Why is "Confusing $y = |f(x)|$ with $y = f(|x|)$" a common mistake in Graph transformations?

Why it happens: Both involve modulus bars; students don't check whether they wrap x or f(x). How to avoid it: Bars around f(x) act on the OUTPUT (reflect up); bars around x act on the INPUT (mirror left↔right). Say which before sketching. Source: 9231 Paper 1 Examiner Reports

Why is "Drawing $y^2 = f(x)$ where $f(x) < 0$" a common mistake in Graph transformations?

Why it happens: Forgetting that a square cannot be negative. How to avoid it: First shade the region where f(x) ≥ 0; the curve exists only there. Mark f < 0 as 'no curve'. Source: 9231 Paper 1 Examiner Reports

Why is "Flipping the sign when taking a reciprocal (e.g. a min of $f$ below the axis becoming a min above)" a common mistake in Graph transformations?

Why it happens: Assuming reciprocal reflects the curve in the x-axis. How to avoid it: Reciprocal PRESERVES sign: f<0 \frac1f < 0. A negative minimum of f becomes a negative MAXIMUM of \frac1f. Source: 9231 Paper 1 Examiner Reports

Why is "Drawing a smooth curve through the $x$-intercepts of $|f(x)|$" a common mistake in Graph transformations?

Why it happens: Reflecting visually but rounding off the join. How to avoid it: Reflection in the x-axis creates a sharp corner (cusp) at each x-intercept — draw it pointed, not rounded. Source: 9231 Paper 1 mark schemes

Why is "Not turning zeros of $f$ into asymptotes of $\frac1f$" a common mistake in Graph transformations?

Why it happens: Treating the reciprocal curve as just a vertical stretch. How to avoid it: Each zero of f becomes a vertical asymptote of \frac1f; each asymptote of f becomes a zero of \frac1f. Map these first. Source: 9231 Paper 1 Examiner Reports

Why is "Drawing $y^2 = f(x)$ meeting the $x$-axis at a smooth/horizontal point" a common mistake in Graph transformations?

Why it happens: Copying the shape of y = √(f) without the lower branch. How to avoid it: At f(x) = 0 the two branches √(f) meet with a VERTICAL tangent — draw them meeting the axis at right angles. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Graph transformations

Q: Why is "Drawing $y^2 = f(x)$ where $f(x) < 0$" a common mistake in Graph transformations?

Why it happens: Forgetting that a square cannot be negative. How to avoid it: First shade the region where f(x) ≥ 0; the curve exists only there. Mark f < 0 as 'no curve'. Source: 9231 Paper 1 Examiner Reports

Q: Why is "Flipping the sign when taking a reciprocal (e.g. a min of $f$ below the axis becoming a min above)" a common mistake in Graph transformations?

Why it happens: Assuming reciprocal reflects the curve in the x-axis. How to avoid it: Reciprocal PRESERVES sign: f<0 \frac1f < 0. A negative minimum of f becomes a negative MAXIMUM of \frac1f. Source: 9231 Paper 1 Examiner Reports

Q: Why is "Drawing a smooth curve through the $x$-intercepts of $|f(x)|$" a common mistake in Graph transformations?

Why it happens: Reflecting visually but rounding off the join. How to avoid it: Reflection in the x-axis creates a sharp corner (cusp) at each x-intercept — draw it pointed, not rounded. Source: 9231 Paper 1 mark schemes

Q: Why is "Not turning zeros of $f$ into asymptotes of $\frac1f$" a common mistake in Graph transformations?

Why it happens: Treating the reciprocal curve as just a vertical stretch. How to avoid it: Each zero of f becomes a vertical asymptote of \frac1f; each asymptote of f becomes a zero of \frac1f. Map these first. Source: 9231 Paper 1 Examiner Reports

Q: Why is "Drawing $y^2 = f(x)$ meeting the $x$-axis at a smooth/horizontal point" a common mistake in Graph transformations?

