A Level · Cambridge (CIE) · 9709

Transform graph inputs and combine horizontal moves

Solve for the new input coordinate instead of copying a sign, and preserve the transformed domain.

Section 1.2 · Functions

Pure Mathematics 1 · Functions · 1.2

Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Written practice is self-checked; show your reasoning and domain restrictions.

Before you begin

Solve a linear equation for its input variable.
Understand vertical graph transformations and declared domains.

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

Quick readiness check

Which input x makes x + 4 equal to an original input 2?

What you will learn

Describe y = f(x + a) and y = f(ax) using horizontal transformations.
Map points and domains under a simple combination of input and output changes.

Ask which new input produces the old one

If (u, v) lies on y = f(x), then y = f(x + a) needs x + a = u to produce that same output. Thus the new point is (u − a, v): translation by vector (−a, 0). A plus inside f shifts left; a minus inside shifts right. For example, f(x + 4) moves (2, −1) to (−2, −1).

For y = f(ax), with a ≠ 0, solve ax = u to get the new point (u/a, v). A positive a gives a stretch parallel to the x-axis with factor 1/a. For a < 0, include reflection in the y-axis and use the positive horizontal factor 1/|a|. Thus f(−2x) reflects in the y-axis and halves horizontal distances; it does not double them. The exceptional a = 0 gives f(0) as a constant only if 0 is in f's domain.

Factor the input expression to read a combination

For y = f(2x − 6) + 1, write the input as 2(x − 3). An original point (u, v) moves to ((u + 6)/2, v + 1). One valid description is a horizontal stretch factor 1/2, then translation right by 3, then up by 1. Original point (4, 5) therefore becomes (5, 6). Moving right by 6 and then halving horizontal distances gives the same horizontal result; moving right by 3 and then halving does not.

Always impose the original function's domain on the transformed argument. For f(u) = (u − 1)² with u ≥ 1, the rule f(2x − 6) + 1 becomes (2x − 7)² + 1, but is defined only when 2x − 6 ≥ 1, hence x ≥ 7/2. Its vertex is (7/2, 1) and range y ≥ 1. Keeping the old domain x ≥ 1 would admit inputs the original function cannot accept.

For y = A f(B(x − H)) + K, B ≠ 0: (u, v) → (H + u/B, Av + K)

Pause and explain

What horizontal scale factor describes y = f(2x)?

Put the idea to work

Worked example

Describe y = f(2x − 6) + 1 and give the image of an original point (4, 5).

Show the worked solution

Factor 2x − 6 as 2(x − 3), giving horizontal scale factor 1/2 followed by a shift right by 3.
The outside +1 translates the result upward by 1.
Solve 2x − 6 = 4 to get x = 5; the new output is 5 + 1 = 6.

Answer Horizontal stretch factor 1/2, then translation (3, 0), then (0, 1); point (5, 6).

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

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

A point (2, −1) lies on y = f(x). Describe y = f(x + 4) and give that point's image.

Give me a hint

Set the new argument x + 4 equal to the old input 2.

Compare my reasoning

Solve x + 4 = 2 to get x = −2.
The original output −1 is unchanged.
Every point moves left by 4, so the translation vector is (−4, 0).

Translation (−4, 0); point (−2, −1).

Look for these in your work

Solve the argument equation to determine direction.
Keep the output coordinate unchanged.

2 · Independent

A point (6, 3) lies on y = f(x). Describe y = f(−2x) and give the point's image.

Give me a hint

Solve −2x = 6 and separate reflection from positive scaling.

Compare my reasoning

The new input is x = −3 and the output remains 3.
The negative multiplier reflects the input positions in the y-axis.
Its magnitude 2 gives a horizontal stretch factor 1/2, so the image is (−3, 3).

Reflection in the y-axis, horizontal stretch factor 1/2; point (−3, 3).

Look for these in your work

Use the reciprocal horizontal scale.
Name the reflection axis and preserve y.

3 · Transfer

Let f(u) = (u − 1)² on u ≥ 1. For h(x) = f(2x − 6) + 1, find its rule, domain, range and vertex.

Give me a hint

Impose u = 2x − 6 ≥ 1 before interpreting the completed square.

Compare my reasoning

Substitute the full argument: h(x) = (2x − 7)² + 1 = 4(x − 7/2)² + 1.
The original input restriction requires 2x − 6 ≥ 1, so the new domain is x ≥ 7/2.
The allowed vertex is (7/2, 1); the square gives every output at least 1 on this right branch.

h(x) = (2x − 7)² + 1, domain x ≥ 7/2, range y ≥ 1, vertex (7/2, 1).

Look for these in your work

Substitute correctly and transform the original domain.
State the range and full vertex coordinates.

Common mistakes

A positive constant inside f moves the graph left, not right.
The horizontal scale factor for f(ax) is 1/|a|, together with a reflection if a < 0.
Transform the input domain as well as points and the algebraic rule.

Recall without your notes

What can you explain now?

An original point is (6, 3). Find its image on y = f(−2x) and describe the transformation.

Compare with the explanation

(−3, 3); reflection in the y-axis and horizontal stretch factor 1/2.

Solve −2x = 6 to get −3; the output stays 3. The negative input multiplier reflects and its magnitude 2 gives reciprocal horizontal scale 1/2.

After trying it yourself, choose your next review. This is your self-assessment.

Your review choice appears on Today. Sign in to sync it across devices.

Back to Today

Make it stick

Derive the point-mapping rule for f(B(x − H)) from an argument equation. Use it to explain horizontal signs, reciprocal scales and changed input restrictions.

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.