Mathematics Advanced • Year 11 • Module 1 • Lesson 9
Translations of Functions
Practise HSC-style writing on translations — describe, predict, write equations, and produce a structured proof that translations preserve features.
1. Short-answer questions
1.1 Explain in 1–2 sentences why y = f(x − 3) represents a shift 3 units to the right, even though the number −3 appears inside the function. Use the idea of inputs and outputs in your explanation. 2 marks Band 3
1.2 The vertex of y = f(x) is at (−1, 2). Write the coordinates of the vertex after each transformation:
(a) y = f(x) + 4.
(b) y = f(x − 3).
(c) y = f(x + 2) − 5. 3 marks Band 3
1.3 A parabola is translated so that its vertex moves from (0, 0) to (4, −2).
(a) Write the equation of the translated parabola in vertex form.
(b) State the domain and range of the translated parabola in interval notation.
(c) Explain in 1–2 sentences why the domain did or did not change, and why the range did or did not change. 4 marks Band 4
2. Extended response
2.1 A parabola has equation y = (x + 1)² − 4.
(a) State the vertex by reading off the vertex form.
(b) State the domain and range in interval notation.
(c) Find the x-intercepts in exact form.
(d) The graph of y = x² is translated by some vector (h, k) to obtain y = (x + 1)² − 4. Write (h, k), describe the translation in words, and prove that this translation maps the y-intercept of y = x² (which is at (0, 0)) onto the y-intercept of y = (x + 1)² − 4, by direct calculation.
(e) Hence describe how the axis of symmetry, vertex, and roots of the parent y = x² (axis x = 0; vertex (0, 0); single root x = 0) each transform under this translation, naming the corresponding feature of y = (x + 1)² − 4. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — vertex (−1, −4) (correct sign on h, correct k).
Part (b) — 1 mark
• 1 mark — domain (−∞, ∞); range [−4, ∞).
Part (c) — 1 mark
• 1 mark — sets (x + 1)² − 4 = 0, gets x + 1 = ±2, so x = 1 and x = −3 (or in factorised form (x − 1)(x + 3) = 0).
Part (d) — 2 marks
• 1 mark — identifies (h, k) = (−1, −4); describes as "1 unit left and 4 units down".
• 1 mark — shows that (0, 0) on the parent ↦ (−1, −4) on the translated graph (by direct calculation: (−1 + 1)² − 4 = −4 ✓) and shows that the y-intercept of the translated graph (substitute x = 0) is at (0, (0 + 1)² − 4) = (0, −3), making the explicit point that the y-intercept is NOT in general the image of the parent's y-intercept — these are two different points unless the translation has h = 0.
Part (e) — 2 marks
• 1 mark — axis of symmetry x = 0 ↦ x = −1; vertex (0, 0) ↦ (−1, −4) (each with the (h, k) = (−1, −4) translation).
• 1 mark — single root x = 0 ↦ two distinct roots x = 1 and x = −3, noting that the number of roots changes because the vertical shift k = −4 pushes the parabola below the x-axis, creating two real intersections (a feature that does not transform by the simple translation rule).
Your response:
Stuck on (e)? Most features (axis, vertex, individual points) transform by adding (h, k). Number-of-roots is NOT such a feature — discuss why.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Why f(x − 3) shifts right (2 marks)
Sample response. To get the same output from f(x − 3) that we would get from f(0), the input x must satisfy x − 3 = 0, i.e. x = 3. So the point that produced y-value f(0) at x = 0 in the original graph now produces the same y-value at x = 3 in the translated graph — the entire graph has moved 3 units in the positive x direction (right).
Marking notes. 1 mark — uses the input/output framing (or equivalent "set the bracket = 0" reasoning). 1 mark — explicit conclusion "shift right by 3", correctly addressing the apparent sign paradox. Common error: stating "right by 3 because the rule says so" without explanation — loses the reasoning mark.