Why it happens: Copying the shape of y = √(f) without the lower branch. How to avoid it: At f(x) = 0 the two branches √(f) meet with a VERTICAL tangent — draw them meeting the axis at right angles. Source: 9231 Paper 1 Examiner Reports

0% Complete

Detailed Study Notes

Detailed notes on Rational Functions and Graphs for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Graph Transformations of Rational and Other Curves — Cambridge 9231 Further Mathematics Paper 1 (2026-2027 syllabus)

Starting from the graph of $y = f(x)$ , sketch the four key related curves: $y = |f(x)|$ , $y = f(|x|)$ , $y^2 = f(x)$ and $y = \frac{1}{f(x)}$ . Each has a precise, memorable rule.

At a glance

$y = |f(x)|$ : reflect any part below the $x$ -axis up to above it.
$y = f(|x|)$ : keep $x \geq 0$ , then mirror it in the $y$ -axis (even function).
$y^2 = f(x)$ : only where $f(x) \geq 0$ ; symmetric about the $x$ -axis.
$y = \frac{1}{f(x)}$ : zeros ↔ asymptotes swap; maxima ↔ minima swap.
Always start from a clear sketch of the original $y = f(x)$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.2d — Understand and use relationships between the graphs of $y = f(x)$ , $y^2 = f(x)$ , $y = \frac{1}{f(x)}$ , $y = f(|x|)$ and $y = |f(x)|$ .
1.2e — Sketch the transformed curve showing how key features (intercepts, asymptotes, turning points, symmetry) map across.

y = |f(x)| — reflect the negative part up

Wherever f(x) is below the x-axis, flip it above. Sharp corners appear at the x-intercepts.

Rule. $|f(x)|$ is never negative, so:

Where $f(x) \geq 0$ : the graph is unchanged.
Where $f(x) < 0$ : reflect that part in the $x$ -axis (flip it up).

What to watch:

The $x$ -axis crossings of $f$ become sharp corners (cusps) of $|f|$ .
Maxima that were below the axis become minima turned upward.
The whole curve sits on or above the $x$ -axis.

|f(x)| flips the below-axis portion upward, creating a corner at each x-intercept of f.

Cambridge tip. Mark the corners clearly — a smooth curve where there should be a cusp loses the feature mark.

Keep $f(x) \geq 0$ parts; reflect $f(x) < 0$ parts up.
$x$ -intercepts of $f$ → sharp corners.
Curve lies entirely on/above the $x$ -axis.

See the full worked example for graph transformations →

y = f(|x|) — make it symmetric in the y-axis

Take only the x ≥ 0 part of f and mirror it into x < 0. Always an even function.

Rule. $f(|x|)$ uses $f$ evaluated at $|x|$ , which is always $\geq 0$ . So:

For $x \geq 0$ : $f(|x|) = f(x)$ — keep the right half unchanged.
For $x < 0$ : $f(|x|) = f(-x)$ — reflect the right half in the $y$ -axis.

The left half of the original graph is discarded and replaced by the mirror image of the right half. The result is always symmetric about the $y$ -axis (an even function).

What to watch:

Any features of $f$ for $x < 0$ are irrelevant — only the $x \geq 0$ behaviour matters.
The point at $x = 0$ is shared by both halves, so the graph is continuous there (often a corner).

Contrast the two modulus transforms:

Transform	Acts on	Effect
$y = \lvert f(x)\rvert$	the output $y$	reflect below-axis parts up
$y = f(\lvert x\rvert)$	the input $x$	mirror right half into left

Cambridge tip. Decide which modulus first: bars around $f(x)$ act on $y$ ; bars around $x$ act on $x$ . Mixing them up is the single most common error in this subtopic.

Keep $x \geq 0$ part; mirror it into $x < 0$ .
Result is even — symmetric about the $y$ -axis.
$|f(x)|$ acts on output; $f(|x|)$ acts on input.

See the full worked example for graph transformations →

y² = f(x) — symmetric about the x-axis

Only exists where f(x) ≥ 0; for each such x there are two y-values, ±√f(x).

