Graphs of Functions · Study Notes · Edexcel IGCSE 4MA1 Higher

Sketch and recognise graphs: linear, quadratic, cubic, reciprocal, exponential. Identify roots, vertices, asymptotes.

At a glance

Linear: $y = mx + c$ . Straight.
Quadratic: $y = ax^{2} + bx + c$ . Parabola.
Cubic: $y = ax^{3} + ...$ . S-shape.
Reciprocal: $y = k/x$ . Two branches, asymptotes at axes.
Exponential: $y = a^{x}$ . Through $(0, 1)$ , asymptote $y = 0$ .
Recognise key features: roots, vertex, intercepts, asymptotes.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.4 — Recognise and sketch graphs of common functions.
3.4 — Identify intercepts, roots, vertices, asymptotes.
3.4 — Use graphs to solve equations and inequalities.

Linear and quadratic — review

▼

Linear: straight. Quadratic: parabola.

Linear $y = mx + c$ .

Straight line. Gradient $m$ , y-intercept $c$ .

Quadratic $y = ax^{2} + bx + c$ .

Parabola.

$a > 0$ : opens UP. Minimum vertex.
$a < 0$ : opens DOWN. Maximum vertex.

A positive coefficient gives a U-shape with a minimum; a negative one gives an n-shape with a maximum.

Roots = where parabola crosses $x$ -axis.

Discriminant	Roots	Graph
$b^{2} - 4ac > 0$	2 distinct	Crosses $x$ -axis twice
$b^{2} - 4ac = 0$	1 repeated	Touches (tangent)
$b^{2} - 4ac < 0$	no real	Doesn't cross

Sketching:

Find $y$ -intercept: at $x = 0$ .
Find roots: solve $ax^{2} + bx + c = 0$ .
Find vertex: $x_{v} = -b/(2a)$ , then compute $y_{v}$ .
Sketch.

Example. $y = x^{2} - 4x - 5$ .

Roots: $(x-5)(x+1) = 0$ → $-1, 5$ .
y-intercept: $-5$ .
Vertex: $x = 2$ , $y = -9$ .

Edexcel tip. Always show 3-4 key points: roots, y-intercept, vertex. Mark schemes credit each.

Linear: straight, gradient + intercept.
Quadratic: parabola, $a > 0$ up, $a < 0$ down.
Roots from factoring or formula.
Vertex at $x = -b/2a$ .

Cubic graphs

▼

S-shape. Up to 3 roots. Double roots are tangent.

Cubic $y = ax^{3} + bx^{2} + cx + d$ .

S-shape (or 'wave' shape).

$a > 0$ : bottom-left to top-right.
$a < 0$ : top-left to bottom-right.

Roots: up to $3$ .

Single root: graph CROSSES $x$ -axis.
Double root: graph is TANGENT (touches but doesn't cross).
Triple root: graph has 'inflection' at the axis.

A single root crosses the axis; a double root touches it and turns back without crossing.

Example. $y = x^{3} - 3x^{2}$ .

Factor: $x^{2}(x - 3)$ .
Roots: $x = 0$ (double), $x = 3$ (single).
At $x = 0$ : tangent to axis.
At $x = 3$ : crosses axis.

Other cubics:

Function	Shape
$y = x^{3}$	Through origin, S-shape, 1 root
$y = x^{3} - x$	$x(x^{2}-1)$ , three roots $-1, 0, 1$
$y = -x^{3} + 4x$	$x(4 - x^{2})$ , three roots, opens opposite

Worked qualitative. What's different about a double root?

The graph TOUCHES the $x$ -axis without crossing.
Both 'sides' of the root are on the same side of the axis.
E.g. $y = x^{2}(x - 3)$ at $x = 0$ : $y$ goes from positive (left) to positive (right) of $0$ , touching at $0$ .

Edexcel tip. When a question gives a factored form like $y = x(x - 1)(x + 2)$ , identify the roots immediately: $0, 1, -2$ . Sketch using these.

Cubic: S-shape.
Up to 3 roots.
Double root: tangent.
Single root: crosses.

Reciprocal graphs

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$y = k/x$ . Two branches. Asymptotes at axes.

Reciprocal $y = \dfrac{k}{x}$ ( $k \ne 0$ ).

