Mathematics Advanced • Year 11 • Module 1 • Lesson 14
Further Transformations & Synthesis
Build fluency with the three "further" transformations: y = |f(x)|, y = f(|x|), and y = 1/f(x).
1. Quick recall
Answer each part in the space provided. 1 mark each
Q1.1 The transformation y = |f(x)| reflects ____________________ of the graph above the x-axis, leaving the rest unchanged.
Q1.2 The transformation y = f(|x|) keeps the graph for x ____ 0 and replaces the other side with ____________________________________.
Q1.3 For y = 1/f(x), complete the table:
An x-intercept of f(x) (where f = 0) becomes a ________________________ on 1/f(x).
A vertical asymptote of f(x) becomes an ________________________ on 1/f(x).
A maximum of f(x) becomes a ________________________ on 1/f(x).
2. Worked example — sketching y = |x² − 4|
Each step matches the lesson's worked example.
Problem. Sketch y = |x² − 4|, identifying intercepts and turning points.
Step 1 — Sketch the parent f(x) = x² − 4.
Upward parabola, vertex (0, −4), x-intercepts at x = ±2, y-intercept (0, −4).
Step 2 — Reflect the below-axis part.
The arc between x = −2 and x = 2 dips below the x-axis. Reflect that arc above. The vertex (0, −4) becomes (0, +4).
Step 3 — Identify the new features.
x-intercepts at (−2, 0) and (2, 0) become cusps: the graph touches the x-axis but does not cross. New local max at (0, 4). Arms continue upward outside [−2, 2].
Conclusion. Range ⊆ [0, ∞). Curve has a cusp at ±2 and a local maximum at (0, 4).
3. Faded example — sketching y = 1/(x² − 4)
Fill in each blank using the lesson's reciprocal strategy. 4 marks
Step 1 — Find vertical asymptotes (zeros of f).
x² − 4 = 0 ⇒ x = ____ or x = ____. So VAs are x = ____ and x = ____.
Step 2 — Find invariant points (where f = ±1).
x² − 4 = 1 ⇒ x² = ____ ⇒ x = ±____. So invariant at (±____, 1).
x² − 4 = −1 ⇒ x² = ____ ⇒ x = ±____. So invariant at (±____, −1).
Step 3 — Image of the minimum.
f has minimum at (0, −4); reciprocal has __________ at (0, ____).
Step 4 — Horizontal asymptote of the reciprocal.
As x → ±∞, f → ±∞, so 1/f → ____. So HA: y = ____.
Conclusion. Three regions: between asymptotes the curve is negative with a maximum at (0, ____); outside the asymptotes the curve is positive, approaching y = 0.
4. Graduated practice
Foundation — describe the effect (4 questions)
| Q | Transformation | Effect on the graph in one phrase |
|---|---|---|
| 4.1 1 | y = |f(x)| | |
| 4.2 1 | y = f(|x|) | |
| 4.3 1 | y = 1/f(x) | |
| 4.4 1 | y = −f(x) vs y = f(−x) |
Standard — sketch and label (6 questions)
Sketch each (use the boxes for description if no diagram), giving all intercepts, turning points and asymptotes.
4.5 Parent f(x) = x − 2. Sketch y = |f(x)|. State the cusp coordinate. 2 marks
4.6 Parent f(x) = (x − 1)². Sketch y = f(|x|). State the resulting vertices. 2 marks
4.7 Parent f(x) = x² − 1. Sketch y = 1/f(x). State the asymptotes and invariant points where f = ±1. 3 marks
4.8 Parent f(x) = √x. Sketch y = f(|x|). State the domain and any new turning point. 2 marks
4.9 Given y = f(x) has x-intercepts at x = 1, 5, a maximum at (3, 4), and y-intercept (0, −5), state the images on y = |f(x)|: x-intercepts, "cusp" condition, image of (3, 4), image of (0, −5). 2 marks
4.10 Let g(x) = 1/(x² − 9). State the vertical asymptotes, any turning point(s), and explain why g has no x-intercepts. 3 marks
Extension — synthesise (2 questions)
4.11 Sketch y = f(|x|) where f(x) = (x − 2)² − 1. Show x-intercepts, y-intercept, any turning points, and explain why the graph is symmetric about the y-axis. 3 marks
4.12 A function f has a single minimum at (2, −3) and is positive for x < 1 and x > 5 (zeros at 1 and 5). Sketch the reciprocal 1/f, labelling asymptotes, the image of the minimum, and the sign in each region. 3 marks
5. Self-check the easy 3
Tick once you've checked your method on the foundation row.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — y = |f(x)|
Reflects any part below the x-axis above the x-axis; on-axis and above-axis parts are unchanged.
