Mathematics Advanced • Year 11 • Module 1 • Lesson 10
Reflections of Functions
Apply reflection rules to game-sprite design, mirrored data, symmetry diagnostics and a coastline survey problem.
Problem 1 — Game sprite mirrored to face the other way
A game character's silhouette is described by y = f(x) = (1/4)(x − 6)(8 − x), defined on 0 ≤ x ≤ 8. When the character turns around, the artist re-uses the same silhouette but mirrored across the vertical centre line of the screen (the y-axis).
Set up: What are we solving for?
(i) Write the equation of the mirrored sprite y = f(−x), in factored form. State its new domain. 3 marks
(ii) The "feet" of the original sprite (where y = 0) are at x = 6 and x = 8. Where are the feet of the mirrored sprite? 1 mark
(iii) Explain in one sentence why the y-axis reflection f(−x) — rather than the x-axis reflection −f(x) — is the correct transformation for "turning around". 1 mark
Stuck on (i)? Replace each x by (−x) inside the factored form, then tidy the signs.Problem 2 — Mirrored data table (modelling)
An experiment measures temperature T as a function of distance x (in m) from a hot wire. The measurements are taken on one side only (x > 0) and the lab assumes symmetry about the wire (x = 0). The recorded data is shown.
| x (m) | T (°C) |
|---|---|
| 1 | 72 |
| 2 | 48 |
| 3 | 30 |
Set up: What are we solving for?
(i) Assuming T(x) is an even function of x (so its graph is symmetric about the y-axis), write the predicted T values at x = −1, −2, −3 and justify in one sentence. 2 marks
(ii) A student proposes the model T(x) = 72/x². Confirm this is even by computing T(−x), and confirm it produces T(1) = 72. 2 marks
(iii) A different student proposes T(x) = 72/x. Test whether this model is even, odd, or neither, and explain in one sentence why the assumption of left-right symmetry rules it out. 2 marks
Problem 3 — Reflection diagnostics for a quadratic
A parabola is given by f(x) = x² − 4x + 3, with vertex at (2, −1) (from completing the square).
Set up: What are we solving for?
(i) Write the equation of y = f(−x) in expanded form, and state the vertex of the reflected parabola. 3 marks
(ii) Write the equation of y = −f(x) in expanded form, and state the vertex of the reflected parabola. 3 marks
(iii) Without further calculation, write the equation of y = −f(−x), and state its vertex. Hence summarise the (x, y) mapping for each of the three reflections. 2 marks
Stuck on (iii)? Apply both single reflections in succession — flip x then flip y, or use −f(−x) directly.Problem 4 — Identify the transformation from a pair of equations
For each pair (original → transformed), state the single reflection involved. If the relationship is not a single reflection (e.g. it includes translation), say so and describe what additional move is needed.
Set up: What are we solving for?
(i) y = x² + 5 → y = −x² − 5. 1 mark
(ii) y = (x − 3)² → y = (x + 3)². 2 marks
(iii) y = √x → y = −√(−x). 2 marks
(iv) y = x³ → y = x³ + 4. 1 mark
Stuck on (iii)? Apply −f(−x) to f(x) = √x and check.Problem 5 — Coastline survey (sign-convention reasoning)
A surveyor maps a coastline by measuring the height h (above sea level) of the cliff as a function of horizontal distance d east of a reference post. She models the data with h(d) = 12 − 0.5 d² for −4 ≤ d ≤ 4 m. Her colleague repeats the measurements but defines distance as "metres west of the post" — i.e. with the positive direction reversed.
Set up: What are we solving for?
(i) Write the colleague's equation hW(d) by reflecting h(d) in the y-axis (i.e. d → −d), and simplify. 2 marks
(ii) Compare hW(d) with h(d). Explain in one sentence why an even function is "blind" to a reversal of the positive direction — and check by computing h(2) and hW(−2). 2 marks
(iii) Suppose, instead, the colleague is interested in the depth below sea level rather than the height, i.e. she wants the graph of y = −h(d). Write this equation and state the new range. 2 marks
Stuck on (ii)? Try substituting d = −2 into h(d) and then into hW(d).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Game sprite mirrored
Set up. Apply f(−x) to a factored profile, then read off the mirrored "feet" (zeros) and explain the rationale.
