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IB Maths AA HL · Unit 2: Functions

IB Maths AA HL Transformations and Modulus Questions

Exam-style IB Maths AA HL transformations and modulus questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.

Practise Transformations and Modulus questions → AA HL formula booklet

What you need to know

HL AA Paper 1 tests |x-2| < 3 or (x-1)/(x+2) ≥ 0. Sign diagrams are your friend — never multiply through by a variable expression. Inequalities — rational and modulus overview →

What's examined in AA HL transformations and modulus

The question bank covers these transformations and modulus question types (number of questions in brackets):

Key formulas

Modulus & argument
\(r = |z| = \sqrt{a^2 + b^2},\ \theta = \arg z\)

In the same notation as the IB formula booklet. All AA HL formulas →

Transformations and Modulus worked examples

Worked example 1: Applying basic translations · easy

The graph of $f(x) = x^2 - 4$ is translated by the vector $\binom{-2}{3}$ to form the graph of $g(x)$. Find an expression for $g(x)$ in the form $ax^2 + bx + c$.

Solution

1. Interpret the translation vector $\binom{-2}{3}$: this shifts the graph $2$ units to the left and $3$ units up.

2. Apply the transformations to the function notation: $g(x) = f(x - (-2)) + 3 = f(x + 2) + 3$.

3. Substitute $(x + 2)$ into the original function: $g(x) = (x + 2)^2 - 4 + 3$.

4. Expand the binomial square: $g(x) = (x^2 + 4x + 4) - 4 + 3$.

5. Simplify the constant terms: $g(x) = x^2 + 4x + 4 - 1$.

6. State the final expanded expression: $\mathbf{g(x) = x^2 + 4x + 3}$.

Examiner tip: Translating left by $c$ units means adding $c$ inside the function bracket (i.e., $f(x+c)$). Reversing this sign and writing $f(x-2)$ for a leftward shift is an incredibly common and costly mistake.

Worked example 2: Solving absolute value equations · medium

Consider the function $f(x) = |2x - 6|$. Solve the equation $f(x) = x$ algebraically.

Solution

1. Set up the equation: $|2x - 6| = x$.

2. Form Case 1, assuming the argument is positive (so the modulus does nothing): $2x - 6 = x$.

3. Solve Case 1: Subtract $x$ and add $6$ to get $x = 6$.

4. Form Case 2, assuming the argument is negative (so the modulus multiplies it by $-1$): $-(2x - 6) = x \implies -2x + 6 = x$.

5. Solve Case 2: Add $2x$ to get $3x = 6 \implies x = 2$.

6. Verify both solutions in the original equation to check for extraneous roots. For $x=6$: $|12 - 6| = 6$ (Valid). For $x=2$: $|4 - 6| = |-2| = 2$ (Valid). Solutions: $\mathbf{x = 2, x = 6}$.

Examiner tip: When solving equations with a modulus on one side and a variable on the other, you must check your final answers to ensure they don't produce a negative output for the absolute value, which would be mathematically impossible.

Worked example 3: Advanced composite transformations · hard

The graph of $y = \cos x$ is transformed to the graph of $g(x) = 3\cos(2x - \pi) + 1$. Describe fully the sequence of four geometric transformations required.

Solution

1. Factorise the argument of the trigonometric function so the coefficient of $x$ is separated: $2x - \pi = 2(x - \frac{\pi}{2})$.

2. Identify the horizontal stretch from the multiplier $2$: A horizontal stretch by scale factor $\frac{1}{2}$.

3. Identify the horizontal translation from the $(x - \frac{\pi}{2})$ term: A horizontal translation by $\frac{\pi}{2}$ units to the right.

4. Identify the vertical stretch from the multiplier $3$ outside the function: A vertical stretch by scale factor $3$.

5. Identify the vertical translation from the $+1$ at the end: A vertical translation by $1$ unit upwards.

6. State the full sequence. (Note: stretches should generally be applied before translations unless bracketed otherwise).

Examiner tip: Always factorise the coefficient inside the trigonometric argument (i.e. changing $bx-c$ into $b(x - c/b)$) before reading off the horizontal translation. Otherwise, your phase shift will be completely incorrect!

Try these IB Maths AA HL transformations and modulus questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.

Question 1 · easy · 2 marks · Paper 1

The graph of \(y = f(x)\) is translated by the vector \(\begin{pmatrix} -3 \\ 4 \end{pmatrix}\). Write down the equation of the transformed graph in terms of \(f(x)\).

Attempt it and see the mark scheme →

All 20 transformations and modulus questions with mark schemes →

FAQ

How many IB Maths AA HL transformations and modulus questions are there?

There are 20 exam-style transformations and modulus questions in the AA HL question bank (Paper 1: 16 · Paper 2: 4), graded 7 easy, 7 medium, 6 hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is transformations and modulus on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 16 · Paper 2: 4. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA HL Unit 2 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

More AA HL Unit 2 topics

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