What are the key characteristics of a function in Pure Mathematics 3?

In Edexcel International A Levels Pure Mathematics 3, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Key characteristics include domain, range, and the function's behavior such as continuity, limits, and differentiability. Understanding these characteristics is crucial for analyzing and graphing functions.

How do you determine the domain and range of a function?

To determine the domain of a function, identify all possible input values (x-values) for which the function is defined. For the range, determine all possible output values (y-values) the function can produce. In Pure Mathematics 3, this often involves analyzing the function's equation and considering any restrictions such as division by zero or square roots of negative numbers.

What is the significance of asymptotes in graphing functions?

Asymptotes are lines that a graph approaches but never touches. They are significant in graphing functions as they indicate the behavior of the function at extreme values or near undefined points. In Edexcel International A Levels Pure Mathematics 3, understanding vertical, horizontal, and oblique asymptotes helps in sketching accurate graphs and analyzing the function's behavior at infinity or near discontinuities.

How can transformations be applied to functions and their graphs?

Transformations such as translations, reflections, stretches, and compressions can be applied to functions to alter their graphs. In Pure Mathematics 3, these transformations are used to shift the graph horizontally or vertically, reflect it across axes, or change its shape. Understanding these transformations is essential for graphing functions and solving related problems.

What role do derivatives play in understanding the graphs of functions?

Derivatives provide information about the slope of a function at any given point, which is crucial for understanding the graph's behavior. In Edexcel International A Levels Pure Mathematics 3, derivatives are used to find critical points, determine intervals of increase and decrease, and identify points of inflection. This analysis helps in sketching accurate graphs and understanding the function's overall behavior.

Why is "Confusing $y = |f(x)|$ with $y = f(|x|)$" a common mistake in Functions and graphs?

Why it happens: Both involve modulus but they transform the graph in opposite directions (x-axis reflection vs y-axis reflection). How to avoid it: Rule of thumb: |f(x)| = 'modulus AFTER f' so reflect below-x-axis parts UP (in x-axis). f(|x|) = 'modulus BEFORE f' so reflect the right-half across the y-axis. Source: WMA13/01 examiner reports — flagged annually

Why is "Drawing the modulus graph but not labelling the vertex" a common mistake in Functions and graphs?

Why it happens: Treating the question as 'just a sketch'. How to avoid it: Every modulus / parabola sketch in P3 must have key coordinates labelled: intercepts AND vertex (or turning point). Examiners deduct marks for missing labels even when the shape is correct.

Why is "Applying outside-$y$ transformations before inside-$x$" a common mistake in Functions and graphs?

Why it happens: Reading left-to-right. How to avoid it: y = a f(b(x - h)) + k: inside-x first (the x closer to f), outside-y last (applied to f(x)). Compute the new vertex/key-point step-by-step.

Why is "Calling a stretch a 'shift' or a translation a 'scale'" a common mistake in Functions and graphs?

Why it happens: Casual vocabulary. How to avoid it: Use the exact mark-scheme words. 'Translation by pmatrix a b pmatrix' for shifts; 'stretch parallel to the -axis, scale factor c' for stretches. Mark schemes will not accept 'shift' or 'scale'. Source: WMA13/01 mark scheme requirements

Why is "Squaring $|f(x)| = |g(x)|$ but not solving $f(x) = \pm g(x)$ directly" a common mistake in Functions and graphs?

Why it happens: Both methods work but students sometimes mix them. How to avoid it: For |f(x)| = |g(x)|: either square both sides (one quadratic) or split into f(x) = g(x) and f(x) = -g(x). The squaring method does NOT introduce extraneous roots here because both sides are non-negative.

