Mathematics Advanced • Year 11 • Module 1 • Lesson 12
Combined Transformations
Practise HSC-style writing on combined transformations — including a structured extended response with marking criteria.
1. Short-answer questions
1.1 Describe in full the sequence of transformations applied to y = f(x) to produce y = −3 f(2x + 6) − 1. Be specific about axes, directions and factors. 3 marks Band 3-4
1.2 The point (1, 2) lies on the graph of y = f(x). Find the corresponding point on the graph of y = −2 f(3x − 6) + 4. Show all working, including the factorisation step. 3 marks Band 4
1.3 A student claims: "y = f(2x − 4) represents a horizontal dilation by factor ½ followed by a translation 4 units to the left." Evaluate this claim. If incorrect, explain the error and state the correct transformations. 3 marks Band 4-5
Stuck on 1.3? Always factor the coefficient of x before reading the translation.2. Extended response
2.1 Let f(x) = x². Consider the transformed function g(x) = −½ f(2(x − 1)) + 3.
(a) Describe all transformations mapping f to g, in order.
(b) Find the equation of g in expanded polynomial form y = Ax² + Bx + C.
(c) State the coordinates of the vertex of g and explain how the standard form makes this immediate.
(d) The point (3, 9) lies on y = f(x). Use the point-mapping formula to find its image on y = g(x), then verify by substituting the image x-value back into your expanded equation from (b). 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies all four transformations: horizontal dilation by ½, translation 1 right, vertical dilation by ½ with reflection in x-axis, translation 3 up.
• 1 mark — correct axes/directions/factors (e.g. names the reflection separately, not "dilation by −½").
Part (b) — 2 marks
• 1 mark — writes g(x) = −½ · (2(x − 1))² + 3 = −½ · 4(x − 1)² + 3 = −2(x − 1)² + 3.
• 1 mark — expands fully to g(x) = −2x² + 4x + 1.
Part (c) — 1 mark
• States vertex (1, 3) and notes that the form y = a(x − h)² + k places (h, k) directly at the vertex.
Part (d) — 2 marks
• 1 mark — computes image with x_new = 3/2 + 1 = 5/2, y_new = (−½)(9) + 3 = −3/2, giving (5/2, −3/2).
• 1 mark — substitutes x = 5/2 into expanded equation and verifies y = −2(5/2)² + 4(5/2) + 1 = −25/2 + 10 + 1 = −3/2 ✓.
Your response:
Stuck on (d) verification? Use exact fractions throughout; converting to decimals will obscure the agreement.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Describe y = −3 f(2x + 6) − 1 (3 marks)
Sample response. Factor inside: 2x + 6 = 2(x + 3) = 2(x − (−3)). So y = −3 f(2(x − (−3))) − 1. The transformations are: (1) horizontal dilation by factor ½ from the y-axis (b = 2), (2) translation 3 units left (h = −3), (3) vertical dilation by factor 3 from the x-axis combined with a reflection in the x-axis (a = −3), and (4) translation 1 unit down (k = −1).
Marking notes. 1 mark — correct factorisation 2(x + 3). 1 mark — identifies all four transformations. 1 mark — correct axes/directions (loses if "reflection in y-axis" or "translation 6 left"). Common error: failing to factor, then claiming "translation 6 left", which scores 0/3 for transformation direction.
1.2 — Image of (1, 2) on y = −2 f(3x − 6) + 4 (3 marks)
Sample response. Factor inside: 3x − 6 = 3(x − 2). So y = −2 f(3(x − 2)) + 4. Read off a = −2, b = 3, h = 2, k = 4.
x_new = x/b + h = 1/3 + 2 = 7/3. y_new = a·y + k = (−2)(2) + 4 = 0. Image: (7/3, 0).
Marking notes. 1 mark — correct factorisation. 1 mark — correct x_new with working shown. 1 mark — correct y_new and final point. A naked answer with no factor step scores 1/3 at most.
1.3 — Critique student claim (3 marks)
Sample response. The student's claim is incorrect. The error is failing to factorise the 2 out of 2x − 4 before reading the translation. Correctly, 2x − 4 = 2(x − 2), so the translation inside the function is 2 units to the right, not 4 units to the left. The correct transformations are: horizontal dilation by factor ½ from the y-axis, then translation 2 units right. There is no reflection (a = 1 and b = 2, both positive) and no vertical change (k = 0).
Marking notes. 1 mark — identifies the claim is incorrect with the specific error named ("did not factor"). 1 mark — shows the factorisation 2(x − 2). 1 mark — states correct transformations (dilation factor ½, right 2). Common slip: students agree the dilation is correct but mis-state the translation as "right 4" (still wrong, because b = 2 means the inside translation is h, not bh).
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — transformations. The equation g(x) = −½ f(2(x − 1)) + 3 has the standard form with a = −½, b = 2, h = 1, k = 3. Reading inside-out: [1 mark — all four identified]
• b = 2: horizontal dilation by factor 1/2 from the y-axis.
• h = 1: translation 1 unit right.
• a = −½: vertical dilation by factor ½ from the x-axis and reflection in the x-axis.
• k = 3: translation 3 units up. [1 mark — specific axes/directions]
Part (b) — expanded polynomial form.
g(x) = −½ (2(x − 1))² + 3 = −½ · 4(x − 1)² + 3 = −2(x − 1)² + 3. [1 mark — correct vertex form]
Expanding: g(x) = −2(x² − 2x + 1) + 3 = −2x² + 4x − 2 + 3 = −2x² + 4x + 1. [1 mark — correct expansion]
Part (c) — vertex. The vertex is at (1, 3). In the form y = a(x − h)² + k, the vertex sits at (h, k), so h = 1 and k = 3 are read directly off the standard form. [1 mark]
Part (d) — image of (3, 9). Using the point-mapping formula:
x_new = x/b + h = 3/2 + 1 = 5/2. y_new = a·y + k = (−½)(9) + 3 = −9/2 + 6/2 = −3/2. So the image is (5/2, −3/2). [1 mark]
Verification by substitution. g(5/2) = −2(5/2)² + 4(5/2) + 1 = −2(25/4) + 10 + 1 = −25/2 + 11 = −25/2 + 22/2 = −3/2. ✓ Agrees with the image y-value. [1 mark]
Total: 7/7. ▮
Band descriptors for marker.
Band 3: Identifies some transformations correctly but misses the reflection or names a wrong direction; expanded form may have algebra slips; vertex stated without justification. ≈ 2−3 marks.
Band 4: All four transformations identified, expansion mostly correct, vertex correct. Point-mapping attempted but no verification or arithmetic slip. ≈ 4−5 marks.
Band 5: All parts correct, point-mapping and verification both shown, but uses approximate decimals (e.g. −1.5 instead of −3/2) so the verification "agreement" is less convincing. ≈ 5−6 marks.
Band 6: Complete proof with exact fractions throughout, explicit invocation of "the form y = a(x − h)² + k places (h, k) at the vertex" in (c), and verification that uses common denominators (not decimal approximations). 7/7.