Rule. $y^2 = f(x)$ means $y = \pm\sqrt{f(x)}$ , so:

The curve only exists where $f(x) \geq 0$ (you cannot square-root a negative).
For each valid $x$ there are two $y$ -values ( $+\sqrt{f}$ and $-\sqrt{f}$ ) → the graph is symmetric about the $x$ -axis.

Key mapping points:

Where $f(x) = 0$ : $y = 0$ — the curve meets the $x$ -axis (often at a sharp point/vertical tangent).
Where $f(x) = 1$ : $y = \pm 1$ .
Large $f(x)$ → the $\pm\sqrt{f}$ values grow more slowly (square-root flattening).
Where $f(x) < 0$ : no curve at all (a gap).

Behaviour near roots. Near a simple zero of $f$ , $y^2 = f(x)$ has a vertical tangent — the two branches meet the axis at right angles, like the left tip of a sideways parabola.

y² = f(x) draws two mirror branches above and below the x-axis, starting where f(x) = 0 with a vertical tangent.

Cambridge tip. Shade or note the region where $f(x) < 0$ as "no curve" — examiners look for awareness that the graph is undefined there.

Exists only where $f(x) \geq 0$ .
Symmetric about the $x$ -axis ( $y = \pm\sqrt{f}$ ).
Vertical tangent where $f(x) = 0$ .

See the full worked example for graph transformations →

y = 1/f(x) — swap zeros and asymptotes

Roots of f become vertical asymptotes; turning points invert; sign is preserved.

Rule. Taking reciprocals swaps "small" and "large" and keeps the sign.

Key mappings:

Where $f(x) = 0$ → $\frac{1}{f(x)}$ has a vertical asymptote.
Where $f(x)$ has a vertical asymptote → $\frac{1}{f(x)} = 0$ (curve meets the axis).
Where $f(x) = 1$ → $\frac{1}{f} = 1$ ; where $f = -1$ → $\frac{1}{f} = -1$ (fixed points).
A maximum of $f$ (positive) becomes a minimum of $\frac{1}{f}$ , and vice versa.
Sign is preserved: $f > 0 \Rightarrow \frac1f > 0$ ; $f < 0 \Rightarrow \frac1f < 0$ .

Behaviour:

As $f(x) \to \pm\infty$ , $\frac{1}{f(x)} \to 0$ .
As $f(x) \to 0^+$ , $\frac{1}{f(x)} \to +\infty$ ; as $f(x) \to 0^-$ , $\frac1{f(x)} \to -\infty$ .

Cambridge tip. Plot the lines $y = 1$ and $y = -1$ lightly — the original and reciprocal curves cross exactly where $f = \pm 1$ , which anchors your sketch.

Zeros of $f$ → vertical asymptotes of $\frac1f$ (and vice versa).
Max ↔ min; sign preserved.
$f \to \pm\infty \Rightarrow \frac1f \to 0$ .

See the full worked example for graph transformations →

Quick recap

$y = |f(x)|$ : reflect below-axis parts up; corners at $x$ -intercepts.
$y = f(|x|)$ : mirror the $x \geq 0$ half into $x < 0$ ; even function.
$y^2 = f(x)$ : only where $f \geq 0$ ; symmetric about the $x$ -axis.
$y = \frac{1}{f(x)}$ : zeros ↔ asymptotes; max ↔ min; sign preserved.
Always sketch the original $y = f(x)$ first.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$|f(x)|$ acts on the output; $f(|x|)$ acts on the input.
$y^2 = f(x)$ needs $f(x) \geq 0$ and is $x$ -axis symmetric.
Reciprocal swaps zeros ↔ asymptotes, max ↔ min, keeps sign.
Anchor points $f = 0, \pm 1$ and the asymptotes.