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Functions and graphs

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Functions and Graphs Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

The visual side of P3 function theory: sketches of $y = |f(x)|$ vs $y = f(|x|)$ , combined transformations of standard graphs, and solving modulus equations and inequalities graphically as well as algebraically. The exam expects fluent sketches with labelled key coordinates and precise transformation language.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P3 2.1 — State the domain and range of a function; identify one-to-one and many-to-one mappings.
P3 2.2 — Sketch $y = |f(x)|$ and $y = f(|x|)$ from $y = f(x)$ , marking key points.
P3 2.3 — Solve modulus equations $|f(x)| = a$ and $|f(x)| = |g(x)|$ both algebraically and graphically.
P3 2.4 — Combine transformations (translation, stretch, reflection) of standard graphs.

Domain and range

Inputs in, outputs out. State both for any function question.

Domain = set of allowable input values $x$ . Range = set of resulting output values $f(x)$ .

Notation: $f : x \mapsto f(x), \;\; x \in D$ .

Worked example. $f(x) = \sqrt{x - 2}$ .

Domain: $x \geq 2$ (so that $x - 2 \geq 0$ ).
Range: $f(x) \geq 0$ (square root is non-negative).

Mapping types.

One-to-one (injective): different inputs give different outputs. Every horizontal line meets graph at most once. Has an inverse.
Many-to-one: at least two inputs share an output. E.g. $f(x) = x^{2}$ over $\mathbb{R}$ : $f(2) = f(-2) = 4$ .
One-to-many: NOT a function — one input maps to multiple outputs.

Restricting the domain to make one-to-one. $f(x) = x^{2}$ over $\mathbb{R}$ is many-to-one. Restricted to $x \geq 0$ , it becomes one-to-one with inverse $f^{-1}(x) = \sqrt{x}$ .

Stating domain / range in P3 exams. Use proper notation: $x \geq 2$ or $x \in [2, \infty)$ . Use interval notation or inequalities; avoid 'all positive numbers' style phrases.

Domain = inputs. Range = outputs.
One-to-one ⇔ inverse exists.
Many-to-one needs domain restriction for an inverse.
Always quote ranges in inequality or interval notation.

Sketching $y = |f(x)|$

Reflect below- $x$ -axis parts UPWARD in the $x$ -axis.

Rule. To sketch $y = |f(x)|$ from $y = f(x)$ :

Keep any part of $y = f(x)$ that lies on or above the $x$ -axis.
Reflect any part below the $x$ -axis upward (in the $x$ -axis).

All output values become non-negative. $|f(x)| \geq 0$ .

Worked example. $f(x) = x^{2} - 4x - 5 = (x - 5)(x + 1)$ .

$f$ has roots $x = -1, 5$ and minimum $(2, -9)$ .
Between $x = -1$ and $x = 5$ , $f(x) < 0$ — reflect upward.
$|f(x)|$ has same roots $-1, 5$ (on the $x$ -axis they touch) and a NEW maximum at $(2, 9)$ .
Outside $[-1, 5]$ , $f \geq 0$ — graph unchanged.

Cusps. Where $f(x) = 0$ (root), $|f(x)|$ has a sharp corner (cusp) — the gradients on the two sides are negatives of each other.

Solving $|f(x)| = k$ graphically. Draw $y = |f(x)|$ and $y = k$ ; count intersections. For $k = 9$ above, three intersections; for $k = 0$ , two intersections; for $k = 10$ , two intersections.

Solving $|f(x)| < k$ graphically. Find $x$ where the graph of $|f|$ is BELOW $y = k$ .

Reflect below-axis parts upward.
Roots become cusps (sharp corners).
Local min below axis becomes local max above.
Graphical intersections give equation solutions.

Sketching $y = f(|x|)$

Keep right half; reflect across $y$ -axis to make left half. Always even.

Rule. To sketch $y = f(|x|)$ from $y = f(x)$ :

Keep the part of $y = f(x)$ for $x \geq 0$ .
Replace the $x < 0$ part with the reflection of the kept part in the $y$ -axis.

Result: an EVEN function — symmetric about the $y$ -axis. The graph satisfies $f(|-x|) = f(|x|)$ .

