Mathematics Advanced • Year 11 • Module 1 • Lesson 13
Sketching & Modelling Transformed Functions
Apply transformations to build real-world models: bridges, projectiles, profit curves, population decay and tank drainage.
Problem 1 — Suspension-bridge cable (geometric model)
The main cable of a small suspension bridge sags between two towers that are 80 m apart. The cable's lowest point is 5 m above the deck. At the towers (40 m from centre), the cable is 25 m above the deck. The cable is modelled by a parabola in the form
h(x) = a·x² + 5, for −40 ≤ x ≤ 40
Set up: What are we solving for?
(i) Find the value of a using the tower condition h(40) = 25. 2 marks
(ii) Identify the parent function and the four transformations that map y = x² to h(x). 2 marks
(iii) A vertical strut is to be installed 20 m from the centre. Find the height of the cable at that point and explain in one line how the model could be re-shifted to put x = 0 at one of the towers. 3 marks
Stuck? Revisit lesson § Worked Example 3 — Modelling with Transformations.Problem 2 — Vertical projectile (motion model)
A ball is thrown straight up from the edge of a cliff. Its height above the ground (in metres) at time t seconds is modelled by
h(t) = −5(t − 2)² + 25
Set up: What are we solving for?
(i) State the maximum height and the time at which it occurs. Justify by reading the vertex form. 2 marks
(ii) Find the launch height and the time the ball hits the ground. 3 marks
(iii) State the domain and range of h in the context of this problem (i.e. as a physical model, not as an algebraic function). 2 marks
Problem 3 — Revenue curve (business model)
A coffee shop's daily revenue R (in hundreds of dollars) when it sells x hundred cups is modelled by
R(x) = −0.5(x − 4)² + 8, x ≥ 0
Set up: What are we solving for?
(i) Find the maximum revenue and the number of cups that produce it. 2 marks
(ii) Find the break-even quantities (R = 0) and interpret them in context. 2 marks
(iii) The shop projects daily revenue of $500 (R = 5). Show that this is achievable for two different sales quantities and find both. 3 marks
Stuck on (ii)? Set R(x) = 0 and solve the resulting quadratic.Problem 4 — Bacterial growth (exponential-style model)
A bacterial colony grows so that the population P (thousands) after t hours follows the transformed exponential
P(t) = 5·2^(t − 1) + 10
Set up: What are we solving for?
(i) State the parent function and the four transformations applied. 2 marks
(ii) Find the population at t = 0 (study start) and the horizontal asymptote of P(t). Interpret each in context. 3 marks
(iii) Explain why this transformed exponential is appropriate when the parent y = 2^t alone would not be — relating your answer to "the model selector" rules in the lesson. 2 marks
Problem 5 — Tank drainage (rational/square-root model)
A water tank drains so that the height of water (in metres) after t minutes is given by
H(t) = 2√(16 − t) for 0 ≤ t ≤ 16
Set up: What are we solving for?
(i) Rewrite H(t) in the form a√(b(t − h)) + k by factorising the inside. State a, b, h, k and describe the transformations applied to y = √x. 3 marks
(ii) Find the initial height H(0) and the time the tank empties (H = 0). Confirm the domain matches the physical situation. 2 marks
(iii) Sketch a labelled key-feature diagram showing initial point, end point, and the general shape (square-root curve, decreasing in t). Use 4 plotted points only. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bridge cable
Set up. We are fitting a quadratic model to a real geometric situation, then using it for predictions and rephrasing the coordinate system.
(i) 25 = a(40)² + 5 ⇒ 1600a = 20 ⇒ a = 1/80 = 0.0125.
(ii) Parent y = x². Transformations: (1) vertical dilation by factor 1/80 (very flat), (2) k = 5 means translation 5 up. No reflection (a > 0), no horizontal translation, no horizontal dilation. The cable opens upward as expected for a sag.
