Why is "Drawing $\cosh x$ with minimum at the origin $(0, 0)$" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Confusing cosh with a parabola or with sinh, or forgetting 0 = 1. How to avoid it: Always plot the minimum at (0, 1) and label it; the range is y ≥ 1, not y ≥ 0. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Giving $\sinh x$ horizontal asymptotes" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Carrying over the bounded behaviour of to . How to avoid it: x — it is unbounded and has NO asymptotes. Only (and the reciprocals) have horizontal asymptotes. Source: 9231 Paper 2 mark schemes

Why is "Stating the range of $\tanh x$ as $-1 \leq y \leq 1$ (closed interval)" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Treating the asymptotes as attainable values. How to avoid it: x approaches but never equals 1, so the range is the OPEN interval -1 < y < 1. Source: 9231 Paper 2 Examiner Reports

Why is "Turning a minimum of cosh into a minimum of sech (instead of a maximum)" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Forgetting that reciprocation flips minima to maxima and sends to 0. How to avoid it: Reciprocal-graph rule: a minimum of f at value m>0 becomes a maximum of 1/f at value 1/m; where f , 1/f 0. Source: 9231 Paper 2 Examiner Reports

Why is "Omitting the vertical asymptote $x = 0$ of $\coth$ and $\text{cosech}$" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Not checking where the denominator x vanishes. How to avoid it: Both x = x/ x and cosech\,x = 1/ x are undefined at x = 0 — mark the vertical asymptote. Source: 9231 Paper 2 mark schemes

Why is "Getting the direction/factor of a horizontal stretch wrong (e.g. $\cosh\frac{x}{2}$ stretched by $\frac{1}{2}$)" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Confusing f(x/2) (stretch factor 2) with f(2x) (stretch factor 1/2). How to avoid it: Replacing x by x/k stretches horizontally by factor k. So /2 is WIDER (factor 2). Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Sketching Graphs of Hyperbolic Functions

Q: Why is "Turning a minimum of cosh into a minimum of sech (instead of a maximum)" a common mistake in Sketching Graphs of Hyperbolic Functions?

Why it happens: Forgetting that reciprocation flips minima to maxima and sends to 0. How to avoid it: Reciprocal-graph rule: a minimum of f at value m>0 becomes a maximum of 1/f at value 1/m; where f , 1/f 0. Source: 9231 Paper 2 Examiner Reports

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Detailed Study Notes

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

Sketching Graphs of Hyperbolic Functions — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (2026-2027 syllabus)

Master the shapes of $\sinh$ , $\cosh$ , $\tanh$ and their reciprocals: domains, ranges, odd/even symmetry, and the asymptotic behaviour that flows directly from $e^x$ and $e^{-x}$ .

At a glance

$\sinh x$ — odd, through the origin, flattened-cubic shape, no asymptotes, range all reals.
$\cosh x$ — even, minimum $(0, 1)$ , the catenary (hanging chain), range $y \geq 1$ .
$\tanh x$ — odd, S-shaped, horizontal asymptotes $y = \pm 1$ , range $-1 < y < 1$ .
Reciprocals ( $\text{sech}, \text{cosech}, \coth$ ) — get their shapes by reciprocating cosh/sinh/tanh.
Everything follows from $e^x$ and $e^{-x}$ : which term dominates tells you the limiting behaviour.

What you’ll learn

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

2.1d — Sketch the graphs of $\sinh x$ , $\cosh x$ and $\tanh x$ , identifying key features (intercepts, symmetry, asymptotes).
2.1e — Sketch the reciprocal hyperbolic functions $\text{sech}\,x$ , $\text{cosech}\,x$ and $\coth x$ and state their domains and ranges.
2.1f — Relate the asymptotic behaviour of each curve to the exponential terms $e^x$ and $e^{-x}$ .

The graphs of sinh and cosh

sinh: odd, through the origin. cosh: even, minimum 1 — the catenary.

Both graphs come straight from the definitions $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$ .

$y = \sinh x$ :

Odd: $\sinh(-x) = -\sinh x$ , so it has rotational symmetry about the origin.
Passes through $(0, 0)$ with gradient $1$ (since $\frac{d}{dx}\sinh x = \cosh x$ and $\cosh 0 = 1$ ).
Increasing everywhere, unbounded: $\sinh x \to +\infty$ as $x \to +\infty$ and $\to -\infty$ as $x \to -\infty$ .
No asymptotes. Shape: a 'flattened cubic'.