How it’s examined

Graph transformations are tested as sketch sub-questions within the Topic 1.2 rational-functions question (2-4 marks each). Cambridge gives you $y = f(x)$ (or a rational function to analyse) and asks you to sketch one or two of the related curves. Marks are awarded for the correct overall shape, the right treatment of asymptotes/intercepts, and symmetry. The classic error is confusing $|f(x)|$ with $f(|x|)$ .

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Graph transformations

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

Question type:

1Sketching y = |f(x)|
Getting startedGraph or diagram• modulus
Question
The line $y = 2x - 4$ is given. Describe the graph of $y = |2x - 4|$ .
Step-by-step solution
1. Step 1
  Find where $f(x) < 0$ . $2x - 4 < 0$ for $x < 2$ — this part is reflected up.
2. Step 2
  For $x \geq 2$ : $y = 2x - 4$ unchanged (already $\geq 0$ ).
3. Step 3
  For $x < 2$ : reflect, giving $y = -(2x - 4) = 4 - 2x$ . A V-shape with vertex at $(2, 0)$ .
Answer
A V-shape with its vertex (corner) at $(2, 0)$ , both arms rising; never below the $x$ -axis.
Examiner tip
The corner sits exactly at the $x$ -intercept of the original line. Mark it clearly.
2Sketching y = f(|x|)
Getting startedGraph or diagram• modulus
Question
Given $f(x) = x^2 - 3x$ , describe the graph of $y = f(|x|)$ .
Step-by-step solution
1. Step 1
  Keep the $x \geq 0$ part of $y = x^2 - 3x$ (a parabola through $0$ and $3$ , minimum at $x = 1.5$ ).
2. Step 2
  Discard $x < 0$ and replace it with the mirror image of the right half in the $y$ -axis.
3. Step 3
  Result: symmetric about the $y$ -axis, with minima at $x = \pm 1.5$ (value $-2.25$ ) and a local maximum/corner at $(0, 0)$ .
Answer
An even (y-axis symmetric) 'W-like' curve: two minima at $(\pm 1.5, -2.25)$ meeting at a corner $(0,0)$ .
Examiner tip
Only the $x \geq 0$ behaviour of $f$ is used; the original $x < 0$ branch is irrelevant.
3Sketching y² = f(x)
Building confidenceGraph or diagram• y-squared
Question
Given $f(x) = x - 1$ , sketch $y^2 = x - 1$ .
Step-by-step solution
1. Step 1
  Domain: need $f(x) \geq 0$ , i.e. $x \geq 1$ . No curve for $x < 1$ .
2. Step 2
  Two branches $y = \pm\sqrt{x - 1}$ , symmetric about the $x$ -axis.
3. Step 3
  At $x = 1$ : $y = 0$ with a vertical tangent. At $x = 2$ : $y = \pm 1$ ; at $x = 5$ : $y = \pm 2$ .
Answer
A sideways parabola opening right, vertex $(1, 0)$ , symmetric about the $x$ -axis, existing only for $x \geq 1$ .
Examiner tip
$y^2 = x - 1$ is the standard sideways parabola — recognising it confirms the vertical tangent at the vertex.
4Sketching y = 1/f(x)
Building confidenceGraph or diagram• reciprocal
Question
Given $f(x) = x^2 - 1$ , describe the key features of $y = \dfrac{1}{x^2 - 1}$ .
Step-by-step solution
1. Step 1
  Zeros of $f$ become vertical asymptotes: $x^2 - 1 = 0 \Rightarrow x = \pm 1$ .
2. Step 2
  Horizontal asymptote: as $x \to \pm\infty$ , $f \to \infty$ so $\frac1f \to 0$ , giving $y = 0$ .
3. Step 3
  Turning point: $f$ has a minimum $-1$ at $x = 0$ , so $\frac1f$ has a maximum $-1$ at $(0, -1)$ (sign preserved, max↔min).
Answer
Vertical asymptotes $x = \pm 1$ , horizontal asymptote $y = 0$ , local maximum at $(0, -1)$ ; three branches.
Examiner tip
Between $x = -1$ and $1$ , $f < 0$ so $\frac1f < 0$ — the middle branch lies entirely below the axis with a maximum at $(0,-1)$ .
5Modulus of a rational function
StretchGraph or diagram• modulus, rational
Question
The curve $y = \dfrac{x}{x - 2}$ has asymptotes $x = 2$ , $y = 1$ . Describe $y = \left|\dfrac{x}{x - 2}\right|$ .
Step-by-step solution
1. Step 1
  Where is $f < 0$ ? $\frac{x}{x-2} < 0$ for $0 < x < 2$ (numerator and denominator have opposite signs).
2. Step 2
  Reflect that region up. Elsewhere ( $x < 0$ or $x > 2$ ) the curve is unchanged.
3. Step 3
  Asymptotes: $x = 2$ stays; the horizontal asymptote $y = 1$ stays (it was already positive). The branch for $0 < x < 2$ now rises from $(0,0)$ up to $+\infty$ near $x = 2$ .
Answer
All branches on/above the $x$ -axis; vertical asymptote $x = 2$ , horizontal asymptote $y = 1$ ; the $0 < x < 2$ portion is reflected upward with a corner at $(0, 0)$ .
Examiner tip
Identify the sign of $f$ region-by-region using the critical points $x = 0$ and $x = 2$ before reflecting.
6y² = f(x) for a quadratic f
StretchGraph or diagram• y-squared
Question
Sketch $y^2 = x(4 - x)$ , identifying where the curve exists and its symmetry.
Step-by-step solution
1. Step 1
  Domain: need $x(4 - x) \geq 0$ , i.e. $0 \leq x \leq 4$ .
2. Step 2
  Symmetry: about the $x$ -axis ( $y = \pm\sqrt{x(4-x)}$ ). The curve also turns out to be symmetric about $x = 2$ .
3. Step 3
  Key points: $y = 0$ at $x = 0$ and $x = 4$ (vertical tangents); maximum $|y|$ at $x = 2$ , where $y^2 = 4$ , so $y = \pm 2$ .
Answer
A closed oval-like loop between $x = 0$ and $x = 4$ , symmetric about the $x$ -axis, widest at $x = 2$ ( $y = \pm 2$ ).
Examiner tip
$y^2 = x(4-x)$ is actually the circle/ellipse-type loop $\;(x-2)^2 + y^2 = 4$ — a complete closed curve, a common 'show that' follow-up.

