What is A10: Gradients and intercepts?

The gradient of a linear function is the change in the vertical quantity per unit change in the horizontal quantity. Its sign shows whether the line rises or falls.

How many marks are A10 questions worth?

In verified past-paper appearances, A10 questions on Edexcel GCSE Maths 1MA1 have been worth 2–4 marks.

Which past exam papers has A10 appeared in?

A10 has appeared in 8 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A10

Identify and interpret gradients and intercepts of linear functions graphically and algebraically

Gradients and intercepts

Worked answers, methods and verified real exam appearances for A10 on Edexcel GCSE Maths 1MA1.

Explanation

The gradient of a linear function is the change in the vertical quantity per unit change in the horizontal quantity. Its sign shows whether the line rises or falls.
The $y$ -intercept is the output when $x=0$ ; the $x$ -intercept is where the output is zero.
Read axis scales and units before calculating or interpreting either feature.
In a context, a gradient is a rate and an intercept is often an initial value or fixed charge.
Examiners require a value, its unit and a sentence explaining what it represents.

The intercept is the output at zero; the gradient is rise divided by run.

Worked example

Water volume follows $V=120-6t$ , where $V$ is litres and $t$ is minutes. Interpret the gradient and intercept.

1.The coefficient of $t$ is $-6$ , so volume changes by $-6$ litres per minute.
2.Therefore the volume decreases by $6$ litres each minute.
3.When $t=0$ , $V=120$ , so the intercept is the initial volume.

Answer: The tank starts with $120$ litres and loses $6$ litres per minute.

Common mistakes

Don't calculate run divided by rise for the gradient.
Don't read the wrong intercept because the axes have been confused.
Don't give a contextual gradient as a bare number without units or meaning.

Exam tip

For “interpret”, state what happens per horizontal-axis unit and what the intercept means at zero.

Worked practice

Tier 1 · Easy

State the gradient and

y

-intercept of

y=-3x+8

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Gradient $-3$ $y$ -intercept $8$	2	Compare with $y=mx+c$ . The coefficient of $x$ is $m=-3$ , and the constant is $c=8$ .

Tier 2 · Standard

A straight line crosses the axes at

(0,12)

and

(6,0)

. Find its gradient and both intercepts.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Gradient $-2$ $y$ -intercept $12$ $x$ -intercept $6$	3	Using the two points, $m=(0-12)/(6-0)=-12/6=-2$ . The point with $x=0$ gives the $y$ -intercept $12$ , and the point with $y=0$ gives the $x$ -intercept $6$ .

Tier 3 · Hard

A straight-line graph of water volume

V

litres against time

t

minutes passes through

(0,120)

and

(8,72)

. Find and interpret its gradient and

V

-intercept, then write

V

in terms of

t

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Gradient $-6$ litres per minute $V$ -intercept $120$ litres, the initial volume $V=120-6t$	4	The gradient is $(72-120)/(8-0)=-48/8=-6$ , meaning the volume decreases by $6$ litres each minute. At $t=0$ , $V=120$ , so the initial volume is $120$ litres. Therefore $V=-6t+120$ .

Tier 1 · Easy

Write down the

y

-intercept of

y=7-2x

and write down whether the line rises or falls as

x

increases.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$y$ -intercept $7$ The line falls	2	Rewrite the equation as $y=-2x+7$ . The intercept is $7$ , and the negative gradient means the line falls as $x$ increases.

Tier 2 · Standard

A straight-line graph shows a cyclist's distance

d

kilometres from home

t

hours after noon. The graph has gradient

12

and

d

-intercept

5

. Explain what the gradient represents. Explain what the

d

-intercept represents.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	The cyclist moves away from home at $12$ kilometres per hour At noon, the cyclist is $5$ kilometres from home	2	The gradient is the change in distance per hour, so the cyclist moves away from home at $12$ kilometres per hour. The $d$ -intercept is the distance when $t=0$ , so at noon the cyclist is $5$ kilometres from home.

