Edexcel International GCSE Mathematics A 4MA1: Sequences, functions and graphs – Revision Notes

For full explanations and worked examples, use the Sequences, functions and graphs study guide. These notes condense Topic 3 (sections 3.1–3.4) of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Sections 3.2 and 3.4, and the extra Higher statements in 3.1 and 3.3, are Higher tier only; the rest is on both tiers. A calculator may be used on every paper.

Course hub: Edexcel IGCSE Mathematics. Tick off each statement on the printable checklist, then test yourself with the practice questions or a free diagnostic.

Key terms

• Term-to-term rule: how to get the next term from the one before (“add 4”, “multiply by 2”).
• Position-to-term rule: a formula for the nth term in terms of n.
• Arithmetic sequence: constant difference between terms. First term a, common difference d (Higher tier only notation).
• Function: a mapping that sends each input to exactly one output. Written f(x) = … or f : x ↦ … (Higher tier only).
• Domain: allowed inputs. Range: resulting outputs.
• Composite fg: do g first, then f. Inverse f⁻¹: reverses f.
• Gradient: (increase in y) ÷ (increase in x). Tangent: straight line touching a curve at one point; its gradient is the gradient of the curve there.
• Stationary (turning) point: where dy/dx = 0.

Formulas

Result	Formula	On the Higher formulae sheet?
nth term (arithmetic)	a + (n − 1)d	No, learn it
Sum of first n terms	Sₙ = (n/2)[2a + (n − 1)d]	Yes
Midpoint	((x₁ + x₂)/2, (y₁ + y₂)/2)	No
Gradient	(y₂ − y₁)/(x₂ − x₁)	No
Straight line	y = mx + c	No
Perpendicular gradients (Higher)	m₁ × m₂ = −1	No
Differentiation	y = axⁿ → dy/dx = naxⁿ⁻¹	No
Kinematics	v = ds/dt, a = dv/dt	No

Method in steps

nth term of an arithmetic sequence (both tiers)

1. Find the difference between terms.
2. Write (difference) × n.
3. Compare with the first term and add or subtract the adjustment.
4. Check by substituting n = 2.

Is a number a term? Set the nth term equal to it and solve. A whole-number n means yes.

Finding a and d from two terms (Higher)

1. Write each term as a + (n − 1)d.
2. Subtract the two equations to find d.
3. Substitute back for a.

Inverse function (Higher)

1. Write y = f(x).
2. Rearrange to make x the subject. If x appears twice, collect x terms on one side and factorise.
3. Replace y with x and write f⁻¹(x) = ….

Equation of a line through a given point (Higher)

1. Find the gradient m (from two points, or from a parallel or perpendicular line).
2. Substitute a known point into y = mx + c to find c.
3. Write y = mx + c; rearrange to ax + by = c only if asked.

Turning points (Higher)

1. Differentiate.
2. Set dy/dx = 0 and solve.
3. Substitute each x into the original equation for y.
4. Decide maximum or minimum from the shape of the graph (for example, a cubic with positive x³ coefficient has its maximum on the left and its minimum on the right).

Graphs to recognise

• Linear: y = mx + c. x = k vertical; y = c horizontal.
• Quadratic: U-shape if the x² coefficient is positive, ∩-shape if negative; symmetrical about a vertical line through the turning point.
• Cubic (Higher): S-shaped curve; the sign of the x³ coefficient decides whether it ends top right (+) or bottom right (−).
• Reciprocal (Higher): y = k/x has two branches and never meets the axes. y = k/x² stays on one side of the x-axis for k ≠ 0.
• Trig (Higher): sin and cos repeat every 360° and stay between −1 and 1; tan repeats every 180° with asymptotes at 90°, 270°, ….

Reading and drawing graphs (both tiers)

• Plotting: complete the table of values first, plot each point with a small cross, then join a quadratic with one smooth curve through every point.
• Travel graphs: on a distance–time graph, a steeper line means faster; a horizontal line means stopped; a line sloping back down means returning to the start. Convert minutes to hours before calculating speed in km/h (40 minutes = 40/60 hour).
• Conversion graphs: read from one axis to the line, then across to the other axis. For values beyond the graph, use a value you can read and scale it up.
• Coordinates from geometry: use properties of the shape. For example, if three vertices of a square are (1, 1), (5, 1) and (5, 5), the fourth is (1, 5).

Solving equations with graphs (Higher tier only)

If y₁ is a straight line and y₂ is a curve, the x-coordinates of their intersection points are the solutions of y₂ − y₁ = 0.

Curve y = x² + 2x − 1 and line y = x + 5
x² + 2x − 1 − (x + 5) = 0   →   x² + x − 6 = 0
So the graphs meet where x = −3 and x = 2.

