A Level · Cambridge (CIE) · 9709

Interpret modulus and sketch a linear absolute-value graph

Read modulus as distance from zero and construct both straight branches of y = |ax + b|.

Pure Mathematics 2 & 3 · Shared Algebra foundations · 2.1 / 3.1

Exam scope and assessment notes

Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Shared Algebra foundations, Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. These modulus and polynomial requirements occur in both sections 2.1 and 3.1. Paper 3 additionally requires partial fractions and rational-power binomial expansions; follow its separate lessons for those methods. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Both sections use linear modulus graphs; nonlinear y = |f(x)| and y = f(|x|) graphs are excluded. Polynomial dividends have degree at most 4 and divisors are linear or quadratic. Written practice is self-checked.

Before you begin

Solve a linear equation and sketch a straight line.
Interpret signed coordinates and distance from zero.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What is |−5|?

What you will learn

Use |u| = u for u ≥ 0 and |u| = −u for u < 0.
Find the vertex, intercepts and branch gradients of a linear modulus graph.

The graph y = |2x − 6| has vertex (3, 0), y-intercept (0, 6), left branch y = −2x + 6 and right branch y = 2x − 6. Its branch gradients are −2 and 2. The labelled point (5, 4) checks the right branch. The plotted window is −1 ≤ x ≤ 7, and both branches continue beyond it unless a domain restriction is supplied. The horizontal and vertical drawing scales differ, so use coordinates to calculate gradients rather than the visual angle.

Modulus keeps magnitude and removes a sign

For a real number u, |u| is its non-negative distance from zero. It equals u when u ≥ 0 and −u when u < 0. Thus |−5| = 5 and |5| = 5, but |u| is not always equal to −u. For an expression such as 2x − 6, first locate where the inner expression changes sign; the modulus is applied to its whole value, not separately to each term.

For y = |2x − 6|, the inner expression is zero at x = 3. When x ≤ 3 the branch is y = −2x + 6, and when x ≥ 3 it is y = 2x − 6. The branches meet at the vertex (3, 0), with gradients −2 on the left and 2 on the right. The graph is a V, has y-intercept (0, 6) and never takes a negative output.

Use the inner root to locate the join

For a ≠ 0, y = |ax + b| has its vertex at (−b/a, 0). The left and right gradients are −|a| and |a| respectively, regardless of the sign of a. The y-intercept is |b|. If a = 0, the graph is the horizontal line y = |b| rather than a V. A sketch should show the correct join, labelled intercept and both branches continuing through the intended domain.

For y = |−3x + 6|, the inner expression is non-negative for x ≤ 2, so the left branch is −3x + 6. For x ≥ 2 the branch is 3x − 6. The vertex is (2, 0). Check the sign of the inner expression before selecting a branch, especially when its coefficient is negative. This section covers linear inner expressions; nonlinear transformed modulus graphs are outside its stated requirements.

Pause and explain

For y = |−3x + 6|, what are the vertex and left-branch gradient?

Put the idea to work

Worked example

Sketch y = |2x − 6|, showing its vertex, y-intercept and branch equations.

Show the worked solution

Solve 2x − 6 = 0 to find vertex (3, 0).
For x ≤ 3 use y = −2x + 6; for x ≥ 3 use y = 2x − 6.
Mark y-intercept (0, 6), with gradients −2 and 2 and non-negative outputs.

Answer Vertex (3, 0); y-intercept (0, 6); left branch −2x + 6, right branch 2x − 6.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Sketch y = |x + 2| with its vertex, y-intercept and both branch equations.

Give me a hint

The join occurs where x + 2 = 0; test an input on either side.

Compare my reasoning

The vertex is (−2, 0) and the y-intercept is (0, 2).
For x ≤ −2 the branch is −x − 2; for x ≥ −2 it is x + 2.
The branch gradients are −1 and 1, giving a V with non-negative outputs.

Vertex (−2, 0); y-intercept (0, 2); branches −x − 2 and x + 2.

Look for these in your work

Locate the inner zero correctly.
Label both branches and retain non-negative outputs.

2 · Independent

Sketch y = |−3x + 6| and give its value at x = 0 and 4.

Give me a hint

A negative inner gradient changes which sign case applies to each side.

Compare my reasoning

The vertex is (2, 0). For x ≤ 2, y = −3x + 6; for x ≥ 2, y = 3x − 6.
At x = 0 the value is 6, and at x = 4 the value is also 6.
The symmetric V has left gradient −3 and right gradient 3, despite the inner coefficient being negative.

Vertex (2, 0); branch gradients −3 and 3; y(0) = y(4) = 6.

