Why is "Stating an oblique asymptote without showing the polynomial division" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Guessing the slant line or using a calculator/graph. How to avoid it: Always write the division result as 'quotient +remainderdenominator'. The working earns the method mark. Source: 9231 Paper 1 Examiner Reports

Why is "Sketching a branch that crosses a vertical asymptote" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Drawing the curve as one continuous line through the asymptote. How to avoid it: Treat the regions between asymptotes as separate branches; test the sign just either side of each vertical asymptote. Source: 9231 Paper 1 Examiner Reports

Why is "Multiplying a rational inequality by the denominator" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Treating P/Q > k like a linear equation and multiplying by Q whose sign is unknown. How to avoid it: Bring all terms to one side, combine into a single fraction, then use a sign table on the critical values. Source: 9231 Paper 1 mark schemes

Why is "Giving the excluded interval as the range (or vice versa)" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Reading the discriminant inequality the wrong way round. How to avoid it: A positive leading coefficient with discriminant ≥ 0 gives the OUTSIDE region (y ≤ a or y ≥ b). State which region is the range. Source: 9231 Paper 1 Examiner Reports

Why is "Assuming every rational function has a horizontal asymptote" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Not checking the relative degrees of numerator and denominator. How to avoid it: Compare degrees first: lower numerator → y=0; equal → ratio; numerator one higher → oblique (no horizontal). Source: 9231 Paper 1 Examiner Reports

Why is "Sketch missing labelled intercepts or turning points" a common mistake in Sketching graphs of simple rational functions?

Why it happens: Rushing the sketch once the asymptotes are drawn. How to avoid it: Run the checklist: asymptotes, intercepts, turning points, branch directions — label coordinates on each. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Sketching graphs of simple rational functions

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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.

Sketching Graphs of Simple Rational Functions — Cambridge 9231 Further Mathematics Paper 1 (2026-2027 syllabus)

Find vertical, horizontal and oblique asymptotes, locate intercepts and turning points, and use the discriminant to pin down the range. A complete, labelled sketch is worth the bulk of the marks.

At a glance

Vertical asymptote: where the denominator $= 0$ (and numerator $\neq 0$ ).
Horizontal vs oblique: compare the degrees of numerator and denominator.
Oblique asymptote: do polynomial (long) division — the quotient is the slant line.
Range: set $y = f(x)$ , form a quadratic in $x$ , require discriminant $\geq 0$ .
A sketch must show: asymptotes (dashed), axis intercepts, turning points, end behaviour.

What you’ll learn

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

1.2a — Determine the asymptotes (vertical, horizontal, oblique) of a rational function with numerator/denominator of degree at most 2.
1.2b — Find axis intercepts and stationary points, and sketch the curve showing all key features.
1.2c — Use the discriminant to find the set of values the function can take (its range), or to solve a related inequality.

Finding the asymptotes

Vertical: denominator = 0. Horizontal/oblique: compare degrees, divide if needed.

Asymptotes are the skeleton of the sketch. Get them first, draw them dashed, and the curve almost draws itself.

Vertical asymptotes. Set the denominator $= 0$ (provided the numerator is non-zero there). The curve shoots to $\pm\infty$ near these lines.

$y = \dfrac{x+1}{x-3}$ has a vertical asymptote at $x = 3$ .

Horizontal or oblique? Compare the degree of the numerator ( $N$ ) with the denominator ( $D$ ):

Degrees	Asymptote	How to find it
$\deg N < \deg D$	Horizontal $y = 0$	The x-axis
$\deg N = \deg D$	Horizontal $y = \frac{\text{lead }N}{\text{lead }D}$	Ratio of leading coefficients
$\deg N = \deg D + 1$	Oblique (slant)	Polynomial division

Oblique asymptote — the division step. When the numerator is exactly one degree higher, divide: $y = \frac{x^2 + 1}{x - 2} = x + 2 + \frac{5}{x-2}$ As $x \to \pm\infty$ the $\frac{5}{x-2}$ term vanishes, so the curve approaches the line $y = x + 2$ . That line is the oblique asymptote.

Cambridge tip. Always write the result of the division as "linear part $+\;\frac{\text{remainder}}{\text{denominator}}$ ". The linear part is the asymptote — no further work needed.

Vertical: solve denominator $= 0$ .
Equal degrees → horizontal $y = \frac{\text{lead}}{\text{lead}}$ .
Numerator one degree higher → divide; quotient is the oblique asymptote.

