Edexcel IGCSE

Graphs

A graph is the picture of an equation: every point on the line or curve is a pair of coordinates that makes the equation true, and every pair that makes it true lies on it. It is section 3.3 of the Edexcel International GCSE Mathematics A specification, and almost every question on it is one of two moves — read a fact off the picture, or turn a fact into the equation $y = m x + c$ .

Edexcel IGCSE 4MA1Specification 3.393 approved practice questions

Set up your class — free Practise on your own

No card needed. Free for individual students and teachers.

Before you start

What you need to be able to do already

If any of these is shaky, fix it first — every worked example below leans on all three.

Add and subtract negative numbers without losing a sign
Work out $5 - (- 3)$ , and then $2 - 4$ . They are $8$ and $- 2$ . Gradients are built out of exactly those two subtractions, and sign slips while subtracting negative coordinates are catalogued as a misconception on this subtopic.
Simplify a fraction and find its negative reciprocalFractions
Simplify $\frac{4}{6}$ , then write the number you multiply $2$ by to get $- 1$ . They are $\frac{2}{3}$ and $- \frac{1}{2}$ . A gradient left as $\frac{4}{6}$ is usually still accepted, but you cannot tell from the question whether it will be, and perpendicular lines are entirely a negative-reciprocal question.
Rearrange an equation to make $y$ the subject, and expand a bracketAlgebraic manipulation
Rearrange $2 y - 6 x = 4$ into the form $y = m x + c$ . It is $y = 3 x + 2$ : add $6 x$ , then divide every term by $2$ . Gradient and intercept can only be read off once $y$ is alone, so this rearrangement comes before almost every answer here.

The idea

What is graphs?

A pair of coordinates fixes a point by two numbers: how far across, then how far up, both measured from the origin $(0, 0)$ . The graph of an equation is the set of every point whose coordinates satisfy it. Any straight line other than a vertical one can be written as $y = m x + c$ , where $m$ is the gradient — how much $y$ rises for each $1$ that $x$ moves right — and $c$ is the value of $y$ where the line crosses the $y$ -axis. Curves obey the same rule; only their shapes differ.

The notation, and how to read it

Every worked solution below is written in these five symbols.

SymbolReads asMeans
$(x, y)$ the point x, yAcross first, then up. $(3, 7)$ and $(7, 3)$ are different points, and swapping them over is a misconception recorded on this subtopic.
$y = m x + c$ y equals m x plus cThe equation of a straight line: $m$ is the gradient and $c$ the $y$ -intercept. So $y = 3 x - 6$ has gradient $3$ and crosses the $y$ -axis at $(0, - 6)$ .
$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ the gradient between two pointsRise over run. Subtract the two points in the same order on the top and on the bottom, or the sign of the gradient comes out reversed.
$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ the midpoint of a line segmentThe average of the two $x$ -coordinates and the average of the two $y$ -coordinates. Both halves need the division by $2$ .
$y = f (x)$ y equals f of xFunction notation for a curve. $f (x) + a$ slides it up by $a$ and $f (x + a)$ slides it left by $a$ . Higher tier.

The one idea underneath all of it

A point lies on a graph exactly when its coordinates satisfy the equation, so anything readable off the picture is also obtainable from the algebra, and the other way round. That is why a gradient can be counted off the squares or calculated from two coordinates and must agree; why the point where two graphs meet is the solution of their equations taken together; and why substituting a point back into an equation is always available as a check on your own answer.

The smallest possible example

A straight line passes through $(2, 7)$ and $(5, 16)$ . Find its gradient.

$m = 3$

$y$ rises by $16 - 7 = 9$ while $x$ runs across by $5 - 2 = 3$ , so $m = \frac{9}{3} = 3$ . Subtract in the same order on both lines — later minus earlier — and keep rise on top. Turning the fraction upside down, and subtracting the two pairs in opposite orders, are each catalogued as misconceptions on this subtopic.

Key terms

The words a question will use

Exam wording is precise. These five carry the meaning a question depends on.

