Mathematics Advanced • Year 11 • Module 1 • Lesson 11
Dilations of Functions
Practise HSC-style writing on dilations, describe, compute, and produce a structured analysis of when vertical and horizontal dilations coincide.
1. Short-answer questions
1.1 Explain in 1–2 sentences why y = f(2x) represents a horizontal compression by factor 1/2, not a stretch by factor 2. Use the idea of inputs and outputs in your explanation. 2 marks Band 3
1.2 The graph of y = f(x) passes through (1, 2), (3, 5), and (6, −1). Find the corresponding points on:
(a) y = 2 f(x). (b) y = f(3x). (c) y = 2 f(3x). 4 marks Band 4
1.3 The graph of y = sin x has period 2π and amplitude 1. The graph of y = 4 sin(3x) is a dilated version.
(a) State the amplitude and period of y = 4 sin(3x).
(b) Hence find the x-coordinate of the first positive maximum of y = 4 sin(3x). 3 marks Band 4-5
2. Extended response
2.1 A student conjectures: "For any function f, applying a vertical dilation by factor k gives the same graph as applying a horizontal dilation by factor 1/k."
(a) Test the conjecture for f(x) = x by writing each dilated equation explicitly and comparing.
(b) Test the conjecture for f(x) = x² by writing each dilated equation explicitly and comparing. State whether the two graphs are the same and, if not, give the relationship between their leading coefficients in terms of k.
(c) Show that for a general power function f(x) = xⁿ (n ≥ 1), a vertical dilation by factor k coincides with a horizontal dilation by factor 1/k iff k = kⁿ. Hence find the values of n for which the student's conjecture is true for all k > 0.
(d) Generalise: for which classes of functions is the conjecture true? Justify in 1–2 sentences. 7 marks Band 5-6
Explicit marking criteria
Part (a), 1 mark
• 1 mark Vertical dilation by k: y = kf(x) = kx. Horizontal dilation by 1/k: y = f(kx) = kx. Both give y = kx.
Part (b), 2 marks
• 1 mark vertical: y = k x²; horizontal (factor 1/k): y = f(kx) = (kx)² = k² x².
• 1 mark compares leading coefficients k and k² and gives their ratio k²/k = k; the graphs generally differ, and coincide when k = 1.
Part (c), 3 marks
• 1 mark vertical: y = k xⁿ; horizontal (factor 1/k): y = (kx)ⁿ = kⁿ xⁿ.
• 1 mark equal iff k xⁿ = kⁿ xⁿ for all x, iff k = kⁿ.
• 1 mark for the equation k = kⁿ to hold for all k > 0, we need n = 1. So the conjecture is true for all k only for the linear case f(x) = x (or more generally any linear f through the origin).
Part (d), 1 mark
• 1 mark for stating f(kx) = kf(x) for every x in a domain closed under positive scaling and every k > 0 (positive homogeneity of degree 1). Linear functions through the origin and the nonlinear function f(x) = |x| are examples.
Your response:
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What I'll revisit before next class:
1.1, Why f(2x) compresses (2 marks)
Sample response. For f(2x) to produce the same output as f(1) would, the input x must satisfy 2x = 1, i.e. x = 1/2. So the feature that occurred at x = 1 on y = f(x) now occurs at x = 1/2 on y = f(2x), every x-coordinate is halved, which compresses the graph toward the y-axis by factor 1/2 (rather than stretching by factor 2).
Marking notes. 1 mark, input/output framing (or "set 2x = original x and solve" reasoning). 1 mark, explicit conclusion "compression by factor 1/2". Common error: writing "stretched by 2", the answer must include the words "compressed" or "factor 1/2" to score the second mark.
1.2, Three dilations applied to three points (4 marks)
Sample response.
(a) y = 2 f(x): y × 2. (1, 2) ↦ (1, 4); (3, 5) ↦ (3, 10); (6, −1) ↦ (6, −2).