(iii) h(20) = (1/80)(400) + 5 = 5 + 5 = 10 m. To put x = 0 at one tower, shift right 40 units: replace x with (x − 40), giving h(x) = (1/80)(x − 40)² + 5 with valid domain 0 ≤ x ≤ 80.
Problem 2 — Vertical projectile
Set up. We are reading a quadratic projectile model in vertex form and extracting physical quantities (max height, ground impact, domain).
(i) Vertex form: h = −5(t − 2)² + 25 has vertex at (h, k) = (2, 25), so maximum height 25 m at t = 2 s.
(ii) Launch height: h(0) = −5(4) + 25 = 5 m (the cliff edge). Hits ground when h = 0: −5(t − 2)² + 25 = 0 ⇒ (t − 2)² = 5 ⇒ t = 2 ± √5. Taking t > 0 and t > 2 (after the peak): t = 2 + √5 ≈ 4.24 s.
(iii) Domain: 0 ≤ t ≤ 2 + √5 (ball exists from launch until impact). Range: 0 ≤ h ≤ 25 (cannot go below ground or above peak).
Problem 3 — Revenue curve
Set up. We are using a vertex-form quadratic to find optimum and break-even points, and inverting to solve for a target revenue.
(i) Vertex at (4, 8), so maximum revenue is $800 when 400 cups are sold.
(ii) 0 = −0.5(x − 4)² + 8 ⇒ (x − 4)² = 16 ⇒ x = 4 ± 4, so x = 0 or x = 8 hundred cups. Break-even at 0 cups (no revenue if no sales) and 800 cups (price drops swamp volume gains, in this model).
(iii) 5 = −0.5(x − 4)² + 8 ⇒ (x − 4)² = 6 ⇒ x = 4 ± √6 ≈ 1.55 or 6.45 hundred cups, i.e. about 155 or 645 cups. Two solutions because the parabola is symmetric about x = 4.
Problem 4 — Bacterial growth
Set up. We are identifying transformations on a parent exponential and reading the asymptote as a baseline population.
(i) Parent y = 2^t. Transformations: horizontal translation 1 right (replace t with t − 1), vertical dilation by factor 5 (multiply by 5), vertical translation 10 up (+10). No reflection. So a = 5, h = 1, k = 10.
(ii) P(0) = 5·2^(−1) + 10 = 5/2 + 10 = 12.5 thousand bacteria at study start. Horizontal asymptote of 2^t is y = 0, which transforms to y = 5·0 + 10 = 10 thousand. Interpretation: as t → −∞ (which is not physical here), the population approaches 10 000, suggesting a baseline minimum population the model "sits above" historically.
(iii) The lesson's selector says exponential is appropriate when growth approaches a limit or grows multiplicatively. A plain y = 2^t starts at 1 and is too small for thousands of bacteria; the transformations (vertical dilation by 5 and shift up by 10) calibrate the parent shape to fit the observed numerical scale and baseline. Without the transformations the shape is right but the values are wrong.
Problem 5 — Tank drainage
Set up. We are recognising a square-root model in disguise, factorising the inside to identify a horizontal reflection, and sketching key features.
(i) 16 − t = −(t − 16) = −1·(t − 16). So H(t) = 2√(−1(t − 16)) + 0. a = 2, b = −1, h = 16, k = 0. Transformations of y = √x: reflection in y-axis (b = −1), translation 16 right (h = 16), vertical dilation by factor 2 (a = 2). No vertical translation.
(ii) H(0) = 2√16 = 8 m. H = 0 when 16 − t = 0, i.e. t = 16 min. Domain 0 ≤ t ≤ 16 matches: water height must be non-negative and the model is undefined for t > 16.
(iii) Plot four labelled points: (0, 8), (7, 6), (12, 4), (16, 0). Curve is decreasing and concave down (steepest at t = 16). Label starting point (0, 8) and ending point (16, 0). Shape is a reflected square-root curve "running into" the t-axis at t = 16.