Worked example. $f(x) = x^{2} + 3x - 4$ .

Right half (for $x \geq 0$ ): $y = x^{2} + 3x - 4$ . Roots: $x^{2} + 3x - 4 = 0 \Rightarrow x = 1$ or $x = -4$ . Only $x = 1$ is in the kept region. $y$ -intercept: $-4$ .
Reflect this right-half in the $y$ -axis: left half is the mirror image. Root at $x = -1$ .

So $y = f(|x|)$ has $x$ -intercepts $\pm 1$ , $y$ -intercept $-4$ , and two minima at $(\pm \tfrac{3}{2}, -\tfrac{25}{4})$ (symmetric reflection of the right-half minimum).

Algebraic check. For $x < 0$ , $f(|x|) = f(-x) = (-x)^{2} + 3(-x) - 4 = x^{2} - 3x - 4$ . Differs from $f(x) = x^{2} + 3x - 4$ for $x < 0$ — good.

Crucial distinction.

Graph	Rule
$y = \\|f(x)\\|$	Reflect parts BELOW $x$ -axis UP into $x$ -axis
$y = f(\\|x\\|)$	Reflect right half across $y$ -axis (replace left half)

$f(|x|)$ is even — symmetric in $y$ -axis.
Right half stays; left half is the mirror.
Test by computing $f(-x)$ on the left: should match the mirror.
$|f(x)|$ vs $f(|x|)$ — different transformations.

Solving modulus equations graphically

Sketch both sides; count and read off intersection $x$ -values.

General principle. To solve $|f(x)| = g(x)$ or $|f(x)| = a$ :

Sketch $y = |f(x)|$ and $y = g(x)$ (or $y = a$ ) on the same axes.
Each intersection corresponds to a solution.
Algebraically find the $x$ -coordinates of those intersections (split into cases).

Worked example. Solve $|x - 2| = x + 1$ graphically AND algebraically.

Graphically: $|x - 2|$ is a V at $x = 2$ with arms going up to the left and right; $y = x + 1$ is a line through $(0, 1)$ , gradient $1$ . They cross once.

Algebraically:

Case 1: $x \geq 2$ . $|x - 2| = x - 2$ . Equation: $x - 2 = x + 1 \Rightarrow -2 = 1$ , contradiction. NO solution here.
Case 2: $x < 2$ . $|x - 2| = 2 - x$ . Equation: $2 - x = x + 1 \Rightarrow x = \tfrac{1}{2}$ . Check: $\tfrac{1}{2} < 2$ ✓; $|0.5 - 2| = 1.5 = 0.5 + 1$ . ✓

So $x = \tfrac{1}{2}$ is the unique solution — consistent with the one graphical intersection.

Solving $|f(x)| < a$ graphically. Shade where the $|f|$ -graph is below the horizontal $y = a$ .

Worked example. Solve $|x - 2| \leq 3$ . Graph is V with vertex $(2, 0)$ . $y = 3$ is a horizontal line. They meet at $x = -1$ and $x = 5$ . The V is below the line BETWEEN these intersections.

So $-1 \leq x \leq 5$ .

Sketch both sides on the same axes.
Number of intersections = number of solutions.
Algebraic case-split AFTER you know how many to expect.
Inequality: shade above/below the appropriate boundary.

Combined transformations

Inside- $x$ first; outside- $y$ last. Use exact mark-scheme language.

Order of transformations. $y = a f(b(x - h)) + k$ :

Replace $x$ with $x - h$ : translation by $\begin{pmatrix} h \\ 0 \end{pmatrix}$ — right $h$ units.
Replace $x$ with $bx$ (inside): stretch parallel to $x$ -axis, scale factor $\tfrac{1}{b}$ . (Or reflect in $y$ -axis if $b < 0$ .)
Multiply $f$ by $a$ : stretch parallel to $y$ -axis, scale factor $a$ . (Or reflect in $x$ -axis if $a < 0$ .)
Add $k$ to result: translation by $\begin{pmatrix} 0 \\ k \end{pmatrix}$ — up $k$ units.

