Mathematics Advanced • Year 11 • Module 1 • Lesson 10
Reflections of Functions
Practise HSC-style writing on reflections — identify, compute, and prove the link between reflections and even/odd symmetry.
1. Short-answer questions
1.1 The graph of y = f(x) passes through the points (−2, 1), (0, 3), and (4, −2). Write the coordinates of the corresponding points on:
(a) y = −f(x). (b) y = f(−x). (c) y = −f(−x). 3 marks Band 3
1.2 Let f(x) = x² − 4x + 3.
(a) Write the equation of the graph after reflection in the y-axis.
(b) Simplify your answer from part (a) by expanding any brackets.
(c) Determine whether the reflected graph is the same as the original graph, justifying with reference to the vertex of each. 4 marks Band 3-4
1.3 The point (−5, 8) lies on the graph of y = f(x), where f is an odd function.
(a) State the coordinates of two other points that must lie on the graph, justifying each from the definition of an odd function.
(b) Hence write down the value of −f(−5). 3 marks Band 4
2. Extended response
2.1 A student claims: "Reflecting y = f(x) in the x-axis and then in the y-axis gives back the original graph — the two reflections cancel out."
(a) Write the equation produced after reflecting y = f(x) first in the x-axis and then in the y-axis, in terms of f.
(b) Evaluate the student's claim. Decide whether it is always, sometimes, or never true, and justify in 1–2 sentences.
(c) Identify the special class of functions for which the claim is true, and prove your characterisation.
(d) Provide a single concrete counterexample function f, compute the composite reflection explicitly, and show that the result is not equal to f. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — composite is y = −f(−x), with the order of operations shown clearly (first y → −y, then x → −x).
Part (b) — 1 mark
• 1 mark — selects sometimes with the explanation "the composite −f(−x) equals f(x) only when f is odd; in general it is not equal to f".
Part (c) — 3 marks
• 1 mark — identifies the special class as odd functions.
• 1 mark — proves (⇒): if −f(−x) = f(x), rearrange to f(−x) = −f(x), the definition of odd.
• 1 mark — proves (⇐): if f is odd, then f(−x) = −f(x), so −f(−x) = −(−f(x)) = f(x).
Part (d) — 2 marks
• 1 mark — chooses any non-odd f (e.g. f(x) = x + 1 or f(x) = x²) and substitutes correctly: e.g. −f(−x) = −(−x + 1) = x − 1, OR −f(−x) = −((−x)²) = −x².
• 1 mark — shows that this composite is not equal to f(x), with a concluding sentence that the student's general claim therefore fails (and ideally noting whether the chosen f is even — and explaining why even is not sufficient: for an even function −f(−x) = −f(x), still not equal to f).
Your response:
Stuck on (c) (⇐)? Start from "f odd ⇒ f(−x) = −f(x)" and substitute into −f(−x).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Point images under three reflections (3 marks)
Sample response.
(a) y = −f(x) sends (x, y) ↦ (x, −y): (−2, 1) ↦ (−2, −1); (0, 3) ↦ (0, −3); (4, −2) ↦ (4, 2).
(b) y = f(−x) sends (x, y) ↦ (−x, y): (−2, 1) ↦ (2, 1); (0, 3) ↦ (0, 3); (4, −2) ↦ (−4, −2).
(c) y = −f(−x) sends (x, y) ↦ (−x, −y): (−2, 1) ↦ (2, −1); (0, 3) ↦ (0, −3); (4, −2) ↦ (−4, 2).
Marking notes. 1 mark per part (each part is "all three points correct"). Allow ½ if exactly one point is wrong but the mapping rule is clearly correct. Common error in (b): forgetting that (0, 3) is invariant under the y-axis reflection.
1.2 — y-axis reflection of f(x) = x² − 4x + 3 (4 marks)
Sample response.
(a) y = f(−x) = (−x)² − 4(−x) + 3, i.e. y = (−x)² − 4(−x) + 3.
(b) Expanded: (−x)² = x² and −4(−x) = 4x, so y = x² + 4x + 3.
(c) The reflected graph is not the same as the original. Completing the square: original f(x) = (x − 2)² − 1, vertex (2, −1); reflected (x + 2)² − 1, vertex (−2, −1). The vertices differ (mirrored across the y-axis), so the graphs are different.
Marking notes. (a) 1 mark — substitutes −x with brackets. (b) 1 mark — fully simplified. (c) 2 marks — one for the conclusion "not the same", one for a vertex-based justification. Accept any equivalent justification (e.g. compute f(1) vs f(−1) to show different y-values).
1.3 — Odd function through (−5, 8) (3 marks)
Sample response. (a) Point 1: (5, −8), because odd ⇒ f(−(−5)) = −f(−5) ⇒ f(5) = −8. Point 2: (0, 0), because for an odd function with 0 in its domain, f(0) = f(−0) = −f(0) ⇒ 2f(0) = 0 ⇒ f(0) = 0.
(b) −f(−5) = −(8) = −8 (the y-value of (5, −8) confirms).
Marking notes. (a) 1 mark per point with justification (2 marks total). (b) 1 mark — value with brief working. Accept either the algebraic chain or the geometric "180° rotation about the origin" framing.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Reflecting y = f(x) in the x-axis first gives y = −f(x). Reflecting this in the y-axis (replace x by −x) gives y = −f(−x). [1 mark — order of operations shown clearly.]
Part (b). The student's claim is sometimes true. The composite −f(−x) equals the original f(x) only when f is odd; for a general function the two reflections do not cancel. [1 mark.]
Part (c). The composite returns the original iff f is an odd function.
(⇒). Suppose −f(−x) = f(x) for every x in the domain. Multiplying both sides by −1 gives f(−x) = −f(x), which is the definition of an odd function. [1 mark — class identified.] [1 mark — forward direction.]
(⇐). Suppose f is odd, so f(−x) = −f(x). Then −f(−x) = −(−f(x)) = f(x). Hence the composite reflection returns f, as required. [1 mark — reverse direction.]
Part (d). Take the non-odd function f(x) = x + 1. Then −f(−x) = −((−x) + 1) = −(−x + 1) = x − 1. So the composite reflection produces y = x − 1, while the original was y = x + 1. [1 mark — explicit counterexample with computation.]
Since x + 1 ≠ x − 1 (they differ by 2 for every x), the composite reflection is not the original graph. The student's general claim therefore fails. (Note: choosing an even function such as f(x) = x² would also be a counterexample, because for even f, −f(−x) = −f(x) ≠ f(x) unless f is identically zero — even is not sufficient; only odd is.) [1 mark — concluding sentence + recognition that "even" is not the right characterisation.] ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Writes −f(−x) for (a) and states "false in general" for (b), but does not name the special class in (c) or only gives one direction of the iff. ≈ 2-3 marks.
Band 4: Identifies the class as odd functions and gives one direction of the proof in (c); provides a counterexample in (d) but does not address why "even" is the wrong characterisation. ≈ 4-5 marks.
Band 5: Both directions of the iff in (c); counterexample with computation but without the closing note on the even-vs-odd distinction. ≈ 5-6 marks.
Band 6: Full response with explicit iff in both directions, an explicit non-odd counterexample with computation, and the principled remark distinguishing "even" (insufficient) from "odd" (the exact characterisation). 7/7.