What is A14: Real-life graphs?

A graph represents the ordered pairs satisfying a rule. Use intercepts, turning points and asymptotes to describe its behaviour.

How many marks are A14 questions worth?

In verified past-paper appearances, A14 questions on Edexcel GCSE Maths 1MA1 have been worth 3–5 marks.

Which past exam papers has A14 appeared in?

A14 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A14

Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

Real-life graphs

Worked answers, methods and verified real exam appearances for A14 on Edexcel GCSE Maths 1MA1.

Explanation

A graph represents the ordered pairs satisfying a rule. Use intercepts, turning points and asymptotes to describe its behaviour.
Reciprocal graphs such as $y=k/x$ have two branches and approach the axes without meeting them.
Higher tier: exponential graphs change by a constant multiplier for equal changes in $x$ .
To solve equations graphically, plot both relations on the same axes and read every intersection accurately.
In a real context, the examiner expects the coordinate to be interpreted using the quantities and units on the axes, not reported as an unexplained pair of numbers.

For

y=k/x

with

k>0

, reciprocal branches lie in opposite quadrants and approach both axes.

Worked example

The journey time $t$ hours is modelled by $t=24/v$ , where $v$ is the average speed in km/h. Find and interpret the point when $v=16$ .

1.Substitute $v=16$ into the model: $t=24\div16$ .
2.Evaluate to obtain $t=1.5$ , so the coordinate is $(16,1.5)$ .
3.Interpret the axes: at $16$ km/h, the journey takes $1.5$ hours.

Answer: $(16,1.5)$ ; travelling at $16$ km/h gives a journey time of $1.5$ hours.

Common mistakes

Don't read the $y$ -coordinate when the question asks for the solution in $x$ .
Don't give an intersection such as $(16,1.5)$ without interpreting either coordinate or its unit.

Exam tip

For an 'estimate' question, show the plotted curves and read the intersection to a precision justified by the graph scale.

Worked practice

Tier 1 · Easy

The time

t

hours for a fixed journey is modelled by

t=24/v

, where

v

is the average speed in km/h. Work out the point on this graph when

v=16

and interpret it.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$(16,1.5)$ At an average speed of $16$ km/h, the journey takes $1.5$ hours.	2	Substitute $v=16$ : $t=24/16=1.5$ . The coordinates are speed then time, so the graph contains $(16,1.5)$ and this represents a $1.5$ -hour journey at $16$ km/h.

Tier 2 · Standard

Higher only: Plot

y=2^x

and

y=9-2x

for

1\le x\le3

. Use the intersection to estimate the solution of

2^x=9-2x

to one decimal place.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x\approx2.2$	4	Draw the increasing exponential curve $y=2^x$ and the decreasing straight line $y=9-2x$ on the same axes. Their intersection has $x$ -coordinate about $2.2$ (the numerical value is about $2.201$ ), so the graphical estimate to one decimal place is $2.2$ .

Tier 3 · Hard

For

0\le t\le5

, two moving objects have distances from a marker modelled by

d=3t^2+2

and

d=14t

, with

d

in metres and

t

in seconds. Draw both graphs and estimate the later time when the objects are equally far from the marker.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$t\approx4.5$ seconds	5	Plot the quadratic $d=3t^2+2$ and the line $d=14t$ . Equal distances occur at intersections. The later intersection has $t\approx4.52$ , so an appropriate graph gives about $4.5$ seconds; the earlier intersection near $0.15$ seconds is not requested.

Tier 1 · Easy

The graph of the height

h

cm of a candle against time

t

hours passes through

(9,0)

. Explain what this point represents.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	After $9$ hours, the candle has height $0$ cm, so it has burned completely.	1	The first coordinate represents time and the second represents height. Therefore $(9,0)$ means that after $9$ hours the candle has height $0$ cm and has burned completely.

Tier 2 · Standard

For

x=2,3,6,9

, work out the corresponding values of

y

for

y=18/x

. Plot the four points and draw the branch for

x>0

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	The points are $(2,9)$ , $(3,6)$ , $(6,3)$ and $(9,2)$ A decreasing reciprocal curve through the four points, approaching both positive axes	3	Substitution gives $18/2=9$ , $18/3=6$ , $18/6=3$ and $18/9=2$ . Plot $(2,9)$ , $(3,6)$ , $(6,3)$ and $(9,2)$ , then draw a smooth decreasing reciprocal branch that approaches but does not meet either axis.

