OxfordAQA International GCSE Mathematics: Functions, coordinates and straight-line graphs (9260) – Revision Notes

Quick recall for the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, section 3.2.2, statements A9, A10 and A11. The study guide has the full explanations. Core content can also be assessed on the Extension papers. Content that only Extension candidates need is marked Extension only.

Also useful: the practice questions, the course hub, the printable checklist and the free diagnostics.

Key facts at a glance

Idea	Statement	Level
Function machine = expression	A9	Core
f(x) notation, domain, range	A9	Extension only
Composite fg(x) = f(g(x))	A9	Extension only
Inverse f⁻¹ undoes f	A9	Extension only
Coordinates in four quadrants	A10	Core
Plot lines; y = mx + c; parallel lines	A11	Core
Line through two points, or point and gradient	A11	Extension only
Perpendicular gradients	A11	Extension only

A9 Core: inputs and outputs

• Input → rule → output. Write the steps as an expression.
• “Add 3 then multiply by 5” is 5(x + 3). “Multiply by 5 then add 3” is 5x + 3.
• To find an input from an output, undo the steps in reverse order.
• Negative inputs go in brackets: for x² − 3x with x = −2, the output is 4 + 6 = 10.

A9 Extension only: function toolkit

Term	Meaning	Quick example
f(x)	Output of f for input x	f(x) = 2x + 5, f(3) = 11
Domain	Allowed inputs	1/(x − 4) has x ≠ 4
Range	Resulting outputs	x² + 2 on −3 ≤ x ≤ 2 gives 2 ≤ f ≤ 11
fg(x)	Do g first, then f	f = 2x + 3, g = x² − 5: fg = 2x² − 7
f⁻¹(x)	Undoes f	h = (3x − 4)/5 gives h⁻¹ = (5x + 4)/3

Method: range of a quadratic on a restricted domain

1. Sketch or think about the shape.
2. Find the output at each end of the domain.
3. Find the output at the turning point if it lies inside the domain.
4. The range runs from the smallest to the largest of these.

Method: composite function

1. Write the inner function in brackets.
2. Substitute it into the outer function wherever x appears.
3. Expand and simplify.
4. fg and gf are usually different.

Method: inverse function

1. Set y equal to the rule for f.
2. Get x on its own. If x appears twice, collect the x terms on one side and factorise.
3. Swap y for x in the result.
4. Check: f(f⁻¹(x)) should simplify to x.

Worked reminder: k(x) = (x + 1)/(x − 2). Then y(x − 2) = x + 1, so x(y − 1) = 2y + 1, giving k⁻¹(x) = (2x + 1)/(x − 1).

A10 Core: four quadrants

Quadrant	Signs (x, y)
1st, top right	(+, +)
2nd, top left	(−, +)
3rd, bottom left	(−, −)
4th, bottom right	(+, −)

• Write (x, y): across first, then up or down.
• y = b is horizontal and x = a is vertical.
• y = x passes through quadrants 1 and 3. y = −x passes through quadrants 2 and 4.
• To complete a parallelogram ABCD, use the same step from A to B as from D to C.

A11 Core: straight lines

y = mx + c: m = gradient, c = y-intercept.

Skill	How
Plot a line	Table of values, at least three points, one ruled line
Gradient	(change in y) ÷ (change in x)
y-intercept	Set x = 0
x-intercept	Set y = 0
Rearrange	3x + 4y = 12 becomes y = −(3/4)x + 3
Parallel	Equal gradients

• Positive gradient: the line slopes up to the right. Negative: down to the right.
• Worked reminder: y = 1.5x − 3 has gradient 1.5, y-intercept −3 and x-intercept 2.

A11 Extension only: equations and perpendiculars

Task	Method
Point (x₁, y₁) and gradient m	y − y₁ = m(x − x₁)
Two points	Find m first, then use either point
Perpendicular gradient	m₁ × m₂ = −1, so −1/m

Worked reminder: through (−3, −4) and (5, 8), the gradient is 12/8 = 3/2 and the line is y = (3/2)x + 1/2.

Worked reminder: take A(1, 3) and B(5, 6). Then gradient AB = 3/4. The line through B perpendicular to AB has gradient −4/3 and equation 4x + 3y = 38.

Method: perpendicular through a point

1. Rearrange the given line to y = mx + c and read m.
2. Flip it and change its sign to get −1/m.
3. Substitute the point into y − y₁ = (−1/m)(x − x₁).
4. Tidy up, and check by substituting the point.