Two-branch hyperbola.

$k > 0$ : branches in upper-right (positive $x, y$ ) and lower-left (negative $x, y$ ).
$k < 0$ : opposite (upper-left and lower-right).

Asymptotes:

VERTICAL: $x = 0$ (the $y$ -axis). Graph never touches.
HORIZONTAL: $y = 0$ (the $x$ -axis).

The two branches hug the axes (dashed) ever closer but never reach them.

Behaviour:

As $x \to \infty$ , $y \to 0$ .
As $x \to 0^{+}$ , $y \to +\infty$ (or $-\infty$ for $k < 0$ ).
As $x \to 0^{-}$ , $y \to -\infty$ (or $+\infty$ ).

Example. $y = 6/x$ .

Test: $(1, 6), (2, 3), (3, 2), (6, 1)$ for positive branch.
Test: $(-1, -6), (-2, -3), (-3, -2)$ for negative branch.
Sketch curve approaching but never touching axes.

Worked qualitative. Why does $y = 1/x$ never cross the $y$ -axis?

$x = 0$ would mean dividing by zero — undefined.
Graph has a 'gap' at $x = 0$ .
The vertical asymptote.

Edexcel tip. Always draw asymptotes as DASHED lines. Mark schemes credit recognising them.

Two branches.
Asymptotes at $x = 0$ and $y = 0$ .
$k > 0$ : 1st & 3rd quadrants.
$k < 0$ : 2nd & 4th quadrants.

Exponential graphs

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$y = a^{x}$ . Always positive. Through $(0, 1)$ . Asymptote $y = 0$ .

Exponential $y = a^{x}$ for $a > 0, a \ne 1$ .

Always positive (for real $x$ ): $a^{x} > 0$ .

Through $(0, 1)$ : $a^{0} = 1$ .

Behaviour:

$a > 1$ : GROWTH. As $x \to \infty$ , $y \to \infty$ . As $x \to -\infty$ , $y \to 0^{+}$ .
$0 < a < 1$ : DECAY. As $x \to \infty$ , $y \to 0^{+}$ . As $x \to -\infty$ , $y \to \infty$ .

Asymptote: $y = 0$ for all exponentials.

Every exponential passes through (0, 1) and flattens towards the x-axis without touching it.

Common exponentials:

Function	Type	Through
$y = 2^{x}$	Growth	$(0,1), (1,2), (2,4), (-1, 0.5)$
$y = 3^{x}$	Growth	$(0,1), (1,3), (2,9)$
$y = (1/2)^{x}$	Decay	$(0,1), (1, 0.5), (2, 0.25)$

Real-world contexts:

Compound interest: exponential growth.
Radioactive decay.
Population growth (initially).
Cooling.

Worked qualitative. Why is $a^{x}$ always positive?

$a > 0$ : any positive number raised to any power is positive.
Even if $x$ is negative, $a^{-x} = 1/a^{x}$ which is still positive.
Graph never crosses the $x$ -axis.

Edexcel tip. Sketch with at least 3 points and the asymptote. Show the shape changes between $x < 0$ and $x > 0$ .

Always positive.
Through $(0, 1)$ .
Asymptote $y = 0$ .
$a > 1$ growth, $0 < a < 1$ decay.

Quick recap

Linear: $y = mx + c$ , straight.
Quadratic: parabola.
Cubic: S-shape, up to 3 roots.
Reciprocal: two branches, asymptotes at axes.
Exponential: through $(0,1)$ , asymptote $y = 0$ .
Identify key features: roots, vertices, intercepts, asymptotes.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Linear: $y = mx + c$ .
Quadratic: parabola; up if $a > 0$ .
Cubic: S-shape; double root = tangent.
Reciprocal: $y = k/x$ , asymptotes at axes.
Exponential: through $(0,1)$ , never zero.

