Mathematics Advanced • Year 11 • Module 1 • Lesson 12
Combined Transformations
Apply the general form y = af(b(x − h)) + k to modelling, game graphics, and point-mapping problems.
Problem 1 — Placing a game character on screen
A 2D game holds a sprite outline as a function y = f(x). The on-screen image is produced by the equation
y_screen = −1.5 f( 0.5 (x − 200) ) + 300
(positive y is downward on screen; coordinates are in pixels).
Set up: What are we solving for?
(i) State a, b, h and k for this transformation, and describe each as a screen action (e.g. "stretched", "flipped"). 3 marks
(ii) The original sprite has a marker dot at (4, 6) on y = f(x). Find its on-screen pixel position using the point-mapping formula. 2 marks
(iii) The engine processes one frame per millisecond. Explain in one line why "applying translation first then scaling" would not place the marker at the same screen position. Give the resulting (wrong) pixel coordinates for the marker if translation were applied before scaling. 3 marks
Stuck? Revisit lesson § Order Matters (Sometimes) and § Transforming Individual Points.Problem 2 — Tide model (modelling)
A harbour engineer represents the standard "ideal" tide profile by y = f(t), where t is hours after midnight and y is in metres above mean sea level. A particular harbour's tide is then modelled by
y = 2 f( ½ (t − 3) ) − 1
Set up: What are we solving for?
(i) Describe the four transformations applied to f, in plain words a council planner could read. 3 marks
(ii) The ideal profile has a high tide of height 1 m at t = 6. Use the point-mapping formula to predict when this harbour reaches its corresponding high tide and how high the water reaches. 2 marks
(iii) The harbour master wants to know whether the horizontal dilation factor ½ "speeds up" or "slows down" the tidal cycle relative to f. Answer in one sentence and justify using the formula. 2 marks
Problem 3 — Instrument calibration (real data)
An audio engineer measures the response curve y = f(x) of a microphone in a lab. To match a new venue, she applies the transformation
y_venue = 3 f( 2x + 8 ) − 4
Set up: What are we solving for?
(i) Rewrite y_venue in the standard form y = a f(b(x − h)) + k by factorising the inside, then state a, b, h, k. 2 marks
(ii) The lab curve passes through (−6, 2). Find the corresponding venue point. 2 marks
(iii) The engineer notices that this transformation does not include a reflection (since a > 0 and the inside coefficient 2 is positive). Explain in one sentence why she can still claim "the new curve has the same shape, just stretched and moved". 2 marks
Stuck on (i)? Lesson § Common Mistakes — factor out the coefficient of x before reading h.Problem 4 — Recovering the equation from a description
An architect describes the geometry of a roof span as: "the basic parabola y = x² is reflected in the x-axis, dilated vertically by factor 2, shifted 3 units right and 4 units up".
Set up: What are we solving for?
(i) Write the equation of the roof span in vertex form. 2 marks
(ii) State the coordinates of the apex of the span and explain how you read them off straight from the equation. 2 marks
(iii) Expand your equation into standard polynomial form y = Ax² + Bx + C. 2 marks
Problem 5 — Does order matter?
A student is convinced that applying transformations in any order gives the same final equation.
Set up: What are we solving for?
(i) Starting from y = f(x) and using f(x) = x, apply: (a) translation 3 right, then vertical dilation by factor 2 | (b) vertical dilation by factor 2, then translation 3 right. Write both final equations and compare. 2 marks
(ii) Repeat (a) and (b) but using horizontal translation 3 right and horizontal dilation by factor ½. Do they agree? 2 marks
(iii) State which combinations of transformations commute and which do not, citing your evidence from (i) and (ii). 2 marks
Stuck? Revisit lesson § Order Matters (Sometimes).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Game character
Set up. We are reading a, b, h, k from the equation and using the point-mapping formula to place a sprite marker on screen.
(i) a = −1.5 (vertical stretch by 1.5 and flip upside down on screen). b = 0.5 (horizontal stretch by factor 1/0.5 = 2). h = 200 (shift right 200 pixels). k = 300 (shift down 300 pixels on screen, because positive y is downward here).