More worked reminders

Solving with a composite (Extension only). f(x) = 2x + 3 and g(x) = x² − 5, so fg(x) = 2x² − 7. Solve fg(x) = 25:

2x² − 7 = 25
x² = 16
x = 4 or x = −4

Linear range (Extension only). g(x) = 2x − 1 for 1 ≤ x ≤ 6. The rule only rises, so g(1) = 1 and g(6) = 11. Range: 1 ≤ g(x) ≤ 11.

Gradient from two points. For (−2, −6) and (4, 3): (3 + 6)/(4 + 2) = 9/6 = 1.5. Subtract the first y from the second y, and the first x from the second x, in the same order.

Quadrants from a rule. Points with x + y = 3 include (5, −2) in the 4th quadrant, (0, 3) on the y-axis and (−1, 4) in the 2nd quadrant. A straight line usually passes through three quadrants.

Describing a pair of lines. Rearrange both into y = mx + c, compare the gradients, then decide:

Gradients	Relationship
Equal	Parallel
Product is −1 (Extension only)	Perpendicular
Anything else	Neither, so the lines cross at one point

Equal gradients with equal intercepts mean the same line.

Exam technique for this unit

• Show the substitution in a function question, such as f(−3) = 3(−3) − 2. Do not write only the answer.
• When asked for a line “in the form y = mx + c”, finish in that form even if an earlier line is neater.
• If a question says “find the equation”, give an equation with y in it. A bare number for the gradient answers only part of the question.
• Use the written steps of a method as a checklist, and glance at the sign of every gradient before moving on.
• Check that a line you have drawn passes through the axis crossings you calculated. A mismatch tells you to recheck the table of values.
• For an inverse function, test one value: put a number into f, then put the result into your f⁻¹. You should get the number back.

Must-know distinctions

• Domain vs range: domain is the inputs, range is the outputs.
• fg vs gf: the function nearest to x acts first.
• f⁻¹(x) vs 1/f(x): the first is the inverse function. The second is a reciprocal.
• Parallel vs perpendicular: equal gradients vs product of gradients equal to −1.
• x-intercept vs y-intercept: set y = 0 for the first and x = 0 for the second.

Quick self-test

1. A machine multiplies by 6 then subtracts 5. What input gives output 43?
2. In which quadrant is the point (−7, −1)?
3. Find the gradient and y-intercept of 2y + 6x = 10.
4. (Extension only) f(x) = 3x − 2 and g(x) = x² + 4. Find fg(x).
5. (Extension only) Find f⁻¹(x) for f(x) = (x − 1)/4.
6. (Extension only) g(x) = x² for −2 ≤ x ≤ 5. State the range.
7. Is the line y = 5 − 2x parallel to 4x + 2y = 9?
8. (Extension only) Find the equation of the line with gradient 2 through (3, 1).
9. (Extension only) A line has gradient −7/3. What is the gradient of a perpendicular line?
10. (Extension only) Find the equation of the line through (0, 4) and (6, 1).

Answers

1. 6x − 5 = 43, so x = 8.
2. Both coordinates are negative, so the 3rd quadrant.
3. 2y = −6x + 10, so y = −3x + 5. Gradient −3, y-intercept 5.
4. fg(x) = 3(x² + 4) − 2 = 3x² + 10.
5. y = (x − 1)/4 gives x = 4y + 1, so f⁻¹(x) = 4x + 1.
6. The smallest output is g(0) = 0 and the largest is g(5) = 25. Range 0 ≤ g(x) ≤ 25.
7. 4x + 2y = 9 gives y = −2x + 4.5. Both gradients are −2, so yes.
8. y − 1 = 2(x − 3), so y = 2x − 5.
9. 3/7.
10. Gradient = (1 − 4)/6 = −1/2, and the y-intercept is 4, so y = −x/2 + 4.

Where marks are usually lost

• Substituting only the endpoints of a domain and missing a turning point.
• Reading a gradient straight from an equation that is not yet in the form y = mx + c.
• Losing a negative sign when squaring a negative input.
• Forgetting the second root when solving a quadratic that comes from a composite function.
• Giving the inverse as a function of y rather than x, or leaving it unchecked.
• Dropping the sign when finding a perpendicular gradient.
• Plotting points carelessly and then drawing a line that bends to fit them.
• Leaving a final line equation without checking it passes through a given point.
• Giving a line as “y = mx + c” without the numbers substituted.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.2, statements A9, A10 and A11.