Mini-Lesson

Algebra and functions

This mini-lesson covers AQA section B — Algebra and functions: surds and indices, quadratics and the discriminant, polynomials and the factor theorem, partial fractions, the modulus function, and the machinery of functions — domain and range, composite and inverse functions, and graph transformations.

Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.

Algebra · surds and indices

Surds, indices and rationalising

A quick but essential foundation — A-level answers are expected in exact form, so surd manipulation must be automatic.

Laws of indices: a^m × aⁿ = a^m+n · a^m ÷ aⁿ = a^m−n · (a^m)ⁿ = a^mn · a⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^1/n = ⁿ√a.
Simplifying surds: √48 = √(16 × 3) = 4√3 — always pull out the largest square factor.
Rationalising: multiply by the conjugate. 1/(√5 − 2) × (√5 + 2)/(√5 + 2) = (√5 + 2)/(5 − 4) = √5 + 2.

Watch out: √(a + b) is not √a + √b. Test it: √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7. Never split a surd over a sum.

Algebra · quadratics

Completing the square and the discriminant

Every quadratic can be written in completed-square form, which hands you the turning point straight away:

ax² + bx + c = a(x + b/2a)² + (c − b²/4a)the vertex is at x = −b/2a

The discriminant Δ = b² − 4ac tells you how many real roots the quadratic has:

Δ > 0 — two distinct real roots (the curve cuts the x-axis twice).
Δ = 0 — one repeated root (the curve touches the x-axis).
Δ < 0 — no real roots (the curve never meets the x-axis).

Worked example — a condition on k

For what k does 3x² + kx + 12 = 0 have equal roots?

Equal roots ⇒ Δ = 0 ⇒ k² − 4(3)(12) = 0 ⇒ k² = 144 ⇒ k = ±12.

Both signs are valid — a very common place to drop a mark.

Quick check

Equal roots

?The equation 3x² + kx + 12 = 0 has a repeated root. What is k?

Algebra · polynomials

The factor theorem and algebraic division

The factor theorem is the fastest way into a cubic:

(x − a) is a factor of f(x) ⇔ f(a) = 0and more generally, f(a) is the remainder when f(x) is divided by (x − a)

Worked example — factorise fully

f(x) = 2x³ − 3x² − 11x + 6. Try small factors of the constant term 6.

f(3) = 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6 = 0, so (x − 3) is a factor.

Dividing: 2x³ − 3x² − 11x + 6 = (x − 3)(2x² + 3x − 2) = (x − 3)(2x − 1)(x + 2).

Roots: x = 3, x = ½, x = −2.

Method: test the factors of (constant ÷ leading coefficient). Once one root is found, divide out and factorise the remaining quadratic — do not keep guessing.

Sort it

Classify the transformation

Tap a transformed equation, then tap the type of transformation it describes.

➡️ Translation

↕️ Stretch

🪞 Reflection

Algebra · partial fractions

Partial fractions

Partial fractions split a single algebraic fraction into a sum of simpler ones. You will need them again for integration and for binomial expansions, so they matter well beyond this section.

(5x + 1)/((x − 1)(x + 2)) ≡ A/(x − 1) + B/(x + 2)the identity must hold for all x

Worked example — the cover-up method

Multiply through: 5x + 1 ≡ A(x + 2) + B(x − 1).

Substitute x = 1 (killing B): 6 = 3A, so A = 2.

Substitute x = −2 (killing A): −9 = −3B, so B = 3.

Answer: (5x + 1)/((x − 1)(x + 2)) ≡ 2/(x − 1) + 3/(x + 2). Check at x = 0: LHS = 1/(−2) = −0.5; RHS = −2 + 1.5 = −0.5 ✓

Repeated factors need an extra term: f(x)/((x − 1)²(x + 3)) ≡ A/(x − 1) + B/(x − 1)² + C/(x + 3). Missing the B term is the standard error.

Quick check

Find the domain

?What is the largest possible domain of the real function f(x) = √(x − 4)?

Algebra · modulus

The modulus function

|x| is the distance of x from zero, so it is never negative: |−4| = 4. The graph of y = |f(x)| takes everything below the x-axis and reflects it up.

|x| = a ⇔ x = a or x = −a (for a ≥ 0)always two branches — solve both

Worked example — solving a modulus equation

Solve |x − 4| = 3.

Branch 1: x − 4 = 3 ⇒ x = 7. Branch 2: x − 4 = −3 ⇒ x = 1.

Read it as "x is 3 away from 4" — which gives 7 and 1 immediately.

Careful with y = f(|x|): that is a different transformation — it keeps the part of the curve for x ≥ 0 and reflects it in the y-axis. |f(x)| reflects in the x-axis; f(|x|) reflects in the y-axis.

Calculate

Your turn — the discriminant

1Calculate the discriminant b² − 4ac of the quadratic 2x² − 5x + 4. (Your sign tells you how many real roots there are.)