(i) f(−x) = (1/4)((−x) − 6)(8 − (−x)) = (1/4)(−x − 6)(8 + x) = (1/4)(−1)(x + 6)(8 + x) = −(1/4)(x + 6)(x + 8). Domain shifts to 0 ≤ −x ≤ 8 ⇒ −8 ≤ x ≤ 0.
(ii) Feet of original: x = 6, 8 ⇒ feet of mirrored: x = −6 and x = −8 (each x-coordinate negated).
(iii) "Turning around" means swapping left and right, which is a left-right mirror — exactly what reflection in the y-axis (f(−x)) does. The x-axis reflection (−f(x)) would flip the sprite upside down, not turn it around.
Problem 2 — Mirrored data
Set up. Use even-function symmetry to predict missing left-side values, and test two candidate models for consistency with that symmetry.
(i) T(−1) = 72 °C; T(−2) = 48 °C; T(−3) = 30 °C. Justification: an even function satisfies T(−x) = T(x), so values at −x match values at +x.
(ii) T(−x) = 72/(−x)² = 72/x² = T(x) ✓ even. T(1) = 72/1 = 72 °C ✓.
(iii) T(−x) = 72/(−x) = −72/x, which is neither T(x) nor +T(x) — it is −T(x), so 72/x is odd. The odd-function symmetry would give T(−1) = −72 °C, contradicting the assumed mirror symmetry of the temperature data (you cannot have negative absolute Kelvin or a negative-relative offset in this model).
Problem 3 — Reflection diagnostics for a quadratic
Set up. Compute both reflections of a quadratic and track the vertex under each.
(i) y = f(−x) = (−x)² − 4(−x) + 3 = x² + 4x + 3. Reflected vertex: (x, y) ↦ (−x, y), so (2, −1) ↦ (−2, −1). (Check: completing the square on x² + 4x + 3 = (x + 2)² − 1 confirms vertex at (−2, −1).)
(ii) y = −f(x) = −(x² − 4x + 3) = −x² + 4x − 3. Reflected vertex: (x, y) ↦ (x, −y), so (2, −1) ↦ (2, 1).
(iii) y = −f(−x) = −(x² + 4x + 3) = −x² − 4x − 3. Vertex (x, y) ↦ (−x, −y), so (2, −1) ↦ (−2, 1). Summary: −f(x) flips y only; f(−x) flips x only; −f(−x) flips both.
Problem 4 — Identify the transformation
Set up. Recognise which single reflection (if any) connects each pair, with extra moves named where needed.
(i) y = x² + 5 → y = −(x² + 5) = −x² − 5: reflection in the x-axis (a single reflection — every y becomes −y).
(ii) y = (x − 3)² → y = (x + 3)² = (−x − 3)² = ((−(x − (−3))))² i.e. replacing x by −x in (x − 3)² gives (−x − 3)² = (x + 3)². So this is a reflection in the y-axis (a single reflection — the vertex (3, 0) maps to (−3, 0)).
(iii) y = √x → y = −√(−x): apply −f(−x). This is reflection in both axes (equivalently, 180° rotation about the origin). The original domain [0, ∞) maps to (−∞, 0] and the original non-negative outputs map to non-positive outputs.
(iv) y = x³ → y = x³ + 4: not a reflection — this is a translation 4 units up.
Problem 5 — Coastline survey
Set up. Reflect a measurement model in the y-axis (sign-convention swap) and check that the answer is the same when the model is already even; then handle a separate x-axis reflection for "depth below sea level".
(i) hW(d) = h(−d) = 12 − 0.5(−d)² = 12 − 0.5 d² (identical to h(d)).
(ii) h is even (only even powers of d), so h(−d) = h(d) for every d — reversing the positive direction has no effect on the height profile. Check: h(2) = 12 − 0.5(4) = 10 and hW(−2) = 12 − 0.5(4) = 10 ✓.
(iii) y = −h(d) = −12 + 0.5 d². Original range was [4, 12] (since at d = ±4, h = 12 − 8 = 4); negating flips the range to [−12, −4] (depths between 4 m and 12 m below sea level).