How it’s examined

Function graphs every Higher Tier paper (4-7 marks). Sketching, intercepts, identifying types. Examiner reports flag (1) wrong direction for parabola, (2) double root crossing instead of tangent, (3) reciprocal crossing axes.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

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

1Sketch a quadratic
Higher• quadratic, sketch
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Question
Sketch $y = x^{2} - 4x - 5$ . Show roots and y-intercept.
Step-by-step solution
1. Step 1
  Roots: solve $x^{2} - 4x - 5 = 0$ . Factor: $(x - 5)(x + 1) = 0$ . Roots: $x = 5, -1$ .
2. Step 2
  y-intercept: at $x = 0$ , $y = -5$ .
3. Step 3
  Coefficient of $x^{2}$ is positive → opens UP. Minimum between roots.
4. Step 4
  Vertex: midway between roots. $x = (5 + (-1))/2 = 2$ . $y = 4 - 8 - 5 = -9$ . Vertex $(2, -9)$ .
Answer
Parabola through $(-1, 0)$ , $(5, 0)$ , y-intercept $(0, -5)$ , vertex $(2, -9)$ , opens up.
2Sketch a cubic
Higher• Adapted from 4MA1/2H Jan 2024 Q17• cubic
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Question
Sketch $y = x^{3} - 3x^{2}$ . Show key features.
Step-by-step solution
1. Step 1
  Factor: $y = x^{2}(x - 3)$ . Roots: $x = 0$ (double), $x = 3$ .
2. Step 2
  $x = 0$ is a TANGENT (double root) — graph touches axis at origin.
3. Step 3
  Cubic with positive leading coefficient: bottom-left to top-right.
4. Step 4
  Test values: at $x = 2$ , $y = 8 - 12 = -4$ . At $x = 4$ , $y = 64 - 48 = 16$ .
Answer
Cubic, touches $x$ -axis at origin (tangent), crosses at $(3, 0)$ , increasing for large $x$ .
Examiner tip
Double root: graph TOUCHES (tangent). Single root: graph CROSSES. Memorise the difference.
3Reciprocal graph
Higher• reciprocal
▼
Question
Sketch $y = \dfrac{6}{x}$ . Show asymptotes.
Step-by-step solution
1. Step 1
  Two branches: one in upper-right (positive $x, y$ ), one in lower-left (negative $x, y$ ).
2. Step 2
  Asymptotes: $x = 0$ (vertical) and $y = 0$ (horizontal).
3. Step 3
  Test points: $(1, 6), (2, 3), (3, 2), (6, 1)$ and $(-1, -6), (-2, -3)$ , etc.
Answer
Hyperbola in 1st and 3rd quadrants, asymptotes at axes.
4Exponential graph
Higher• exponential
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Question
Sketch $y = 2^{x}$ .
Step-by-step solution
1. Step 1
  Always POSITIVE. Passes through $(0, 1)$ since $2^{0} = 1$ .
2. Step 2
  As $x \to \infty$ , $y \to \infty$ (rapid growth).
3. Step 3
  As $x \to -\infty$ , $y \to 0^{+}$ (asymptote $y = 0$ ).
4. Step 4
  Test points: $(0, 1), (1, 2), (2, 4), (3, 8), (-1, 0.5), (-2, 0.25)$ .
Answer
Exponential growth curve, through $(0, 1)$ , asymptote $y = 0$ for negative $x$ .
5Counting solutions of a sine equation over two cycles [1 mark]
Getting started• Paper 1H/2H style — 1 mark• trigonometric graphs, sine
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Question
Write down the number of solutions of the equation $\sin x = 0.2$ for $0^{\circ} \leqslant x \leqslant 720^{\circ}$ [1]
Step-by-step solution
1. Step 1
  Between $0^{\circ}$ and $360^{\circ}$ the graph of $y = \sin x$ rises to $1$ and comes back down through positive values until $180^{\circ}$ , so the line $y = 0.2$ cuts it twice in each cycle.
2. Step 2
  B1 The graph repeats every $360^{\circ}$ , and $0^{\circ}$ to $720^{\circ}$ is two full cycles, so there are $2 \times 2$ solutions (at about $11.5^{\circ}$ , $168.5^{\circ}$ , $371.5^{\circ}$ and $528.5^{\circ}$ ).
Answer
$4$
Examiner tip
Examiner tip: Sketch two cycles of the sine curve and the line $y = 0.2$ , then count the crossings. You do not need to find the angles.