See the full worked example for sketching graphs of simple rational functions →

Range using the discriminant

Rearrange y = f(x) into a quadratic in x; real x needs discriminant ≥ 0.

A favourite Cambridge sub-question: "find the set of values $y$ can take." The discriminant method handles it cleanly.

The method.

Write $y = \dfrac{\text{quadratic}}{\text{quadratic}}$ (or similar).
Multiply out to get a quadratic equation in $x$ with $y$ as a parameter.
For a real $x$ to exist, the discriminant $b^2 - 4ac \geq 0$ .
Solve that inequality for $y$ — the solution set is the range.

Worked outline. For $y = \dfrac{x^2 - 4x + 3}{x - 2}$ (or a $\frac{\text{quad}}{\text{quad}}$ form), cross-multiply: $y(x - 2) = x^2 - 4x + 3 \;\Rightarrow\; x^2 - (4 + y)x + (3 + 2y) = 0$ Discriminant $\geq 0$ : $(4 + y)^2 - 4(3 + 2y) \geq 0 \;\Rightarrow\; y^2 + 4 \geq 0$ Here it is true for all $y$ , so the function takes all real values except at any value excluded by the asymptote/turning-point structure. (When the discriminant gives e.g. $y \leq a$ or $y \geq b$ , those bounds are the turning-point $y$ -values.)

Link to turning points. The boundary values of $y$ from the discriminant are exactly the $y$ -coordinates of the stationary points. This is a quick alternative to differentiating.

Cambridge tip. State the range with correct inequality signs and say whether the endpoints are included. " $y \leq -1$ or $y \geq 7$ " is a complete answer; " $-1 \leq y \leq 7$ " would be the gap, not the range — read which region your inequality gives.

Form a quadratic in $x$ with $y$ as parameter.
Real $x$ ⟺ discriminant $\geq 0$ → solve for $y$ .
Discriminant boundary values = turning-point $y$ -values.

See the full worked example for sketching graphs of simple rational functions →

Putting the sketch together

Asymptotes, intercepts, turning points, then the branches and end behaviour.

A mark-scheme sketch is judged on features, not artistic accuracy. Work through this checklist every time.

Sketch checklist:

Asymptotes — draw all vertical and the horizontal/oblique as dashed lines, labelled with their equations.
Axis intercepts — $x$ -intercepts where numerator $= 0$ ; $y$ -intercept by putting $x = 0$ .
Turning points — from the discriminant range (or by differentiating); mark coordinates.
Behaviour near asymptotes — does each branch go to $+\infty$ or $-\infty$ ? Test a value just left/right of each vertical asymptote.
End behaviour — how the curve approaches the horizontal/oblique asymptote as $x \to \pm\infty$ (from above or below).

Draw asymptotes first (dashed), then sketch each branch hugging them. Label intercepts and turning points.

Cambridge tip. Examiners frequently penalise sketches where a branch crosses an asymptote it should approach, or where the curve is drawn touching the horizontal axis at a turning point that should be off it. Plot one extra test point per branch to fix the shape.

Asymptotes dashed and labelled first.
Mark intercepts and turning-point coordinates.
Check each branch goes the right way to $\pm\infty$ .

See the full worked example for sketching graphs of simple rational functions →

Quick recap

Vertical asymptote: denominator $= 0$ .
Equal degrees → horizontal $y = \frac{\text{lead}}{\text{lead}}$ ; numerator one higher → oblique via division.
Range: quadratic in $x$ , discriminant $\geq 0$ , solve for $y$ .
Discriminant boundaries = turning-point $y$ -values.
Sketch = asymptotes + intercepts + turning points + correct branch directions.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Degree-comparison table for horizontal vs oblique asymptotes.
Oblique asymptote = quotient of polynomial division.
Discriminant method for the range.
Sketch checklist (asymptotes → intercepts → turning points → branches).

How it’s examined

Topic 1.2 appears every Paper 1 (6-9 marks). Part (a) asks for the asymptotes (algebraically — a calculator answer scores nothing); part (b) is a full sketch marked on features; part (c) often asks for the range or to solve an inequality. The biggest mark-losers are (i) reading off an oblique asymptote without showing the division, (ii) sketches with branches crossing asymptotes, and (iii) range answers with the wrong inequality region.