It also works backwards: to solve x³ − 4x + 1 = 0 using a drawn graph of y = x³ − 3x, rewrite it as x³ − 3x = x − 1 and draw the line y = x − 1.

Transformations (Higher tier only)

Equation	Transformation	A point (p, q) moves to
y = f(x) + a	translate up a	(p, q + a)
y = f(x + a)	translate left a	(p − a, q)
y = af(x)	stretch in y-direction, factor a	(p, aq)
y = f(ax)	stretch in x-direction, factor 1/a	(p/a, q)

Changes outside the bracket act on y as you expect. Changes inside the bracket act on x the opposite way.

Must-know distinctions

• Term-to-term vs position-to-term: “add 7” does not tell you the 100th term directly; 7n − 4 does.
• nth term vs Sₙ: the 20th term is one number in the list; S₂₀ adds the first twenty.
• fg vs gf: order matters. fg(x) means f(g(x)).
• f⁻¹(x) vs 1/f(x): the inverse is not the reciprocal.
• Distance–time vs speed–time: gradient gives speed on the first, acceleration on the second.
• Parallel vs perpendicular (Higher): same gradient vs negative reciprocal gradient.
• Gradient of a straight line vs gradient of a curve: one value everywhere vs a value that changes; for a curve use a tangent or differentiate.
• v = 0 vs a = 0: “at rest” means v = 0; “maximum velocity” means a = 0.

Small worked reminders

Sequence 13, 9, 5, 1, …   difference −4  →  nth term = 17 − 4n
(Higher) Line through (1, −2) parallel to y = 4x + 9:  y = 4x − 6
y = 3x² − 12x:  dy/dx = 6x − 12 = 0 → x = 2, y = −12, minimum (U-shape)

Quick self-test

1. Find the nth term of 9, 5, 1, −3, …
2. Find the sum of the first 15 terms of 2 + 5 + 8 + 11 + … (Higher)
3. f(x) = 4 − x². Find f(−3). (Higher)
4. f(x) = x + 2 and g(x) = 3x. Find fg(x) and gf(x). (Higher)
5. f(x) = (x − 7)/4. Find f⁻¹(x). (Higher)
6. Find the gradient of the line through (−2, 5) and (4, −7). (Higher)
7. Find the equation of the line parallel to y = 3x − 1 that passes through (2, 9). (Higher)
8. A line has gradient −1/4. Write down the gradient of a line perpendicular to it. (Higher)
9. Differentiate y = 5x⁴ − 3/x. (Higher)
10. Find the coordinates of the turning point of y = x² − 6x + 1 and say whether it is a maximum or minimum. (Higher)
11. The graph of y = f(x) has a maximum at (2, 5). Write down the maximum point of (a) y = f(x − 3), (b) y = 3f(x). (Higher)
12. Find the midpoint of (−1, −6) and (7, 2).

Answers

1. Difference −4, so −4n + 13 (check: n = 1 gives 9). 13 − 4n
2. a = 2, d = 3: S₁₅ = (15/2)(4 + 14 × 3) = (15/2)(46) = 345
3. 4 − 9 = −5
4. fg(x) = 3x + 2 (g first: 3x, then add 2); gf(x) = 3(x + 2) = 3x + 6
5. y = (x − 7)/4 → x = 4y + 7, so f⁻¹(x) = 4x + 7
6. (−7 − 5) ÷ (4 − (−2)) = −12 ÷ 6 = −2
7. 9 = 3(2) + c, c = 3: y = 3x + 3
8. Negative reciprocal of −1/4: 4
9. y = 5x⁴ − 3x⁻¹, so dy/dx = 20x³ + 3/x²
10. dy/dx = 2x − 6 = 0, x = 3, y = 9 − 18 + 1 = −8: (3, −8), minimum
11. (a) (5, 5) (b) (2, 15)
12. (3, −2)

Where marks are usually lost

• Writing the nth term with the wrong sign when the sequence decreases (writing 4n + 13 for 9, 5, 1, …).
• Using n instead of (n − 1) in a + (n − 1)d, which shifts every term by one place.
• Rounding n to the nearest whole number when it should be tested: n = 22.4 means “not a term”, or “23 terms needed” in a sum problem.
• Leaving an inverse as “x = …” in terms of y, without writing f⁻¹(x).
• Giving the domain exclusion as the value of the denominator (0) rather than the value of x that makes it zero.
• Taking the gradient as (change in x) ÷ (change in y), or losing a negative sign when subtracting coordinates.
• Using the gradient of a perpendicular line without changing its sign.
• Drawing a tangent that cuts the curve, or choosing points on the tangent too close together to read accurately.
• Substituting the turning-point x into dy/dx instead of y to find the y-coordinate.
• In kinematics, differentiating once when the question asks for acceleration.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 3, Sequences, functions and graphs, sections 3.1–3.4 (Foundation and Higher tier content).