Mathematics Advanced • Year 11 • Module 1 • Lesson 13
Sketching & Modelling Transformed Functions
Practise HSC-style sketches with full key-feature labelling and a structured modelling response.
1. Short-answer questions
1.1 The graph of y = f(x) has a turning point at (2, −3) and x-intercepts at x = −1 and x = 5. Find the corresponding turning point and x-intercept images of (−1, 0) and (5, 0) on the graph of y = 2 f(x − 1) + 4. Show all working. 3 marks Band 3-4
1.2 Sketch the graph of y = 2/(x + 1) − 3, clearly showing the asymptotes (dashed, labelled with equations), the x-intercept, the y-intercept and at least one other point. 4 marks Band 4
1.3 The cables of a small suspension bridge form a parabola with its lowest point 4 m above the deck, and rises to 20 m at the towers, which are 60 m apart. Find the equation of the cable in vertex form, taking the origin at the centre of the deck (so the lowest point is on the y-axis). 3 marks Band 4
Stuck on 1.3? Vertex is at (0, 4); use the tower condition (±30, 20) to find a.2. Extended response
2.1 A small parachute company is designing a drop ride. The height of the rider above the ground after t seconds is modelled (during the free-fall portion only) by h(t) = −5(t − 1)² + 45, where h is in metres.
(a) State the parent function and list the four transformations applied.
(b) Find the height at t = 0 (start of the modelled portion) and the time the rider would hit the ground if the chute did not deploy.
(c) Sketch h(t) on the physically meaningful domain, labelling the vertex, the starting point, and the ground-impact time.
(d) For safety the chute must deploy at h = 25 m. Find both algebraic times at which h = 25 and explain which is the physical "deploy" time. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — parent y = t².
• 1 mark — lists all four transformations (vertical dilation factor 5, reflection in t-axis, translation 1 right, translation 45 up).
Part (b) — 2 marks
• 1 mark — h(0) = −5(1) + 45 = 40 m.
• 1 mark — solves h = 0 to get t = 1 + 3 = 4 s (rejecting t = −2).
Part (c) — 2 marks
• 1 mark — downward parabola with vertex (1, 45) labelled.
• 1 mark — starting point (0, 40) and ground impact (4, 0) labelled, sensible domain shown (0 ≤ t ≤ 4).
Part (d) — 2 marks
• 1 mark — solves (t − 1)² = 4 to get t = −1 or t = 3.
• 1 mark — identifies t = 3 s as the physical deploy time (after the peak, on descent) and rejects t = −1.
Your response:
Stuck on (d)? Both algebraic solutions are valid in the model; choose the one that fits the physical timeline (rider falls, then chute deploys after vertex).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Transformed turning point and intercepts (3 marks)
Sample response. a = 2, b = 1, h = 1, k = 4. Turning point (2, −3) → (2/1 + 1, 2·(−3) + 4) = (3, −2). Image of (−1, 0) → (−1/1 + 1, 2·0 + 4) = (0, 4). Image of (5, 0) → (5 + 1, 0 + 4) = (6, 4). Note these images of the original x-intercepts are not x-intercepts of the new graph (because the original y = 0 maps to y = 4, not 0).
Marking notes. 1 mark — correct identification of a, b, h, k. 1 mark — correct turning-point image with point-mapping formula shown. 1 mark — correct images of (−1, 0) and (5, 0). Bonus credit for explicitly noting that these are not new x-intercepts (a common Band-5 omission).
1.2 — Sketch y = 2/(x + 1) − 3 (4 marks)
Sample response (description). Parent y = 1/x. Read a = 2, h = −1, k = −3.
VA: x = 0/1 + (−1) = x = −1 (dashed). HA: y = 2·0 + (−3) = y = −3 (dashed).
x-intercept: 0 = 2/(x + 1) − 3 ⇒ 2/(x + 1) = 3 ⇒ x + 1 = 2/3 ⇒ x = −1/3, plot (−1/3, 0).
y-intercept: y(0) = 2/1 − 3 = −1, plot (0, −1).