$y = \cosh x$ :

Even: $\cosh(-x) = \cosh x$ , symmetric about the $y$ -axis.
Minimum point $(0, 1)$ — the value $\cosh 0 = 1$ , and $\cosh x \geq 1$ always (because $e^x + e^{-x} \geq 2$ ).
Range $y \geq 1$ ; unbounded above.
This U-shaped curve is the catenary, the shape of a chain hanging under its own weight.

cosh (blue): even, minimum (0,1). sinh (red): odd, through the origin. Both ≈ ½eˣ for large positive x, so they converge on the right.

Cambridge tip. The single most common mark lost here is drawing $\cosh$ with its minimum at the origin. Always plot and label the minimum at $(0, 1)$ .

$\sinh$ : odd, origin, gradient $1$ , no asymptotes.
$\cosh$ : even, minimum $(0, 1)$ , range $y \geq 1$ (the catenary).
Both $\approx \frac{1}{2}e^x$ as $x \to +\infty$ , so they converge.

See the full worked example for sketching graphs of hyperbolic functions →

tanh and its asymptotes

An odd S-curve trapped between y = 1 and y = −1.

$y = \tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^{2x} - 1}{e^{2x} + 1}$ .

Key features:

Odd, through the origin with gradient $\text{sech}^2 0 = 1$ .
S-shaped and increasing.
Horizontal asymptotes $y = 1$ and $y = -1$ . As $x \to +\infty$ , $\tanh x \to 1$ ; as $x \to -\infty$ , $\tanh x \to -1$ .
Range $-1 < y < 1$ (open interval — the asymptotes are never reached).

Why the asymptotes appear. Divide the exponential form top and bottom by $e^{2x}$ : $\tanh x = \frac{1 - e^{-2x}}{1 + e^{-2x}}$ As $x \to +\infty$ , $e^{-2x} \to 0$ , giving $\tanh x \to \frac{1}{1} = 1$ . As $x \to -\infty$ , $e^{2x} \to 0$ in the original form gives $\frac{0 - 1}{0 + 1} = -1$ .

A deeper way to see it: for large $x$ , $\sinh x$ and $\cosh x$ both behave like $\frac{1}{2}e^x$ , so their ratio tends to $1$ .

tanh: odd, through the origin, squeezed between the asymptotes y = ±1 (dashed) which it approaches but never reaches.

Cambridge tip. Always draw the asymptotes as dashed lines and state the range with strict inequalities: $-1 < y < 1$ . Writing $-1 \leq y \leq 1$ is a marked error.

$\tanh$ : odd S-curve through the origin.
Horizontal asymptotes $y = \pm 1$ (dashed, never reached).
Range is the OPEN interval $-1 < y < 1$ .

See the full worked example for sketching graphs of hyperbolic functions →

The reciprocal functions: sech, cosech, coth

Reciprocate cosh, sinh, tanh — minima become maxima, infinities become zeros.

Sketch each reciprocal by applying the reciprocal-graph rules to the function underneath:

A minimum at value $m > 0$ becomes a maximum at $\frac{1}{m}$ .
Where the original $\to \pm\infty$ , the reciprocal $\to 0$ .
Where the original $= 0$ , the reciprocal has a vertical asymptote.

$y = \text{sech}\,x = \dfrac{1}{\cosh x}$ :

$\cosh x$ has minimum $1$ at $x = 0$ , so $\text{sech}$ has a maximum $(0, 1)$ .
Even (cosh is even), bell-shaped.
As $x \to \pm\infty$ , $\cosh x \to \infty$ , so $\text{sech}\,x \to 0^+$ . Horizontal asymptote $y = 0$ . Range $0 < y \leq 1$ .

$y = \text{cosech}\,x = \dfrac{1}{\sinh x}$ :

$\sinh x = 0$ at $x = 0$ , so there is a vertical asymptote $x = 0$ . Domain $x \neq 0$ .
Odd; two branches.
As $x \to \pm\infty$ , $\text{cosech}\,x \to 0$ . Horizontal asymptote $y = 0$ . Range $y \neq 0$ .

$y = \coth x = \dfrac{\cosh x}{\sinh x} = \dfrac{1}{\tanh x}$ :

Vertical asymptote $x = 0$ (where $\sinh x = 0$ ). Domain $x \neq 0$ .
Odd. Since $\tanh x \to \pm 1$ , $\coth x \to \pm 1$ : horizontal asymptotes $y = \pm 1$ .
Range $|y| > 1$ (i.e. $y > 1$ or $y < -1$ ).

sech x: even, maximum (0,1), with the x-axis (y = 0) as a horizontal asymptote on both sides.