Hint: a = 2, b = −5, c = 4. So b² − 4ac = (−5)² − 4(2)(4) = 25 − 32. A negative discriminant means no real roots.

Calculate

Your turn — the factor theorem

2For f(x) = 2x³ − 3x² − 11x + 6, evaluate f(3). (If you get zero, (x − 3) is a factor.)

Hint: f(3) = 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6. The full factorisation is (x − 3)(2x − 1)(x + 2).

Functions · domain and range

Functions: domain, range and inverses

A function is a rule with a domain (the allowed inputs) and a range (the outputs it actually produces). Change the domain and you change the function.

f(x) = √(x − 4): domain x ≥ 4 (you cannot square-root a negative); range f(x) ≥ 0.
f(x) = 1/(x − 2): domain x ≠ 2; range f(x) ≠ 0.

An inverse f⁻¹ exists only if f is one-to-one. To find it: write y = f(x), rearrange for x, then swap letters. The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x, and the domain and range swap over.

Worked example — an inverse

f(x) = (2x + 5)/(x − 3). Set y = (2x + 5)/(x − 3) ⇒ y(x − 3) = 2x + 5 ⇒ xy − 2x = 3y + 5 ⇒ x(y − 2) = 3y + 5.

So f⁻¹(x) = (3x + 5)/(x − 2). Check: f⁻¹(7) = 26/5 = 5.2, and f(5.2) = 15.4/2.2 = 7 ✓

Quick check

Order of composition

?For functions f and g, what does the composite fg(x) mean?

Calculate

Your turn — partial fractions

3Write (5x + 1)/((x − 1)(x + 2)) ≡ A/(x − 1) + B/(x + 2). Find the value of A.

Hint: Multiply up: 5x + 1 ≡ A(x + 2) + B(x − 1). Substituting x = 1 kills the B term: 6 = 3A. (For the record, B = 3.)

Functions · composites

Composite functions

A composite function applies one function to the output of another. The notation is read right to left:

fg(x) means f(g(x)) — do g first, then fgf(x) means g(f(x)) — a different function

Worked example — order matters

f(x) = 3x − 1 and g(x) = x² + 2.

fg(4) = f(g(4)) = f(4² + 2) = f(18) = 3(18) − 1 = 53.

gf(4) = g(f(4)) = g(11) = 11² + 2 = 123.

53 ≠ 123 — composition is not commutative.

Domain of a composite: fg(x) is only defined where g(x) is defined and g(x) lies in the domain of f. And note ff⁻¹(x) = x — inverses undo each other.

Match it

Match the expression to its exact value

Tap an item on the left, then its partner on the right.

Expression

Exact value

Functions · transformations

Graph transformations

Six transformations, and the ones inside the bracket always behave in the opposite way to how they look.

y = f(x) + a — translation up by a (vector [0, a]).
y = f(x + a) — translation left by a (vector [−a, 0]) — not right.
y = a f(x) — vertical stretch, scale factor a.
y = f(ax) — horizontal stretch, scale factor 1/a — a squash if a > 1.
y = −f(x) — reflection in the x-axis.
y = f(−x) — reflection in the y-axis.

inside the bracket → horizontal → oppositeoutside the bracket → vertical → as expected

Order matters in combined transformations. For y = 2f(x + 3) − 1, apply the translation left 3 first, then the stretch ×2, then the translation down 1.

Quick check

Name the transformation

?The graph of y = f(x) is transformed to y = f(x + 3). What has happened?

Calculate

Your turn — a composite

4f(x) = 3x − 1 and g(x) = x² + 2. Calculate fg(4).

Hint: fg(4) = f(g(4)). First g(4) = 4² + 2 = 18, then f(18) = 3(18) − 1. (Compare gf(4) = 123 — the order really does matter.)

Calculate

Your turn — an inverse

5f(x) = (2x + 5)/(x − 3), with x ≠ 3. Find the value of f⁻¹(7).

Hint: f⁻¹(7) is the x with f(x) = 7. So 2x + 5 = 7(x − 3) ⇒ 2x + 5 = 7x − 21 ⇒ 26 = 5x. Check: f(5.2) = 15.4/2.2 = 7 ✓

Recap

The big ideas to know

Surds: √48 = 4√3 · rationalise with the conjugate · √(a + b) ≠ √a + √b

Discriminant: b² − 4ac > 0 two roots · = 0 repeated · < 0 none

Factor theorem: (x − a) is a factor ⇔ f(a) = 0

Partial fractions: cover-up method; repeated factors need A/(x−1) + B/(x−1)²

Modulus: |x − 4| = 3 has two branches: x = 7 and x = 1

Functions: domain and range swap under f⁻¹ · fg means "g first" · inside the bracket = horizontal and opposite

That is AQA section B — Algebra and functions. Press Finish to see your score.

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