Mathematics Advanced • Year 11 • Module 3 • Lesson 8
Stationary Points
Practise HSC-style writing on stationary points — including a structured "find and classify" extended response.
1. Short-answer questions
1.1 Find and classify the stationary points of f(x) = x³ − 3x² + 4. 4 marks Band 4
1.2 Show that f(x) = x³ has a horizontal point of inflection at x = 0. Justify using a sign table for f′(x), not just by stating f″(0) = 0. 3 marks Band 4
1.3 The curve y = 2x³ + ax² + b has a local minimum at the point (1, −4). Find a and b. 3 marks Band 4
Stuck on 1.3? You have two conditions: y(1) = −4 and y′(1) = 0.2. Extended response
2.1 The curve y = ax³ + bx² + cx + d has a local maximum at (1, 4) and a local minimum at (3, 0).
(a) Write down two equations from the curve passing through the named points, and two from the curve having zero gradient there.
(b) Solve the system to determine a, b, c and d.
(c) Confirm using the second derivative test that (1, 4) is indeed a local maximum and (3, 0) is a local minimum.
(d) Hence sketch the cubic, labelling the y-intercept and any x-intercepts found by solving y = 0. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — two value equations: a + b + c + d = 4 and 27a + 9b + 3c + d = 0.
• 1 mark — two gradient equations from y′(x) = 3ax² + 2bx + c: 3a + 2b + c = 0 and 27a + 6b + c = 0.
Part (b) — 3 marks
• 1 mark — uses subtraction of the gradient equations to isolate b in terms of a (b = −6a).
• 1 mark — substitutes to find c = 9a.
• 1 mark — uses value equations to determine a = 1, hence b = −6, c = 9, d = 0.
Part (c) — 2 marks
• 1 mark — computes y″(1) = 6(1) + 2(−6) = −6 < 0 (local max confirmed).
• 1 mark — computes y″(3) = 6(3) + 2(−6) = 6 > 0 (local min confirmed).
Part (d) — 1 mark
• 1 mark — sketch shows the correct cubic shape with y-intercept (0, 0), local max (1, 4) and local min (3, 0); notes the touching x-intercept at (3, 0) from the repeated root.
Your response:
Stuck on (d)? You should already have y = x³ − 6x² + 9x. Factor as x(x − 3)² to find the intercepts.How did this worksheet feel?
What I'll revisit before next class:
1.1 — f(x) = x³ − 3x² + 4 (4 marks)
Sample response. f′(x) = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2.
f(0) = 4; f(2) = 8 − 12 + 4 = 0.
f″(x) = 6x − 6. f″(0) = −6 < 0 ⇒ local maximum at (0, 4). f″(2) = 6 > 0 ⇒ local minimum at (2, 0).
Marking notes. 1 — solves f′(x) = 0 for both x-values. 1 — finds both y-coordinates by substitution. 1 — uses f″ to classify (0, 4). 1 — uses f″ to classify (2, 0). Stopping at x-values only (no y, no nature) loses 2 marks.
1.2 — f(x) = x³ at x = 0 (3 marks)
Sample response. f′(x) = 3x². f′(0) = 0, so (0, 0) is a stationary point. f″(x) = 6x, f″(0) = 0, so the second derivative test is inconclusive.
Sign table for f′(x) = 3x² around x = 0:
x: −0.1 0 +0.1
f′(x): 0.03 0 0.03
sign: + 0 +
Since f′(x) does not change sign at x = 0, the curve does not turn — it has a horizontal point of inflection at (0, 0).
Marking notes. 1 — identifies stationary point. 1 — recognises second derivative test fails and constructs a sign table for f′. 1 — concludes "horizontal inflection" with the reason that f′ keeps the same sign on both sides. A response that just says "f″(0) = 0 so it's an inflection" earns 0/3 for the conclusion; the sign argument is essential.
1.3 — y = 2x³ + ax² + b with local min at (1, −4) (3 marks)
Sample response. The point (1, −4) lies on the curve: 2 + a + b = −4 ⇒ a + b = −6 [eqn 1].
The gradient at x = 1 is zero: y′ = 6x² + 2ax, so y′(1) = 6 + 2a = 0 ⇒ a = −3 [eqn 2].
Substituting into eqn 1: −3 + b = −6 ⇒ b = −3.
(Optional check: y″ = 12x + 2a = 12 − 6 = 6 > 0, confirming a local minimum at x = 1.)
Marking notes. 1 — uses the point condition correctly. 1 — uses the zero-gradient condition correctly. 1 — solves the system to find a = −3 and b = −3. Confusing the "local min" condition with y″ > 0 (instead of y′ = 0) loses the second mark.
2.1 — Cubic with given features (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Curve through (1, 4): a + b + c + d = 4 [P1]. Curve through (3, 0): 27a + 9b + 3c + d = 0 [P2]. [1 mark — value equations.]
y′(x) = 3ax² + 2bx + c. y′(1) = 0: 3a + 2b + c = 0 [G1]. y′(3) = 0: 27a + 6b + c = 0 [G2]. [1 mark — gradient equations.]
Part (b). Subtract [G1] from [G2]: 24a + 4b = 0 ⇒ b = −6a. [1 mark.]
Substitute into [G1]: 3a + 2(−6a) + c = 0 ⇒ c = 12a − 3a = 9a. [1 mark.]
Substitute into [P1]: a + (−6a) + 9a + d = 4 ⇒ 4a + d = 4 ⇒ d = 4 − 4a.
Substitute into [P2]: 27a + 9(−6a) + 3(9a) + d = 0 ⇒ 27a − 54a + 27a + d = 0 ⇒ d = 0.
So 0 = 4 − 4a ⇒ a = 1, b = −6, c = 9, d = 0. The cubic is y = x³ − 6x² + 9x. [1 mark.]
Part (c). y″(x) = 6x + 2b = 6x − 12.
y″(1) = 6 − 12 = −6 < 0 ⇒ (1, 4) is a local maximum. ✓ [1 mark.]
y″(3) = 18 − 12 = 6 > 0 ⇒ (3, 0) is a local minimum. ✓ [1 mark.]
Part (d). Sketch: y = x³ − 6x² + 9x = x(x² − 6x + 9) = x(x − 3)². Intercepts at (0, 0) (single root, curve crosses) and (3, 0) (double root, curve touches and turns). Local max at (1, 4) between the two intercepts; local min at (3, 0) which coincides with the touching x-intercept. The cubic rises from −∞, passes through the origin, peaks at (1, 4), descends to touch the x-axis at (3, 0), then rises again to +∞. [1 mark — correct sketch with labels.]
Total: 8/8.
Band descriptors for marker.
Band 3: Sets up two equations only (typically the point conditions); does not use the zero-gradient conditions; cannot solve for all unknowns. ≈ 2-3 marks.
Band 4: Writes all four equations but does not solve cleanly; arrives at three of four unknowns; partial sketch. ≈ 4-5 marks.
Band 5: Solves the system in full; classifies both stationary points via y″; sketch present but missing the touching-versus-crossing distinction at x = 3. ≈ 6-7 marks.
Band 6: All four equations clearly labelled; system solved by elimination; both stationary points classified algebraically using y″; sketch shows the correct cubic shape with the (x − 3)² touching behaviour explicitly noted. 8/8.