 \implies (0, -6)$ (A1)
Part (c) [2 Marks]:
Since the horizontal asymptote is $y = 2$ and $2x + 6 \ne 2(x - 1)$ for any real $x$:
$$\text{Range: } \{y \in \mathbb{R} \mid y \ne 2\} \quad (\text{or } (-\infty, 2) \cup (2, \infty)) \quad \text{\bf (A2)}$$Question 6: Inverse and Intersection with $y = x$
Let $f(x) = \frac{3x + 4}{x - 2}$, for $x \ne 2$.
(a) Find an expression for $f^{-1}(x)$. [3 marks]
(b) Solve $f(x) = x$ to find the exact coordinates of the points of intersection between the graph of $f(x)$ and the line $y = x$. [4 marks]
View Official Markscheme & Model Solution
Part (a) [3 Marks]:
$$x = \frac{3y + 4}{y - 2} \implies x(y - 2) = 3y + 4 \quad \text{\bf (M1)}$$ $$xy - 2x = 3y + 4 \implies xy - 3y = 2x + 4 \quad \text{\bf (M1)}$$ $$y(x - 3) = 2x + 4 \implies f^{-1}(x) = \frac{2x + 4}{x - 3} \quad (\text{for } x \ne 3) \quad \text{\bf (A1)}$$Part (b) [4 Marks]:
$$\frac{3x + 4}{x - 2} = x \implies 3x + 4 = x(x - 2) \quad \text{\bf (M1)}$$ $$3x + 4 = x^2 - 2x \implies x^2 - 5x - 4 = 0 \quad \text{\bf (A1)}$$ $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-4)}}{2(1)} = \frac{5 \pm \sqrt{25 + 16}}{2} = \frac{5 \pm \sqrt{41}}{2} \quad \text{\bf (M1)}$$Points of intersection lie on $y = x$:
$$\left(\frac{5 + \sqrt{41}}{2}, \frac{5 + \sqrt{41}}{2}\right) \quad \text{and} \quad \left(\frac{5 - \sqrt{41}}{2}, \frac{5 - \sqrt{41}}{2}\right) \quad \text{\bf (A1)}$$Topic 2 Curriculum
All Functions Study Guides
About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.