Worked example. Describe the transformations mapping $y = f(x) = |x|$ onto $y = 2|x - 3| - 1$ .

Step 1: $|x| \to |x - 3|$ . Translation right by $3$ .
Step 2: $|x - 3| \to 2|x - 3|$ . Stretch parallel to $y$ -axis, scale factor $2$ .
Step 3: $2|x - 3| \to 2|x - 3| - 1$ . Translation down by $1$ .

Vertex tracks: $(0, 0) \to (3, 0) \to (3, 0)$ (stretch through axis doesn't move vertex) $\to (3, -1)$ .

Mark-scheme language. Exam mark schemes are strict:

'Translation by $\begin{pmatrix} a \\ b \end{pmatrix}$ ' — NOT 'shift'.
'Stretch parallel to the $y$ -axis, scale factor $a$ ' — NOT 'scale' or 'vertical stretch'.
'Reflection in the $x$ -axis' — NOT 'flip'.

Mixed transformations. $y = -f(x + 2) - 3$ : translation $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$ , reflection in $x$ -axis, translation $\begin{pmatrix} 0 \\ -3 \end{pmatrix}$ .

Inside- $x$ first (the $x$ closest to $f$ ).
Translation = shift; stretch = scale.
Use exact mark-scheme wording.
Track key coordinates step-by-step.

How it’s examined

Functions and Graphs is a $6$ - $10$ -mark mid-paper question on WMA13/01. Typical setups: 'sketch $y = |f(x)|$ ' or 'sketch $y = f(|x|)$ ' with intercepts and turning points labelled; describe a sequence of transformations linking two graphs (with strict mark-scheme language); solve modulus equation/inequality graphically and algebraically. The grading: M for the algebraic step, B for each correctly-labelled key feature. Examiners commonly note that 'students lose marks for unlabelled sketches' — make labels habitual.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Functions and graphs

Step-by-step solutions to past-paper-style questions on functions and graphs, written exactly the way a tutor would explain them at the board.

