A Algebra · graphs — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in A, read straight through: 4 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 1MA1 section A. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

How this checking works

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

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.

A15 · Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Higher tier only. Gradient is change in the vertical coordinate divided by change in the horizontal coordinate, with units formed from the axis units.
For a curve, draw a tangent at the required point and use two well-separated points on that tangent, not two points on the curve.
Estimate an area under a non-linear graph by dividing it into strips and applying the trapezium rule.
On a distance-time graph gradient represents speed; on a velocity-time graph gradient represents acceleration and area represents displacement.
The examiner expects a numerical result, correct units and an interpretation of its sign where relevant.

Worked example

A distance-time graph is a straight line from $(12,150)$ to $(32,510)$ , with time in seconds and distance in metres. Calculate and interpret its gradient.

1.Use change in distance divided by change in time: $\dfrac{510-150}{32-12}$ .
2.Evaluate $\dfrac{360}{20}=18$ .
3.Attach the units metres per second and interpret the constant straight-line gradient.

Answer: $18$ m/s; the object travels at a constant speed of $18$ m/s.

Common mistakes

Don't use two points on the curve instead of two points on the tangent when estimating a gradient.
Don't add trapezium heights without multiplying by half the strip width.
Don't report a velocity-time area in m/s instead of metres.

Exam tip. For an estimate, leave the tangent or trapezia visible because the method marks depend on the construction.

A16 · Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Higher tier only. A circle centred at $(0,0)$ with radius $r$ has equation $x^2+y^2=r^2$ .
Check that a given point lies on the circle by substituting its coordinates. The radius from the origin to a point $(a,b)$ has gradient $b/a$ when $a\ne0$ .
The tangent at that point is perpendicular to the radius, so its gradient is the negative reciprocal, and a point-gradient equation gives the tangent.
The examiner expects an exact equation, normally simplified, and it should pass through the stated point.
Horizontal and vertical radii produce vertical and horizontal tangents respectively.

The tangent at a point on a circle is perpendicular to the radius.

Worked example

The point $P(5,12)$ lies on $x^2+y^2=169$ . Find the equation of the tangent at $P$ .

1.The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ .
2.Use point-gradient form: $y-12=-\dfrac{5}{12}(x-5)$ .
3.Multiply by $12$ and rearrange to obtain $5x+12y=169$ .

Answer: $5x+12y=169$ .

Common mistakes

Don't use the radius gradient $12/5$ as the tangent gradient.
Don't find the negative reciprocal correctly but write a line that does not pass through the given point.

Exam tip. Substitute the contact point into your final tangent equation for a quick accuracy check.

A17 · Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Definition. Specification A17: solve linear equations in one unknown algebraically, including those with the unknown on both sides of the equation; find approximate solutions using a graph.

Do the same thing to both sides until the unknown stands alone. With a fraction, multiply every term on both sides by the denominator, writing the line $5-x=2(2x-7)$ before expanding; the method mark needs the whole of each side multiplied, not just one term.
Expand brackets, then collect the $x$ terms on one side and the numbers on the other in a single visible step such as $4x+x=14+5$ .
Divide last, and leave the answer as an exact fraction or decimal; $\dfrac{19}{5}$ and $3.8$ both score, but $\dfrac{-19}{-5}$ does not.
Check by substituting into the original equation.
A graphical solution is the $x$ -coordinate where the two graphs meet, read to the accuracy of the grid.

Worked example

Solve

\dfrac{3(x-2)}{4}-\dfrac{x+1}{3}=\dfrac{5}{6}

(3)

(Total for Question 1 is 3 marks)

Working

1.Multiply every term by $12$ , the lowest common multiple of $4$ , $3$ and $6$ : $9(x-2)-4(x+1)=10$ .
2.Expand the brackets: $9x-18-4x-4=10$ .
3.Collect the $x$ terms and the numbers: $5x=10+18+4=32$ .
4.Divide by $5$ : $x=\dfrac{32}{5}$ .

Answer: $x=\dfrac{32}{5}$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a correct first step, eg $9(x-2)-4(x+1)=10$ or $9x-18-4x-4=10$	M1	Method must show every term multiplied by 12 (or a common multiple)
	(dep) for isolating terms in $x$ , eg $9x-4x=10+18+4$ or $5x=32$	M1
	for $\dfrac{32}{5}$	A1	oe; accept 6.4 or $6\dfrac{2}{5}$

Mark-scheme wording to use

“for a correct first step, eg 5 – x = 2(2x − 7) or 5 – x = 4x − 14 or 5/2 – x/2 = 2x – 7” — Pearson 1MA1, June 2018 Paper 3H, Q7
“(dep) for isolating terms in x … Accept 19/5 , 3 4/5 oe but not −19/−5 oe” — Pearson 1MA1, June 2018 Paper 3H, Q7
“for correct expansion of the bracket or dividing all terms by 3 as a first step … for isolating terms in x on one side of an equation eg 5x – 6 – 3x = −3” — Pearson 1MA1, November 2017 Paper 2H, Q1

What the examiner wants. Every term treated the same. The June 2018 Paper 3H scheme states that the method must show the left side and both terms on the right multiplied by $2$ ; the November 2024 Paper 3H report says many knew to multiply both sides by the denominator but too often did not carry it out correctly, and then lost the marks for isolating the terms.

Common mistakes

Don't multiply only the fractional terms, not every term, when clearing denominators.
Don't change the sign of a term when moving it without performing the same addition or subtraction on both sides.

Exam tip. For a multi-mark 'solve' question, keep each balancing step visible so an arithmetic slip does not lose the method marks.