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 — Sketching graphs of simple rational functions

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

Question type:

1Vertical and horizontal asymptotes
Getting startedDirect calculation• asymptote
Question
State the equations of the asymptotes of $y = \dfrac{2x + 1}{x - 3}$ .
Step-by-step solution
1. Step 1
  Vertical asymptote: set the denominator to zero. $x - 3 = 0$ .
  $x = 3$
2. Step 2
  Horizontal asymptote: degrees of numerator and denominator are equal (both $1$ ), so use the ratio of leading coefficients $\frac{2}{1}$ .
  $y = 2$
Answer
Vertical: $x = 3$ . Horizontal: $y = 2$ .
Examiner tip
Equal degrees always give a horizontal asymptote at the leading-coefficient ratio — no division needed.
2Oblique asymptote by division
Getting startedDirect calculation• oblique, division
Question
Find the asymptotes of $y = \dfrac{x^2 + 1}{x - 2}$ .
Step-by-step solution
1. Step 1
  Vertical asymptote: $x - 2 = 0$ .
  $x = 2$
2. Step 2
  Numerator degree (2) is one more than denominator (1) → oblique asymptote. Divide.
  $\frac{x^2 + 1}{x - 2} = x + 2 + \frac{5}{x - 2}$
3. Step 3
  As $x \to \pm\infty$ , the $\frac{5}{x-2}$ term $\to 0$ , leaving the line.
  $y = x + 2$
Answer
Vertical: $x = 2$ . Oblique: $y = x + 2$ .
Examiner tip
Show the division line $x + 2 + \frac{5}{x-2}$ — examiners award the method mark for the working, not the stated asymptote alone.
3Range using the discriminant
Building confidenceMulti-step problem• range, discriminant
Question
Find the set of values taken by $y = \dfrac{x^2 - 3}{x - 1}$ .
Step-by-step solution
1. Step 1
  Cross-multiply to form a quadratic in $x$ .
  $y(x - 1) = x^2 - 3 \;\Rightarrow\; x^2 - yx + (y - 3) = 0$
2. Step 2
  Real $x$ requires discriminant $\geq 0$ .
  $(-y)^2 - 4(1)(y - 3) \geq 0 \;\Rightarrow\; y^2 - 4y + 12 \geq 0$
3. Step 3
  Discriminant of this quadratic in $y$ : $(-4)^2 - 4(12) = 16 - 48 = -32 < 0$ , and the $y^2$ coefficient is positive, so $y^2 - 4y + 12 > 0$ for all $y$ .
Answer
$y$ can take all real values ( $y \in \mathbb{R}$ ).
Examiner tip
When the $y$ -inequality holds for all $y$ , the function's range is all of $\mathbb{R}$ — there is no gap and no turning point to exclude.
4Range with an excluded interval
Building confidenceMulti-step problem• range, discriminant, turning-points
Question
Show that $y = \dfrac{x^2 + x + 1}{x}$ cannot take values strictly between $-1$ and $3$ .
Step-by-step solution
1. Step 1
  Form a quadratic in $x$ .
  $yx = x^2 + x + 1 \;\Rightarrow\; x^2 + (1 - y)x + 1 = 0$
2. Step 2
  Discriminant $\geq 0$ for real $x$ .
  $(1 - y)^2 - 4 \geq 0$
3. Step 3
  Solve the inequality. $(1-y)^2 \geq 4 \Rightarrow |1 - y| \geq 2$ .
  $1 - y \geq 2 \;\text{ or }\; 1 - y \leq -2$
4. Step 4
  Rearrange: $y \leq -1$ or $y \geq 3$ . So no values lie strictly between.
Answer
$y \leq -1$ or $y \geq 3$ ; the interval $-1 < y < 3$ is excluded.
Examiner tip
The boundary values $y = -1$ and $y = 3$ are the $y$ -coordinates of the two turning points — a free by-product of the discriminant method.
5A complete labelled sketch
StretchGraph or diagram• sketch
Question
For $y = \dfrac{x^2 + 1}{x - 2}$ , state the asymptotes, find the coordinates of the stationary points, and describe the sketch.
Step-by-step solution
1. Step 1
  Asymptotes (from earlier): $x = 2$ and $y = x + 2$ .
2. Step 2
  Stationary points via the range. $y(x-2) = x^2 + 1 \Rightarrow x^2 - yx + (2y + 1) = 0$ . Discriminant $\geq 0$ : $y^2 - 4(2y+1) \geq 0 \Rightarrow y^2 - 8y - 4 \geq 0$ .
  $y = \frac{8 \pm \sqrt{64 + 16}}{2} = 4 \pm 2\sqrt{5}$
3. Step 3
  Turning-point $y$ -values are $y = 4 - 2\sqrt5 \approx -0.47$ and $y = 4 + 2\sqrt5 \approx 8.47$ . Substitute back ( $x = \frac{y}{2}$ at the boundary) to get the $x$ -coordinates.
4. Step 4
  Sketch: two branches. Lower-left branch rises to a maximum at $y \approx -0.47$ ; upper-right branch falls to a minimum at $y \approx 8.47$ ; both hug $x = 2$ and the slant $y = x + 2$ .
Answer
Asymptotes $x = 2$ , $y = x + 2$ ; stationary $y$ -values $4 \pm 2\sqrt5$ ; two branches as described.
Examiner tip
Reaching the stationary values via the discriminant avoids the messy quotient-rule differentiation of $\frac{x^2+1}{x-2}$ .
6Solving an inequality from the graph
StretchMulti-step problem• inequality
Question
Using a sketch of $y = \dfrac{x+4}{x-1}$ , solve $\dfrac{x+4}{x-1} > 2$ .
Step-by-step solution
1. Step 1
  Find where equality holds. $\frac{x+4}{x-1} = 2 \Rightarrow x + 4 = 2(x-1) \Rightarrow x = 6$ .
2. Step 2
  Vertical asymptote at $x = 1$ ; horizontal asymptote $y = 1$ . The curve exceeds $2$ between the asymptote and the crossing point.
3. Step 3
  Test a value, say $x = 2$ : $\frac{6}{1} = 6 > 2$ ✓ (inside). $x = 7$ : $\frac{11}{6} \approx 1.83 < 2$ (outside).
Answer
$1 < x < 6$ .
Examiner tip
Solving rational inequalities by sketch + critical points avoids the sign-error trap of multiplying both sides by $(x-1)$ , whose sign is unknown.