TermWhat it meansNotation
Gradient
How steep a line is: the change in $y$ for every $1$ that $x$ increases. A positive gradient rises to the right, a negative one falls, and a gradient of $0$ is a horizontal line. $m = 3$
Intercept
Where a graph crosses an axis. Put $x = 0$ for the $y$ -intercept, which is the $c$ in $y = m x + c$ ; put $y = 0$ for the $x$ -intercepts, which are the roots of the equation. $(0, c)$
Midpoint
The point exactly halfway along a line segment: average the two $x$ -coordinates, then average the two $y$ -coordinates. The midpoint of $(2, 5)$ and $(8, 9)$ is $(5, 7)$ ; adding without dividing by $2$ is a recorded misconception here. $(5, 7)$
Parallel and perpendicular
Parallel lines have equal gradients. Perpendicular gradients multiply to $- 1$ , so a line at right angles to $y = 2 x - 5$ has gradient $- \frac{1}{2}$ . Higher tier. $m_{1} m_{2} = - 1$
Tangent
A straight line that touches a curve at one point and matches its direction there. Drawing one and finding its gradient is how the gradient of a curve at a point is measured. Higher tier. $\frac{rise}{run}$

The method

How to work out a graphs

Five steps, in this order. The worked examples below point back to these numbers.

Plot or read every point as across first, then up, from the origin. $(3, 7)$ is $3$ right and $7$ up; $(7, 3)$ is somewhere else entirely. Negative coordinates go left or down, so the four quadrants are simply the four combinations of signs. Everything below depends on this being automatic, because a misread coordinate corrupts every later line.
For a straight line, work out the gradient as the change in $y$ divided by the change in $x$ .From $(2, 7)$ to $(5, 16)$ that is $\frac{16 - 7}{5 - 2} = 3$ . Subtract the points in the same order top and bottom. On a drawn line you may instead count squares, provided you read the scale on each axis rather than assuming both are $1$ .
Find $c$ by putting the gradient and one known point into $y = m x + c$ .Gradient $3$ through $(2, 7)$ gives $7 = 3 \times 2 + c$ , so $c = 1$ and the line is $y = 3 x + 1$ . From a drawing, $c$ is simply where the line crosses the $y$ -axis — but only if that axis is drawn at $x = 0$ , which is worth checking.
For a curve, complete a table of values, plot the points, and join them with a smooth freehand curve.Substitute each $x$ into the equation to get its $y$ ; a quadratic or cubic is never joined with a ruler. To find the gradient at one point on a curve you draw a tangent there and apply step 2 to the tangent — draw it long, because a short tangent makes the two coordinates hard to read accurately.
Check by substituting a point back into the equation.Is $(5, 16)$ on $y = 3 x + 1$ ? $3 \times 5 + 1 = 16$ , so yes. The same check settles intersections: a point where two graphs meet must satisfy both equations at once, which is exactly why solving them together finds it.

Use it when

The question involves coordinates, a drawn line or curve, or an equation you are asked to draw, read or find. ‘Work out the gradient’, ‘find the equation of the line’ and ‘write down the coordinates’ are all these five steps.

Do not use it when

The line is vertical. $x = 4$ has no defined gradient — the run is zero, so $\frac{rise}{run}$ does not exist, and that is not the same as a gradient of $0$ , which is a horizontal line — and it cannot be written as $y = m x + c$ ; vertical lines are $x =$ a number and horizontal ones are $y =$ a number. Nor should you find the gradient of a curve by taking two points on the curve itself: that gives the average gradient between them, not the gradient at a point, and only a tangent answers the question actually asked.

Worked examples

6 IGCSE Maths practice questions on graphs, fully worked

One per objective, so the range is covered rather than the same question repeated. Every step is written out, and the answer is shown — this is study material, not a test.

Example 1

The graph of y = x² - 4x + 3 is a parabola. What is the value of y when x = 5? [1 mark]

Plotting quadratic graphs

Worked solution

Step 1: Start with the equation y = x² - 4x + 3.
Step 2: Substitute x = 5 into the equation.
Step 3: Calculate 5² = 25.
Step 4: Calculate - $4 \times 5 = - 20$ .
Step 5: Add the terms: $25 - 20 + 3 = 8$ .
Step 6: So y = 8 when x = 5.

Answer8

Example 2

A cyclist travels at a constant speed of 15 mph for 2 hours. Work out the distance travelled by the cyclist. [2 marks]

Distance-time graphs

Worked solution

Step 1: Recall the formula: distance = speed × time.
Step 2: Speed is 15 miles per hour and time is 2 hours.
Step 3: Multiply the speed by the time: $15 \times 2 = 30$ .
Step 4: So the distance travelled is 30 miles.

