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.