Common mistake: Answering $2$ , by thinking only of $0^{\circ}$ to $360^{\circ}$ . The range here is two full cycles.
6Table of values for y = 16/x² [2 marks]
Getting started• Paper 1H/2H style — 2 marks• reciprocal graphs, table of values
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Question
Complete the table of values for $y = \dfrac{16}{x^{2}}$

$\begin{array}{lcccccc} x & 0.5 & 1 & 2 & 4 & 5 & 8 \\ \hline y & \square & 16 & \square & 1 & \square & \square \end{array}$

[2]
Step-by-step solution
1. Step 1
  Square $x$ first, then divide $16$ by the result.
  $x = 0.5:\ \frac{16}{0.25} = 64$
2. Step 2
  B2 The other three values. (B1 for two or three correct.)
  $x = 2:\ \frac{16}{4} = 4 \qquad x = 5:\ \frac{16}{25} = 0.64 \qquad x = 8:\ \frac{16}{64} = 0.25$
Answer
$64$ , $4$ , $0.64$ , $0.25$ (for $x = 0.5, 2, 5, 8$ )
Examiner tip
Examiner tip: Dividing by a number less than $1$ makes the answer bigger, so $x = 0.5$ gives the largest value in the table. The values fall quickly as $x$ increases, which is the shape of a $\dfrac{k}{x^{2}}$ graph.

Common mistake: Dividing by $x$ instead of $x^{2}$ , giving $32$ , $8$ , $3.2$ and $2$ . Check against the given value: $x = 4$ must give $1$ , and $\dfrac{16}{4} = 4$ does not.
7Finding k for a reciprocal graph [2 marks]
Building confidence• Paper 1H/2H style — 2 marks• reciprocal graphs, sketching
▼
Question
The graph of $y = \dfrac{k}{x}$ , where $k$ is a constant, passes through the point $(-2, 3)$

(a) Find the value of $k$ [1]

(b) Does the graph cross the $x$ -axis? Give a reason for your answer. [1]
Step-by-step solution
1. Step 1
  B1 (a) Substitute $x = -2$ and $y = 3$ .
  $3 = \frac{k}{-2} \quad\Rightarrow\quad k = -6$
2. Step 2
  B1 (b) On the $x$ -axis $y = 0$ , but $\dfrac{-6}{x}$ can never equal $0$ because the numerator is not $0$ . So the graph never meets the $x$ -axis; its two branches lie top-left and bottom-right and approach the axes without touching them.
Answer
(a) $k = -6$ (b) No, because $\dfrac{-6}{x} = 0$ has no solution, so $y$ is never $0$ .
Examiner tip
Examiner tip: A reciprocal graph $y = \dfrac{k}{x}$ never crosses either axis: $x = 0$ is not allowed, and $y$ cannot be $0$ . A negative $k$ puts the branches in the top-left and bottom-right.

Common mistake: In (a), writing $k = -1.5$ from $3 \div -2$ . Multiply both sides by $x$ : $k = xy = (-2)(3)$ .
8Where a line meets y = 12/x [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• reciprocal graphs, intersection
▼
Question
The line $y = x - 1$ meets the curve $y = \dfrac{12}{x}$ at two points.

Find the coordinates of both points.

Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  M1 Set the two expressions for $y$ equal and multiply through by $x$ (allowed, as $x \neq 0$ ).
  $\frac{12}{x} = x - 1 \quad\Rightarrow\quad 12 = x^{2} - x \quad\Rightarrow\quad x^{2} - x - 12 = 0$
2. Step 2
  M1 Factorise and solve.
  $(x - 4)(x + 3) = 0 \quad\Rightarrow\quad x = 4 \ \text{or}\ x = -3$
3. Step 3
  A1 Use $y = x - 1$ for each $y$ -coordinate.
  $x = 4:\ y = 3 \qquad x = -3:\ y = -4$
Answer
$(4, 3)$ and $(-3, -4)$
Examiner tip
Examiner tip: Check each point in the curve's equation: $\dfrac{12}{4} = 3$ and $\dfrac{12}{-3} = -4$ . One point is on each branch of the reciprocal graph.