Tier 3 · Hard

The depth

h

metres of water at times

t=0,1,2,3,4,5,6

hours is

2,5,8,10,8,5,2

respectively. Plot the points, join consecutive points with straight lines and use the graph to work out for how long the depth is greater than

6

metres.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	About $3.3$ hours (accept $3.2$ to $3.4$ hours from an accurate graph)	4	Plot $(0,2)$ , $(1,5)$ , $(2,8)$ , $(3,10)$ , $(4,8)$ , $(5,5)$ and $(6,2)$ . The line $h=6$ meets the graph at about $t=1.33$ and $t=4.67$ . The depth is more than $6$ metres between these times, for about $4.67-1.33=3.34$ hours, which is about $3.3$ hours; a graph reading from $3.2$ to $3.4$ hours is acceptable.

Tier 2 · Standard

For each integer value of

x

from

-4

3

, work out the corresponding value of

y

for

y=x^2+x-4

. Plot the points and draw a smooth curve. Use your graph to estimate the two solutions of

x^2+x-4=3

, giving each solution to one decimal place.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x\approx-3.2$ or $x\approx2.2$ (accept $-3.3$ to $-3.1$ and $2.1$ to $2.3$ )	4	The $y$ -values for $x=-4,-3,-2,-1,0,1,2,3$ are $8,2,-2,-4,-4,-2,2,8$ . Plot these points and draw the parabola. Where the curve crosses $y=3$ , the $x$ -coordinates are about $-3.2$ and $2.2$ (the numerical values are $-3.1925\ldots$ and $2.1925\ldots$ ).

Tier 3 · Hard

For

1\le x\le6

, draw the graphs of

y=20/x

and

y=2x-1

on the same axes. Use the intersection to estimate the positive solution of

20/x=2x-1

, giving your answer to one decimal place.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x\approx3.4$ (accept $3.3$ to $3.5$ from an accurate graph)	4	Draw the decreasing reciprocal curve $y=20/x$ and the increasing line $y=2x-1$ . Their intersection has $x$ -coordinate about $3.4$ (the numerical value is $3.4221\ldots$ ), so the graphical estimate to one decimal place is $3.4$ .

Tier 3 · Hard

The table shows values of

y=x^3-4x

for

x=-2.5,-2,-1.5,-1,-0.5,0,0.5,1,1.5,2,2.5

. The corresponding values of

y

are

-5.625,0,2.625,3,1.875,0,-1.875,-3,-2.625,0,5.625

. Plot the points and draw a smooth curve. Add the line

y=1

and estimate all three solutions of

x^3-4x=1

to one decimal place.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x\approx-1.9$ , $x\approx-0.3$ or $x\approx2.1$ (accept $-2.0$ to $-1.8$ , $-0.4$ to $-0.2$ , and $2.0$ to $2.2$ from an accurate graph)	5	Plot the eleven supplied points, join them with a smooth cubic curve and draw the horizontal line $y=1$ . The three intersection $x$ -coordinates are approximately $-1.8608$ , $-0.2541$ and $2.1149$ , so the graphical estimates to one decimal place are $-1.9$ , $-0.3$ and $2.1$ . The smallest distance from a rounding boundary is $0.0041$ , or $0.041$ final-digit units.

Q10

Tier 3 · Hard

Higher only: For integer values of

x

from

0

5

, work out values for

y=2^x

and

y=x^2+1

. Draw both graphs on the same axes. Write down the two integer solutions of

2^x=x^2+1

and estimate the remaining solution to one decimal place.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$x=0$ , $x=1$ and $x\approx4.3$ (accept $4.2$ to $4.4$ for the non-integer solution from an accurate graph)	5	For $x=0,1,2,3,4,5$ , the exponential values are $1,2,4,8,16,32$ and the quadratic values are $1,2,5,10,17,26$ . The graphs meet at $x=0$ and $x=1$ , then again between $4$ and $5$ . The final intersection is at $x=4.2574\ldots$ , giving a graphical estimate of $4.3$ to one decimal place. It is $0.074$ final-digit units from the nearest rounding boundary.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	2F	Q16	3	Allowed	Foundation	QP MS
2022-11	3F	Q14	3	Allowed	Foundation	QP MS
2023-06	2F	Q10	4	Allowed	Foundation	QP MS
2019-11	2F	Q11	3	Allowed	Foundation	QP MS
2019-11	1F	Q16	3	Non-calculator	Foundation	QP MS
2021-11	2F	Q14	3	Allowed	Foundation	QP MS
2024-11	2F	Q14	3	Allowed	Foundation	QP MS
2022-11	3F	Q19	4	Allowed	Foundation	QP MS
2021-11	2F	Q23	5	Allowed	Foundation	QP MS
2021-11	2H	Q3	5	Allowed	Higher	QP MS

Other points in A Algebra · graphs

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