Mathematics
05802025–2027 syllabus

MATHEMATICS · CHAPTER 2

Algebra and graphs

Use symbols to express general relationships, solve unknowns and interpret how quantities change.

Core + Extended12 connected sectionsSyllabus-aligned guide

LEARNING OBJECTIVES

What you will be able to do

  • translate statements into algebra and substitute values correctly
  • expand, factorise and complete the square
  • manipulate algebraic fractions and expressions with indices
  • solve linear, fractional, simultaneous and quadratic equations
  • rearrange formulas, including when the new subject appears more than once
  • solve inequalities and describe regions on coordinate grids
  • find rules for linear, quadratic, cubic, exponential and combined sequences
  • model direct and inverse proportion
  • interpret practical graphs and the main function families
  • differentiate simple polynomials and analyse stationary points
  • use function, inverse-function and composite-function notation

AT A GLANCE

Syllabus0580Coverage2025–2027Sections12LevelCore + Extended

INTRODUCTION · THE BIG IDEA

Use symbols to express general relationships, solve unknowns and interpret how quantities change.

Algebra expresses a pattern once instead of calculating it separately for every value. A letter may represent an unknown, a changing quantity or any member of a number set.

A graph is the visual form of a relationship. Its intercepts, gradient, turning points and asymptotes reveal information that may be difficult to see from an equation alone.

01

SECTION 01

Algebraic language and substitution

Core concept

An expression has no equals sign; an equation states that two expressions are equal; a formula links several quantities; and an identity is true for every permitted value.

A term is separated by + or −. Its numerical factor is the coefficient. Substitute by replacing each letter with its value, using brackets around negative values before applying powers.

Construct algebra in the order described. For example, the product of two consecutive even numbers can be written 2n(2n + 2).

DETAILED EXPLANATION

  • If x = −4, then x² = (−4)² = 16.
  • A solution must make the original equation true.
Original worked example

Substituting into a formula

  1. Use v² = u² + 2as with u = −3, a = 4 and s = 5.
  2. Insert brackets: v² = (−3)² + 2(4)(5).
  3. v² = 9 + 40 = 49.
  4. If v is a speed, take the non-negative root.

Answer: v = 7.

02

SECTION 02

Expansion, factorisation and completing the square

Core concept

Expansion multiplies every term in one factor by every term in the other factors. Factorisation reverses this process and should be continued until every factor is fully simplified.

First extract a common factor. Then check for grouping, a difference of two squares, a perfect square or a quadratic. Cubic expressions in this syllabus can be factorised by taking out a common factor before handling the remaining quadratic.

Completing the square rewrites ax² + bx + c as a(x − p)² + q. The form reveals the axis of symmetry x = p and turning point (p, q).

DETAILED EXPLANATION

  • Difference of squares: a²x² − b²y² = (ax − by)(ax + by).
  • Perfect squares: a² ± 2ab + b² = (a ± b)².
  • For ax² + bx + c, split the middle term using numbers with product ac and sum b.
RULE 1
(a + b)(c + d) = ac + ad + bc + bd
RULE 2
a² − b² = (a − b)(a + b)
RULE 3
x² + 2px + p² = (x + p)²
RULE 4
x² + bx = (x + b/2)² − (b/2)²
Original worked example

Completing the square for 2x² − 12x + 5

  1. Factor 2 from the x-terms: 2(x² − 6x) + 5.
  2. Complete the square inside: x² − 6x = (x − 3)² − 9.
  3. Substitute: 2[(x − 3)² − 9] + 5.
  4. Simplify the constant: 2(x − 3)² − 13.

Answer: 2x² − 12x + 5 = 2(x − 3)² − 13, with minimum point (3, −13).

03

SECTION 03

Algebraic fractions and indices

Core concept

Factor numerators and denominators before cancelling. Only common factors may be cancelled; separate terms joined by + or − cannot.

For addition or subtraction, use the lowest common denominator and combine the resulting numerator. For division, multiply by the reciprocal of the second fraction.

Index laws apply to matching bases. Fractional indices represent roots and negative indices represent reciprocals. Simple exponential equations can be solved by writing both sides with the same base.

DETAILED EXPLANATION

  • aᵐaⁿ = aᵐ⁺ⁿ, aᵐ/aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ.
  • If 5ˣ⁺¹ = 25ˣ, write 25ˣ = 5²ˣ, then equate indices.
RULE 1
a⁻ⁿ = 1/aⁿ
RULE 2
a¹⁄ⁿ = ⁿ√a
RULE 3
aᵐ⁄ⁿ = ⁿ√(aᵐ)
Original worked example

Adding 2/x and 3/(x + 1)

  1. The common denominator is x(x + 1).
  2. Rewrite: 2(x + 1)/[x(x + 1)] + 3x/[x(x + 1)].
  3. Combine numerators: 2x + 2 + 3x = 5x + 2.
  4. State the original restrictions.

Answer: (5x + 2)/[x(x + 1)], where x ≠ 0 and x ≠ −1.