Cambridge tip. Never lose the vertical asymptotes of $\text{cosech}$ and $\coth$ at $x = 0$ — they appear because the denominator $\sinh x$ vanishes there.

$\text{sech}$ : even, maximum $(0, 1)$ , asymptote $y = 0$ .
$\text{cosech}$ , $\coth$ : vertical asymptote $x = 0$ (where $\sinh = 0$ ).
$\coth$ also has horizontal asymptotes $y = \pm 1$ ; range $|y| > 1$ .

See the full worked example for sketching graphs of hyperbolic functions →

Quick recap

$\sinh$ : odd, origin, gradient $1$ , no asymptotes, range $\mathbb{R}$ .
$\cosh$ : even, minimum $(0, 1)$ , range $y \geq 1$ (the catenary).
$\tanh$ : odd S-curve, asymptotes $y = \pm 1$ , range $-1 < y < 1$ (strict).
$\text{sech}$ : even, maximum $(0, 1)$ , asymptote $y = 0$ .
$\text{cosech}$ and $\coth$ : vertical asymptote $x = 0$ ; $\coth$ also has $y = \pm 1$ .
All asymptotic behaviour comes from which of $e^x$ , $e^{-x}$ dominates.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\cosh$ minimum is $(0, 1)$ , NOT the origin.
$\tanh$ range is $-1 < y < 1$ (strict) with asymptotes $y = \pm 1$ .
$\sinh$ has no asymptotes — it is unbounded.
Reciprocal rule: minimum $m \to$ maximum $\frac{1}{m}$ ; $\infty \to 0$ ; zero $\to$ vertical asymptote.
$\text{cosech}$ and $\coth$ have a vertical asymptote at $x = 0$ .

How it’s examined

Sketching questions appear in Paper 2 (Further Pure 2), often as the opening part of a longer hyperbolic-functions question (3-5 marks), or attached to a transformation (e.g. sketch $y = a\cosh\frac{x}{a}$ ). Marks reward correct symmetry, the labelled minimum $(0,1)$ for cosh, asymptotes drawn as dashed lines, and strict ranges. The most penalised errors are a cosh minimum at the origin, missing/extra asymptotes, and a closed range for tanh.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2); 9231/22 & 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

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Step-by-step worked examples — Sketching Graphs of Hyperbolic Functions

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

Question type:

1Key features of y = sinh x
Getting startedGraph or diagram• sinh, sketch
Question
State the domain, range and symmetry of $y = \sinh x$ , and describe its behaviour as $x \to \pm\infty$ .
Step-by-step solution
1. Step 1
  Domain. $\sinh x = \frac{e^x - e^{-x}}{2}$ is defined for every real $x$ .
  $\text{domain: } x \in \mathbb{R}$
2. Step 2
  Range and symmetry. $\sinh$ is odd ( $\sinh(-x) = -\sinh x$ ), passes through the origin, and takes every real value.
  $\text{range: } y \in \mathbb{R}$
3. Step 3
  Limits. For large $x$ , $e^{-x} \to 0$ so $\sinh x \approx \frac{1}{2}e^x \to +\infty$ ; for large negative $x$ , $\sinh x \to -\infty$ .
  $\sinh x \to +\infty \text{ as } x \to +\infty, \quad \sinh x \to -\infty \text{ as } x \to -\infty$
Answer
Domain $\mathbb{R}$ , range $\mathbb{R}$ , odd (origin symmetry), shaped like a flattened cubic with no asymptotes.
Examiner tip
'Odd' and 'through the origin' are the two phrases the mark scheme wants. $\sinh$ has NO horizontal asymptotes (unlike tanh).
2Key features of y = cosh x
Getting startedGraph or diagram• cosh, sketch
Question
State the domain, range, symmetry and minimum point of $y = \cosh x$ .
Step-by-step solution
1. Step 1
  Domain. Defined for all real $x$ .
  $\text{domain: } x \in \mathbb{R}$
2. Step 2
  Minimum. $\cosh 0 = \frac{1 + 1}{2} = 1$ , and $\cosh x \geq 1$ for all $x$ because $e^x + e^{-x} \geq 2$ (AM-GM).
  $\text{minimum point } (0, 1), \quad \text{range: } y \geq 1$
3. Step 3
  Symmetry. $\cosh$ is even: $\cosh(-x) = \cosh x$ , so the curve is symmetric about the $y$ -axis. This U-shape is the catenary (a hanging chain).
  $\cosh(-x) = \cosh x$
Answer
Domain $\mathbb{R}$ , range $y \geq 1$ , even (symmetric about the $y$ -axis), minimum at $(0, 1)$ — the catenary.
Examiner tip
The minimum value is $1$ , NOT $0$ . Stating the range as $y \geq 1$ and the point $(0,1)$ are both marked.
3Asymptotes of y = tanh x
Building confidenceGraph or diagram• tanh, asymptote
Question
Show that $y = \tanh x$ has horizontal asymptotes $y = 1$ and $y = -1$ , and state its domain and range.
Step-by-step solution
1. Step 1
  Use the exponential form $\tanh x = \dfrac{e^{2x} - 1}{e^{2x} + 1}$ .
  $\tanh x = \frac{e^{2x} - 1}{e^{2x} + 1}$
2. Step 2
  As $x \to +\infty$ , $e^{2x} \to \infty$ , so dividing top and bottom by $e^{2x}$ gives $\frac{1 - e^{-2x}}{1 + e^{-2x}} \to \frac{1}{1} = 1$ .
  $\tanh x \to 1 \text{ as } x \to +\infty$
3. Step 3
  As $x \to -\infty$ , $e^{2x} \to 0$ , so $\tanh x \to \frac{-1}{1} = -1$ .
  $\tanh x \to -1 \text{ as } x \to -\infty$
4. Step 4
  Domain and range. $\cosh x \neq 0$ ever, so the domain is all reals; the range is the open interval between the asymptotes.
  $\text{domain: } x \in \mathbb{R}, \quad \text{range: } -1 < y < 1$
Answer
Asymptotes $y = \pm 1$ ; domain $\mathbb{R}$ ; range $-1 < y < 1$ ; odd, through the origin.
Examiner tip
Dividing through by $e^{2x}$ is the rigorous way to find the limits. The range uses strict inequalities — $\tanh x$ never actually reaches $\pm 1$ .
4Sketching y = sech x
Building confidenceGraph or diagram• sech, reciprocal
Question
By considering $\text{sech}\,x = \dfrac{1}{\cosh x}$ , describe the graph of $y = \text{sech}\,x$ (maximum, symmetry, asymptote).
Step-by-step solution
1. Step 1
  Reciprocal of cosh. Since $\cosh x \geq 1$ with minimum $1$ at $x = 0$ , its reciprocal has a MAXIMUM at $x = 0$ .
  $\text{sech}\,0 = \frac{1}{\cosh 0} = \frac{1}{1} = 1$
2. Step 2
  Symmetry. $\cosh$ is even, so $\text{sech}$ is even too — symmetric about the $y$ -axis.
  $\text{sech}(-x) = \text{sech}\,x$
3. Step 3
  Asymptote. As $x \to \pm\infty$ , $\cosh x \to \infty$ , so $\text{sech}\,x \to 0^{+}$ . The $x$ -axis is a horizontal asymptote.
  $0 < \text{sech}\,x \leq 1$
Answer
A bell-shaped even curve with maximum $(0, 1)$ , range $0 < y \leq 1$ , and the $x$ -axis as a horizontal asymptote.
Examiner tip
Reciprocal-graph reasoning: a minimum of cosh becomes a maximum of sech; cosh→∞ becomes sech→0. The curve never touches the x-axis.
5Sketching y = coth x
StretchGraph or diagram• coth, reciprocal, asymptote
Question
Describe the graph of $y = \coth x = \dfrac{\cosh x}{\sinh x}$ , including all asymptotes, domain and range.
Step-by-step solution
1. Step 1
  Vertical asymptote. $\coth x = \frac{\cosh x}{\sinh x}$ is undefined where $\sinh x = 0$ , i.e. at $x = 0$ . So $x = 0$ is a vertical asymptote.
  $\text{domain: } x \neq 0$
2. Step 2
  Horizontal asymptotes. $\coth x = \frac{1}{\tanh x}$ , and since $\tanh x \to \pm 1$ , $\coth x \to \pm 1$ as $x \to \pm\infty$ .
  $\coth x \to 1 \text{ as } x \to +\infty, \quad \coth x \to -1 \text{ as } x \to -\infty$
3. Step 3
  Behaviour near $0$ . As $x \to 0^{+}$ , $\sinh x \to 0^{+}$ while $\cosh x \to 1$ , so $\coth x \to +\infty$ . As $x \to 0^{-}$ , $\coth x \to -\infty$ . The function is odd.
  $|\coth x| > 1$
Answer
Odd function; vertical asymptote $x = 0$ ; horizontal asymptotes $y = \pm 1$ ; domain $x \neq 0$ ; range $|y| > 1$ (i.e. $y > 1$ or $y < -1$ ).
Examiner tip
coth has THREE asymptotes: one vertical ( $x=0$ ) and two horizontal ( $y=\pm1$ ). The range excludes $-1 \leq y \leq 1$ entirely.
6Why sinh resembles a flattened cubic
StretchGraph or diagram• sinh, behaviour
Question
Explain, using the exponential definition, why $y = \sinh x$ behaves like $\frac{1}{2}e^x$ for large positive $x$ but passes through the origin with gradient $1$ .
Step-by-step solution
1. Step 1
  Large positive $x$ . In $\sinh x = \frac{e^x - e^{-x}}{2}$ , the term $e^{-x} \to 0$ , so the curve is dominated by $\frac{1}{2}e^x$ .
  $\sinh x \approx \tfrac{1}{2}e^x \text{ for large } x$
2. Step 2
  At the origin. $\sinh 0 = 0$ , so the curve passes through $(0,0)$ .
  $\sinh 0 = \frac{1 - 1}{2} = 0$
3. Step 3
  Gradient at origin. $\frac{d}{dx}\sinh x = \cosh x$ , and $\cosh 0 = 1$ , so the gradient at the origin is $1$ — the curve crosses with the same slope as $y = x$ .
  $\left.\frac{d}{dx}\sinh x\right|_{x=0} = \cosh 0 = 1$
Answer
Near $x = 0$ the curve looks like $y = x$ (gradient $1$ through the origin); for large $x$ it grows like $\frac{1}{2}e^x$ — giving the flattened-cubic shape.
Examiner tip
Linking the shape to the dominant exponential term is exactly the kind of reasoning examiners reward. The gradient $\cosh 0 = 1$ explains the near-linear behaviour at the origin.

