Mathematics Advanced • Year 11 • Module 1 • Lesson 11
Dilations of Functions
Apply dilations to photo zoom, sound-wave periods, scaled-up parabolas in design, sensor-units conversions, and a reverse-engineering task.
Problem 1 — Pinch-to-zoom on a graphics overlay
An app overlays a graphic whose silhouette is y = f(x), with a known point (3, 4) on it. The user pinches to zoom the image to 200% horizontally and 150% vertically.
Set up: What are we solving for?
(i) Write the equation of the zoomed silhouette in the form y = a · f(x/b), giving the values of a and b. 2 marks
(ii) Find the image of (3, 4) under this dilation. 2 marks
(iii) Explain in one sentence why the horizontal-200% dilation is written with x/2 inside the function (not 2x). 1 mark
Stuck on (iii)? f(x/b) dilates horizontally by factor b — so 200% (= factor 2) means x/2 inside.Problem 2 — Sound wave: period change under horizontal dilation
A pure tone is modelled by y = sin x (with x in radians; period 2π). A second tone is generated by y = sin(3x).
Original: y = sin x. New: y = sin(3x).
Set up: What are we solving for?
(i) Describe the transformation from y = sin x to y = sin(3x) in words (dilation type, factor, axis). 2 marks
(ii) Find the period of y = sin(3x). Hence find its frequency (in cycles per 2π) and explain in one sentence why a compressed graph has a higher frequency. 2 marks
(iii) A third tone has equation y = 4 sin(3x). Does the period change? Does the amplitude change? Justify each in one line. 2 marks
Stuck on (ii)? Period of sin(bx) is 2π/b.Problem 3 — Design parabola scaled up for a feature wall
An artist designs a parabolic motif y = x² on a 1 m grid, with key points (0, 0), (1, 1), (2, 4), (3, 9). The motif is enlarged to fit a feature wall: each x-coordinate is to be doubled, and each y-coordinate is to be tripled.
Set up: What are we solving for?
(i) Write the equation of the enlarged motif as y = a · f(x/b) where a, b are constants and f(x) = x². Simplify to a single polynomial in x. 3 marks
(ii) Find the image of the four key points (0, 0), (1, 1), (2, 4), (3, 9) under the dilation. Verify (numerically) that one of them satisfies your simplified equation in (i). 3 marks
(iii) The motif must fit within a 6 m × 9 m frame. Does the enlarged motif fit if the horizontal extent in the original was from x = 0 to x = 3? Justify. 2 marks
Problem 4 — Sensor unit-conversion as a dilation
A sensor outputs voltage V (in volts) as a function of pressure p (in kilopascals): V(p) = 0.05 p (for 0 ≤ p ≤ 1000). The engineer wants to express V in terms of pressure measured in pascals instead (1 kPa = 1000 Pa).
Set up: What are we solving for?
(i) If pPa is pressure in Pa, then pkPa = pPa/1000. Write Vnew(pPa) by substituting and simplifying. 2 marks
(ii) Describe the transformation from V(p) to Vnew(pPa) as a horizontal dilation, stating the dilation factor and the axis. 2 marks
(iii) Find Vnew(500 000) (i.e. the voltage when the pressure is 500 000 Pa) and check it agrees with V(500), the same physical pressure expressed in kPa. 2 marks
Stuck on (ii)? Going from kPa to Pa stretches every x by a factor of 1000.Problem 5 — Reverse-engineer a dilation from two data points
A student suspects that a measured graph is a single vertical or horizontal dilation of the parent y = √x. She has two data points on the measured graph: (4, 6) and (16, 12).
Set up: What are we solving for?
(i) Test the vertical-dilation hypothesis y = a √x. Substitute (4, 6) to find a, then verify (or rule out) the second point. 2 marks
(ii) Test the horizontal-dilation hypothesis y = √(x/b). Substitute (4, 6) to find b, then verify (or rule out) the second point. 2 marks
(iii) State which hypothesis fits the data and give the final equation. In one sentence, explain how the algebraic structure of √x makes this particular reverse-engineering possible. 2 marks
Stuck? √(x/b) = (1/√b) · √x — so any horizontal dilation of √x is algebraically a vertical dilation as well.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Photo zoom
Set up. Translate a "zoom by 200% horizontal, 150% vertical" instruction into a dilation equation, and apply to a single point.