Answer30 miles

Example 3

A straight line has equation y = 3x - 5. Work out the gradient of this line. [1 mark]

y = mx + c form

Worked solution

Step 1: The equation y = 3x - 5 is in the form y = mx + c.
Step 2: In y = mx + c, m represents the gradient of the line.
Step 3: Look at the number in front of x (the coefficient of x). The coefficient is 3.
Step 4: Therefore, the gradient (m) is 3.

Answer3

Show 3 more worked examples

Example 4

A car accelerates from rest at a constant rate. Its speed-time graph is a straight line from (0,0) to (10,20). What is the acceleration of the car in m/s²? [1 mark]

Speed-time graphs

Worked solution

Step 1: The car starts from rest, so initial speed v=0 m/s.
Step 2: After 10 seconds, the speed is 20 m/s (from the point (10,20)).
Step 3: Acceleration is the gradient (change in speed / change in time).
Step 4: Change in speed = $20 - 0 = 20$ m/s.
Step 5: Change in time = $10 - 0 = 10$ s.
Step 6: Acceleration = $\frac{20}{10} = 2$ m/s².

Answer2 m/s²

Example 5

A point has coordinates (3, 7). Work out the coordinates of the point after it is moved 4 units left and 2 units down. [2 marks]

Understand and use conventions for rectangular Cartesian coordinates

Worked solution

Step 1: Start with the original coordinates (3, 7).
Step 2: Moving 4 units left means subtracting 4 from the x-coordinate: 3 − 4 = -1.
Step 3: Moving 2 units down means subtracting 2 from the y-coordinate: 7 − $2 = 5$ .
Step 4: The new coordinates are (-1, 5).

Answer(-1, 5)

Example 6

Complete the table of values for y = 2x + 1. Then draw the graph of y = 2x + 1 for x from 0 to 4. [2 marks]

Plotting straight-line graphs

Worked solution

Step 1: Start with the equation y = 2x + 1.
Step 2: Here the gradient is 2 and the y-intercept is 1.
Step 3: To find y when x = 0: y = $2 \times 0 + 1 = 0 + 1 = 1$ .
Step 4: To find y when x = 1: y = $2 \times 1 + 1 = 2 + 1 = 3$ .
Step 5: To find y when x = 2: y = $2 \times 2 + 1 = 4 + 1 = 5$ .
Step 6: To find y when x = 3: y = $2 \times 3 + 1 = 6 + 1 = 7$ .
Step 7: To find y when x = 4: y = $2 \times 4 + 1 = 8 + 1 = 9$ .
Step 8: Plot these points and draw a straight line through them.

AnswerA graph showing a line through (0,1), (1,3), (2,5), (3,7), (4,9)

Common mistakes

Where marks actually get lost

Learnly tags 13 distinct misconceptions on this subtopic. These are errors that show up in real student work — not generic revision advice.

Examiner traps

Averaging x and y separately
What goes wrong: Students correctly find the average of x-coordinates and y-coordinates but may misapply the formula by adding and dividing by 2 incorrectly.
What to do instead: To find the midpoint, add the x-coordinates and divide by 2, then do the same for y-coordinates. Example: $(5 + 9) /2 = 7$ , $(3 + 11) /2 = 7$ .
Adding coordinates instead of averaging
What goes wrong: Students add the x-coordinates and y-coordinates but forget to divide by 2, giving the sum as the midpoint.
What to do instead: Remember: midpoint is the average, not the sum. Add the x's and divide by 2, add the y's and divide by 2.
Swapping x and y coordinates
What goes wrong: Students mistakenly average the x-coordinate of one point with the y-coordinate of the other, or vice versa.
What to do instead: Keep x's with x's and y's with y's. Average the x-coordinates together, then average the y-coordinates together.
Subtracting coordinates instead of adding
What goes wrong: Students subtract the coordinates instead of adding them before dividing by 2.
What to do instead: Midpoint uses addition: (x1+x2)/2 and (y1+y2)/2. Do not subtract; add the coordinates first.
Dividing by number of points incorrectly
What goes wrong: Students divide the sum of coordinates by the number of coordinates (4) instead of by 2.
What to do instead: There are 2 points, so divide each sum by 2, not 4. For x: (x1+x2)/2, for y: (y1+y2)/2.
Tangent not at correct point
What goes wrong: Students draw the tangent line at a different x-value, such as x=0 or x=2, instead of the specified point.
What to do instead: Always check that your tangent touches the curve exactly at the given x-coordinate. Mark the point first.