1Sketch $y = |f(x)|$ for a quadratic (5 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q5• modulus graph, transformation
Question
$f(x) = x^{2} - 4x - 5$ . (a) Sketch $y = f(x)$ , marking intercepts and the minimum. (b) Hence sketch $y = |f(x)|$ on the same axes.
Step-by-step solution
1. Step 1
  (a) Factorise: $x^{2} - 4x - 5 = (x - 5)(x + 1)$ . $x$ -intercepts: $x = 5, -1$ . $y$ -intercept: $f(0) = -5$ . Completed square: $f(x) = (x - 2)^{2} - 9$ . Minimum at $(2, -9)$ .
2. Step 2
  Sketch the parabola: upward-opening, passing through $(-1, 0)$ , $(5, 0)$ , $(0, -5)$ , vertex $(2, -9)$ .
3. Step 3
  (b) For $y = |f(x)|$ , reflect ANY portion of the graph where $y < 0$ in the $x$ -axis. The portion between $x = -1$ and $x = 5$ (where $f(x) < 0$ ) flips up. The minimum $(2, -9)$ becomes the maximum $(2, 9)$ of $|f(x)|$ .
4. Step 4
  Outside $[-1, 5]$ the graph is unchanged. $x$ -intercepts of $y = |f(x)|$ are still $-1$ and $5$ ; $y$ -intercept is $|{-5}| = 5$ .
Answer
$y = |f(x)|$ has intercepts $(-1, 0), (5, 0), (0, 5)$ and local maximum $(2, 9)$ .
Examiner tip
(a) M1 correct shape and roots. A1 vertex labelled. (b) M1 reflection of below-axis portion. A1 correct intercepts on $|f|$ . A1 vertex $(2, 9)$ marked. Sketches MUST label key coordinates — examiners deduct marks for unlabelled vertices and intercepts even when the shape is correct.
2Sketch $y = f(|x|)$ vs $y = |f(x)|$ (5 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q4• modulus graph, even function
Question
Given $f(x) = x^{2} + 3x - 4$ , sketch $y = f(|x|)$ , marking all axis intercepts.
Step-by-step solution
1. Step 1
  $y = f(|x|)$ replaces $x$ with $|x|$ . For $x \geq 0$ , $f(|x|) = f(x) = x^{2} + 3x - 4$ . For $x < 0$ , $f(|x|) = f(-x) = x^{2} - 3x - 4$ .
2. Step 2
  Graphical rule: keep the $x \geq 0$ part of $y = f(x)$ and reflect it in the $y$ -axis to get the $x < 0$ part. The graph is even (symmetric about $y$ -axis).
3. Step 3
  For $x \geq 0$ : roots of $x^{2} + 3x - 4 = (x + 4)(x - 1)$ are $x = -4$ (rejected, negative) and $x = 1$ . So the right-half graph passes through $(0, -4)$ and $(1, 0)$ .
4. Step 4
  Reflecting in the $y$ -axis gives a root at $x = -1$ on the left, also passing through $(0, -4)$ . So $y = f(|x|)$ has $x$ -intercepts $\pm 1$ and $y$ -intercept $-4$ .
Answer
$y = f(|x|)$ is even with $x$ -intercepts $(\pm 1, 0)$ and $y$ -intercept $(0, -4)$ , with minima at $(\pm\tfrac{3}{2}, -\tfrac{25}{4})$ .
Examiner tip
M1 for stating the right-half-reflect-into-left rule (or symmetry about $y$ -axis). A1 intercepts $(\pm 1, 0)$ . A1 $y$ -intercept $(0, -4)$ . A1 symmetry. A1 minima coordinates. Students often draw $y = |f(x)|$ instead — completely different. Test by checking $f(|-2|) = f(2)$ : if $y = f(|x|)$ and $y = f(x)$ differ on the left of the $y$ -axis, you have the right graph.
3Solve $|2x + 1| = |x - 3|$ (5 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q4• modulus equation, two moduli
Question
Solve $|2x + 1| = |x - 3|$ .
Step-by-step solution
1. Step 1
  Two moduli on opposite sides ⇒ square both sides (the modulus does nothing on a non-negative value).
  $(2x + 1)^{2} = (x - 3)^{2}$
2. Step 2
  Expand both sides:
  $4x^{2} + 4x + 1 = x^{2} - 6x + 9$
3. Step 3
  Rearrange:
  $3x^{2} + 10x - 8 = 0$
4. Step 4
  Factorise: $(3x - 2)(x + 4) = 0$ . So $x = \tfrac{2}{3}$ or $x = -4$ .
5. Step 5