Q1.2 — y = f(|x|)
Keeps the graph for x ≥ 0 and replaces the other side with a mirror image of the right side across the y-axis. Always produces an even graph.
Q1.3 — reciprocal table
x-intercept of f → vertical asymptote of 1/f. VA of f → x-intercept of 1/f. Maximum of f → minimum of 1/f.
Q3 — Faded example y = 1/(x² − 4)
Step 1: x² = 4 ⇒ x = −2 or x = 2; VAs x = −2 and x = 2.
Step 2: x² = 5, x = ±√5; invariant (±√5, 1). x² = 3, x = ±√3; invariant (±√3, −1).
Step 3: f min at (0, −4), reciprocal has maximum at (0, −1/4).
Step 4: 1/f → 0; HA y = 0.
Conclusion: max at (0, −1/4).
Q4.1−4.4 — Effects
4.1 |f(x)|: fold below-axis part above the x-axis (range ≥ 0). 4.2 f(|x|): keep x ≥ 0 side, mirror it across the y-axis — always even. 4.3 1/f(x): zeros → vertical asymptotes; max ↔ min; sign preserved. 4.4 −f(x): reflect in x-axis (negate y). f(−x): reflect in y-axis (negate x). Different unless f is odd.
Q4.5 — y = |x − 2|
Original line has zero at x = 2 and is below the axis for x < 2. Reflecting gives a V-shape with cusp at (2, 0), two arms with slopes ±1.
Q4.6 — y = (|x| − 1)²
For x ≥ 0: y = (x − 1)² with vertex (1, 0). Mirror across y-axis adds vertex (−1, 0). Local max at (0, 1) between them. Resulting "W"-style shape with two minima.
Q4.7 — y = 1/(x² − 1)
VAs at x = ±1. Invariant: f = 1 when x² = 2 ⇒ x = ±√2; f = −1 when x² = 0 ⇒ x = 0, giving (0, −1). Min of f at (0, −1) → max of 1/f at (0, −1). Between −1 and 1 the curve is negative with max at (0, −1); outside the asymptotes the curve is positive, approaching y = 0.
Q4.8 — y = √|x|
For x ≥ 0: same as √x starting at (0, 0). Mirror across y-axis. Domain all real x. New "cusp-like" turning point at (0, 0). Even function.
Q4.9 — |f(x)| images
x-intercepts unchanged at x = 1, 5 (now cusps, graph touches axis without crossing). Maximum (3, 4) above axis: unchanged at (3, 4). y-intercept (0, −5) below axis: reflected to (0, 5).
Q4.10 — g(x) = 1/(x² − 9)
VAs at x = ±3 (zeros of x² − 9). Min of f at (0, −9) → max of g at (0, −1/9). No x-intercepts: 1/(x² − 9) = 0 has no solution because the numerator is the constant 1, which is never 0. Range: y > 0 or y ≤ −1/9.
Q4.11 — y = (|x| − 2)² − 1
For x ≥ 0: y = (x − 2)² − 1, vertex (2, −1), x-intercepts at (x − 2)² = 1 ⇒ x = 1 or 3. y-intercept y(0) = (0 − 2)² − 1 = 3. Mirror across y-axis: vertex (−2, −1), x-intercepts at −1 and −3. Local max at (0, 3) (the y-intercept, common to both halves). Symmetric about the y-axis because |−x| = |x|, so f(|−x|) = f(|x|), making the function even by construction.
Q4.12 — Reciprocal of a function with min at (2, −3), zeros at 1, 5
Reciprocal has VAs at x = 1 and x = 5. Min of f at (2, −3) → max of 1/f at (2, −1/3). Sign matching: f > 0 for x < 1 and x > 5 ⇒ 1/f > 0 there; f < 0 between 1 and 5 (where f passes through its min −3) ⇒ 1/f < 0 there with max at (2, −1/3). HA y = 0 as |x| → ∞.