Model Answers — Sketching graphs of simple rational functions

High-scoring sample answers for sketching graphs of simple rational functions on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Write down the equations of the asymptotes of the curve $y = \dfrac{3x - 2}{x + 4}$ .
Model answer
Vertical asymptote — where the denominator is zero: $x + 4 = 0 \;\Rightarrow\; x = -4$

Horizontal asymptote — numerator and denominator both have degree $1$ , so the asymptote is the ratio of leading coefficients: $y = \frac{3}{1} = 3$
Why this scores
B1 for $x = -4$ , B1 for recognising equal degrees, B1 for $y = 3$ . No division required when degrees are equal.
Question 2
4 marks
Find the equations of the asymptotes of $y = \dfrac{2x^2 - x + 3}{x + 1}$ .
Model answer
Vertical asymptote: $x + 1 = 0 \Rightarrow x = -1$ .

Oblique asymptote — numerator degree ( $2$ ) is one more than denominator degree ( $1$ ), so divide: $\frac{2x^2 - x + 3}{x + 1} = 2x - 3 + \frac{6}{x + 1}$

As $x \to \pm\infty$ , $\dfrac{6}{x+1} \to 0$ , so the oblique asymptote is: $y = 2x - 3$
Why this scores
B1 for $x = -1$ ; M1 for the division; A1 for the correct quotient; A1 for stating $y = 2x - 3$ . The division working must be shown.
Question 3
6 marks
Find the set of values that $y = \dfrac{x^2 + 2}{x + 1}$ cannot take.
Model answer
Form a quadratic in $x$ : $y(x + 1) = x^2 + 2 \;\Rightarrow\; x^2 - yx + (2 - y) = 0$

For real $x$ , the discriminant must be $\geq 0$ : $(-y)^2 - 4(1)(2 - y) \geq 0$ $y^2 + 4y - 8 \geq 0$

Solve $y^2 + 4y - 8 = 0$ : $y = \frac{-4 \pm \sqrt{16 + 32}}{2} = -2 \pm 2\sqrt{3}$

Since the coefficient of $y^2$ is positive, $y^2 + 4y - 8 \geq 0$ outside the roots. Therefore real $x$ exists only when $y \leq -2 - 2\sqrt3$ or $y \geq -2 + 2\sqrt3$ .