04

SECTION 04

Equations and simultaneous equations

Core concept

Keep an equation balanced by applying the same operation to both sides. For fractional equations, multiply every term by the lowest common denominator and reject any value that makes an original denominator zero.

Two linear simultaneous equations may be solved by elimination or substitution. For a linear and non-linear pair, substitute the linear expression into the non-linear equation, solve the resulting quadratic, then find both matching values.

Solve quadratics by factorisation, completing the square or the quadratic formula. The discriminant b² − 4ac indicates two, one or no real roots.

RULE 1
x = (−b ± √(b² − 4ac))/(2a)
RULE 2
discriminant = b² − 4ac
Original worked example

Solving y = x + 1 and y = x² − 5

  1. Equate the two expressions for y: x + 1 = x² − 5.
  2. Rearrange: x² − x − 6 = 0.
  3. Factorise: (x − 3)(x + 2) = 0.
  4. For x = 3, y = 4. For x = −2, y = −1.

Answer: The solutions are (3, 4) and (−2, −1).

05

SECTION 05

Changing the subject of a formula

Core concept

To isolate a subject that appears once, undo operations in reverse order while keeping the equation balanced. Deal with an outer square or root only after isolating the expression it acts on.

If the subject appears in several terms, move all of those terms to one side, factor out the subject and divide by its remaining factor.

Original worked example

Making x the subject of y = (ax + b)/(cx + d)

  1. Multiply by cx + d: y(cx + d) = ax + b.
  2. Expand: cxy + dy = ax + b.
  3. Collect x-terms: cxy − ax = b − dy.
  4. Factor: x(cy − a) = b − dy.
  5. Divide by cy − a.

Answer: x = (b − dy)/(cy − a), provided cy ≠ a.

06

SECTION 06

Inequalities and regions

Core concept

Solve a linear inequality like an equation, but reverse the inequality sign when multiplying or dividing by a negative number. A compound inequality sets both a lower and upper restriction.

On a number line, strict boundaries use open circles and inclusive boundaries use closed circles. On a coordinate grid, strict boundaries are broken and inclusive boundaries are solid.

For a two-variable inequality, test a point such as (0, 0) when it is not on the boundary. Cambridge questions normally shade the unwanted side unless told otherwise.

Original worked example

Solving −3 ≤ 2x − 5 < 7

  1. Add 5 to all three parts: 2 ≤ 2x < 12.
  2. Divide all three parts by 2: 1 ≤ x < 6.
  3. Use a closed circle at 1 and an open circle at 6.

Answer: 1 ≤ x < 6.

07

SECTION 07

Sequences

Core concept

A term-to-term rule produces the next value; an nth-term rule gives a value directly from its position n. Check both the first term and at least one later term.

Constant first, second and third differences indicate linear, quadratic and cubic rules. For a quadratic sequence, half the constant second difference is the coefficient of n².

An exponential sequence has a constant multiplier. Combined sequences may need a familiar polynomial pattern plus an alternating or exponential part.

RULE 1
linear: an + b
RULE 2
quadratic: an² + bn + c
RULE 3
cubic: an³ + bn² + cn + d
RULE 4
exponential: arⁿ
Original worked example

Finding the nth term of 2, 9, 28, 65, …

  1. Compare with cubes: 1, 8, 27, 64, …
  2. Every given term is one more than n³.
  3. Test n = 4: 4³ + 1 = 65.

Answer: The nth term is n³ + 1.

08

SECTION 08

Direct and inverse proportion

Core concept

Direct proportion means y is a constant multiple of a stated power of x. In inverse proportion, y is a constant divided by that power. The syllabus includes powers, squares, square roots, cubes and cube roots.

Replace ∝ with = k, use one known pair to calculate k, then use the equation for new values. Keep the same units throughout.

RULE 1
y ∝ xⁿ ⇒ y = kxⁿ
RULE 2
y ∝ 1/xⁿ ⇒ y = k/xⁿ
Original worked example

Inverse square-root proportion

  1. y is inversely proportional to √x, so y = k/√x.
  2. When x = 9, y = 8: 8 = k/3, so k = 24.
  3. When x = 64, y = 24/8.

Answer: y = 3.

09

SECTION 09

Practical graphs and rates of change

Core concept

On a distance–time graph, gradient is speed. A horizontal section means stationary. On a speed–time graph, gradient is acceleration and area beneath the graph is distance travelled.

A negative speed–time gradient represents deceleration. A tangent estimates the instantaneous gradient of a curve; choose two well-separated points on the tangent, not on the curve.

Conversion graphs translate between units or currencies. Read from one axis to the line and across to the other axis using the stated scale.

RULE 1
gradient = vertical change ÷ horizontal change
RULE 2
distance = area under a speed–time graph
RULE 3
acceleration = change in speed ÷ time
Original worked example

Distance from a speed–time graph

  1. A vehicle accelerates from 0 to 18 m/s in 6 s, travels at 18 m/s for 10 s, then slows to rest in 4 s.
  2. Triangle area = 1/2 × 6 × 18 = 54 m.
  3. Rectangle area = 10 × 18 = 180 m.
  4. Final triangle area = 1/2 × 4 × 18 = 36 m.

