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.
A quick but essential foundation — A-level answers are expected in exact form, so surd manipulation must be automatic.
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.
Every quadratic can be written in completed-square form, which hands you the turning point straight away:
The discriminant Δ = b² − 4ac tells you how many real roots the quadratic has:
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.
The factor theorem is the fastest way into a cubic:
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.
Tap a transformed equation, then tap the type of transformation it describes.
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.
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.
|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.
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.
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.
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.
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 ✓
A composite function applies one function to the output of another. The notation is read right to left:
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.
Tap an item on the left, then its partner on the right.
Six transformations, and the ones inside the bracket always behave in the opposite way to how they look.
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.
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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