The function cannot take values in $\;-2 - 2\sqrt3 < y < -2 + 2\sqrt3$ .
Why this scores
M1 for the quadratic in $x$ , M1 for discriminant $\geq 0$ , A1 for $y^2 + 4y - 8 \geq 0$ , M1 for solving, A1 A1 for the excluded interval with exact endpoints. Surd form is expected — do not round.
Question 4
6 marks
The curve $y = \dfrac{x - 1}{(x + 1)(x - 2)}$ has vertical asymptotes $x = -1$ and $x = 2$ . State the horizontal asymptote, find the axis intercepts, and describe the behaviour of the three branches.
Model answer
Horizontal asymptote: the numerator has degree $1$ and the denominator degree $2$ , so $\deg N < \deg D$ and the asymptote is: $y = 0$

Axis intercepts:

$x$ -intercept: numerator $= 0 \Rightarrow x = 1$ , giving $(1, 0)$ .

$y$ -intercept: put $x = 0$ : $y = \dfrac{-1}{(1)(-2)} = \dfrac{1}{2}$ , giving $\left(0, \tfrac12\right)$ .

Three branches (separated by $x = -1$ and $x = 2$ ):

For $x < -1$ : as $x \to -1^-$ the curve $\to -\infty$ ; as $x \to -\infty$ it approaches $y = 0$ from below.

For $-1 < x < 2$ : passes through $(0, \tfrac12)$ and $(1, 0)$ , running from $+\infty$ (near $x=-1^+$ ) to $-\infty$ (near $x = 2^-$ ).

For $x > 2$ : comes from $+\infty$ (near $x = 2^+$ ) and approaches $y = 0$ from above as $x \to \infty$ .
Why this scores
B1 horizontal asymptote, B1 B1 intercepts, B1 B1 B1 for correct branch behaviour. Sign-testing near each asymptote is the reliable way to get the branch directions right.
Question 5
7 marks
Find the coordinates of the stationary points of $y = \dfrac{x^2 - 3x + 4}{x - 1}$ .
Model answer
Form a quadratic in $x$ with $y$ as parameter: $y(x - 1) = x^2 - 3x + 4 \;\Rightarrow\; x^2 - (3 + y)x + (4 + y) = 0$

Stationary points occur where $y$ is at a boundary value, i.e. the discriminant is zero: $(3 + y)^2 - 4(4 + y) = 0$ $9 + 6y + y^2 - 16 - 4y = 0 \;\Rightarrow\; y^2 + 2y - 7 = 0$ $y = \frac{-2 \pm \sqrt{4 + 28}}{2} = -1 \pm 2\sqrt{2}$

At a repeated root, $x = \dfrac{3 + y}{2}$ :

$y = -1 + 2\sqrt2 \;\Rightarrow\; x = \dfrac{2 + 2\sqrt2}{2} = 1 + \sqrt2$

$y = -1 - 2\sqrt2 \;\Rightarrow\; x = \dfrac{2 - 2\sqrt2}{2} = 1 - \sqrt2$

Stationary points: $\left(1 + \sqrt2,\; -1 + 2\sqrt2\right)$ and $\left(1 - \sqrt2,\; -1 - 2\sqrt2\right)$ .
Why this scores
M1 quadratic in $x$ , M1 discriminant $= 0$ , A1 $y^2 + 2y - 7 = 0$ , A1 the two $y$ -values, M1 using $x = \frac{3+y}{2}$ at the repeated root, A1 A1 the coordinates. Setting discriminant $=0$ (not $\geq 0$ ) pinpoints the turning points.
Question 6
7 marks
Solve the inequality $\dfrac{x^2 - 4}{x - 3} \leq x$ .
Model answer
Bring everything to one side (do not multiply across by $x - 3$ , whose sign is unknown): $\frac{x^2 - 4}{x - 3} - x \leq 0 \;\Rightarrow\; \frac{x^2 - 4 - x(x - 3)}{x - 3} \leq 0$

Simplify the numerator: $\frac{x^2 - 4 - x^2 + 3x}{x - 3} = \frac{3x - 4}{x - 3} \leq 0$

Critical values: numerator zero at $x = \dfrac{4}{3}$ ; denominator zero at $x = 3$ (excluded).

Sign analysis of $\dfrac{3x - 4}{x - 3}$ :

$x < \tfrac43$ : $(-)/(-) = +$

$\tfrac43 < x < 3$ : $(+)/(-) = -$ ✓

$x > 3$ : $(+)/(+) = +$

The expression is $\leq 0$ on $\tfrac43 \leq x < 3$ (equality at $x = \tfrac43$ , asymptote excluded at $x = 3$ ).

Solution: $\dfrac{4}{3} \leq x < 3$ .
Why this scores
M1 for moving to one side, M1 for the single combined fraction, A1 for $\frac{3x-4}{x-3}$ , M1 critical values, M1 sign table, A1 A1 for the solution set with correct open/closed endpoints. Multiplying through by $(x-3)$ loses marks for sign errors.