Model Answers — Graph transformations

High-scoring sample answers for graph transformations on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Sketch the graph of $y = |3 - x|$ , stating the coordinates of the vertex and the $y$ -intercept.
Model answer
$f(x) = 3 - x$ is zero at $x = 3$ and negative for $x > 3$ .

For $x \leq 3$ : $y = 3 - x$ (unchanged, positive).

For $x > 3$ : reflect, $y = -(3 - x) = x - 3$ .

This is a V-shape with vertex at $(3, 0)$ . The $y$ -intercept (put $x = 0$ ) is $y = |3| = 3$ , i.e. $(0, 3)$ . Both arms have gradient $\pm 1$ .
Why this scores
B1 vertex $(3,0)$ , B1 $y$ -intercept $(0,3)$ , B1 correct V-shape never dipping below the axis.
Question 2
4 marks
The graph of $y = f(x)$ where $f(x) = x^3 - x$ is given. Describe how to obtain, and the symmetry of, the graph of $y = f(|x|)$ .
Model answer
For $y = f(|x|)$ :

Keep the portion of $y = x^3 - x$ for $x \geq 0$ (which passes through $(0,0)$ and $(1,0)$ with a minimum between).

Replace the $x < 0$ portion with the reflection of the $x \geq 0$ part in the $y$ -axis.