Model Answers — Sketching Graphs of Hyperbolic Functions

High-scoring sample answers for sketching graphs of hyperbolic functions 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 $y = \sinh x$ , indicating clearly any points where it crosses the axes and describing its symmetry.
Model answer
Crossing point. $\sinh 0 = 0$ , so the curve passes through the origin $(0, 0)$ only.

Symmetry. $\sinh(-x) = -\sinh x$ , so the function is odd — it has rotational symmetry of order $2$ about the origin.

Shape. For large positive $x$ , $\sinh x \approx \frac{1}{2}e^x \to +\infty$ ; for large negative $x$ , $\sinh x \to -\infty$ . There are no asymptotes. The curve is monotonically increasing, resembling a flattened cubic, with gradient $\cosh 0 = 1$ at the origin.
Why this scores
B1 for through the origin, B1 for odd/rotational symmetry, B1 for the correct increasing flattened-cubic shape with no asymptotes. A common error is drawing asymptotes that belong to tanh.
Question 2
4 marks
Sketch $y = \cosh x$ , stating its minimum point and range, and explain why $\cosh x \geq 1$ for all $x$ .
Model answer
Minimum point. $\cosh 0 = \frac{e^0 + e^0}{2} = \frac{1+1}{2} = 1$ , giving a minimum at $(0, 1)$ .

Why $\cosh x \geq 1$ . By the AM-GM inequality (or completing the square), for positive quantities $e^x$ and $e^{-x}$ : $\frac{e^x + e^{-x}}{2} \geq \sqrt{e^x \cdot e^{-x}} = \sqrt{e^0} = 1$ with equality only when $e^x = e^{-x}$ , i.e. $x = 0$ .

Range. $y \geq 1$ .

Shape. An even, U-shaped curve (the catenary), symmetric about the $y$ -axis, increasing for $x > 0$ and decreasing for $x < 0$ , with no upper bound.
Why this scores
B1 minimum $(0,1)$ , B1 range $y \geq 1$ , M1 for the AM-GM (or equivalent) argument, A1 for a correct even U-shape. Stating minimum $0$ instead of $1$ is the classic slip.
Question 3
5 marks
Show that $y = \tanh x$ has asymptotes $y = \pm 1$ , and hence sketch the curve stating its range.
Model answer
Exponential form: $\tanh x = \frac{e^{2x} - 1}{e^{2x} + 1}$

As $x \to +\infty$ , divide top and bottom by $e^{2x}$ : $\tanh x = \frac{1 - e^{-2x}}{1 + e^{-2x}} \to \frac{1 - 0}{1 + 0} = 1$ so $y = 1$ is a horizontal asymptote.

As $x \to -\infty$ , $e^{2x} \to 0$ : $\tanh x \to \frac{0 - 1}{0 + 1} = -1$ so $y = -1$ is a horizontal asymptote.

