Mathematics Advanced • Year 11 • Module 1 • Lesson 10

Reflections of Functions

Build procedural fluency in identifying −f(x) vs f(−x), reflecting points, and writing the equation of a reflected graph.

Build · Skill Drill

1. Quick recall — the reflection rules

Answer each question in the space provided. 1 mark each

Q1.1 Complete the point mappings:

y = −f(x) sends (x, y) to ( ____, ____ ). This is a reflection in the ________-axis.

y = f(−x) sends (x, y) to ( ____, ____ ). This is a reflection in the ________-axis.

Q1.2 Where is the negative sign? Match each to its reflection.

(a) Negative outside the function ⇒ reflection in the ________-axis.

(b) Negative inside (with x) ⇒ reflection in the ________-axis.

Q1.3 Complete: y = −f(−x) is equivalent to a ________° rotation about the ________. A function is unchanged by this transformation iff it is ________.

Stuck? Revisit lesson § Formula Reference and § Misconceptions.

2. Worked example — write the equation after reflecting f(x) = x² − 4x + 3 in the y-axis

Follow each line. The reason is given on the right.

Problem. Let f(x) = x² − 4x + 3. Write the equation after reflection in the y-axis, in expanded form.

Step 1 — Apply the f(−x) rule: replace every x with (−x), using brackets.

y = f(−x) = (−x)² − 4(−x) + 3

Reason: replace each x by (−x); brackets prevent sign errors on the −4x term.

Step 2 — Simplify each power separately.

(−x)² = x² and −4(−x) = 4x

∴ y = x² + 4x + 3

Step 3 — Quick check on vertex.

Original vertex (axis x = 2) ↔ reflected vertex (axis x = −2). ✓

Conclusion. Reflected equation: y = x² + 4x + 3.

3. Faded example — write equations for both reflections of f(x) = x³ − 2x

Reflect in the x-axis and then (separately) reflect the original in the y-axis. Compare the two answers. 4 marks

Step 1 — Reflection in the x-axis: multiply the entire function by ____.

y = −f(x) = −(x³ − 2x) = ________________

Step 2 — Reflection in the y-axis: replace every x with ____ (in brackets).

y = f(−x) = ( ____ )³ − 2( ____ ) = ________________

Step 3 — Compare: the two answers are (the same / different) because f is an ________ function. Verify by writing the algebraic test f(−x) = ________ · f(x).

Conclusion. For f(x) = x³ − 2x, both reflections yield ________________ — a hallmark of an ________ function.

Stuck? Revisit lesson § Worked Example 3 — Equation of a Reflected Graph.

4. Graduated practice — identify, reflect, write

Foundation — name the reflection (4 questions)

Q	Equation	Reflection (x-axis / y-axis / both / neither)
4.1 1	y = −f(x)
4.2 1	y = f(−x)
4.3 1	y = −f(−x)
4.4 1	y = f(x) + 2

Standard — apply to coordinates (6 questions)

For Q4.5–4.7 the original point on y = f(x) is (3, 2). For Q4.8–4.10, apply each reflection to f(x) = x² + 3x and write the simplified result.

4.5 Image of (3, 2) on y = −f(x) is ( ____, ____ ). 1 mark

4.6 Image of (3, 2) on y = f(−x) is ( ____, ____ ). 1 mark

4.7 Image of (3, 2) on y = −f(−x) is ( ____, ____ ). 1 mark

4.8 Write y = −f(x) for f(x) = x² + 3x, expanded. 2 marks

4.9 Write y = f(−x) for f(x) = x² + 3x, expanded. 2 marks

4.10 For g(x) = √x, write the equation of the graph after reflection in the y-axis, and state the new domain in interval notation. 2 marks

Extension — link reflections to symmetry (2 questions)

4.11 Let f(x) = 2x⁴ − 5x² + 1. (a) Find f(−x) in simplified form. (b) Hence state whether reflecting f in the y-axis leaves the graph unchanged, and justify in one line. 3 marks

4.12 Let f(x) = x³ − 4x. (a) Find −f(−x) in simplified form. (b) Hence state whether reflecting f in both axes (equivalently, 180° rotation about the origin) leaves the graph unchanged, and name the symmetry type. 3 marks

Stuck on 4.12(b)? An odd function satisfies −f(−x) = f(x).

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.5 (y = −f(x)) I flipped only the y-coordinate, giving (3, −2).

For 4.6 (y = f(−x)) I flipped only the x-coordinate, giving (−3, 2).

For 4.7 (y = −f(−x)) I flipped both coordinates, giving (−3, −2).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Point mappings

y = −f(x): (x, y) ↦ (x, −y). Reflection in the x-axis. y = f(−x): (x, y) ↦ (−x, y). Reflection in the y-axis.

Q1.2 — Where's the negative?

(a) Outside ⇒ x-axis. (b) Inside (with x) ⇒ y-axis.

Q1.3 — Both reflections = rotation

y = −f(−x) is equivalent to a 180° rotation about the origin. A function is unchanged by this iff it is odd (satisfies −f(−x) = f(x)).

Q3 — Faded example: f(x) = x³ − 2x

Step 1: y = −(x³ − 2x) = −x³ + 2x.
Step 2: y = f(−x) = (−x)³ − 2(−x) = −x³ + 2x = −x³ + 2x.
Step 3: The two answers are the same, because f is an odd function. Algebraic test: f(−x) = −1 · f(x).
Conclusion: both reflections yield y = −x³ + 2x — a hallmark of an odd function.

Q4.1–4.4 — Name the reflection

4.1: x-axis. 4.2: y-axis. 4.3: both (≡ 180° rotation about the origin). 4.4: neither — it's a vertical translation (2 up).

Q4.5–4.7 — Image of (3, 2)

4.5: (3, −2). 4.6: (−3, 2). 4.7: (−3, −2).

Q4.8 — y = −f(x) for f(x) = x² + 3x

y = −(x² + 3x) = −x² − 3x.

Q4.9 — y = f(−x) for f(x) = x² + 3x

y = (−x)² + 3(−x) = x² − 3x.

Q4.10 — y-axis reflection of g(x) = √x

g(−x) = √(−x). Need −x ≥ 0 ⇒ x ≤ 0. Domain: (−∞, 0].

Q4.11 — f(x) = 2x⁴ − 5x² + 1

(a) f(−x) = 2(−x)⁴ − 5(−x)² + 1 = 2x⁴ − 5x² + 1 = f(x).
(b) Reflecting in the y-axis leaves the graph unchanged, because f(−x) = f(x) — f is an even function (only even powers of x).

Q4.12 — f(x) = x³ − 4x

(a) −f(−x) = −[(−x)³ − 4(−x)] = −[−x³ + 4x] = x³ − 4x = f(x).
(b) Reflecting in both axes leaves the graph unchanged, because −f(−x) = f(x). f is odd — the symmetry is 180° rotational symmetry about the origin.