What is A12: Graphs of functions?

Recognise graph families from their defining shapes and features: linear graphs have constant gradient, quadratics are symmetric parabolas, simple cubics have an S-shape, and y=1/x has two reciprocal branches with asymptotes x=0 and y=0.

How many marks are A12 questions worth?

In verified past-paper appearances, A12 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has A12 appeared in?

A12 has appeared in 23 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A12

Recognise, sketch and interpret graphs of linear, quadratic and simple cubic functions, the reciprocal y = 1/x (x ≠ 0), exponential y = k^x (k > 0), and y = sin x, cos x, tan x for angles of any size

Graphs of functions

Worked answers, methods and verified real exam appearances for A12 on Edexcel GCSE Maths 1MA1.

Explanation

Recognise graph families from their defining shapes and features: linear graphs have constant gradient, quadratics are symmetric parabolas, simple cubics have an S-shape, and $y=\dfrac{1}{x}$ has two reciprocal branches with asymptotes $x=0$ and $y=0$ .
Sketch by marking intercepts, roots, turning points, asymptotes and representative values.
Higher tier: also recognise exponentials and sine, cosine and tangent graphs for angles of any size, using their periods and standard values.
Examiners expect a sketch to show correct shape and position, not merely a collection of plotted points.

The reciprocal graph

y=1/x

has branches in quadrants I and III and approaches, but never reaches, both axes.

Worked example

For $y=\dfrac{6}{x}$ , state both asymptotes and the quadrants containing its branches.

1. $x=0$ is excluded, so the vertical asymptote is $x=0$ .
2.As $|x|$ grows, $\dfrac{6}{x}$ approaches $0$ , so the horizontal asymptote is $y=0$ .
3. $x$ and $y$ have the same sign, so the branches lie in quadrants I and III.

Answer: Asymptotes $x=0$ and $y=0$ ; branches in quadrants I and III.

Common mistakes

Don't draw a reciprocal branch touching or crossing an axis.
Don't sketch a cubic as a parabola rather than an S-shaped curve.
Don't treat exponential growth as a straight line (Higher tier).

Exam tip

Before sketching, list the intercepts, turning points and asymptotes that fix the graph's shape.

Worked practice

Tier 1 · Easy

A graph passes through

(0,1)

and its

y

-value doubles whenever

x

increases by

1

. Name the function

y=2^x

as linear, quadratic, cubic, reciprocal or exponential.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Exponential	1	Equal increases in $x$ multiply the output by a constant factor, so $y=2^x$ is an exponential function.

Tier 2 · Standard

For the graph

y=\frac6x

, state both asymptotes and the two quadrants containing its branches.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Asymptotes $x=0$ and $y=0$ Branches in quadrants I and III	3	The expression is undefined at $x=0$ , giving vertical asymptote $x=0$ . As $\|x\|$ grows, $6/x$ approaches $0$ , giving horizontal asymptote $y=0$ . Since $6/x$ has the same sign as $x$ , the branches lie in quadrants I and III.

Tier 3 · Hard

For

y=\cos x

-180^\circ\leq x\leq360^\circ

, list the

x

-intercepts and the coordinates of every maximum and minimum needed for an accurate sketch.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$x$ -intercepts at $x=-90^\circ,90^\circ,270^\circ$ Maxima $(0^\circ,1)$ and $(360^\circ,1)$ Minima $(-180^\circ,-1)$ and $(180^\circ,-1)$	4	Cosine is zero at odd multiples of $90^\circ$ , giving $-90^\circ$ , $90^\circ$ and $270^\circ$ in the interval. It reaches $1$ at multiples of $360^\circ$ , here $0^\circ$ and $360^\circ$ , and reaches $-1$ at odd multiples of $180^\circ$ , here $-180^\circ$ and $180^\circ$ .

Tier 1 · Easy

Which graph,

y=x^2

y=x^3

, has rotational symmetry about the origin?

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$y=x^3$ (it has rotational symmetry of order $2$ about the origin)	1	The graph of $y=x^3$ is unchanged by a half-turn about the origin, so it has rotational symmetry of order $2$ about the origin.

Tier 2 · Standard

A quadratic graph has

y

-values

8,3,0,-1,0

x=-1,0,1,2,3

respectively. Write down the coordinates of its turning point. Write down the equation of its line of symmetry.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Turning point $(2,-1)$ Line of symmetry $x=2$	2	The least value is $-1$ at $x=2$ , so the turning point is $(2,-1)$ . The equal values at $x=1$ and $x=3$ show symmetry about $x=2$ , so the line of symmetry is $x=2$ .

Tier 3 · Hard

The equation of a quadratic graph is

y=-2(x+3)^2+11

. State whether the graph opens upwards or downwards and give a reason. Write down the maximum value of

y

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	The graph opens downwards because the coefficient of $(x+3)^2$ is negative Maximum value of $y$ is $11$	3	The coefficient of $(x+3)^2$ is $-2$ . Since it is negative, the graph opens downwards. Its turning point is $(-3,11)$ , so the maximum value of $y$ is $11$ .