  Substitute back to confirm. $|2(\tfrac{2}{3}) + 1| = |\tfrac{7}{3}| = \tfrac{7}{3}$ and $|\tfrac{2}{3} - 3| = \tfrac{7}{3}$ . ✓ Similarly $x = -4$ gives $|-7| = |-7|$ . ✓
Answer
$x = \tfrac{2}{3}$ or $x = -4$ .
Examiner tip
M1 for squaring (a perfectly valid step when both sides are moduli). A1 for the correct quadratic. M1 for solving. A1 each root. Squaring is the cleanest method here because both sides are guaranteed non-negative. The case-split method (4 cases) also works but takes longer.
4State domain and range of a piecewise modulus function (5 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q3• domain, range, piecewise
Question
$f(x) = |x - 2| + 3$ , $x \in \mathbb{R}$ . (a) Sketch $y = f(x)$ , marking the vertex. (b) State the range. (c) Determine whether $f$ has an inverse.
Step-by-step solution
1. Step 1
  (a) $|x - 2| \geq 0$ with minimum $0$ at $x = 2$ . So $f$ is a V-shape with vertex $(2, 3)$ , gradient $+1$ for $x \geq 2$ and gradient $-1$ for $x \leq 2$ .
2. Step 2
  (b) Range: $f(x) \geq 3$ .
3. Step 3
  (c) $f$ is NOT one-to-one: $f(0) = 5 = f(4)$ , so the horizontal line $y = 5$ crosses twice. No inverse exists over $\mathbb{R}$ .
4. Step 4
  If we restrict to $x \geq 2$ (one branch only), then $f(x) = (x - 2) + 3 = x + 1$ is one-to-one and $f^{-1}(x) = x - 1$ , $x \geq 3$ .
Answer
(a) V-shape, vertex $(2, 3)$ . (b) Range $f(x) \geq 3$ . (c) Not one-to-one over $\mathbb{R}$ ; no inverse without domain restriction.
Examiner tip
(a) B1 V-shape with correct vertex. (b) B1 range. (c) M1 horizontal-line test or example. A1 conclusion. B1 (bonus) the restricted inverse. The V-shape is a recurring shape in P3 — always state the vertex coordinates as a labelled point on the sketch.
5Combined transformation involving modulus (6 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q7• transformations, modulus, combined
Question
$f(x) = |x|$ . Describe the transformations that map $y = f(x)$ onto $y = 2 |x - 3| - 1$ . Sketch the final graph.
Step-by-step solution
1. Step 1
  Write $g(x) = 2 f(x - 3) - 1$ . Apply transformations to $f(x) = |x|$ in the order: inside- $x$ first, outside- $y$ last.
2. Step 2
  Step (i): replace $x$ with $x - 3$ — translation by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ (right $3$ ). New: $|x - 3|$ , vertex at $(3, 0)$ .
3. Step 3
  Step (ii): multiply by $2$ — vertical stretch, scale factor $2$ . New: $2|x - 3|$ , vertex unchanged at $(3, 0)$ ; gradients are $\pm 2$ .
4. Step 4
  Step (iii): subtract $1$ — translation by $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$ (down $1$ ). Final vertex: $(3, -1)$ .
5. Step 5
  $x$ -intercepts where $2|x - 3| - 1 = 0 \Rightarrow |x - 3| = \tfrac{1}{2} \Rightarrow x = 3 \pm \tfrac{1}{2}$ . So $(\tfrac{5}{2}, 0)$ and $(\tfrac{7}{2}, 0)$ .
Answer
Transformations: right $3$ , vertical stretch $\times 2$ , down $1$ . Final graph: V-shape, vertex $(3, -1)$ , $x$ -intercepts $\tfrac{5}{2}$ and $\tfrac{7}{2}$ .
Examiner tip
M1 for identifying right-translation. A1 'translation $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ ' (exact wording). M1 vertical stretch. A1 'scale factor $2$ parallel to the $y$ -axis' (exact wording). M1 down-translation. A1 'translation $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$ '. Mark schemes are extremely strict on the language — say 'translation' (not 'shift') and 'stretch' (not 'scale').