The resulting graph is symmetric about the $y$ -axis (an even function). It passes through $(\pm 1, 0)$ and $(0, 0)$ , with matching minima at $x = +\tfrac{1}{\sqrt3}$ and $x = -\tfrac{1}{\sqrt3}$ .
Why this scores
B1 keep right half, B1 reflect into left, B1 state even/y-axis symmetry, B1 correct mapped features. Note the original odd symmetry of $x^3 - x$ is destroyed — only $x \geq 0$ matters.
Question 3
5 marks
Sketch $y^2 = 2x + 4$ , stating the domain of $x$ and the points where the curve meets the axes.
Model answer
Rearrange: $y = \pm\sqrt{2x + 4}$ , which requires $2x + 4 \geq 0$ , i.e. $x \geq -2$ (domain).

Meets the $x$ -axis where $y = 0$ : $2x + 4 = 0 \Rightarrow x = -2$ , the point $(-2, 0)$ — a vertical tangent here.

Meets the $y$ -axis where $x = 0$ : $y^2 = 4 \Rightarrow y = \pm 2$ , the points $(0, 2)$ and $(0, -2)$ .

The curve is a sideways parabola opening to the right, vertex $(-2, 0)$ , symmetric about the $x$ -axis, existing only for $x \geq -2$ .
Why this scores
B1 domain $x \geq -2$ , B1 vertex/x-intercept with vertical tangent, B1 B1 the two $y$ -intercepts, B1 correct symmetric shape.
Question 4
4 marks
Given $f(x) = x - 2$ , sketch $y = \dfrac{1}{x - 2}$ and state its asymptotes.
Model answer
Vertical asymptote where $f = 0$ : $x = 2$ .

Horizontal asymptote: as $x \to \pm\infty$ , $\dfrac{1}{x-2} \to 0$ , so $y = 0$ .

Sign behaviour:

For $x > 2$ : $f > 0$ so $y > 0$ ; as $x \to 2^+$ , $y \to +\infty$ .

For $x < 2$ : $f < 0$ so $y < 0$ ; as $x \to 2^-$ , $y \to -\infty$ .

The graph is a rectangular hyperbola with branches in the regions $x > 2$ (above axis) and $x < 2$ (below axis). The $y$ -intercept is $\left(0, -\tfrac12\right)$ .
Why this scores
B1 $x = 2$ , B1 $y = 0$ , B1 correct branch signs, B1 $y$ -intercept. Reciprocal of a zero gives the asymptote — the core mapping.
Question 5
6 marks
The curve $y = f(x)$ where $f(x) = (x + 1)(x - 3)$ is a parabola with minimum $(1, -4)$ . Sketch $y = \dfrac{1}{f(x)}$ , giving all asymptotes and turning points.
Model answer
Vertical asymptotes at the zeros of $f$ : $x = -1$ and $x = 3$ .

Horizontal asymptote: as $x \to \pm\infty$ , $f \to +\infty$ , so $y = \dfrac1f \to 0$ , i.e. $y = 0$ .

Turning point: $f$ has a minimum $-4$ at $x = 1$ , so $\dfrac1f$ has a maximum at $x = 1$ with value $\dfrac{1}{-4} = -\dfrac14$ , i.e. the point $\left(1, -\tfrac14\right)$ .

Three branches:

$x < -1$ : $f > 0$ , so $y > 0$ , decreasing from $0^+$ up to $+\infty$ near $x = -1^-$ .

$-1 < x < 3$ : $f < 0$ , so $y < 0$ , with the maximum $\left(1, -\tfrac14\right)$ ; $y \to -\infty$ at both asymptotes.

$x > 3$ : $f > 0$ , so $y > 0$ , falling from $+\infty$ towards $0^+$ .
Why this scores
B1 B1 the two vertical asymptotes, B1 horizontal asymptote, B1 the max↔min inversion giving $(1,-\tfrac14)$ , B1 B1 correct sign of each branch. The minimum of $f$ becoming a maximum of $\frac1f$ (sign preserved) is the key A* discriminator.
Question 6
7 marks
The curve $y = f(x)$ where $f(x) = \dfrac{x - 1}{x + 1}$ has asymptotes $x = -1$ and $y = 1$ , and passes through $(1, 0)$ and $(0, -1)$ . (a) Sketch $y = |f(x)|$ . (b) Sketch $y^2 = f(x)$ , stating where it exists.
Model answer
(a) $y = |f(x)|$ . Determine where $f < 0$ : $\dfrac{x-1}{x+1} < 0$ for $-1 < x < 1$ . Reflect that region above the axis; elsewhere unchanged.