See 7 more examiner traps

Tangent drawn as a curve
What goes wrong: Students draw a curved line that follows the graph instead of a straight line that just touches at one point.
What to do instead: A tangent must be a straight line. Use a ruler to draw it so it only touches the curve at one point.
Incorrect gradient calculation
What goes wrong: Students misread coordinates or use the wrong formula (e.g., using Δx/Δy instead of Δy/Δx).
What to do instead: Gradient = rise over run = (change in y) ÷ (change in x). Pick two points on the tangent and subtract carefully.
Tangent line too short for accuracy
What goes wrong: Students draw a very short tangent, making it hard to read coordinates accurately for gradient calculation.
What to do instead: Extend your tangent line well beyond the point of contact so you can pick two clear points far apart.
Inconsistent coordinate order
What goes wrong: Students subtract the coordinates in the wrong order, e.g., (y2 - y1)/(x1 - x2) instead of (y2 - y1)/(x2 - x1).
What to do instead: Always subtract the first point's coordinates from the second point's coordinates: (y2 - y1)/(x2 - x1).
Inverted numerator and denominator
What goes wrong: Students compute (x2 - x1)/(y2 - y1) instead of (y2 - y1)/(x2 - x1), swapping rise and run.
What to do instead: Remember gradient = change in y divided by change in x. Think 'rise over run'.
Sign error in subtraction
What goes wrong: Students make sign mistakes when subtracting negative coordinates, e.g., $(5 - (- 3)) / (2 - 4)$ becomes $(5 - 3) / (2 - 4)$ .
What to do instead: Be careful with signs: subtracting a negative is the same as adding. Double-check each subtraction.
Fraction not simplified
What goes wrong: Students leave the gradient as an unsimplified fraction like $\frac{4}{6}$ instead of $\frac{2}{3}$ .
What to do instead: Always simplify your fraction to its lowest terms by dividing numerator and denominator by their greatest common factor.

Exam technique

How the marks are actually awarded

Method marks are given for working a marker can follow. This is what to put on the page.

Where the marks are: Questions in this section run from $1$ to $4$ marks. Writing down a coordinate or a $y$ -intercept is $1$ . A gradient from two points is $2$ , one mark for the correct difference fraction and one for its value — so write $\frac{16 - 7}{5 - 2}$ on the page before you divide. Finding the equation of a line is $3$ , split across the gradient, the value of $c$ , and the final answer written as $y = m x + c$ . On a tangent question one mark is for the tangent itself, so draw it long and straight before any arithmetic. Higher-tier intersection questions are $4$ , and most of the credit is in the quadratic you get after substituting, so write that equation down even if solving it then fails.
Command words to expect: Write down the coordinatesWork out the gradientFind the equation of the lineGive your answer in the form y = mx + cComplete the table of valuesDraw the graph ofUse the graph to estimateBy drawing a tangentShow thatFind the coordinates of the points of intersection
Accuracy and rounding: Coordinates are exact and are written in brackets with a comma: $(0, - 5)$ , not a bare $- 5$ . A gradient is a number, so cancel it: write $\frac{2}{3}$ rather than $\frac{4}{6}$ . An uncancelled fraction is often still accepted, but it is refused the moment a question asks for the simplest form, and leaving it uncancelled is a misconception our bank records here. ‘In the form $y = m x + c$ ’ means $y$ stands alone on the left, so $2 y = 6 x + 4$ has not yet answered the question. Values read off a drawn graph are estimates and are marked to a tolerance, so read to the nearest small square; values obtained by algebra are exact and should be left exact rather than rounded.
Calculator: A calculator is allowed in every paper of this qualification, on both Foundation and Higher tiers, and it will neither draw nor read a graph for you. Use it for the arithmetic of $\frac{16 - 7}{5 - 2}$ and for filling a table of values at speed — then plot the points yourself and confirm that one of them satisfies the equation.