Sketch. An odd, increasing S-shaped curve through the origin, squeezed between $y = 1$ and $y = -1$ . Range: $-1 < y < 1$ (strict — the asymptotes are never reached).
Why this scores
M1 for the exponential form, A1 A1 for the two limits, B1 for the S-shaped sketch through the origin, B1 for range $-1 < y < 1$ . The strict inequality matters.
Question 4
6 marks
By considering $\text{cosech}\,x = \dfrac{1}{\sinh x}$ , describe and sketch $y = \text{cosech}\,x$ , stating its domain, range, symmetry and all asymptotes.
Model answer
Domain. $\text{cosech}\,x = \frac{1}{\sinh x}$ is undefined where $\sinh x = 0$ , i.e. at $x = 0$ . So the domain is $x \neq 0$ and $x = 0$ is a vertical asymptote.

Symmetry. $\sinh$ is odd, so its reciprocal $\text{cosech}$ is also odd: $\text{cosech}(-x) = -\text{cosech}\,x$ .

Behaviour near $0$ . As $x \to 0^{+}$ , $\sinh x \to 0^{+}$ , so $\text{cosech}\,x \to +\infty$ . As $x \to 0^{-}$ , $\text{cosech}\,x \to -\infty$ .

Behaviour at infinity. As $x \to +\infty$ , $\sinh x \to +\infty$ , so $\text{cosech}\,x \to 0^{+}$ ; as $x \to -\infty$ , $\text{cosech}\,x \to 0^{-}$ . The $x$ -axis ( $y = 0$ ) is a horizontal asymptote.

Range. Every value except $0$ : $y \neq 0$ (the curve consists of two branches, one in each of the regions $x>0,\,y>0$ and $x<0,\,y<0$ ).

Sketch. Two disconnected branches, hugging the positive $y$ -axis near the origin (right branch) and the negative $y$ -axis (left branch), each flattening towards the $x$ -axis as $|x|$ grows.
Why this scores
B1 domain $x \neq 0$ with vertical asymptote, B1 odd, B1 behaviour near $0$ , B1 horizontal asymptote $y = 0$ , B1 range $y \neq 0$ , B1 correct two-branch sketch. Reciprocal-graph reasoning earns all the marks.
Question 5
7 marks
On a single diagram, sketch $y = \sinh x$ , $y = \cosh x$ and $y = \tanh x$ . Describe how each relates to the exponential terms $e^x$ and $e^{-x}$ , and identify which curves share which asymptotic behaviour.
Model answer
$y = \cosh x = \frac{e^x + e^{-x}}{2}$ . Even, U-shaped, minimum $(0, 1)$ . For large positive $x$ it follows $\frac{1}{2}e^x$ ; for large negative $x$ it follows $\frac{1}{2}e^{-x}$ . So cosh sits just above sinh for $x > 0$ .

$y = \sinh x = \frac{e^x - e^{-x}}{2}$ . Odd, through the origin, gradient $1$ there. For large positive $x$ it also follows $\frac{1}{2}e^x$ (since $e^{-x} \to 0$ ), so sinh and cosh become indistinguishable as $x \to +\infty$ — both asymptotic to $\frac{1}{2}e^x$ . For large negative $x$ , $\sinh x \to -\frac{1}{2}e^{-x} \to -\infty$ while $\cosh x \to +\frac{1}{2}e^{-x} \to +\infty$ .

$y = \tanh x = \frac{\sinh x}{\cosh x}$ . Odd, S-shaped, through the origin, with horizontal asymptotes $y = \pm 1$ . Because $\sinh x \to \cosh x$ as $x \to +\infty$ , their ratio $\tanh x \to 1$ ; similarly $\tanh x \to -1$ as $x \to -\infty$ .

Shared behaviour.

sinh and cosh both grow like $\frac{1}{2}e^x$ as $x \to +\infty$ (curves converge).

Only tanh has finite asymptotes ( $y = \pm 1$ ); sinh and cosh are unbounded.

All three pass through the origin except cosh (cosh passes through $(0,1)$ ).
Why this scores
B1 for each correctly shaped curve (3 marks), B1 for sinh/cosh both $\to \frac{1}{2}e^x$ , B1 for tanh asymptotes $\pm1$ explained via the ratio, B1 for the cosh-through-(0,1) distinction. The link between the convergence of sinh/cosh and the tanh asymptote is the high-value insight.
Question 6
8 marks
Sketch $y = \cosh x - 1$ and $y = 2\cosh\frac{x}{2}$ , describing in each case how the graph differs from $y = \cosh x$ (minimum point and shape).
Model answer
$y = \cosh x - 1$ . This is $y = \cosh x$ translated down by $1$ unit. The minimum moves from $(0, 1)$ to $(0, 0)$ , so the curve now passes through the origin and touches the $x$ -axis there. Range: $y \geq 0$ . Still even, still U-shaped, with the same unbounded growth as $\cosh x$ .