Common mistake: Multiplying only the $x$ term by $x$ , giving $12 = x^{2} - 1$ . The whole right-hand side, $x - 1$ , must be multiplied.
9Solving tan x = 3 beyond 360 degrees [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• trigonometric graphs, tangent
▼
Question
Solve $\tan x = 3$ for $0^{\circ} \leqslant x \leqslant 540^{\circ}$

Give your answers correct to 1 decimal place. [3]
Step-by-step solution
1. Step 1
  B1 The calculator gives the first solution.
  $x = \tan^{-1} 3 = 71.565\ldots^{\circ} \approx 71.6^{\circ}$
2. Step 2
  M1 The graph of $y = \tan x$ repeats every $180^{\circ}$ , so add $180^{\circ}$ to find further solutions.
  $71.565\ldots^{\circ} + 180^{\circ} = 251.565\ldots^{\circ}$
3. Step 3
  A1 Add $180^{\circ}$ once more; the next one, $611.6^{\circ}$ , is beyond $540^{\circ}$ .
  $251.565\ldots^{\circ} + 180^{\circ} = 431.565\ldots^{\circ}$
Answer
$x = 71.6^{\circ}$ , $251.6^{\circ}$ and $431.6^{\circ}$
Examiner tip
Examiner tip: For tangent, every solution is the first one plus a multiple of $180^{\circ}$ . List them until the next would pass the end of the range.

Common mistake: Adding $360^{\circ}$ (the sine and cosine period) and giving only $71.6^{\circ}$ and $431.6^{\circ}$ . That misses $251.6^{\circ}$ and loses the A mark.
10Finding the line to draw, then the exact intersections [5 marks]
Stretch• Paper 1H/2H style — 5 marks• suitable straight line, quadratic graphs, surds
▼
Question
The graph of $y = x^{2} - 2x - 1$ has been drawn on a grid.

(a) Find the equation of the straight line that should be drawn on the grid to solve $x^{2} - 4x + 2 = 0$ [2]

(b) The line from part (a) meets the curve at the points $A$ and $B$ . Find the exact coordinates of $A$ and $B$ . [3]
Step-by-step solution
1. Step 1
  M1 (a) Subtract the equation from the curve: $(x^{2} - 2x - 1) - (x^{2} - 4x + 2)$ .
  $x^{2} - 4x + 2 = 0 \iff x^{2} - 2x - 1 = 2x - 3$
2. Step 2
  A1 So the line is
  $y = 2x - 3$
3. Step 3
  M1 (b) The $x$ -coordinates of $A$ and $B$ are the solutions of $x^{2} - 4x + 2 = 0$ . Complete the square (or use the quadratic formula).
  $(x - 2)^{2} - 4 + 2 = 0 \quad\Rightarrow\quad (x - 2)^{2} = 2$
4. Step 4
  A1 Take both square roots.
  $x = 2 + \sqrt{2} \quad\text{or}\quad x = 2 - \sqrt{2}$
5. Step 5
  A1 Substitute into $y = 2x - 3$ for the $y$ -coordinates.
  $y = 2(2 \pm \sqrt{2}) - 3 = 1 \pm 2\sqrt{2}$
Answer
(a) $y = 2x - 3$ (b) $(2 + \sqrt{2},\ 1 + 2\sqrt{2})$ and $(2 - \sqrt{2},\ 1 - 2\sqrt{2})$
Examiner tip
Examiner tip: "Exact" means leave the answers as surds. As a check, $2 + \sqrt{2} = 3.41\ldots$ and $2 - \sqrt{2} = 0.58\ldots$ , which are the values you would read from the graph.

Common mistake: Taking only the positive square root, $x - 2 = \sqrt{2}$ , and finding just one point. Two points are asked for, so both roots are needed.

Question 1
Paper 1H/2H style — 1 mark1 mark
The curve $y = x^{2} - 6x + 2$ passes through the points $(1, -3)$ and $(5, -3)$

Write down the equation of the line of symmetry of the curve. [1]
Model answer
The two points have the same $y$ -coordinate, so the line of symmetry is halfway between them: $\dfrac{1 + 5}{2} = 3$

Answer: $x = 3$
Why this scores
Mark scheme:
- B1 for $x = 3$ cao
Why this scores full marks: The answer is written as the equation of a line, $x = 3$ , not just the number $3$ . The symmetry of a parabola means any two points at the same height are the same distance from its line of symmetry.
Question 2
Paper 1H/2H style — 2 marks2 marks
Complete the table of values for $y = \cos x$