(ii) x_new = x/b + h = 4/0.5 + 200 = 8 + 200 = 208. y_new = a·y + k = (−1.5)(6) + 300 = −9 + 300 = 291. Screen position: (208 px, 291 px).
(iii) Translation and dilation along the same axis do not commute. If translation were applied first (4 → 4 + 200 = 204) and then dilation (204 / 0.5 = 408), the x-coordinate would be 408, not 208. Y would also differ: dilate first then translate gives 6 · (−1.5) + 300 = 291; translate first then dilate gives (6 + 300) · (−1.5) = −459 (wrong). The standard form enforces the correct order via the parentheses.
Problem 2 — Tide model
Set up. We are interpreting a tide transformation in plain English and using it to predict the timing and height of an analogue event.
(i) Inside: ½(t − 3). So: (1) horizontal dilation by factor 2 from the t-axis → tidal cycle takes twice as long; (2) translation 3 hours right (high tide occurs later); (3) vertical dilation by factor 2 → tides are twice as deep/high; (4) translation 1 m down → mean water level is 1 m lower than the ideal.
(ii) a = 2, b = ½, h = 3, k = −1. t_new = t/b + h = 6/(½) + 3 = 12 + 3 = 15 hours after midnight (= 3 pm). y_new = 2(1) + (−1) = 1 m above mean sea level.
(iii) It slows down the cycle. The horizontal stretch factor is 1/b = 1/(½) = 2, so the period of the new tide is twice that of f — events take longer to recur.
Problem 3 — Instrument calibration
Set up. We are factorising the inside, identifying a, b, h, k, and mapping a data point through the transformation.
(i) 2x + 8 = 2(x + 4) = 2(x − (−4)). So y_venue = 3 f(2(x − (−4))) − 4. a = 3, b = 2, h = −4, k = −4.
(ii) x_new = x/b + h = (−6)/2 + (−4) = −3 − 4 = −7. y_new = 3(2) + (−4) = 2. Venue point: (−7, 2).
(iii) With a > 0 and b > 0 there are no reflections, so the curve is not flipped or mirrored — it is only stretched (factors 3 vertical, ½ horizontal) and translated. The geometric shape (peaks-up, troughs-down) is preserved.
Problem 4 — Recovering the equation
Set up. We are building the equation from a verbal description, reading the apex from it, and then expanding.
(i) Reflect: −x². Vertical dilation 2: −2x². Shift 3 right and 4 up: y = −2(x − 3)² + 4.
(ii) Apex at (3, 4). In vertex form y = a(x − h)² + k the apex is (h, k) — read directly off the equation.
(iii) y = −2(x² − 6x + 9) + 4 = −2x² + 12x − 18 + 4 = −2x² + 12x − 14.
Problem 5 — Commutativity test
Set up. We are comparing the final equations produced by swapping the order of two transformations, to see which combinations commute.
(i) Using f(x) = x.
(a) Right 3 first: y = (x − 3). Then vertical dilation by 2: y = 2(x − 3) = 2x − 6.
(b) Vertical dilation first: y = 2x. Then right 3: y = 2(x − 3) = 2x − 6.
Same result — horizontal translation and vertical dilation commute (they act on different axes).
(ii) (a) Right 3 first: y = (x − 3). Then horizontal dilation by ½: replace x with 2x, giving y = (2x − 3).
(b) Horizontal dilation by ½ first: y = 2x. Then right 3: y = 2(x − 3) = 2x − 6.
Different equations — horizontal translation and horizontal dilation do not commute (they act on the same axis).
(iii) Transformations along different axes commute (e.g. horizontal translation with vertical dilation). Transformations along the same axis (e.g. horizontal translation with horizontal dilation) do not commute. That is why we write the standard form y = af(b(x − h)) + k with parentheses: it fixes the order to read "translate, then dilate" inside, and "dilate, then translate" outside.