Mathematics Advanced • Year 11 • Module 1 • Lesson 11

Dilations of Functions

Build procedural fluency in vertical (af(x)) and horizontal (f(bx)) dilations — naming, factor calculation, coordinate prediction.

Build · Skill Drill

1. Quick recall — the dilation rules

Answer each question in the space provided. 1 mark each

Q1.1 Complete the point mappings:

y = af(x): (x, y) ↦ ( ____, ____ ).   Dilation type: ____________ by factor ____ from the ____-axis.

y = f(bx): (x, y) ↦ ( ____, ____ ).   Dilation type: ____________ by factor ____ from the ____-axis.

Q1.2 The horizontal trap. For y = f(3x), the dilation factor is ____, so the graph is (stretched / compressed) toward the y-axis. For y = f(x/3), the dilation factor is ____, so the graph is (stretched / compressed). Circle one in each.

Q1.3 Effect on intercepts. Tick T or F:

(a) Vertical dilation y = af(x) leaves x-intercepts unchanged.   T / F

(b) Horizontal dilation y = f(bx) leaves y-intercepts unchanged.   T / F

Stuck? Revisit lesson § Formula Reference and § Effect on Key Features.

2. Worked example — image of (2, 3) and (6, −1) under y = 2 f(x/3)

Follow each line. The reason is given on the right.

Problem. The graph of y = f(x) passes through (2, 3) and (6, −1). Find the image of each point under y = 2 f(x/3).

Step 1 — Identify both transformations.

2 outside ⇒ vertical dilation by factor 2 (y → 2y).

x/3 inside ⇒ horizontal dilation by factor 3 (x → 3x).

Reason: f(x/3) means the input is divided by 3, so to get the same value of f we need to feed in 3x — meaning every x-coordinate is multiplied by 3.

Step 2 — Apply the horizontal dilation (×3 on x).

(2, 3) → (6, 3)    (6, −1) → (18, −1)

Step 3 — Apply the vertical dilation (×2 on y).

(6, 3) → (6, 6)    (18, −1) → (18, −2)

Conclusion. Images: (6, 6) and (18, −2).

3. Faded example — dilate y = x² to obtain y = (2x)²

Compare y = 2x² (a vertical dilation) with y = (2x)² (a horizontal dilation). Fill in the blanks. 4 marks

Step 1 — Vertical dilation: y = 2x² is y = ____ · f(x) where f(x) = x². The dilation is ____________ by factor ____ from the ____-axis.

Image of (1, 1) on y = x²: y-coordinate × 2 ⇒ ( ____, ____ ).

Step 2 — Horizontal dilation: y = (2x)² is y = f( ____ · x) where f(x) = x². The dilation is ____________ by factor ____ from the ____-axis.

Image of (1, 1) on y = x²: x-coordinate × (1/____) ⇒ ( ____, ____ ).

Step 3 — Why do both look "narrower"? Expand (2x)² = ____ · x². So y = (2x)² is algebraically the same as y = ____ x², which is also a vertical dilation by factor ____. This explains why two different-looking dilations give the same parabola for f(x) = x².

Conclusion. For f(x) = x², the horizontal dilation y = f(2x) is identical to the vertical dilation y = 4 f(x). This coincidence is specific to ________ functions.

Stuck? Revisit lesson § Vertical and Horizontal Dilations.

4. Graduated practice

Foundation — name the dilation (4 questions)

QEquationType (V/H)Factor
4.1 1y = 4f(x)
4.2 1y = f(3x)
4.3 1y = f(x/2)
4.4 1y = (1/3) f(x)

Standard — apply to coordinates and equations (6 questions)

For Q4.5–4.7 the original point on y = f(x) is (4, 2). For Q4.8–4.10 write the equation after the named dilation, simplifying.