1.2 — Vertex images (3 marks)
Sample response. Starting vertex (−1, 2). (a) y = f(x) + 4: translation 4 up ⇒ vertex (−1, 6). (b) y = f(x − 3): translation 3 right ⇒ vertex (2, 2). (c) y = f(x + 2) − 5: translation 2 left and 5 down ⇒ vertex (−3, −3).
Marking notes. 1 mark per correct vertex. Common error in (c): writing (1, −3) by reading "+2" as "right 2".
1.3 — Translated parabola (4 marks)
Sample response. (a) Translation 4 right, 2 down: y = (x − 4)² − 2.
(b) Domain: (−∞, ∞). Range: [−2, ∞).
(c) The domain did not change because translating left or right does not introduce any new x-restrictions on a polynomial. The range changed because translating 2 units down lowers the minimum y-value from 0 (the original vertex height) to −2 (the new vertex height).
Marking notes. (a) 1 mark — correct vertex-form equation with right signs. (b) 1 mark — both domain and range correct (interval notation). (c) 2 marks — one for the unchanged-domain reasoning, one for the changed-range reasoning. Accept "horizontal translations affect domain but here domain was the whole real line so unchanged; vertical translations affect range, shifting the minimum by k = −2".
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Comparing y = (x − h)² + k with y = (x + 1)² − 4 gives h = −1 and k = −4. Vertex: (−1, −4). [1 mark.]
Part (b). Parabola defined for all real x: domain (−∞, ∞). Opens upward with minimum at the vertex: range [−4, ∞). [1 mark.]
Part (c). (x + 1)² − 4 = 0 ⇒ (x + 1)² = 4 ⇒ x + 1 = ±2 ⇒ x = 1 or x = −3. [1 mark.]
Part (d). The translation vector is (h, k) = (−1, −4): 1 unit left and 4 units down. [1 mark.]
Image of the parent's y-intercept (0, 0): apply the translation, getting (0 + (−1), 0 + (−4)) = (−1, −4). Verify on the translated equation: (−1 + 1)² − 4 = 0 − 4 = −4. ✓ So (0, 0) ↦ (−1, −4) (which is the vertex, not the new y-intercept).
Y-intercept of the translated graph: substitute x = 0: y = (0 + 1)² − 4 = 1 − 4 = −3, so the new y-intercept is (0, −3). Note: (0, −3) is NOT the image of the parent's (0, 0) under the translation — it is the (different) point where the translated graph crosses x = 0. [1 mark — explicit verification of the image AND the deliberate observation that the y-intercept does not transform like a parent point.]
Part (e). Axis of symmetry: parent x = 0 ↦ shifted by h = −1 ⇒ x = −1 (the new axis). ✓ Vertex: (0, 0) ↦ (−1, −4). ✓ [1 mark.]
Roots: the parent y = x² has a single (repeated) root at x = 0 (the parabola sits on the x-axis). The translated parabola has two distinct roots x = 1 and x = −3, from part (c). This is NOT a point-translation rule failure: the vertical shift k = −4 pushes the parabola down so that it now intersects the x-axis in two places, creating roots that did not exist in the parent. The number of real roots is therefore not a feature that transforms by simply adding (h, k); it can change when k is large enough to shift the vertex across the x-axis. [1 mark — both correct image roots AND the principled explanation that "number of roots" is not invariant under translation.] ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Reads off the vertex (a), gives domain/range (b), and finds the roots (c), but does not engage with the conceptual distinction in (d) between "image of parent point" and "y-intercept of translated graph". ≈ 3 marks.
Band 4: Adds the translation vector and verification in (d), but skips part (e) or only restates the rule without addressing the change in number of roots. ≈ 4-5 marks.
Band 5: Completes (e) for the axis and vertex but is unclear on roots — treats the single ↦ two roots transition as either an error or as point-translation. ≈ 5-6 marks.
Band 6: Full response — (d) explicitly distinguishes image of parent y-intercept from y-intercept of translated graph; (e) handles axis, vertex, and gives a principled discussion that number-of-roots is not translation-invariant when k crosses a critical value. 7/7.