Mathematics Advanced • Year 11 • Module 1 • Lesson 9
Translations of Functions
Apply horizontal and vertical translations to real-world models — bridge profiles, cooling curves, asymptote shifts, and inverse-rate reasoning.
Problem 1 — Bridge arch translated to a new site (geometric)
A symmetric parabolic bridge arch has equation y = −0.04 x² + 9, with vertex at (0, 9) m above the deck. To re-use the design at a second site, engineers shift the arch 4 m to the right and lower its highest point by 2 m.
Set up: What are we solving for?
(i) Write the equation of the translated arch in the form y = a(x − h)² + k. 2 marks
(ii) State the new vertex (the new highest point) and the equation of the axis of symmetry. 2 marks
(iii) At what horizontal distances x does the translated arch touch the deck (y = 0)? Give exact values, then to 2 d.p. 3 marks
Stuck on (iii)? Set the translated equation equal to 0 and solve for x — the symmetry of the vertex form does the heavy lifting.Problem 2 — Cooling curve shifted in time (modelling)
A liquid's temperature T (°C) above room temperature follows the model T(t) = 60/(t + 1), where t (minutes) is measured from when the liquid is poured. The experiment is repeated, but this time the thermometer is connected 2 minutes after pouring.
Original: T(t) = 60/(t + 1)
Set up: What are we solving for?
(i) Write the equation of the shifted graph Tnew(t) so that t = 0 now corresponds to the moment the thermometer is connected. 2 marks
(ii) A student answers "Tnew(t) = 60/(t + 3)" (without justification). Explain in 1-2 sentences why this is correct or incorrect, with reference to the inside-the-function shift rule. 2 marks
(iii) Use your answer from (i) to find Tnew(0) and Tnew(4), and interpret each value in one sentence. 2 marks
Problem 3 — Range and asymptotes after translation
The graph of y = 1/x has horizontal asymptote y = 0 and vertical asymptote x = 0. A designer shifts it 3 units right and 2 units down to model the offset of a sensor's response curve.
Set up: What are we solving for?
(i) Write the equation of the translated graph. 1 mark
(ii) State the equations of the two new asymptotes. 2 marks
(iii) State the domain and range of the translated graph in interval notation, and explain in one sentence why the asymptotes act as the boundaries. 3 marks
Stuck? Apply the translation to the asymptotes the same way you do to a key point.Problem 4 — Re-build a graph from three known points
The graph of y = f(x) is known to pass through (0, 1), (2, 5), and (−1, 0).
Set up: What are we solving for?
(i) Find the image of each of the three points under the translation y = f(x − 3) + 4. 3 marks
(ii) Sketch (on the grid in your book) the original three points and their images, and indicate the translation vector with a single arrow. 2 marks
(iii) If the original graph has range [0, ∞), what is the range after this translation? Justify in one sentence. 1 mark
Problem 5 — Reverse-engineer the translation (data analysis)
A scientist has a "parent" curve y = √x. She measures one feature of her translated curve: its leftmost point is at (3, 5). She believes the curve is a horizontal-plus-vertical translation of y = √x (no stretching, no reflection).
Set up: What are we solving for?
(i) Write the equation of the translated curve in the form y = √(x − h) + k. 2 marks
(ii) State the domain of the translated curve in interval notation. 1 mark
(iii) Use the equation from (i) to predict y at x = 7 and at x = 12. 2 marks
Stuck on (i)? The "leftmost point" of y = √x is (0, 0); the same point on the translated graph is (h, k).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bridge arch translated
Set up. Re-write a parabolic profile after a known horizontal and vertical shift, then read off its key features.
(i) Original is y = −0.04 x² + 9. Shift 4 right ⇒ replace x with (x − 4); lower highest point by 2 ⇒ subtract 2 from the constant. Equation: y = −0.04 (x − 4)² + 7.
(ii) New vertex: (h, k) = (4, 7). Axis of symmetry: x = 4.
(iii) Set −0.04 (x − 4)² + 7 = 0 ⇒ 0.04 (x − 4)² = 7 ⇒ (x − 4)² = 175 ⇒ x − 4 = ±√175 ⇒ x = 4 ± 5√7. Numerically: 5√7 ≈ 13.23, so x ≈ 17.23 m or x ≈ −9.23 m.
Problem 2 — Cooling curve shift in time
Set up. Re-zero the time variable by shifting the graph 2 units left along t (so that the new t = 0 corresponds to original t = 2).
(i) If new t = 0 corresponds to original t = 2, then original time = new time + 2. Substitute (t + 2) into the original: Tnew(t) = 60/((t + 2) + 1) = 60/(t + 3).
(ii) Correct. Replacing t with (t + 2) shifts the graph 2 units to the LEFT (the "+ inside" trap acts in the opposite direction), which is exactly what re-zeroing the clock requires: events that happened at original t = 2 now happen at new t = 0.
(iii) Tnew(0) = 60/3 = 20 °C (the moment the thermometer is connected, the liquid is 20 °C above room temperature). Tnew(4) = 60/7 ≈ 8.57 °C above room temperature, 4 minutes after the thermometer was connected.
Problem 3 — Range and asymptotes
Set up. Apply the same translation to both the curve and its asymptotes, then read off domain and range.
(i) y = 1/(x − 3) − 2.
(ii) Vertical asymptote: x = 3 (the original x = 0 shifted 3 right). Horizontal asymptote: y = −2 (the original y = 0 shifted 2 down).
(iii) Domain: (−∞, 3) ∪ (3, ∞). Range: (−∞, −2) ∪ (−2, ∞). The asymptotes act as boundaries because the graph never crosses them: as x → 3 the function blows up to ±∞, and as x → ±∞ the function approaches but never equals −2.
Problem 4 — Re-build from three points
Set up. Apply the same translation vector to every point on the graph.
(i) Translation vector is (h, k) = (3, 4) (3 right, 4 up): (0, 1) → (3, 5). (2, 5) → (5, 9). (−1, 0) → (2, 4).
(ii) Marker discretion — student to sketch original and image points and draw one (3, 4) arrow between any matched pair.
(iii) Translation shifts every y by +4, so range becomes [4, ∞). (Horizontal shift does not affect range.)
Problem 5 — Reverse-engineer translation
Set up. Use the leftmost point of y = √x (which is (0, 0)) as the anchor: it maps to (h, k) under the translation.
(i) The leftmost point of y = √x is at (0, 0). For the translated curve, this anchor moves to (3, 5), so h = 3 and k = 5: y = √(x − 3) + 5.
(ii) Need x − 3 ≥ 0 ⇒ x ≥ 3. Domain: [3, ∞).
(iii) At x = 7: y = √4 + 5 = 7. At x = 12: y = √9 + 5 = 8.