Mini-Lesson

Algebra and functions

Topic 2 is the engine room of 9MA0. This mini-lesson covers surds and indices, the discriminant, the factor and remainder theorems, partial fractions, the modulus function, composite and inverse functions with their domains and ranges, and the four graph transformations.

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.

Algebra · quadratics

The discriminant

For ax² + bx + c = 0 the discriminant is Δ = b² − 4ac. It tells you how the parabola meets the x-axis.

b² − 4ac > 0 → two distinct real roots= 0 → one repeated (equal) root · < 0 → no real roots

Worked example — a “show that” classic

Find k such that x² + kx + 9 = 0 has equal roots, k > 0.

Equal roots ⇒ b² − 4ac = 0 ⇒ k² − 4(1)(9) = 0 ⇒ k² = 36 ⇒ k = ±6, so k = 6.

Check: x² + 6x + 9 = (x + 3)², repeated root x = −3. ✓

Exam trap: “the curve meets the axis” means Δ ≥ 0; “the line is a tangent” means Δ = 0; “the curve is always positive” means a > 0 and Δ < 0. Read the wording, then set up the correct inequality.

Calculate

Your turn — equal roots

1The equation x² + kx + 9 = 0 has equal roots and k > 0. Find k.

Set-up

Δ = k² − 4(1)(9) = k² − 36. Equal roots ⇒ Δ = 0 ⇒ k² = 36 ⇒ k = ±6.

k > 0 so k = 6, giving (x + 3)² = 0. ✓

Hint: Set b² − 4ac = 0: k² − 36 = 0.

Algebra · polynomials

Factor and remainder theorems

For a polynomial f(x):

f(a) = 0 ⇔ (x − a) is a factorRemainder theorem: dividing f(x) by (x − a) leaves remainder f(a)

For a divisor (bx − a), the root is x = a/b, so the remainder is f(a/b).

Worked example — f(x) = 2x³ − 5x² − 4x + 3

f(3) = 2(27) − 5(9) − 12 + 3 = 54 − 45 − 12 + 3 = 0, so (x − 3) is a factor.

Divide: 2x³ − 5x² − 4x + 3 = (x − 3)(2x² + x − 1) = (x − 3)(2x − 1)(x + 1).

Roots: x = 3, x = ½, x = −1.

Exam trap: the remainder on dividing by (x − 2) is f(2), not f(−2). Sign errors here are the single most common lost mark in this topic.

Calculate

Your turn — remainder theorem

2f(x) = 2x³ − 5x² − 4x + 3. Find the remainder when f(x) is divided by (x − 2).

Set-up

f(2) = 2(2)³ − 5(2)² − 4(2) + 3 = 16 − 20 − 8 + 3 = −9.

So the remainder is −9 (and (x − 2) is therefore not a factor).

Hint: Remainder = f(2) = 2(8) − 5(4) − 4(2) + 3.

Quick check

Repeated factors

?Which is the correct form of the partial fractions for (x + 1) ⁄ [(x − 2)²(x + 3)] ?

Algebra · partial fractions

Partial fractions

Split a proper algebraic fraction into simpler pieces. 9MA0 requires denominators with up to three distinct linear factors, or a repeated linear factor.

(9x − 5) ⁄ [(x − 1)(x + 3)] ≡ A/(x − 1) + B/(x + 3)multiply through, then substitute the roots

Worked example

9x − 5 ≡ A(x + 3) + B(x − 1).

Let x = 1: 9 − 5 = 4 = A(4) ⇒ A = 1.

Let x = −3: −27 − 5 = −32 = B(−4) ⇒ B = 8.

So (9x − 5)/[(x − 1)(x + 3)] ≡ 1/(x − 1) + 8/(x + 3).

Check at x = 5: LHS = 40/(4 × 8) = 1.25; RHS = 1/4 + 8/8 = 0.25 + 1 = 1.25 ✓

Why it matters: partial fractions are the gateway to integrating rational functions (Topic 8) and to binomial expansions of them (Topic 4). If the fraction is improper (degree of top ≥ degree of bottom), divide first.

Calculate

Your turn — find B

3(9x − 5) ⁄ [(x − 1)(x + 3)] ≡ A/(x − 1) + B/(x + 3). Find the value of B.

Set-up

Substituting x = −3 kills the A term: 9(−3) − 5 = B(−3 − 1)

−32 = −4B ⇒ B = 8. (And x = 1 gives A = 1.)

Hint: Multiply up: 9x − 5 ≡ A(x + 3) + B(x − 1), then substitute x = −3.

Algebra · modulus

The modulus function

|x| is the distance from zero: |x| = x if x ≥ 0, and |x| = −x if x < 0. The graph of y = |f(x)| reflects anything below the x-axis up above it.

y = |2x − 5| is a V. Solving |2x − 5| = x + 1 means finding both intersections.

Worked example — solve |2x − 5| = x + 1

Positive branch: 2x − 5 = x + 1 ⇒ x = 6. Check: |7| = 7 and 6 + 1 = 7 ✓

Negative branch: 2x − 5 = −(x + 1) ⇒ 3x = 4 ⇒ x = 4/3. Check: |8/3 − 5| = 7/3 and 4/3 + 1 = 7/3 ✓

Exam trap: always check each root back in the original equation — the negative branch can throw up solutions that do not actually work. And |a| = |b| ⇒ a = ±b, so squaring both sides is legitimate but doubles the work.

