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.