Tier 3 · Hard

Two venues charge for hiring

n

chairs. Venue A uses

A=3n+12

and venue B uses

B=4n+5

. Which venue has the greater fixed charge, and which has the greater charge per chair? Give a reason for each answer.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Venue A has the greater fixed charge: £12 Venue B has the greater charge per chair: £4	3	The constant term is the charge when $n=0$ , so the fixed charges are £12 for A and £5 for B; A is greater. The coefficient of $n$ is the charge per chair, so the rates are £3 for A and £4 for B; B is greater.

Tier 2 · Standard

A student's printing credit is modelled by

C=24-0.15p

, where

C

is the credit in pounds after printing

p

pages. Explain what the intercept and the gradient represent. Work out the credit after

60

pages.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	The initial credit is £24 Each page reduces the credit by £0.15 £15	3	At $p=0$ , $C=24$ , so the intercept represents £24 of initial credit. The gradient $-0.15$ means each printed page reduces the credit by £0.15. After $60$ pages, $C=24-0.15(60)=24-9=15$ .

Tier 3 · Hard

A straight-line graph shows the total mass

M

kg of a crate containing

n

identical items. The graph passes through

(3,12.5)

and

(8,22.5)

. Find and interpret the gradient and the

M

-intercept. Write

M

in terms of

n

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Gradient $2$ kg per item $M$ -intercept $6.5$ kg, the mass of the empty crate $M=2n+6.5$	4	The gradient is $\dfrac{22.5-12.5}{8-3}=\dfrac{10}{5}=2$ kg per item. Using $M=2n+c$ and $(3,12.5)$ gives $12.5=6+c$ , so $c=6.5$ . This is the mass when $n=0$ , so the empty crate has mass $6.5$ kg and $M=2n+6.5$ .

Tier 3 · Hard

The value

V

pounds of a machine is modelled by a straight line, where

t

is its age in years. The machine is worth £20 880 when new and its value falls by £145 each month. Find and interpret the gradient and the

V

-intercept. Write

V

in terms of

t

. Work out when the model gives

V=10\,440

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Gradient $-1740$ pounds per year, so the modelled value falls by £1740 each year $V$ -intercept $20\,880$ pounds, the value when the machine is new $V=20\,880-1740t$ $t=6$ years	5	There are $12$ months in a year, so the annual change is $12\times(-145)=-1740$ pounds per year. At $t=0$ , $V=20\,880$ , so the intercept is the new value and $V=20\,880-1740t$ . Setting $V=10\,440$ gives $20\,880-1740t=10\,440$ , so $1740t=10\,440$ and $t=6$ years.

Q10

Tier 3 · Hard

The heights of two saplings are modelled by straight lines. Sapling A's graph passes through

(2,18)

and

(8,36)

, where time is in weeks and height is in centimetres. Sapling B is modelled by

h=2.5t+15

. Which sapling is taller initially, which grows faster, and when do the models give the same height?

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Sapling B is taller initially: $15$ cm compared with $12$ cm Sapling A grows faster: $3$ cm per week compared with $2.5$ cm per week The models give the same height after $6$ weeks, when both give $30$ cm	5	For A, the gradient is $\dfrac{36-18}{8-2}=3$ cm per week. Writing $h=3t+c$ and using $(2,18)$ gives $18=6+c$ , so $c=12$ cm. B has intercept $15$ cm and gradient $2.5$ cm per week, so B is initially taller but A grows faster. Solve $3t+12=2.5t+15$ : $0.5t=3$ , so $t=6$ and the common height is $30$ cm.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	2H	Q16	3	Allowed	Higher	QP MS
2023-11	3F	Q28	3	Allowed	Foundation	QP MS
2024-11	3F	Q27	3	Allowed	Foundation	QP MS
2023-11	3H	Q9	3	Allowed	Higher	QP MS
2024-06	1F	Q25	4	Non-calculator	Foundation	QP MS
2021-11	1F	Q28	2	Non-calculator	Foundation	QP MS
2024-06	1H	Q6	4	Non-calculator	Higher	QP MS
2022-06	3H	Q8	3	Allowed	Higher	QP MS

Other points in A Algebra · equations and reasoning

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