Key Formulae — Functions and graphs

The formulae you need to memorise for functions and graphs on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Graph of $y = |f(x)|$
$y = |f(x)| = \begin{cases} f(x) & \text{if } f(x) \geq 0 \\ -f(x) & \text{if } f(x) < 0 \end{cases}$
$f(x)$
the original function
When to use
Sketching $y = |f(x)|$ : keep any part of the graph on or above the $x$ -axis; reflect any part below the $x$ -axis upward (in the $x$ -axis). NOT in the IAL formula booklet.
Example
For $f(x) = x^{2} - 9$ with minimum $(0, -9)$ , $|f(x)|$ has a local max at $(0, 9)$ between roots $\pm 3$ .
Graph of $y = f(|x|)$
$y = f(|x|) = \begin{cases} f(x) & \text{if } x \geq 0 \\ f(-x) & \text{if } x < 0 \end{cases}$
$f(x)$
the original function
When to use
Sketching $y = f(|x|)$ : keep the $x \geq 0$ portion of $y = f(x)$ ; replace the $x < 0$ portion with the reflection of the $x \geq 0$ portion in the $y$ -axis. The result is always even. NOT in the formula booklet.
Example
For $f(x) = x^{2} + 3x - 4$ , the right half (with root $x = 1$ ) reflects into the left half (with root $x = -1$ ).
Function transformations (summary)
$y = a f\bigl(b(x - h)\bigr) + k$
$h$
horizontal translation (right if $h > 0$ )
$k$
vertical translation (up if $k > 0$ )
$a$
vertical stretch / reflection (in $x$ -axis if $a < 0$ )
$b$
horizontal stretch (factor $1/b$ ) / reflection (in $y$ -axis if $b < 0$ )
When to use
Decomposing a transformed function. Apply inside- $x$ transformations (those affecting $x$ before $f$ ) first; outside- $y$ transformations (those applied to $f(x)$ ) afterwards. NOT in the formula booklet.
Example
$y = 2|x - 3| - 1$ : translation right $3$ , vertical stretch factor $2$ , translation down $1$ .
Domain of a composite (revisited)
$\text{dom}(fg) = \{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$
$f, g$
two functions
When to use
Stating the domain of $fg$ . Check both that $g$ is defined at $x$ AND that $g(x)$ lies in the domain of $f$ . NOT in the formula booklet.
Example
$f(x) = \ln x$ (dom $x > 0$ ), $g(x) = x - 2$ . Then $fg(x) = \ln(x - 2)$ , dom: $x > 2$ .

Key Definitions and Keywords — Functions and graphs

Definitions to memorise and the exact keywords mark schemes credit for functions and graphs answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Function
Examiner keyword
A rule that assigns exactly one output $f(x)$ to each input $x$ in the domain. Notation: $f : x \mapsto f(x)$ or $y = f(x)$ .
Mapping diagram
A diagram with the domain on one side, the range (or codomain) on the other, and arrows showing how each input maps to its output. Used to illustrate one-to-one vs many-to-one mappings.
Graph of $y = |f(x)|$
Examiner keyword
The graph obtained by reflecting all parts of $y = f(x)$ that lie below the $x$ -axis upward in the $x$ -axis. All $y$ -values become non-negative.
Graph of $y = f(|x|)$
Examiner keyword
The graph obtained by keeping the $x \geq 0$ part of $y = f(x)$ and reflecting it in the $y$ -axis to fill in the $x < 0$ part. The result is always an even function.
Translation
Examiner keyword
A rigid shift of a graph by a vector $\begin{pmatrix} a \\ b \end{pmatrix}$ . $y = f(x - a) + b$ shifts $y = f(x)$ right $a$ units and up $b$ units. Mark schemes require the word 'translation' and a column vector.
Challenge
Examiner keyword
A scaling parallel to one axis. $y = a f(x)$ stretches the graph by factor $a$ parallel to the $y$ -axis; $y = f(bx)$ stretches by factor $1/b$ parallel to the $x$ -axis. Mark schemes require the word 'stretch', the factor, and the axis.