4.5 Image of (4, 2) on y = 3 f(x) is ( ____, ____ ).    1 mark

4.6 Image of (4, 2) on y = f(2x) is ( ____, ____ ).    1 mark

4.7 Image of (4, 2) on y = 2 f(x/4) is ( ____, ____ ).    2 marks

4.8 f(x) = x³ − x. Write the equation after vertical dilation by factor 3 (i.e. y = 3 f(x)), simplified.    2 marks

4.9 f(x) = √x. Write the equation after horizontal dilation by factor 1/2 (i.e. y = f(2x)), simplified. State the new domain.    2 marks

4.10 f(x) = sin x. Write y = sin(x/2). What is the period of this dilated graph (recall: sin x has period 2π)?    2 marks

Extension — link to features (2 questions)

4.11 The graph of y = f(x) has x-intercept at x = 6 and y-intercept at y = 4. (a) After y = 3 f(x), what are the new intercepts? (b) After y = f(x/2), what are the new intercepts?    3 marks

4.12 For f(x) = x, prove algebraically that y = 5 f(x) and y = f(5x) produce the same equation. Then explain in one line why this equivalence fails for f(x) = x² (give the two resulting equations side by side).    3 marks

Stuck on 4.12? Compute 5 · x and 5 · x separately, then 5 · x² and (5x)² — and observe the difference.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Point mappings

y = af(x): (x, y) ↦ (x, ay). Vertical, factor a, from x-axis.   y = f(bx): (x, y) ↦ (x/b, y). Horizontal, factor 1/b, from y-axis.

Q1.2 — Horizontal trap

f(3x): factor 1/3, compressed.   f(x/3): factor 3, stretched.

Q1.3 — Intercepts

(a) T — y = af(x) keeps points where y = 0 fixed.   (b) T — y = f(bx) keeps points where x = 0 fixed.

Q3 — Faded example: y = x² → y = 2x² and y = (2x)²

Step 1: y = 2x² is y = 2·f(x). Vertical, factor 2, from x-axis. (1, 1) ↦ (1, 2).
Step 2: y = (2x)² is y = f(2·x). Horizontal, factor 1/2, from y-axis. (1, 1) ↦ (1/2·1, 1) = (1/2, 1).
Step 3: (2x)² = 4·x², so y = (2x)² is the same as y = 4x² (a vertical dilation by factor 4).
Conclusion: For f(x) = x², y = f(2x) ≡ y = 4 f(x). Coincidence specific to quadratic (homogeneous) functions.

Q4.1–4.4 — Name the dilation

4.1: Vertical, factor 4.   4.2: Horizontal, factor 1/3.   4.3: Horizontal, factor 2.   4.4: Vertical, factor 1/3.

Q4.5–4.7 — Image of (4, 2)

4.5 (y = 3 f(x)): y × 3: (4, 6).   4.6 (y = f(2x)): x ÷ 2: (2, 2).   4.7 (y = 2 f(x/4)): x × 4 then y × 2: (16, 4).

Q4.8 — y = 3 f(x) for f(x) = x³ − x

y = 3(x³ − x) = 3x³ − 3x.

Q4.9 — y = f(2x) for f(x) = √x

y = √(2x). Need 2x ≥ 0 ⇒ x ≥ 0.   Domain: [0, ∞). (The domain has been compressed, but [0, ∞) is invariant under compression toward 0.)

Q4.10 — y = sin(x/2)

y = sin(x/2). The factor is 1/(1/2) = 2 (horizontal stretch by 2), so the period is 2 · 2π = 4π.

Q4.11 — Intercepts under dilation

(a) y = 3 f(x): x-intercept unchanged (still x = 6); y-intercept ×3 ⇒ y = 12.
(b) y = f(x/2): y-intercept unchanged (still y = 4); x-intercept ×2 ⇒ x = 12.

Q4.12 — Equivalence for f(x) = x vs f(x) = x²

For f(x) = x: y = 5 f(x) = 5x. y = f(5x) = 5x. Same equation. ✓
For f(x) = x²: y = 5 f(x) = 5x², but y = f(5x) = (5x)² = 25x². Different — equal only if 5 = 25, which is false. The equivalence fails because for non-linear f the input-scale enters with a power higher than 1.