$y = 2\cosh\frac{x}{2}$ . Two transformations of $y = \cosh x$ :

$\frac{x}{2}$ inside stretches the graph horizontally by factor $2$ (a wider U).

The factor $2$ outside stretches it vertically by factor $2$ .

The minimum was at $(0, 1)$ ; the horizontal stretch leaves $x = 0$ fixed, and the vertical stretch takes $y = 1$ to $y = 2$ , so the new minimum is $(0, 2)$ . Range: $y \geq 2$ . Still even and U-shaped, but wider and taller than $\cosh x$ .

Why these matter. $y = 2\cosh\frac{x}{2}$ is, up to constants, the equation of a hanging chain (catenary). Recognising standard transformations applied to $\cosh$ is a common Paper 2 task.
Why this scores
First curve: B1 translation down, B1 new minimum $(0,0)$ /passes through origin, B1 range $y \geq 0$ . Second curve: B1 horizontal stretch factor $2$ , B1 vertical stretch factor $2$ , B1 new minimum $(0,2)$ , B1 range $y \geq 2$ , B1 correct wider/taller even U-shape. Confusing the direction of the $\frac{x}{2}$ stretch (factor $2$ , not $\frac{1}{2}$ ) is the standard error.

Key Formulae — Sketching Graphs of Hyperbolic Functions

The formulae you need to memorise for sketching graphs of hyperbolic functions on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

sinh — domain and range
$y = \sinh x: \;\; \text{domain } \mathbb{R}, \;\; \text{range } \mathbb{R}$
$x$
any real number
When to use
Odd function through the origin, gradient $1$ there, no asymptotes. Increasing everywhere.
cosh — domain and range
$y = \cosh x: \;\; \text{domain } \mathbb{R}, \;\; \text{range } y \geq 1$
$x$
any real number
When to use
Even function (the catenary) with minimum $(0, 1)$ . Symmetric about the $y$ -axis.
tanh — domain, range, asymptotes
$y = \tanh x: \;\; \text{domain } \mathbb{R}, \;\; \text{range } -1 < y < 1, \;\; \text{asymptotes } y = \pm 1$
$x$
any real number
When to use
Odd S-shaped function through the origin, bounded by $\pm 1$ (never reached).
sech — reciprocal of cosh
$y = \text{sech}\,x = \frac{1}{\cosh x}: \;\; \text{domain } \mathbb{R}, \;\; \text{range } 0 < y \leq 1$
$x$
any real number
When to use
Even bell-shaped curve with maximum $(0, 1)$ and the $x$ -axis as a horizontal asymptote.
coth — reciprocal of tanh
$y = \coth x = \frac{1}{\tanh x}: \;\; \text{domain } x \neq 0, \;\; \text{range } |y| > 1, \;\; \text{asymptotes } x = 0,\, y = \pm 1$
$x$
any real number except $0$
When to use
Odd function with a vertical asymptote at $x = 0$ and horizontal asymptotes $y = \pm 1$ .

Key Definitions and Keywords — Sketching Graphs of Hyperbolic Functions

Definitions to memorise and the exact keywords mark schemes credit for sketching graphs of hyperbolic functions answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Even function
Examiner keyword
A function with $f(-x) = f(x)$ , symmetric about the $y$ -axis. $\cosh$ and $\text{sech}$ are even.
Odd function
Examiner keyword
A function with $f(-x) = -f(x)$ , with rotational symmetry of order $2$ about the origin. $\sinh$ , $\tanh$ , $\text{cosech}$ and $\coth$ are odd.
Horizontal asymptote
Examiner keyword
A line $y = c$ that the curve approaches as $x \to \pm\infty$ but never crosses (here). $\tanh$ has $y = \pm 1$ ; $\text{sech}$ and $\text{cosech}$ have $y = 0$ .
Catenary
Examiner keyword
The U-shaped curve traced by a chain hanging under gravity; modelled by $y = a\cosh\frac{x}{a}$ . The graph of $\cosh x$ is the basic catenary.
Range
Examiner keyword
The set of output values a function takes. $\sinh$ : all reals; $\cosh$ : $y \geq 1$ ; $\tanh$ : $-1 < y < 1$ .
Domain
The set of permitted input values. $\sinh$ , $\cosh$ , $\tanh$ , $\text{sech}$ : all reals; $\text{cosech}$ , $\coth$ : $x \neq 0$ .

