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IGCSE · Cambridge (CIE) · 0444

Transform f(x) + k, kf(x) and f(x + k)

Distinguish vertical translation, vertical scaling and horizontal translation · Transform coordinates and domain restrictions · Explain the opposite sign of a horizontal shift

Section E3.12 · Extended · Translations and vertical scaling of graphs

Mathematics (US) 0444 · historical 2025 study · C3.12 / E3.12
Exam scope and assessment notes

Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.

Before you begin

  • Function notation
  • Graph coordinates

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

Quick readiness check

What should you establish before applying the method in this lesson?

What you will learn

  • Distinguish vertical translation, vertical scaling and horizontal translation
  • Transform coordinates and domain restrictions
  • Explain the opposite sign of a horizontal shift
Right by 3, then up by 2-2-10123450246810xyy = x²y = (x − 3)² + 2(0, 0)(3, 2)
The blue graph y = x² has vertex (0, 0). The red graph y = (x − 3)² + 2 has vertex (3, 2): shift right by 3 and up by 2. A corresponding point moves from (1, 1) to (4, 3). Visible curve endpoints are plotting-window limits, not domain restrictions.

Outside changes the output

If (p, q) lies on y = f(x), then (p, q + k) lies on f(x) + k: a vertical translation. For kf(x), the point is (p, kq): vertical scale factor |k|, with reflection in the x-axis when k < 0. At k = 0 the outputs collapse to zero on the original domain.

The domain stays the same for these two outside changes; an algebraically simplified zero expression does not automatically fill excluded original inputs.

Inside changes the input

For y = f(x + k), the old input p occurs when x + k = p, so the new point is (p − k, q). A positive k shifts the graph left, not right. The domain shifts too: if the original requires x ≥ 0, f(x + 2) requires x ≥ −2.

Apply one transformation at a time for combined expressions. This 0444 section lists f(x + k), not an obligation to replace it with f(kx); they are different transformations.

Sources for this lesson

Pause and explain

Which way does f(x + 2) shift the original graph?

Put the idea to work

Worked example

A point (2, 5) is on f. Where does it move for f(x) + 3, −2f(x) and f(x + 3)?

Show the worked solution
  1. Outside addition changes the height to 8.
  2. Outside scaling changes the height to −10.
  3. Inside addition requires x + 3 = 2, so the new x is −1.

Answer (2, 8), (2, −10), (−1, 5), respectively.

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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

A point (−1, 4) lies on f. Give its corresponding points on f(x) − 2 and 3f(x).

Give me a hint

Only change the output coordinate.

Compare my reasoning
  1. Subtracting 2 gives 4 − 2 = 2.
  2. Multiplying by 3 gives 12.

(−1, 2) and (−1, 12).

Look for these in your work

  • Keep the input unchanged
  • Apply the whole vertical change
2 · Independent

For f(x) = x², find the vertex of f(x − 3) + 2 and expand its rule.

Give me a hint

Solve x − 3 = 0 for the shifted vertex input.

Compare my reasoning
  1. The vertex shifts from (0, 0) to (3, 2).
  2. (x − 3)² + 2 = x² − 6x + 11.

Vertex (3, 2); rule x² − 6x + 11.

Look for these in your work

  • Use the inside sign correctly
  • Check the expansion
3 · Transfer

For f(x) = 1/x, find the excluded input of f(x + 2), its output at x = 0, and the domain of 0f(x).

Give me a hint

Keep restrictions from the original evaluation of f.

Compare my reasoning
  1. f(x + 2) = 1/(x + 2), excluding x = −2; at 0 its value is 1/2.
  2. 0f(x) is zero where f is defined, so its domain still excludes 0.

Shifted exclusion −2; output 1/2 at 0. The domain of 0f(x) is x ≠ 0.

Look for these in your work

  • Shift a domain break
  • Do not erase an original undefined input

Common mistakes

  • f(x + 3) shifts left by 3.
  • A vertical multiplier changes all output heights, not input coordinates.
Recall without your notes

What can you explain now?

Which way does f(x + 2) shift the original graph?

Compare with the explanation

Left by 2 input units.

The same original input is reached two units earlier.

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Make it stick

Which way does f(x + 2) shift the original graph?

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 (US)

Course scope and assessment

191 original lessons provide 573 guided, independent and application tasks. All 142 tier entries have linked teaching foundations: 55 Core and 87 Extended entries across 87 unique numbered sections in ten areas. Extended includes Core, so the tier-entry count is not a percentage of teaching time or student mastery. This is not full programme completion: qualified mathematics-teacher review, cumulative and exam-length assessment, and reliable drawn-work review remain unfinished. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.

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