$\begin{array}{lccccccc} x & 0^{\circ} & 60^{\circ} & 120^{\circ} & 180^{\circ} & 240^{\circ} & 300^{\circ} & 360^{\circ} \\ \hline y & 1 & 0.5 & \square & \square & \square & \square & 1 \end{array}$

[2]
Model answer
$\cos 120^{\circ} = -0.5$

$\cos 180^{\circ} = -1$

$\cos 240^{\circ} = -0.5$

$\cos 300^{\circ} = 0.5$

Answer: $-0.5$ , $-1$ , $-0.5$ , $0.5$
Why this scores
Mark scheme:
- B2 for all four values correct; B1 for two or three correct
Why this scores full marks: The values are symmetrical about $x = 180^{\circ}$ ( $120^{\circ}$ and $240^{\circ}$ match, as do $60^{\circ}$ and $300^{\circ}$ ), which is a quick check on the calculator work. A calculator set to radians gives wrong values, so check the mode first.
Question 3
Paper 1H/2H style — 2 marks2 marks
Find the coordinates of the points where the graph of $y = x^{3} - 9x$ crosses the $x$ -axis. [2]
Model answer
On the $x$ -axis, $y = 0$ :

$x^{3} - 9x = 0$

$x(x^{2} - 9) = 0$

$x(x - 3)(x + 3) = 0$

Answer: $(-3, 0)$ , $(0, 0)$ and $(3, 0)$
Why this scores
Mark scheme:
- M1 for $x(x^{2} - 9)$ or $x(x - 3)(x + 3)$ oe
- A1 for $(-3, 0)$ , $(0, 0)$ and $(3, 0)$
Why this scores full marks: Taking out the common factor $x$ first keeps the solution $x = 0$ ; dividing both sides by $x$ would lose it. The difference of two squares gives both $3$ and $-3$ .
Question 4
Paper 1H/2H style — 2 marks2 marks
The graphs of $y = x^{2} - 3x$ and $y = x + 2$ are drawn on the same grid.

The $x$ -coordinates of the points where they intersect are the solutions of the equation $x^{2} + bx + c = 0$

Find the value of $b$ and the value of $c$ [2]
Model answer
$x^{2} - 3x = x + 2$

$x^{2} - 4x - 2 = 0$

Answer: $b = -4$ , $c = -2$
Why this scores
Mark scheme:
- M1 for $x^{2} - 3x = x + 2$ oe
- A1 for $b = -4$ and $c = -2$
Why this scores full marks: At an intersection the two $y$ -values are equal. Moving everything to the left-hand side, with care over signs ( $-3x - x = -4x$ ), gives the equation in the required form.
Question 5
Paper 1H/2H style — 3 marks3 marks
Here is a table of values for $y = x^{3} - 3x^{2} + 4$

$\begin{array}{lccccc} x & -1 & 0 & 1 & 2 & 3 \\ \hline y & 0 & 4 & 2 & 0 & 4 \end{array}$

The graph of $y = x^{3} - 3x^{2} + 4$ has turning points at $(0, 4)$ and $(2, 0)$

(a) Write down the solutions of $x^{3} - 3x^{2} + 4 = 0$ [1]

(b) Write down the number of solutions of $x^{3} - 3x^{2} + 4 = 3$ [1]

(c) The equation $x^{3} - 3x^{2} + 4 = k$ has exactly two solutions when $k = 0$ . Write down the other value of $k$ for which it has exactly two solutions. [1]
Model answer
(a) $y = 0$ at $x = -1$ and $x = 2$ (the curve touches the $x$ -axis at its minimum $(2, 0)$ ).

Answer: $x = -1$ , $x = 2$

(b) $y = 3$ lies between the minimum value $0$ and the maximum value $4$ , so it crosses the curve three times.

Answer: $3$

(c) Exactly two solutions happen when the line $y = k$ touches a turning point and crosses the curve once more. $k = 0$ touches the minimum, so the other is the line through the maximum $(0, 4)$ .