Extra anchor: image of (1, 1) is (1/1 + (−1), 2·1 + (−3)) = (0, −1) (this is the y-intercept, so use (−1, −1) on the parent → (−1/1 + (−1), 2·(−1) + (−3)) = (−2, −5) instead).
Sketch two branches: right of x = −1 the curve runs through (−1/3, 0) and (0, −1), approaching y = −3 from above and x = −1 from the right. Left of x = −1 it runs through (−2, −5), approaching y = −3 from below and x = −1 from the left.
Marking notes. 1 mark — both asymptotes drawn and labelled with equations (dashed lines required). 1 mark — x- and y-intercepts both correct with coordinates. 1 mark — both branches drawn with correct orientation (one above-right of intersection of asymptotes, one below-left). 1 mark — one additional anchor point labelled. Common loss: missing the second branch entirely.
1.3 — Bridge cable (3 marks)
Sample response. Vertex at (0, 4) (lowest point on the y-axis, 4 m above deck). Vertex form: y = a x² + 4. Tower condition: at x = ±30, y = 20. 20 = a(30)² + 4 ⇒ 900 a = 16 ⇒ a = 16/900 = 4/225. Equation: y = (4/225) x² + 4 for −30 ≤ x ≤ 30.
Marking notes. 1 mark — correct vertex form with vertex (0, 4). 1 mark — correct substitution of tower condition. 1 mark — correct a (accept a = 16/900, 4/225, or decimal ≈ 0.0178). Common slip: using x = 60 instead of 30 (full span vs. half-span).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). The model h(t) = −5(t − 1)² + 45 has parent function y = t². [1 mark] Reading a = −5, h = 1, k = 45: (1) vertical dilation by factor 5 from the t-axis, (2) reflection in the t-axis (because a < 0), (3) translation 1 unit right, (4) translation 45 units up. [1 mark]
Part (b). Start of modelled portion: h(0) = −5(0 − 1)² + 45 = −5 + 45 = 40 m. [1 mark]
Ground impact: −5(t − 1)² + 45 = 0 ⇒ (t − 1)² = 9 ⇒ t − 1 = ±3 ⇒ t = −2 or t = 4. Take t > 0: t = 4 s. [1 mark]
Part (c). Sketch a downward parabola with vertex (1, 45), starting at (0, 40) (clearly marked), passing through the vertex, and meeting the t-axis at (4, 0). Domain shown 0 ≤ t ≤ 4. [2 marks — vertex labelled + start and end labelled]
Part (d). Set h = 25: −5(t − 1)² + 45 = 25 ⇒ −5(t − 1)² = −20 ⇒ (t − 1)² = 4 ⇒ t − 1 = ±2 ⇒ t = −1 or t = 3. [1 mark] Rejecting t = −1 (negative time, pre-launch), the chute deploys at t = 3 s — after the peak at t = 1, on the way down, which matches the physical scenario of a free-fall ride descending past the deploy altitude. [1 mark]
Total: 8/8. ▮
Band descriptors for marker.
Band 3: Parent identified, transformations partly listed (often missing reflection); h(0) correct but ground impact misses the negative root or omits the rejection statement. ≈ 3−4 marks.
Band 4: All of (a) and (b) correct; sketch present but missing one of vertex/start/end labels; (d) finds both algebraic times but does not justify the physical choice. ≈ 5−6 marks.
Band 5: All parts correct, sketch fully labelled with all three key points and domain, (d) justifies physical choice but without referencing the timeline (peak then descent). ≈ 6−7 marks.
Band 6: Full response with parent and all transformations named precisely, sketch with vertex (1, 45), start (0, 40), impact (4, 0), domain restricted, and (d) explicitly rejecting t = −1 with the words "after the peak, on descent". 8/8.