Answer: Total distance = 270 m.

10

SECTION 10

Function graphs and graphical solutions

Core concept

Build a table of values before plotting. Include enough x-values to reveal intercepts, symmetry, turning points and asymptotic behaviour, then join with a smooth curve rather than straight segments.

Linear, quadratic, cubic, reciprocal and exponential functions have characteristic shapes. Roots are x-intercepts. Reciprocal graphs approach vertical and horizontal asymptotes; exponential graphs may approach a horizontal asymptote.

Solve f(x) = 0 at the x-intercepts and f(x) = g(x) at intersections. Graphical answers are estimates and must be reported to an accuracy supported by the scale.

Typical linear, quadratic, cubic, reciprocal and exponential graph shapeslinearquadraticcubicreciprocalexponential
Original graph-family guide: use intercepts, symmetry, turning points and asymptotes to identify each family.
Features used to recognise common graph families
FamilyTypical formKey features
Lineary = ax + bconstant gradient; no turning point
Quadraticy = ax² + bx + cparabola; one axis of symmetry; one turning point
Cubicy = ax³ + bx² + cx + dmay have two turning points; opposite end directions
Reciprocaly = a/x + btwo branches; vertical and horizontal asymptotes
Exponentialy = arˣ + brapid growth or decay; horizontal asymptote
Original worked example

Solving x² − 2 = x graphically

  1. Plot y = x² − 2 and y = x on the same axes.
  2. Read the x-coordinates of their two intersections.
  3. The values should be close to −1 and 2; a finer scale gives approximately −1.0 and 2.0.

Answer: The intersections give the solutions to x² − x − 2 = 0: x = −1 and x = 2.

11

SECTION 11

Differentiation

Core concept

Differentiation produces a gradient function. For axⁿ, multiply the coefficient by n and reduce the power by one. Constants differentiate to zero.

A stationary point occurs where dy/dx = 0. Classify it using a sketch, the sign of the gradient on each side, or the second derivative: positive gives a minimum and negative gives a maximum.

The syllabus uses polynomials containing no more than three terms and non-negative integer powers. Points of inflection are not required.

RULE 1
d/dx(axⁿ) = anxⁿ⁻¹
RULE 2
stationary point: dy/dx = 0
Original worked example

Stationary points of y = x³ − 3x² − 9x + 5

  1. Differentiate: dy/dx = 3x² − 6x − 9.
  2. Set to zero and divide by 3: x² − 2x − 3 = 0.
  3. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1.
  4. Substitute into y: y(3) = −22 and y(−1) = 10.
  5. d²y/dx² = 6x − 6: it is positive at 3 and negative at −1.

Answer: Local maximum (−1, 10); local minimum (3, −22).

12

SECTION 12

Functions, inverses and composites

Core concept

A function assigns exactly one output to each input in its domain. The domain is the allowed input set and the range is the resulting output set.

An inverse function reverses a one-to-one function. Write y = f(x), rearrange for x, then interchange x and y. A composite such as gf(x) means apply f first and then g.

A mapping diagram represents each input with an arrow to its output. If one input has two outputs, the relation is not a function.

RULE 1
gf(x) = g(f(x))
RULE 2
f⁻¹(f(x)) = x for permitted x
Original worked example

Inverse and composite functions

  1. Let f(x) = 3x − 5 and g(x) = x² + 1.
  2. For f⁻¹, write y = 3x − 5 and rearrange: x = (y + 5)/3.
  3. Therefore f⁻¹(x) = (x + 5)/3.
  4. gf(x) = g(3x − 5) = (3x − 5)² + 1.

Answer: f⁻¹(x) = (x + 5)/3 and gf(x) = 9x² − 30x + 26.

QUICK CHAPTER SUMMARY

The ideas to carry forward

  • Use brackets when substituting negative values.
  • Factorise fully and verify by expansion.
  • Keep denominator restrictions from the original expression.
  • Choose an equation method that matches its structure and check all solutions.
  • Reverse an inequality only after multiplying or dividing by a negative.
  • Difference patterns and ratios reveal sequence families.
  • Replace proportionality with a constant k.
  • Graph features connect visual behaviour to equations.
  • Differentiation finds gradients and stationary points.
  • Inverse functions undo; composite functions act in order.

QUICK REVISION CHECKLIST

Can you do each of these without your notes?

  • translate statements into algebra and substitute values correctly
  • expand, factorise and complete the square
  • manipulate algebraic fractions and expressions with indices
  • solve linear, fractional, simultaneous and quadratic equations
  • rearrange formulas, including when the new subject appears more than once
  • solve inequalities and describe regions on coordinate grids
  • find rules for linear, quadratic, cubic, exponential and combined sequences
  • model direct and inverse proportion
  • interpret practical graphs and the main function families
  • differentiate simple polynomials and analyse stationary points
  • use function, inverse-function and composite-function notation