Calculate

Your turn — modulus equation

4Solve |2x − 5| = x + 1 and give the larger of the two solutions.

Set-up

2x − 5 = x + 1 ⇒ x = 6 (check: |12 − 5| = 7 = 6 + 1 ✓)

2x − 5 = −(x + 1) ⇒ 3x = 4 ⇒ x = 4/3 (check: |8/3 − 5| = 7/3 = 4/3 + 1 ✓)

Larger solution: x = 6.

x =

Hint: Positive branch: 2x − 5 = x + 1.

Algebra · functions

Composite and inverse functions

A function maps each element of the domain to exactly one element of the range.

Composite fg(x) = f(g(x)) — do g first, then f. Order matters: fg ≠ gf in general.
Inverse f⁻¹ exists only if f is one-to-one. Swap x and y, then rearrange.
Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f.
The graph of y = f⁻¹(x) is the reflection of y = f(x) in y = x.

Worked example — f(x) = 3x − 4, g(x) = x² + 1

fg(2) = f(g(2)) = f(2² + 1) = f(5) = 3(5) − 4 = 11.

gf(2) = g(f(2)): here f(2) = 3(2) − 4 = 2, so gf(2) = g(2) = 2² + 1 = 5. Different answer — order matters.

f⁻¹: let y = 3x − 4 ⇒ x = (y + 4)/3, so f⁻¹(x) = (x + 4)/3. Check: f⁻¹(11) = (11 + 4)/3 = 5, which is exactly the g(2) that f turned into 11. ✓

Exam trap: f(x) = x² on ℝ has no inverse — it is many-to-one (both 3 and −3 map to 9). Restrict the domain to x ≥ 0 and it becomes one-to-one, so f⁻¹(x) = √x exists.

Calculate

Your turn — composite function

5f(x) = 3x − 4 and g(x) = x² + 1. Find fg(2).

Set-up

g(2) = 2² + 1 = 5.

f(5) = 3(5) − 4 = 11.

(Watch the order: gf(2) = g(2) = 5, which is different.)

Hint: fg(2) means f(g(2)) — apply g first.

Quick check

Does it have an inverse?

?Why does f(x) = x², with domain x ∈ ℝ, have no inverse function?

Sort it

Sort the transformations

Tap a transformation of y = f(x), then the family it belongs to.

↔️ Translation

↕️ Stretch

🪞 Reflection

Algebra · transformations

Graph transformations

Changes inside the bracket affect x and do the opposite of what you expect; changes outside affect y and behave normally.

y = f(x + a) → translate by (−a, 0)y = f(x) + a → translate (0, a) · y = af(x) → stretch ×a in y · y = f(ax) → stretch ×1/a in x

y = −f(x) — reflection in the x-axis.
y = f(−x) — reflection in the y-axis.
y = f(x + 2) − 3 — translation by the vector (−2, −3): left 2, down 3.
y = 2f(3x) — stretch scale factor 2 parallel to the y-axis and scale factor ⅓ parallel to the x-axis.

Exam trap: f(2x) squashes the graph towards the y-axis (factor ½), it does not stretch it. And when combining, apply the inside changes in the order that undoes the bracket.

Quick check

Read the transformation

?The curve y = f(x) is transformed to y = f(x + 2) − 3. Describe the transformation.

Match it

Algebra flashcards

Tap a card on the left, then the matching result.

Expression / condition

Meaning

Algebra · surds & indices

Surds, indices and rationalising

These are assumed knowledge but examined constantly inside longer questions.

a^(m/n) = ⁿ√(a^m) · a^(−n) = 1/aⁿ · a⁰ = 1√a × √b = √(ab) · √(a/b) = √a / √b

Rationalise: multiply by the conjugate. 1/(3 − √2) × (3 + √2)/(3 + √2) = (3 + √2)/(9 − 2) = (3 + √2)/7.
Simplify: √50 = √25 × √2 = 5√2.
Index form for calculus: 3/√x = 3x^(−½) — you must convert before differentiating or integrating.

Exam trap: √(a + b) ≠ √a + √b. Test it: √(9 + 16) = 5, but 3 + 4 = 7.

Quick check

Simplify

?Write 1 ⁄ (3 − √2) in the form (a + b√2)/c with integers a, b, c.

Recap

The big ideas to take away

Discriminant: b² − 4ac: > 0 two roots · = 0 equal roots (tangent) · < 0 none

Factor / remainder: f(a) = 0 ⇔ (x − a) is a factor; remainder on ÷(x − a) is f(a)

Partial fractions: substitute the roots; a repeated factor needs A/(x−a) AND B/(x−a)²

Modulus: |f(x)| reflects the negative part upwards; solve both branches and check

Functions: fg means g first; f⁻¹ exists only for one-to-one f; domain and range swap

Transformations: inside the bracket = opposite, outside = as expected

Topic 2 underpins the whole qualification — every calculus question is an algebra question in disguise. Press Finish to see your score.

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