Common Mistakes and Misconceptions — Sketching Graphs of Hyperbolic Functions

The traps other students keep falling into on sketching graphs of hyperbolic functions questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Drawing $\cosh x$ with minimum at the origin $(0, 0)$
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Confusing cosh with a parabola or with sinh, or forgetting $\cosh 0 = 1$ .
How to avoid it
Always plot the minimum at $(0, 1)$ and label it; the range is $y \geq 1$ , not $y \geq 0$ .
✕Giving $\sinh x$ horizontal asymptotes
9231 Paper 2 mark schemes
Why it happens
Carrying over the bounded behaviour of $\tanh$ to $\sinh$ .
How to avoid it
$\sinh x \to \pm\infty$ — it is unbounded and has NO asymptotes. Only $\tanh$ (and the reciprocals) have horizontal asymptotes.
✕Stating the range of $\tanh x$ as $-1 \leq y \leq 1$ (closed interval)
9231 Paper 2 Examiner Reports
Why it happens
Treating the asymptotes as attainable values.
How to avoid it
$\tanh x$ approaches but never equals $\pm 1$ , so the range is the OPEN interval $-1 < y < 1$ .
✕Turning a minimum of cosh into a minimum of sech (instead of a maximum)
9231 Paper 2 Examiner Reports
Why it happens
Forgetting that reciprocation flips minima to maxima and sends $\infty$ to $0$ .
How to avoid it
Reciprocal-graph rule: a minimum of $f$ at value $m>0$ becomes a maximum of $\frac{1}{f}$ at value $\frac{1}{m}$ ; where $f \to \infty$ , $\frac{1}{f} \to 0$ .
✕Omitting the vertical asymptote $x = 0$ of $\coth$ and $\text{cosech}$
9231 Paper 2 mark schemes
Why it happens
Not checking where the denominator $\sinh x$ vanishes.
How to avoid it
Both $\coth x = \frac{\cosh x}{\sinh x}$ and $\text{cosech}\,x = \frac{1}{\sinh x}$ are undefined at $x = 0$ — mark the vertical asymptote.
✕Getting the direction/factor of a horizontal stretch wrong (e.g. $\cosh\frac{x}{2}$ stretched by $\frac{1}{2}$ )
9231 Paper 2 Examiner Reports
Why it happens
Confusing $f(\frac{x}{2})$ (stretch factor $2$ ) with $f(2x)$ (stretch factor $\frac{1}{2}$ ).
How to avoid it
Replacing $x$ by $\frac{x}{k}$ stretches horizontally by factor $k$ . So $\cosh\frac{x}{2}$ is WIDER (factor $2$ ).

Sketching Graphs of Hyperbolic Functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Key features of y = sinh x

2Key features of y = cosh x

3Asymptotes of y = tanh x

4Sketching y = sech x

5Sketching y = coth x

6Why sinh resembles a flattened cubic

sinh — domain and range

cosh — domain and range

tanh — domain, range, asymptotes

sech — reciprocal of cosh

coth — reciprocal of tanh

Even function

Odd function

Horizontal asymptote

Catenary

Range

Domain

✕Drawing cosh⁡x\cosh xcoshx with minimum at the origin (0,0)(0, 0)(0,0)

✕Giving sinh⁡x\sinh xsinhx horizontal asymptotes

✕Stating the range of tanh⁡x\tanh xtanhx as −1≤y≤1-1 \leq y \leq 1−1≤y≤1 (closed interval)

✕Turning a minimum of cosh into a minimum of sech (instead of a maximum)

✕Omitting the vertical asymptote x=0x = 0x=0 of coth⁡\cothcoth and cosech\text{cosech}cosech

✕Getting the direction/factor of a horizontal stretch wrong (e.g. cosh⁡x2\cosh\frac{x}{2}cosh2x​ stretched by 12\frac{1}{2}21​)

✕Drawing $\cosh x$ with minimum at the origin $(0, 0)$

✕Giving $\sinh x$ horizontal asymptotes

✕Stating the range of $\tanh x$ as $-1 \leq y \leq 1$ (closed interval)

✕Omitting the vertical asymptote $x = 0$ of $\coth$ and $\text{cosech}$

✕Getting the direction/factor of a horizontal stretch wrong (e.g. $\cosh\frac{x}{2}$ stretched by $\frac{1}{2}$ )