Answer: $k = 4$
Why this scores
Mark scheme:
- B1 for $-1$ and $2$ (and no others)
- B1 for $3$
- B1 for $4$ cao
Why this scores full marks: The candidate pictures the curve from the table and the turning points: up to $(0, 4)$ , down to touch the axis at $(2, 0)$ , then up again. A horizontal line between $y = 0$ and $y = 4$ meets it three times; above $4$ or below $0$ , only once; exactly at $y = 0$ or $y = 4$ it touches a turning point, giving two solutions. At $k = 4$ these are $x = 0$ and $x = 3$ , which the table confirms.
Question 6
Paper 1H/2H style — 3 marks3 marks
The graph of $y = \dfrac{5}{x^{2}}$ is drawn for $x \neq 0$

(a) Give a reason why no part of the graph is below the $x$ -axis. [1]

(b) Write down the equations of the two lines that the graph gets closer and closer to but never meets. [1]

(c) Find the $x$ -coordinates of the points where the graph meets the line $y = 20$ [1]
Model answer
(a) $x^{2}$ is positive for every $x \neq 0$ , and $5$ is positive, so $\dfrac{5}{x^{2}}$ is always positive.

(b) $x = 0$ and $y = 0$

(c) $\dfrac{5}{x^{2}} = 20 \Rightarrow x^{2} = 0.25 \Rightarrow x = \pm 0.5$

Answer: $x = 0.5$ and $x = -0.5$
Why this scores
Mark scheme:
- B1 for a reason referring to $x^{2}$ (and $5$ ) being positive oe
- B1 for $x = 0$ and $y = 0$
- B1 for $0.5$ and $-0.5$
Why this scores full marks: In (c) the candidate gives both square roots. The June 2024 examiners' report noted students who worked out only the positive value of $x$ after square rooting when the question asked for two points on the curve; here the two branches of the graph each meet $y = 20$ once.
Question 7
Paper 1H/2H style — 4 marks4 marks
Find all the solutions of $\sin x = 0.5$ for $-360^{\circ} \leqslant x \leqslant 360^{\circ}$ [4]
Model answer
$\sin^{-1} 0.5 = 30^{\circ}$

By the symmetry of the sine graph about $90^{\circ}$ : $180^{\circ} - 30^{\circ} = 150^{\circ}$

The graph repeats every $360^{\circ}$ , so subtract $360^{\circ}$ from each:

$30^{\circ} - 360^{\circ} = -330^{\circ}$ and $150^{\circ} - 360^{\circ} = -210^{\circ}$

Answer: $x = -330^{\circ}$ , $-210^{\circ}$ , $30^{\circ}$ , $150^{\circ}$
Why this scores
Mark scheme:
- B1 for $30$
- B1 for $150$
- M1 for subtracting $360$ from a positive solution, or use of the symmetry of the graph for negative $x$
- A1 for $-330$ and $-210$ (and no extra values in the range)
Why this scores full marks: The candidate finds the two solutions in one cycle, then uses the period to move a whole cycle to the left. Four values, in order, with none outside the range; $\sin(-210^{\circ}) = 0.5$ is a quick check.
Question 8
Paper 1H/2H style — 4 marks4 marks
The curve with equation $y = x^{2} + \dfrac{a}{x}$ , where $a$ is a constant, passes through the point $(2, 7)$

(a) Find the value of $a$ [2]

The line $y = 7$ meets the curve at $(2, 7)$ and at two other points.

(b) Show that the $x$ -coordinates of the points where the line $y = 7$ meets the curve satisfy $(x - 2)(x^{2} + 2x - 3) = 0$ , and hence find the $x$ -coordinates of the two other points. [2]
Model answer
(a) $7 = 2^{2} + \dfrac{a}{2}$

$\dfrac{a}{2} = 3$

Answer: $a = 6$

(b) $x^{2} + \dfrac{6}{x} = 7$

$x^{3} - 7x + 6 = 0$

$(x - 2)(x^{2} + 2x - 3) = x^{3} + 2x^{2} - 3x - 2x^{2} - 4x + 6 = x^{3} - 7x + 6$ as required.

$(x - 2)(x - 1)(x + 3) = 0$

Answer: $x = 1$ and $x = -3$
Why this scores
Mark scheme:
- M1 for $7 = 4 + \dfrac{a}{2}$ oe
- A1 for $a = 6$
- M1 for $x^{3} - 7x + 6 = 0$ and a correct expansion of $(x - 2)(x^{2} + 2x - 3)$ oe
- A1 for $1$ and $-3$
Why this scores full marks: The candidate multiplies every term by $x$ to clear the fraction, expands the given product to show it matches, then factorises the quadratic factor. The factor $x - 2$ is the known point $(2, 7)$ . Checking $x = -3$ : $9 + \dfrac{6}{-3} = 7$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
The line $y = x - 3$ meets the curve $y = x^{3} - 3x^{2}$ at three points.