Vertical asymptote $x = -1$ stays.

Horizontal asymptote $y = 1$ stays (positive).

The branch for $-1 < x < 1$ now rises from $+\infty$ (near $x=-1^+$ ) down to a corner at $(1, 0)$ , then the curve continues as $f$ for $x > 1$ approaching $y = 1$ from below. The point $(0,-1)$ becomes $(0, 1)$ .

(b) $y^2 = f(x)$ . Exists only where $f(x) \geq 0$ , i.e. $x \leq -1$ (excluded, asymptote) or $x \geq 1$ . So the curve exists for $x > -1$ is excluded in $(-1,1)$ ; valid for $x < -1$ and $x \geq 1$ .

For $x \geq 1$ : $y = \pm\sqrt{f(x)}$ , starting at $(1, 0)$ with a vertical tangent, the two branches widening towards $y = \pm 1$ as $x \to \infty$ (since $f \to 1$ ).

For $x < -1$ : $f > 1$ there, so $y = \pm\sqrt{f}$ gives two branches with $|y| > 1$ , symmetric about the $x$ -axis, approaching the asymptotic behaviour near $x = -1^-$ .
Why this scores
(a) B1 reflect $-1<x<1$ , B1 asymptotes retained, B1 corner at $(1,0)$ , B1 mapped point $(0,1)$ . (b) B1 domain $f\geq0$ , B1 $x$ -axis symmetry with vertical tangent at $(1,0)$ , B1 limiting behaviour $y\to\pm1$ . Stating the existence region precisely is essential.

Key Formulae — Graph transformations

The formulae you need to memorise for graph transformations on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

y = |f(x)| (modulus of output)
$y = |f(x)| = \begin{cases} f(x) & f(x) \geq 0 \\ -f(x) & f(x) < 0 \end{cases}$
When to use
Reflect the below-axis part of $f$ upward; corners at the $x$ -intercepts.
y = f(|x|) (modulus of input)
$y = f(|x|) = \begin{cases} f(x) & x \geq 0 \\ f(-x) & x < 0 \end{cases}$
When to use
Mirror the $x \geq 0$ half of $f$ into $x < 0$ ; always an even function.
y² = f(x)
$y = \pm\sqrt{f(x)} \quad (f(x) \geq 0)$
When to use
Two branches symmetric about the $x$ -axis; only where $f \geq 0$ ; vertical tangent where $f = 0$ .
y = 1/f(x)
$f(x) = 0 \;\to\; \text{asymptote}; \quad f \to \pm\infty \;\to\; \tfrac1f \to 0$
When to use
Zeros ↔ vertical asymptotes; maxima ↔ minima; sign of $f$ preserved.

Key Definitions and Keywords — Graph transformations

Definitions to memorise and the exact keywords mark schemes credit for graph transformations answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Modulus (absolute value)
Examiner keyword
$|a|$ is the non-negative magnitude of $a$ : $|a| = a$ if $a \geq 0$ , and $-a$ if $a < 0$ .
Even function
Examiner keyword
A function with $f(-x) = f(x)$ , symmetric about the $y$ -axis. $y = f(|x|)$ is always even.
Vertical tangent
Examiner keyword
A point where the tangent line is vertical; occurs on $y^2 = f(x)$ at the points where $f(x) = 0$ .
Reciprocal curve
Examiner keyword
The graph of $y = \frac{1}{f(x)}$ , in which zeros and vertical asymptotes of $f$ are interchanged and the sign of $f$ is preserved.
Invariant (fixed) point
A point unchanged by a transformation. For $y = \frac1{f}$ , the points where $f = \pm 1$ are fixed; for $y = |f|$ , points already on/above the axis are fixed.