Tier 2 · Standard

For the cubic graph

y=x^3-8

, write down the coordinates of the

x

-intercept and the

y

-intercept. Write down whether the graph rises or falls as

x

increases.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x$ -intercept $(2,0)$ $y$ -intercept $(0,-8)$ The graph rises	3	At the $x$ -intercept, $x^3-8=0$ , so $x^3=8$ and $x=2$ , giving $(2,0)$ . At $x=0$ , $y=-8$ , giving $(0,-8)$ . The positive coefficient of $x^3$ means the cubic rises from left to right.

Tier 3 · Hard

The reciprocal graph

y=\dfrac{k}{x}

passes through

(3,-4)

. Work out

k

. Write down both asymptotes and describe where the two branches of the graph lie.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$k=-12$ Asymptotes $x=0$ and $y=0$ One branch where $x<0$ and $y>0$ (top left), the other where $x>0$ and $y<0$ (bottom right).	4	Substitute $(3,-4)$ : $-4=\dfrac{k}{3}$ , so $k=-12$ . A reciprocal graph has asymptotes $x=0$ and $y=0$ . Since $k$ is negative, $x$ and $y$ have opposite signs, so one branch lies where $x<0,\ y>0$ and the other where $x>0,\ y<0$ .

Tier 3 · Hard

Higher only: Ellie says that the graph of

y=\left(\dfrac12\right)^x

rises as

x

increases and crosses the

x

-axis at

(1,0)

. Explain both of Ellie's errors. State the

y

-intercept and the horizontal asymptote.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	The graph falls as $x$ increases because the base lies between $0$ and $1$ It has no $x$ -intercept because $\left(\dfrac12\right)^x$ is always positive $y$ -intercept $(0,1)$ Horizontal asymptote $y=0$	4	Because $0<\tfrac12<1$ , multiplying the output by $\tfrac12$ for each unit increase in $x$ makes the graph fall. Every output is positive, so the curve approaches but never crosses the $x$ -axis. At $x=0$ , the output is $1$ , giving the $y$ -intercept $(0,1)$ , and the horizontal asymptote is $y=0$ .

Q10

Tier 3 · Hard

Three graphs

P

Q

and

R

have equations

y=x^2+4x-5

y=x^3+2

and

y=-\dfrac8x

, in some order. Graph

P

crosses the axes at

(-5,0)

(1,0)

and

(0,-5)

. Graph

Q

passes through

(-1,1)

(0,2)

and

(2,10)

. Graph

R

passes through

(-4,2)

and

(2,-4)

and is undefined when

x=0

. Match each graph to its equation. Show one substitution check for each match.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$P: y=x^2+4x-5$ $Q: y=x^3+2$ $R: y=-\dfrac8x$ $P$ check: $(-5)^2+4(-5)-5=0$ $Q$ check: $2^3+2=10$ $R$ check: $-8\div(-4)=2$	5	For $y=x^2+4x-5$ , substituting $x=-5$ gives $25-20-5=0$ (and $x=1$ also gives $0$ ), so this is $P$ . For $y=x^3+2$ , substituting $x=2$ gives $y=2^3+2=10$ , so this is $Q$ . For $y=-\dfrac8x$ , substituting $x=-4$ gives $y=2$ (and the expression is undefined at $x=0$ ), so this is $R$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	1H	Q22	4	Non-calculator	Higher	QP MS
2021-11	2F	Q24	6	Allowed	Foundation	QP MS
2023-06	2F	Q28	3	Allowed	Foundation	QP MS
2021-11	2H	Q18	4	Allowed	Higher	QP MS
2022-11	1H	Q9	4	Non-calculator	Higher	QP MS
2024-11	2H	Q5	4	Allowed	Higher	QP MS
2024-06	2H	Q5	6	Allowed	Higher	QP MS
2019-11	2H	Q5	2	Allowed	Higher	QP MS
2019-11	1H	Q21	5	Non-calculator	Higher	QP MS
2021-11	3F	Q28	2	Allowed	Foundation	QP MS
2024-06	1H	Q12	3	Non-calculator	Higher	QP MS
2022-06	2H	Q21	3	Allowed	Higher	QP MS
2024-11	2F	Q24	4	Allowed	Foundation	QP MS
2019-11	2F	Q25	2	Allowed	Foundation	QP MS
2021-11	2H	Q4	6	Allowed	Higher	QP MS
2019-06	3H	Q17	2	Allowed	Higher	QP MS
2023-11	2F	Q20	4	Allowed	Foundation	QP MS
2022-06	1H	Q6	6	Non-calculator	Higher	QP MS
2022-11	1H	Q21	2	Non-calculator	Higher	QP MS
2022-06	1F	Q28	6	Non-calculator	Foundation	QP MS
2024-06	2F	Q24	6	Allowed	Foundation	QP MS
2024-11	3H	Q23	5	Allowed	Higher	QP MS
2024-11	3H	Q21	4	Allowed	Higher	QP MS

Other points in A Algebra · equations and reasoning

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