📚 Question 5: Mastering Stationary Points and Curve Sketching | 第5题：掌握驻点与曲线作图
In AQA A-Level Mathematics, Question 5 often appears as a pure maths problem that tests your ability to differentiate, solve equations, and interpret results graphically. This article walks through a typical question step by step, using a cubic function as an example.
在AQA A-Level数学中，第5题通常以纯数学问题的形式出现，考查你的微分、解方程和图像解读能力。本文将以一个三次函数为例，逐步解析这类典型问题。
1. Understanding the Problem | 理解题意
9x + 2 and asks you to find the stationary points, determine their nature, and sketch the curve. You may also be asked to find intercepts or compare values.
第5题的一个常见版本会给出一个三次函数，例如 f(x) = x³ − 6x² + 9x + 2，要求你求驻点、判断其性质并画出曲线。有时还会要求求截距或比较函数值。
To solve this type of problem efficiently, you need to recall:
要高效解决这类问题，你需要回忆以下内容：
- How to differentiate polynomials.
- How to solve quadratic equations.
- How to use the second derivative to determine the nature of a stationary point.
- 如何对多项式求导。
- 如何解二次方程。
- 如何利用二阶导数判断驻点的性质。
2. Differentiate the Function | 对函数求导
The first step is to differentiate f(x) = x³ − 6x² + 9x + 2 with respect to x.
第一步是求 f(x) = x³ − 6x² + 9x + 2 关于 x 的导数。
f'(x) = 3x² − 12x + 9
Notice that each term in the cubic is differentiated using the power rule: multiply by the power, then reduce the power by one.
注意，每一项三次项都使用幂法则求导：先乘以幂数，再将幂数减一。
3. Solve f'(x) = 0 to Find Stationary Points | 解 f'(x) = 0 求驻点
Stationary points occur where the gradient is zero, so set f'(x) = 0.
驻点出现在梯度为零处，因此令 f'(x) = 0。
3x² − 12x + 9 = 0
Divide through by 3 to simplify:
两边同时除以3进行化简：
x² − 4x + 3 = 0
Factorise the quadratic:
对二次式进行因式分解：
(x − 1)(x − 3) = 0
So the stationary points occur at x = 1 and x = 3.
因此驻点出现在 x = 1 和 x = 3 处。
4. Determine the Nature Using the Second Derivative | 用二阶导数判断驻点性质
Differentiate f'(x) to obtain the second derivative.
对 f'(x) 求导得到二阶导数。
f''(x) = 6x − 12
At x = 1:
在 x = 1 处：
f''(1) = 6(1) − 12 = −6 < 0
Since f''(1) < 0, the point at x = 1 is a local maximum.
因为 f''(1) < 0，所以 x = 1 处是局部极大值。
At x = 3:
在 x = 3 处：
f''(3) = 6(3) − 12 = 6 > 0
Since f''(3) > 0, the point at x = 3 is a local minimum.
因为 f''(3) > 0，所以 x = 3 处是局部极小值。
5. Find the y-Coordinates of the Stationary Points | 求驻点的纵坐标
Substitute x = 1 and x = 3 into the original function f(x).
将 x = 1 和 x = 3 代入原函数 f(x)。
| x | f(x) | Result |
| 1 | 1³ − 6(1²) + 9(1) + 2 | 1 − 6 + 9 + 2 = 6 |
| 3 | 3³ − 6(3²) + 9(3) + 2 | 27 − 54 + 27 + 2 = 2 |
Therefore the stationary points are (1, 6) and (3, 2).
因此驻点为 (1, 6) 和 (3, 2)。
6. Find the Intercepts and Behaviour as x → ±∞ | 求截距并分析 x → ±∞ 的行为
For the y-intercept, set x = 0:
求 y 轴截距时，令 x = 0：
f(0) = 0 − 0 + 0 + 2 = 2
So the curve crosses the y-axis at (0, 2).
所以曲线与 y 轴交于点 (0, 2)。
nteger root exists. In an exam, you may not need exact x-intercepts if the question only asks for a sketch.
求 x 轴截距时，令 f(x) = 0。解 x³ − 6x² + 9x + 2 = 0 可能不易分解，但可以尝试检验 2 的整数因子。检验 x = 2 得 8 − 24 + 18 + 2 = 4，不为零。检验 x = −1 得 −1 − 6 − 9 + 2 = −14，因此不存在简单的整数根。在考试中，如果题目只要求画草图，可能不需要精确的 x 截距。
³ term dominates and x³ is positive, so f(x) → ∞. As x → −∞, x³ is negative, so f(x) → −∞.
当 x → ∞ 时，x³ 项起主导作用且为正，因此 f(x) → ∞。当 x → −∞ 时，x³ 为负，因此 f(x) → −∞。
7. Sketch the Graph | 画出曲线草图
Use the information gathered to sketch the curve:
利用已收集的信息画出曲线：
- Local maximum at (1, 6).
- Local minimum at (3, 2).
- y-intercept at (0, 2).
- Curve rises from the left, passes through the maximum, goes down to the minimum, then rises to the right.
- 局部极大值 (1, 6)。
- 局部极小值 (3, 2)。
- y 轴截距 (0, 2)。
- 曲线从左侧上升，经过极大值，下降到极小值，再向右侧上升。
Make sure the sketch is smooth and shows the correct shape. Label the stationary points clearly.
确保草图平滑且形状正确，并清晰标注驻点。
8. Common Errors and How to Avoid Them | 常见错误及避免方法
Students often make mistakes in the following areas:
学生常在以下方面出错：
| Error | Why It Happens | How to Avoid It |
| Forgetting to divide f'(x) by 3 | Rushing the factorisation | Always look for common factors first. |
| Wrong sign of f''(x) | Misremembering the rule | f''(x) < 0 means maximum, f''(x) > 0 means minimum. |
| Sketching without labelling key points | Not reading the question | Always label axes, intercepts and stationary points. |
Double-check your arithmetic when substituting values back into f(x).
在代入 f(x) 计算时，务必再次检查算术。
9. Practice Question | 练习题
Try this similar question on your own:
请独立尝试以下类似问题：
coordinates of the stationary points and determine their nature. Sketch the graph, showing the y-intercept and the stationary points.
已知 g(x) = 2x³ − 3x² − 12x + 7，求驻点坐标并判断其性质。画出草图，标出 y 截距和驻点。
Check your answer: g' ationary points are (2, −13) and (−1, 14), with (−1, 14) a maximum and (2, −13) a minimum.
检查答案：g'(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1)，所以 x = 2 和 x = −1。驻点为 (2, −13) 和 (−1, 14)，其中 (−1, 14) 为极大值，(2, −13) 为极小值。
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