Look for these in your work

Choose each branch from the inner sign.
Check two actual inputs against the original modulus.

3 · Transfer

A supplied signed separation model is d(t) = 3t − 12 metres for 0 ≤ t ≤ 6 seconds. Sketch its distance from the reference, |d(t)|, and find the closest time and the distances at both endpoints.

Give me a hint

Distance is the magnitude of the signed separation; use only the stated time interval.

Compare my reasoning

The inner expression vanishes at t = 4, giving the minimum distance 0 m.
The distance is 12 − 3t for 0 ≤ t ≤ 4 and 3t − 12 for 4 ≤ t ≤ 6.
The initial distance is 12 m and the final distance is 6 m; the closest time is t = 4 s.

Closest at t = 4 s with distance 0 m; endpoint distances 12 m and 6 m.

Look for these in your work

Convert a signed separation into a non-negative distance.
Respect the time domain and label the model's units.

Common mistakes

Apply modulus to the whole expression; |a + b| is not generally |a| + |b|.
Check where the inner expression is negative before reversing its sign.
A zero coefficient gives a constant graph, not a V.

Recall without your notes

What can you explain now?

Find |−7| and |2(1) − 6|.

Compare with the explanation

7 and 4.

The values inside the modulus are −7 and −4; their distances from zero are 7 and 4.

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Make it stick

Explain the two sign cases for a linear modulus, then sketch its join and branches without treating modulus as a term-by-term operation.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics

Check the official syllabus 2026–2027

Course scope and assessment

312 sequenced lessons provide 936 guided, independent and application problems covering the 160 objectives in sections 1.1–1.8, 2.1–2.6, 3.1–3.9, 4.1–4.5, 5.1–5.5 and 6.1–6.5. The map lists all 38 sections across six alternative components; all eight Pure Mathematics 1 sections and all six Pure Mathematics 2 sections, from Paper 2 Algebra through Numerical solution of equations, currently have full teaching sequences. Paper 3 Algebra, section 3.1, links the six shared foundation lessons plus six lessons on partial fractions and rational binomial expansions; Paper 3 logarithms, trigonometry and numerical equations use the shared lessons, with the Paper 1 graph foundation linked into trigonometry. Paper 3 differentiation combines the shared methods with its inverse-tangent derivative lesson. Paper 3 Integration links four shared standard and identity lessons plus seven advanced lessons, covering all six section 3.5 requirements. Eight Vectors lessons cover all six section 3.7 requirements, with 2D and 3D line and solid applications. Nine Differential equations lessons cover all four section 3.8 requirements, including Paper 3 integration techniques, initial conditions, equilibrium solutions and conditional model interpretations. Eleven Complex numbers lessons cover all eight section 3.9 requirements, with seven Argand diagrams and explicit argument and boundary checks. All nine Pure Mathematics 3 sections now have teaching linked to every mapped requirement. Nine Forces and equilibrium lessons cover all seven section 4.1 requirements, with six force diagrams, signed incline-friction intervals, reaction-dependent pull and push limits and separate-body third-law checks. Nine Kinematics lessons cover the four official section 4.2 requirements, with the graph requirement mapped to three relationships. Six motion diagrams support directed quantities, graph slopes and signed areas, Paper 1 time-calculus methods, reversal distances, constant-acceleration stages and shared-clock particle meetings. Six Momentum lessons cover both section 4.3 requirements, including signed system totals, conservation assumptions, outgoing velocities, coalescence, unknown masses and successive impacts. Four diagrams show system momentum through the stipulated events, and supplied kinetic-energy checks distinguish conservation of momentum from changes in kinetic energy. Twelve Newton’s laws lessons cover all four section 4.4 requirements, with vertical support and release stages, rough horizontal and incline motion, inferred parameters, hanging and surface pulley systems, rigid tow-bar thrust and slack-string transitions. Seven force diagrams support the separate-particle balances. Nine Energy, work and power lessons cover all five section 4.5 requirements, with angle-dependent work, kinetic and reference potential energy, signed net-work balances, smooth curved slides, rough and driven slopes, ideal connected-system energy and instantaneous power on hills. Five diagrams distinguish the energy inputs, losses and valid positive-speed force model. All five Mechanics sections now have teaching and staged practice linked to their mapped requirements. Thirteen Representation of data lessons cover all five section 5.1 requirements, including choice and limitations of displays, ordered and back-to-back stems, raw summaries and box plots, unequal-width density histograms, cumulative-frequency interpolation, descriptive moment calculations, grouped estimates, coding, combining two datasets and correcting totals. Six accessible diagrams retain repeated observations and distinguish numerical spread, frequency area and estimated readings. Ten Permutations and combinations lessons cover both official section 5.2 requirements, with selection and ordered roles, repeated objects, adjacency blocks, non-adjacency gaps, leading-zero cases, labelled multi-row seating and mixed restrictions. Six accessible diagrams show outcome trees, ordered versus unordered choices, repeated copies, blocks, gaps and distinct row seats. Eleven Probability lessons cover all four section 5.3 requirements, with equiprobable elementary outcomes, selection-based probabilities, complements and disjoint regions, independence tests, restricted sample spaces, two-way tables, replacement and removal trees, reversed conditions and unknown parameters. Seven accessible diagrams show dice outcomes, random subsets, event regions, a conditioned grid, a count table and two probability trees. Twelve Discrete random variables lessons cover all three official section 5.4 requirements, with complete probability tables, moment and parameter checks, binomial assumptions, exact probabilities and moments, geometric first-success probabilities, tails, expectation and integer reliability thresholds. Six accessible diagrams distinguish outcome grouping, weighted moments, fixed trials, discrete probability bars, waiting conventions and integer tail deadlines. Twelve Normal distribution lessons cover all three official section 5.5 requirements, with continuous-area interpretation, MF19 cumulative tables, full standardisation, tail and interval regions, conditional probabilities, percentiles, one- and two-parameter inference, binomial suitability, continuity corrections and approximate integer thresholds. Seven accessible diagrams show model areas, table reading, aligned scales, intervals, quantiles, matched binomial and normal shapes and whole integer cells. All five Statistics 1 sections now have teaching and staged practice linked to their mapped requirements. Fourteen Poisson distribution lessons cover all five official section 6.1 requirements, with homogeneous event-count assumptions, exposure scaling, exact masses and tails, conditional count events, moments, parameter inference, probability duration targets, disjoint and overlapping windows, rare-count binomial approximations, matched normal approximations and continuity corrections. Nine accessible diagrams show event exposures, unit conversions, discrete bars, complete tail complements, population moments, shared windows, approximate count models and integer cells. Exact checks distinguish approximate and exact close-target decisions. Fourteen Linear combinations lessons cover the six listed results in official section 6.2, with affine moments, signed scales and constant cases, expectation without independence, independent squared variance contributions, fixed and random fees, independent copies versus reused values, affine normal events, independent normal sums and differences, weighted and conditional areas, coefficient inference, exact independent Poisson sums and overlapping exposure limits. Nine accessible diagrams distinguish scale changes, variance contributions, component costs, copy mechanisms, aligned normal scales, summed areas, Poisson count pairs and unsupported distribution claims. Fourteen Continuous random variables lessons cover both official section 6.3 requirements, with density validity and units, support-aware intervals, complements and conditional regions, first and second moments, percentiles by direct density areas, symmetry checks, parameter inference, infinite-support power and exponential integrals, divergent-moment limits and combined applications. Nine accessible diagrams distinguish density height from probability, normalisation, negative-density failures, central and conditional areas, mean versus median, percentile locations and partial infinite tails. Explicit cumulative distribution function knowledge is not required. Eighteen Sampling and estimation lessons cover all eight official section 6.4 requirements, with population and frame definitions, random-number selection, sampling-quality limits, independent sample-mean moments, normal means, CLT conditions, SE scaling, raw, summarised and coded unbiased estimates, known-variance and large-sample mean intervals, repeated-sampling interpretation, precision planning, proportion intervals, sparse-outcome limits and contextual applications. Ten accessible diagrams distinguish selection frames, number rejection, repeated sample means, population versus mean density, CLT models, variance divisors, interval targets, illustrative coverage, integer size checks and proportion margins. Eighteen Hypothesis tests lessons cover all five official section 6.5 requirements, with pre-specified alternatives, inclusive count tails, explicit equal-tail conventions, integer critical regions and actual significance, exposure-aware Poisson tests, continuity-corrected count approximations, normal and large-sample mean tests, confidence-interval connections, error classifications, fixed-region binomial, Poisson and normal error probabilities and sample-size power comparisons. Nine accessible diagrams show the testing procedure, exact count regions, unequal Poisson tails, discrete significance jumps, corrected count boundaries, matching mean intervals and alternative error distributions. All shared lessons are counted once in the course total. All 38 sections now link teaching and practice. Eight earlier overview lessons remain available, but do not count as full section coverage. This is a content-availability measure, not full programme completion or exam readiness. Written practice is self-checked, not officially marked. Check the 2028–2030 syllabus separately if sitting an examination from 2028; this teaching map uses 2026–2027 requirements.