(b) y = f(3x): x ÷ 3. (1, 2) ↦ (1/3, 2); (3, 5) ↦ (1, 5); (6, −1) ↦ (2, −1).
(c) y = 2 f(3x): x ÷ 3 then y × 2. (1, 2) ↦ (1/3, 4); (3, 5) ↦ (1, 10); (6, −1) ↦ (2, −2).
Marking notes. 1 mark each for parts (a) and (b) (all three points correct). 2 marks for part (c), 1 for correct horizontal step, 1 for correct vertical step. Allow ½ if one point in any part is wrong but the mapping rule is clearly correct.
1.3, y = 4 sin(3x): amplitude, period, first max (3 marks)
Sample response. (a) Amplitude: 4 (outside multiplier). Period: 2π/3 (period of sin(bx) is 2π/b, here b = 3).
(b) sin θ has its first positive maximum at θ = π/2. Set 3x = π/2 ⇒ x = π/6. First positive maximum at x = π/6, with y = 4. So the point is (π/6, 4).
Marking notes. (a) 1 mark, amplitude; 1 mark, period (must be 2π/3, not 2π). (b) 1 mark, correctly sets 3x = π/2 and solves to π/6. Common error: forgetting the 1/3 factor and giving x = π/2.
2.1, Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). For f(x) = x:
Vertical dilation by k: y = k · f(x) = kx.
Horizontal dilation by 1/k: y = f(x ÷ (1/k)) = f(kx) = kx.
Both give the same equation y = kx, so the two dilations produce the same graph. ✓ [1 mark.]
Part (b). For f(x) = x²:
Vertical dilation by k: y = k · x² = k x².
Horizontal dilation by 1/k: y = (kx)² = k² x². [1 mark, both equations correct.]
The graphs generally differ because the leading coefficients are k and k². They coincide when k = k², i.e. k = 1 (k = 0 is excluded). The horizontal dilation by factor 1/k acts like a vertical dilation by factor k²/k = k times the original vertical dilation, so "applying vertical by k" and "applying horizontal by 1/k" differ by a further vertical dilation of factor k. [1 mark, graphs different, ratio of leading coefficients explained.]
Part (c). For f(x) = xⁿ (n ≥ 1):
Vertical dilation by k: y = k xⁿ.
Horizontal dilation by 1/k: y = (kx)ⁿ = kⁿ xⁿ. [1 mark, both general equations.]
Equality of these polynomial functions for all x means equality of coefficients: k = kⁿ. [1 mark, equation k = kⁿ.]
For this to hold for all k > 0 (not just for special k), we need kⁿ⁻¹ = 1 to be true for every positive k. The only n for which this is true is n = 1. (For n ≥ 2, take k = 2: kⁿ⁻¹ = 2ⁿ⁻¹ ≥ 2 ≠ 1.) So the conjecture is true for all k only in the linear case f(x) = x. [1 mark, concludes n = 1.]
Part (d). On a domain closed under positive scaling, the general condition is f(kx) = kf(x) for every x and k > 0: f is positively homogeneous of degree 1. This includes linear functions through the origin and nonlinear functions such as f(x) = |x|, since |kx| = k|x|. [1 mark, general class and example.] ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Completes (a) but for (b) writes only one of the two dilated equations or omits the explicit comparison of leading coefficients. ≈ 2-3 marks.
Band 4: Both (a) and (b) complete with comparison; (c) attempts the general power but does not isolate k = kⁿ or does not deduce n = 1. ≈ 4-5 marks.
Band 5: Full (a)-(b)-(c), with n = 1 identified, but (d) omitted or limited to "only linear functions" with no reasoning. ≈ 5-6 marks.
Band 6: All four parts with explicit equation comparisons, the k = kⁿ ⇒ n = 1 deduction, and a generalisation in (d) that states positive homogeneity of degree 1 and gives a brief justification or example. 7/7.