(i) 200% horizontal = factor 2 from y-axis ⇒ inside becomes x/2. 150% vertical = factor 1.5 from x-axis ⇒ outside multiplier 1.5. Equation: y = 1.5 · f(x/2) (so a = 1.5, b = 2).
(ii) Horizontal dilation × 2: (3, 4) ↦ (6, 4). Vertical dilation × 1.5: (6, 4) ↦ (6, 6). Image: (6, 6).
(iii) A factor-2 stretch away from the y-axis means every x-coordinate doubles, which corresponds to dividing the input by 2 (x/2) — not multiplying by 2 (which would compress by half).
Problem 2 — Sound wave
Set up. Read a horizontal dilation from y = sin(bx), then translate to period and frequency.
(i) Horizontal dilation by factor 1/3 from the y-axis (compression toward the y-axis).
(ii) Period = 2π/3. Frequency (cycles per 2π) = 3 cycles per 2π (vs 1 cycle per 2π for sin x). A compressed graph fits more cycles into the same x-interval, so frequency is higher.
(iii) Period: unchanged, still 2π/3 (the outside factor 4 does not affect period; it only scales the y-values). Amplitude: changes from 1 to 4 (vertical dilation by factor 4 takes the maximum from 1 to 4).
Problem 3 — Feature-wall parabola
Set up. Combine horizontal dilation (×2 on x ⇒ x/2 inside) with vertical dilation (×3 on y ⇒ outside multiplier 3), simplify the resulting equation, and check feasibility.
(i) y = 3 · f(x/2) = 3 (x/2)² = 3 · x²/4 = (3/4) x².
(ii) Image map: (x, y) ↦ (2x, 3y). (0, 0) ↦ (0, 0); (1, 1) ↦ (2, 3); (2, 4) ↦ (4, 12); (3, 9) ↦ (6, 27). Verify (2, 3): (3/4)(2)² = (3/4)(4) = 3 ✓.
(iii) Original horizontal extent [0, 3] m ↦ [0, 6] m, so the enlarged motif spans 6 m horizontally — exactly fits the 6 m horizontal dimension. Vertical extent of the enlarged motif at x = 6 m is y = (3/4)(36) = 27 m — this exceeds the 9 m vertical limit. So the motif does not fit; the artist must either crop the top, restrict the horizontal extent, or use a smaller vertical dilation factor.
Problem 4 — Sensor units
Set up. A change of unit on the input variable is a horizontal dilation.
(i) Vnew(pPa) = V(pPa/1000) = 0.05 · (pPa/1000) = 0.00005 · pPa (or equivalently 5 × 10⁻⁵ · pPa).
(ii) Horizontal dilation by factor 1000 from the y-axis (the x-axis is stretched by 1000: a kPa value of 500 corresponds to a Pa value of 500 000).
(iii) Vnew(500 000) = 0.00005 × 500 000 = 25 V. Check: V(500) = 0.05 × 500 = 25 V. ✓ Same physical pressure, same voltage — units of pressure don't change the underlying physical relationship.
Problem 5 — Reverse-engineer dilation
Set up. Test two single-dilation hypotheses against the same two data points and see which is consistent.
(i) y = a √x. At (4, 6): 6 = a √4 = 2a ⇒ a = 3. So y = 3√x. Check (16, 12): 3√16 = 3 · 4 = 12 ✓. Vertical hypothesis fits.
(ii) y = √(x/b). At (4, 6): 6 = √(4/b) ⇒ 36 = 4/b ⇒ b = 4/36 = 1/9. So y = √(9x) = 3√x. Check (16, 12): √(16 · 9) = √144 = 12 ✓. Horizontal hypothesis also fits.
(iii) Both hypotheses fit, with final equation y = 3√x. This works because √(x/b) = (1/√b) · √x — for a square root, a horizontal dilation by factor b is algebraically identical to a vertical dilation by factor 1/√b. The dilation cannot be uniquely identified as "purely vertical" or "purely horizontal" from data alone — only from the context.