Find the coordinates of the three points.

Show clear algebraic working. [5]
Model answer
$x^{3} - 3x^{2} = x - 3$

$x^{3} - 3x^{2} - x + 3 = 0$

$x^{2}(x - 3) - 1(x - 3) = 0$

$(x - 3)(x^{2} - 1) = 0$

$(x - 3)(x - 1)(x + 1) = 0$

$x = 3$ , $x = 1$ , $x = -1$

Using $y = x - 3$ : $y = 0$ , $y = -2$ , $y = -4$

Answer: $(3, 0)$ , $(1, -2)$ and $(-1, -4)$
Why this scores
Mark scheme:
- M1 for $x^{3} - 3x^{2} - x + 3 = 0$ oe
- M1 for $x^{2}(x - 3) - (x - 3)$ or $(x - 3)(x^{2} - 1)$ oe
- M1 for $(x - 3)(x - 1)(x + 1)$ or three values of $x$ from a correct method
- A1 for $x = 3, 1, -1$
- A1 for all three coordinate pairs correct
Why this scores full marks: Grouping the four terms in pairs reveals the common factor $(x - 3)$ , and the difference of two squares finishes the factorisation. The $y$ -coordinates come from the line, which is quicker than the cubic and a good check: $1^{3} - 3(1)^{2} = -2$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
You are given that $\sin 25^{\circ} = 0.423$ , correct to 3 significant figures.

Use the graphs of $y = \sin x$ and $y = \cos x$ to answer these questions. Do not use a calculator.

(a) Write down the value of $\sin 155^{\circ}$ [1]

(b) Write down the value of $\sin 205^{\circ}$ [1]

(c) Write down the value of $\cos 65^{\circ}$ [1]

(d) Solve $\sin x = -0.423$ for $0^{\circ} \leqslant x \leqslant 360^{\circ}$ [2]
Model answer
(a) The sine graph is symmetrical about $x = 90^{\circ}$ , and $155^{\circ} = 180^{\circ} - 25^{\circ}$ .

Answer: $0.423$

(b) $205^{\circ} = 180^{\circ} + 25^{\circ}$ ; from $180^{\circ}$ to $360^{\circ}$ the graph is the first half turned upside down.

Answer: $-0.423$

(c) The cosine graph is the sine graph moved $90^{\circ}$ to the left, so $\cos 65^{\circ} = \sin(65^{\circ} + 90^{\circ}) = \sin 155^{\circ}$ .

Answer: $0.423$

(d) One solution is $205^{\circ}$ from (b). By symmetry about $x = 270^{\circ}$ , the other is $360^{\circ} - 25^{\circ}$ .

Answer: $x = 205^{\circ}$ and $x = 335^{\circ}$
Why this scores
Mark scheme:
- B1 for $0.423$
- B1 for $-0.423$
- B1 for $0.423$
- B1 for $205$
- B1 for $335$
Why this scores full marks: Every answer comes from a symmetry of the graphs, stated briefly, rather than from a calculator. In (c), $\cos 65^{\circ} = \sin 25^{\circ}$ because $25^{\circ} + 65^{\circ} = 90^{\circ}$ , which the candidate reaches by relating the two graphs. In (d), the two solutions lie between $180^{\circ}$ and $360^{\circ}$ , where the sine graph is below the axis.

4MA1 Graphs of Functions

At a glance

What you’ll learn

Linear and quadratic — review

Cubic graphs

Reciprocal graphs

Exponential graphs

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Sketch a quadratic

2Sketch a cubic

3Reciprocal graph

4Exponential graph

5Counting solutions of a sine equation over two cycles [1 mark]

6Table of values for y = 16/x² [2 marks]

7Finding k for a reciprocal graph [2 marks]

8Where a line meets y = 12/x [3 marks]

9Solving tan x = 3 beyond 360 degrees [3 marks]

10Finding the line to draw, then the exact intersections [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10