Key Formulae — Sketching graphs of simple rational functions

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

Vertical asymptote
$\text{denominator} = 0 \;\;(\text{numerator} \neq 0)$
When to use
Locate the vertical lines the curve cannot cross.
Horizontal asymptote (equal degrees)
$y = \frac{\text{leading coeff of numerator}}{\text{leading coeff of denominator}}$
When to use
When numerator and denominator have the same degree.
Oblique asymptote
$y = (\text{quotient}) + \frac{\text{remainder}}{\text{denominator}} \;\to\; y = \text{quotient}$
When to use
When numerator degree is exactly one more than denominator degree; obtain by polynomial division.
Range via discriminant
$b^2 - 4ac \geq 0 \;\;(\text{quadratic in } x,\; y \text{ as parameter})$
$a, b, c$
coefficients of the quadratic in $x$ , each a function of $y$
When to use
Find the set of values $y$ can take; the boundary values are the turning-point $y$ -coordinates.

Key Definitions and Keywords — Sketching graphs of simple rational functions

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

Rational function
Examiner keyword
A function of the form $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials and $Q \neq 0$ . In 9231, each has degree at most $2$ .
Vertical asymptote
Examiner keyword
A line $x = a$ that the curve approaches but never reaches, occurring where the denominator is zero (numerator non-zero).
Oblique (slant) asymptote
Examiner keyword
A non-horizontal straight line the curve approaches as $x \to \pm\infty$ , present when the numerator's degree is one more than the denominator's.
Range (set of values)
Examiner keyword
The set of all $y$ -values the function can output. Found via the discriminant condition for the existence of a real $x$ .
Stationary point
A point where the gradient is zero. Its $y$ -coordinate equals a boundary value of the discriminant range.

Common Mistakes and Misconceptions — Sketching graphs of simple rational functions

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

✕Stating an oblique asymptote without showing the polynomial division
9231 Paper 1 Examiner Reports
Why it happens
Guessing the slant line or using a calculator/graph.
How to avoid it
Always write the division result as 'quotient $+\frac{\text{remainder}}{\text{denominator}}$ '. The working earns the method mark.
✕Sketching a branch that crosses a vertical asymptote
9231 Paper 1 Examiner Reports
Why it happens
Drawing the curve as one continuous line through the asymptote.
How to avoid it
Treat the regions between asymptotes as separate branches; test the sign just either side of each vertical asymptote.
✕Multiplying a rational inequality by the denominator
9231 Paper 1 mark schemes
Why it happens
Treating $\frac{P}{Q} > k$ like a linear equation and multiplying by $Q$ whose sign is unknown.
How to avoid it
Bring all terms to one side, combine into a single fraction, then use a sign table on the critical values.
✕Giving the excluded interval as the range (or vice versa)
9231 Paper 1 Examiner Reports
Why it happens
Reading the discriminant inequality the wrong way round.
How to avoid it
A positive leading coefficient with discriminant $\geq 0$ gives the OUTSIDE region ( $y \leq a$ or $y \geq b$ ). State which region is the range.
✕Assuming every rational function has a horizontal asymptote
9231 Paper 1 Examiner Reports
Why it happens
Not checking the relative degrees of numerator and denominator.
How to avoid it
Compare degrees first: lower numerator → $y=0$ ; equal → ratio; numerator one higher → oblique (no horizontal).
✕Sketch missing labelled intercepts or turning points
9231 Paper 1 Examiner Reports
Why it happens
Rushing the sketch once the asymptotes are drawn.
How to avoid it
Run the checklist: asymptotes, intercepts, turning points, branch directions — label coordinates on each.

Sketching graphs of simple rational 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

1Vertical and horizontal asymptotes

2Oblique asymptote by division

3Range using the discriminant

4Range with an excluded interval

5A complete labelled sketch

6Solving an inequality from the graph

Vertical asymptote

Horizontal asymptote (equal degrees)

Oblique asymptote

Range via discriminant

Rational function

Vertical asymptote

Oblique (slant) asymptote

Range (set of values)

Stationary point

✕Stating an oblique asymptote without showing the polynomial division

✕Sketching a branch that crosses a vertical asymptote

✕Multiplying a rational inequality by the denominator

✕Giving the excluded interval as the range (or vice versa)

✕Assuming every rational function has a horizontal asymptote

✕Sketch missing labelled intercepts or turning points