Check yourself

You should now be able to:

Plot and read points in all four quadrants, across before up, and name the axes and the origin correctly.
Find the midpoint of a line segment by averaging the two $x$ -coordinates and the two $y$ -coordinates.
Use given geometrical information, such as the other vertices of a rectangle, to work out coordinates without measuring.
Find the gradient of a straight line, from a drawing or from two coordinates, as the change in $y$ over the change in $x$ .
Read the gradient and $y$ -intercept straight off $y = m x + c$ , and draw the line from them.
Draw and use a straight-line conversion graph to change between two units in either direction.
Interpret linear and non-linear graphs in context, including distance-time graphs and the speeds they show.
Complete a table of values and plot linear and quadratic graphs, joining a curve smoothly rather than with a ruler.
Plot cubic and reciprocal graphs and recognise their shapes (Higher).
Apply the transformations $f (x) + a$ , $f (x + a)$ , $f (a x)$ and $a f (x)$ to a graph, and write the new function (Higher).
Find a gradient on a curve by drawing a tangent, find where a line meets a curve, and find lines parallel or perpendicular to a given line (Higher).

Specification coverage

The 19 things 4MA1 asks you to do

The assessable objectives for sequences, functions and graphs — graphs, in the order the specification lists them.

Plotting quadratic graphs
Distance-time graphs
y = mx + c form
Speed-time graphs
Understand and use conventions for rectangular Cartesian coordinates
Plotting straight-line graphs

Show 13 more objectives

Plot points (x, y) in any of the four quadrants or locate points with given coordinates
Finding the equation of a line
Determine the coordinates of points identified by geometrical information
Determine the coordinates of the midpoint of a line segment, given the coordinates of the two end points
Draw and interpret straight line conversion graphs
Find the gradient of a straight line
Recognise, plot and draw graphs with equation y = Ax^3 + Bx^2 + Cx + D and reciprocal-type graphs including y = sin x, y = cos x, y = tan x for any size angles in degrees
Apply to the graph of y = f(x) the transformations y = f(x) + a, y = f(ax), y = f(x + a), y = af(x) for linear, quadratic, sine and cosine functions
Interpret and analyse transformations of functions and write the functions algebraically
Find the gradients of non-linear graphs (by drawing a tangent)
Find the intersection points of two graphs (one linear and one non-linear) and recognise the solutions correspond to the solutions of (y2 - y1) = 0
Calculate the gradient of a straight line given the coordinates of two points
Find the equation of a straight line parallel or perpendicular to a given line

See who is stuck on graphs before you teach it.

Create a class, share the 8-character code, and set graphs as practice. Your students work through all 93 approved practice questions — not just the 6 on this page — and you see which objectives the class has not got yet. Free for individual students and teachers.

How Learnly works for teachers

Set up your class — free Practise on your own

What you need to be able to do already

Add and subtract negative numbers without losing a sign

Simplify a fraction and find its negative reciprocalFractions

Rearrange an equation to make \(y\) the subject, and expand a bracketAlgebraic manipulation

What is graphs?

The notation, and how to read it

The words a question will use

Gradient

Intercept

Midpoint

Parallel and perpendicular

Tangent

How to work out a graphs

6 IGCSE Maths practice questions on graphs, fully worked

The graph of y = x² - 4x + 3 is a parabola. What is the value of y when x = 5? [1 mark]

A cyclist travels at a constant speed of 15 mph for 2 hours. Work out the distance travelled by the cyclist. [2 marks]

A straight line has equation y = 3x - 5. Work out the gradient of this line. [1 mark]

A car accelerates from rest at a constant rate. Its speed-time graph is a straight line from (0,0) to (10,20). What is the acceleration of the car in m/s²? [1 mark]

A point has coordinates (3, 7). Work out the coordinates of the point after it is moved 4 units left and 2 units down. [2 marks]

Complete the table of values for y = 2x + 1. Then draw the graph of y = 2x + 1 for x from 0 to 4. [2 marks]

Where marks actually get lost

Averaging x and y separately

Adding coordinates instead of averaging

Swapping x and y coordinates

Subtracting coordinates instead of adding

Dividing by number of points incorrectly

Tangent not at correct point

Tangent drawn as a curve

Incorrect gradient calculation

Tangent line too short for accuracy

Inconsistent coordinate order

Inverted numerator and denominator

Sign error in subtraction

Fraction not simplified

How the marks are actually awarded

You should now be able to:

The 19 things 4MA1 asks you to do

See who is stuck on graphs before you teach it.