Mathematics Advanced • Year 11 • Module 1 • Lesson 14
Further Transformations & Synthesis
Practise HSC-style sketches and analysis of |f|, f(|x|), and 1/f, including a structured extended response.
1. Short-answer questions
1.1 The graph of y = f(x) has x-intercepts at x = 1 and x = 5, a maximum at (3, 4), and a y-intercept at (0, −5). Sketch the graph of y = |f(x)|, labelling the images of these key features. 3 marks Band 3-4
1.2 Sketch the graph of y = f(|x|) where f(x) = (x − 2)² − 1. Show all x-intercepts, the y-intercept and any turning points, and explain in one sentence why the resulting graph is symmetric about the y-axis. 4 marks Band 4
1.3 Let g(x) = 1/(x² − 9). (a) State the equations of the vertical asymptotes. (b) Find the coordinates of any turning points. (c) Explain why g has no x-intercepts. (d) State the range of g. 4 marks Band 4-5
Stuck on (d)? Find the max of g between the asymptotes (image of the parent's min), then describe ranges on each side.2. Extended response
2.1 Let f(x) = x² − 2x − 3.
(a) Find the x-intercepts, y-intercept and turning point of f, then sketch y = f(x).
(b) Sketch y = |f(x)| on a separate axis, labelling the cusps and the image of the turning point. State the range of |f(x)|.
(c) Sketch y = 1/f(x) on a separate axis, labelling the vertical asymptotes, the image of the turning point, and at least one invariant point where f = ±1.
(d) Explain in 2–3 lines what the three sketches (f, |f|, 1/f) reveal about how each "further" transformation reshapes the parent. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — finds x-intercepts (factor (x − 3)(x + 1)): x = −1, 3; y-intercept (0, −3); turning point by completing the square: (1, −4).
• 1 mark — correctly drawn upward parabola with all three features labelled.
Part (b) — 2 marks
• 1 mark — reflects the arc between x = −1 and x = 3 above the axis; cusps at (−1, 0) and (3, 0).
• 1 mark — image of turning point (1, −4) becomes local max (1, 4); range [0, ∞).
Part (c) — 3 marks
• 1 mark — VAs at x = −1, 3.
• 1 mark — turning point (1, −4) of f becomes max at (1, −1/4) of 1/f.
• 1 mark — finds an invariant point (e.g. f = −1 at x² − 2x − 3 = −1 ⇒ x² − 2x − 2 = 0 ⇒ x = 1 ± √3) and labels at least one.
Part (d) — 1 mark
• Synthesises: |f| folds below-axis up (range becomes non-negative); 1/f turns zeros into asymptotes and flips max/min; both preserve domain symmetry but change the shape qualitatively.
Your response:
Stuck on (c) invariant? Solve x² − 2x − 3 = ±1 to find points common to f and 1/f.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sketch y = |f(x)| (3 marks)
Sample response. x-intercepts (1, 0) and (5, 0) become cusps (graph touches the axis without crossing). Maximum (3, 4) is above the axis — unchanged at (3, 4). y-intercept (0, −5) is below the axis — reflected to (0, 5). Sketch a curve that touches the x-axis at 1 and 5, has a local maximum at (3, 4) between them, and runs upward outside [1, 5] passing through (0, 5).
Marking notes. 1 mark — cusps at (1, 0) and (5, 0) explicitly named. 1 mark — (0, −5) reflected to (0, 5). 1 mark — (3, 4) unchanged (above axis). Common error: students reflect everything, including the maximum.
1.2 — Sketch y = f(|x|) for f(x) = (x − 2)² − 1 (4 marks)
Sample response. For x ≥ 0: y = (x − 2)² − 1, vertex (2, −1), x-intercepts at (x − 2)² = 1 ⇒ x = 1 or 3, y-intercept (0, 3). Mirror across y-axis: vertex (−2, −1), x-intercepts at −1 and −3. Both halves share the y-intercept (0, 3) as a local maximum. The graph is symmetric about the y-axis because f(|−x|) = f(|x|) for all x — the composition with |x| is an even function by construction.
Marking notes. 1 mark — correct x ≥ 0 sketch with vertex (2, −1) and intercepts (1, 0), (3, 0). 1 mark — correct mirror with vertex (−2, −1) and intercepts (−1, 0), (−3, 0). 1 mark — y-intercept (0, 3) labelled as a local maximum. 1 mark — symmetry justified by f(|−x|) = f(|x|).
1.3 — g(x) = 1/(x² − 9) (4 marks)
Sample response. (a) Zeros of x² − 9 are x = ±3, so VAs at x = 3 and x = −3. (b) Parent f(x) = x² − 9 has min at (0, −9), so g has max at (0, −1/9). (c) g = 0 would require the numerator 1 to be 0, which is impossible — so no x-intercepts. (d) Between the asymptotes g is negative with max −1/9, so g ≤ −1/9. Outside the asymptotes g is positive, approaching 0 from above. Range: g ≤ −1/9 or g > 0.
Marking notes. 1 mark each for (a), (b), (c), (d). For (b), accept (0, −1/9) with brief working (image of min via reciprocal). For (d), full range required (both pieces, not just "y ≠ 0").
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). f(x) = x² − 2x − 3 = (x − 3)(x + 1). x-intercepts at x = −1, 3. y-intercept: f(0) = −3. Completing the square: f(x) = (x − 1)² − 4, so turning point (1, −4) (minimum since the parabola opens upward). [1 mark] Sketch: upward parabola with these three features labelled. [1 mark]
Part (b). The arc between x = −1 and x = 3 dips below the x-axis; reflect it above. Cusps at (−1, 0) and (3, 0). The turning point (1, −4) reflects to (1, 4), now a local maximum. [1 mark cusps + 1 mark turning-point image] Outside [−1, 3] the curve is unchanged (it was already above the axis). Range: [0, ∞).
Part (c). Zeros of f at x = −1, 3 ⇒ VAs x = −1 and x = 3. [1 mark] Min of f at (1, −4) ⇒ max of 1/f at (1, −1/4). [1 mark] Invariant where f = −1: x² − 2x − 3 = −1 ⇒ x² − 2x − 2 = 0 ⇒ x = 1 ± √3. So invariant points (1 ± √3, −1). [1 mark] Between the asymptotes the curve is negative, peaking at (1, −1/4); outside, the curve is positive, approaching y = 0 as |x| → ∞.
Part (d). The three sketches show: f is the parent — an upward parabola crossing the x-axis. |f| folds the negative arc above the axis, producing cusps at the zeros and converting the minimum into a local maximum; the range is forced non-negative. 1/f flips the picture: zeros become vertical asymptotes (the curve is now forbidden from x = −1, 3), and the minimum becomes a local maximum at the reciprocal y-value (−1/4). Together they illustrate that "further" transformations reshape the parent in qualitatively different ways than translations or dilations — they change which points are present at all. [1 mark synthesis]
Total: 8/8. ▮
Band descriptors for marker.
Band 3: Correctly finds x-intercepts and y-intercept but misses or mis-states the turning point; |f| sketch reflects entire graph (not just the arc); 1/f shows wrong asymptote locations. ≈ 3 marks.
Band 4: All of (a) correct; |f| sketch correct with cusps but does not state range; 1/f shows asymptotes and image of min but no invariant points. ≈ 5 marks.
Band 5: All sketches correct with labels; (c) includes invariant points (1 ± √3, −1); but (d) summary missing or shallow. ≈ 6–7 marks.
Band 6: Full response with all three sketches labelled, invariant points found and labelled, range of |f| explicitly stated, and synthesis paragraph that distinguishes the qualitative effect of |f| from 1/f. 8/8.