Common Mistakes and Misconceptions — Graph transformations

The traps other students keep falling into on graph transformations questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing $y = |f(x)|$ with $y = f(|x|)$
9231 Paper 1 Examiner Reports
Why it happens
Both involve modulus bars; students don't check whether they wrap $x$ or $f(x)$ .
How to avoid it
Bars around $f(x)$ act on the OUTPUT (reflect up); bars around $x$ act on the INPUT (mirror left↔right). Say which before sketching.
✕Drawing $y^2 = f(x)$ where $f(x) < 0$
9231 Paper 1 Examiner Reports
Why it happens
Forgetting that a square cannot be negative.
How to avoid it
First shade the region where $f(x) \geq 0$ ; the curve exists only there. Mark $f < 0$ as 'no curve'.
✕Flipping the sign when taking a reciprocal (e.g. a min of $f$ below the axis becoming a min above)
9231 Paper 1 Examiner Reports
Why it happens
Assuming reciprocal reflects the curve in the $x$ -axis.
How to avoid it
Reciprocal PRESERVES sign: $f<0 \Rightarrow \frac1f < 0$ . A negative minimum of $f$ becomes a negative MAXIMUM of $\frac1f$ .
✕Drawing a smooth curve through the $x$ -intercepts of $|f(x)|$
9231 Paper 1 mark schemes
Why it happens
Reflecting visually but rounding off the join.
How to avoid it
Reflection in the $x$ -axis creates a sharp corner (cusp) at each $x$ -intercept — draw it pointed, not rounded.
✕Not turning zeros of $f$ into asymptotes of $\frac1f$
9231 Paper 1 Examiner Reports
Why it happens
Treating the reciprocal curve as just a vertical stretch.
How to avoid it
Each zero of $f$ becomes a vertical asymptote of $\frac1f$ ; each asymptote of $f$ becomes a zero of $\frac1f$ . Map these first.
✕Drawing $y^2 = f(x)$ meeting the $x$ -axis at a smooth/horizontal point
9231 Paper 1 Examiner Reports
Why it happens
Copying the shape of $y = \sqrt{f}$ without the lower branch.
How to avoid it
At $f(x) = 0$ the two branches $\pm\sqrt{f}$ meet with a VERTICAL tangent — draw them meeting the axis at right angles.

Graph transformations

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Sketching y = |f(x)|

2Sketching y = f(|x|)

3Sketching y² = f(x)

4Sketching y = 1/f(x)

5Modulus of a rational function

6y² = f(x) for a quadratic f

y = |f(x)| (modulus of output)

y = f(|x|) (modulus of input)

y² = f(x)

y = 1/f(x)

Modulus (absolute value)

Even function

Vertical tangent

Reciprocal curve

Invariant (fixed) point

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

✕Drawing y2=f(x)y^2 = f(x)y2=f(x) where f(x)<0f(x) < 0f(x)<0

✕Flipping the sign when taking a reciprocal (e.g. a min of fff below the axis becoming a min above)

✕Drawing a smooth curve through the xxx-intercepts of ∣f(x)∣|f(x)|∣f(x)∣

✕Not turning zeros of fff into asymptotes of 1f\frac1ff1​

✕Drawing y2=f(x)y^2 = f(x)y2=f(x) meeting the xxx-axis at a smooth/horizontal point

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

✕Drawing $y^2 = f(x)$ where $f(x) < 0$

✕Flipping the sign when taking a reciprocal (e.g. a min of $f$ below the axis becoming a min above)

✕Drawing a smooth curve through the $x$ -intercepts of $|f(x)|$

✕Not turning zeros of $f$ into asymptotes of $\frac1f$

✕Drawing $y^2 = f(x)$ meeting the $x$ -axis at a smooth/horizontal point