Common Mistakes and Misconceptions — Functions and graphs

The traps other students keep falling into on functions and graphs questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Confusing $y = |f(x)|$ with $y = f(|x|)$
WMA13/01 examiner reports — flagged annually
Why it happens
Both involve modulus but they transform the graph in opposite directions ( $x$ -axis reflection vs $y$ -axis reflection).
How to avoid it
Rule of thumb: $|f(x)|$ = 'modulus AFTER $f$ ' so reflect below- $x$ -axis parts UP (in $x$ -axis). $f(|x|)$ = 'modulus BEFORE $f$ ' so reflect the right-half across the $y$ -axis.
✕Drawing the modulus graph but not labelling the vertex
Why it happens
Treating the question as 'just a sketch'.
How to avoid it
Every modulus / parabola sketch in P3 must have key coordinates labelled: intercepts AND vertex (or turning point). Examiners deduct marks for missing labels even when the shape is correct.
✕Applying outside- $y$ transformations before inside- $x$
Why it happens
Reading left-to-right.
How to avoid it
$y = a f(b(x - h)) + k$ : inside- $x$ first (the $x$ closer to $f$ ), outside- $y$ last (applied to $f(x)$ ). Compute the new vertex/key-point step-by-step.
✕Calling a stretch a 'shift' or a translation a 'scale'
WMA13/01 mark scheme requirements
Why it happens
Casual vocabulary.
How to avoid it
Use the exact mark-scheme words. 'Translation by $\begin{pmatrix} a \\ b \end{pmatrix}$ ' for shifts; 'stretch parallel to the $\square$ -axis, scale factor $c$ ' for stretches. Mark schemes will not accept 'shift' or 'scale'.
✕Squaring $|f(x)| = |g(x)|$ but not solving $f(x) = \pm g(x)$ directly
Why it happens
Both methods work but students sometimes mix them.
How to avoid it
For $|f(x)| = |g(x)|$ : either square both sides (one quadratic) or split into $f(x) = g(x)$ and $f(x) = -g(x)$ . The squaring method does NOT introduce extraneous roots here because both sides are non-negative.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Functions and graphs — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Functions and graphs

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Sketch y=∣f(x)∣y = |f(x)|y=∣f(x)∣ for a quadratic (5 marks, P3)

2Sketch y=f(∣x∣)y = f(|x|)y=f(∣x∣) vs y=∣f(x)∣y = |f(x)|y=∣f(x)∣ (5 marks, P3)

3Solve ∣2x+1∣=∣x−3∣|2x + 1| = |x - 3|∣2x+1∣=∣x−3∣ (5 marks, P3)

4State domain and range of a piecewise modulus function (5 marks, P3)

5Combined transformation involving modulus (6 marks, P3)

Graph of y=∣f(x)∣y = |f(x)|y=∣f(x)∣

Graph of y=f(∣x∣)y = f(|x|)y=f(∣x∣)

Function transformations (summary)

Domain of a composite (revisited)

Function

Mapping diagram

Graph of y=∣f(x)∣y = |f(x)|y=∣f(x)∣

Graph of y=f(∣x∣)y = f(|x|)y=f(∣x∣)

Translation

Challenge

✕Confusing y=∣f(x)∣y = |f(x)|y=∣f(x)∣ with y=f(∣x∣)y = f(|x|)y=f(∣x∣)

✕Drawing the modulus graph but not labelling the vertex

✕Applying outside-yyy transformations before inside-xxx

✕Calling a stretch a 'shift' or a translation a 'scale'

✕Squaring ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣ but not solving f(x)=±g(x)f(x) = \pm g(x)f(x)=±g(x) directly

Practice questions

Past paper style quiz

Video lesson

Functions and graphs — frequently asked questions

1Sketch $y = |f(x)|$ for a quadratic (5 marks, P3)

2Sketch $y = f(|x|)$ vs $y = |f(x)|$ (5 marks, P3)

3Solve $|2x + 1| = |x - 3|$ (5 marks, P3)

Graph of $y = |f(x)|$

Graph of $y = f(|x|)$

Graph of $y = |f(x)|$

Graph of $y = f(|x|)$

✕Confusing $y = |f(x)|$ with $y = f(|x|)$

✕Applying outside- $y$ transformations before inside- $x$

✕Squaring $|f(x)| = |g(x)|$ but not solving $f(x) = \pm g(x)$ directly