📚 Example 6.10.1: Stationary Points, Tangents and Normals | 例题 6.10.1：驻点、切线与法线
In AQA A-Level Mathematics, Example 6.10.1 is a core worked example on applying differentiation to find stationary points and the equations of tangents and normals. This topic appears frequently in Paper 1 (Pure Mathematics) and forms a foundation for optimisation and curve sketching.
在 AQA A-Level 数学中，例题 6.10.1 是一个关于应用微分求驻点、切线方程和法线方程的核心例题。该主题在试卷一（纯数学）中频繁出现，是优化问题和曲线作图的基础。
1. The Example Statement | 例题陈述
The target function is a cubic polynomial. We will find all stationary points, decide whether each is a local maximum or minimum, and then write down the tangent and normal equations at a specified point.
目标函数是一个三次多项式。我们将求出所有驻点，判断每个驻点是局部极大值还是局部极小值，然后写出指定点处的切线和法线方程。
f(x) = x³ − 6x² + 9x + 1
tandard AQA-style question because it tests several differentiation skills in one example.
我们还需要求 x = 2 处的切线和法线。这是一道标准的 AQA 风格题目，因为它在一个例题中同时考查了多种微分技能。
2. Prerequisite: Basic Differentiation | 预备知识：基本微分
he derivative is given by dy/dx = n xⁿ⁻¹. For a polynomial, we differentiate term by term.
首先回顾 xⁿ 的幂函数求导法则。若 y = xⁿ，则导数为 dy/dx = n xⁿ⁻¹。对于多项式，我们逐项求导。
Geometrically, the derivative f '(x) gives the gradient of the tangent to the curve at any point x. Stationary points occur where the tangent is horizontal, so the gradient is zero.
从几何意义上看，导数 f '(x) 表示曲线在任意点 x 处切线的斜率。驻点出现在切线水平的位置，因此斜率为零。
3. Step 1: Differentiate f(x) | 步骤1：对 f(x) 求导
Differentiate the cubic term by term. The derivative of x³ is 3x², the derivative of −6x² is −12x, and the derivative of 9x is 9. The constant 1 disappears.
对三次多项式逐项求导。x³ 的导数为 3x²，−6x² 的导数为 −12x，9x 的导数为 9，常数 1 的导数为 0。
f '(x) = 3x² − 12x + 9
It is useful to factorise the derivative. Taking out the common factor 3 gives 3(x² − 4x + 3), and the quadratic factorises further to 3(x − 1)(x − 3).
对导数进行因式分解很有帮助。提取公因式 3 得到 3(x² − 4x + 3)，这个二次式可以继续分解为 3(x − 1)(x − 3)。
f '(x) = 3(x − 1)(x − 3)
4. Step 2: Solve f '(x) = 0 | 步骤2：解 f '(x) = 0
Stationary points occur when f ' = 0. Because the derivative is already factorised, we can set each bracket equal to zero.
驻点出现在 f '(x) = 0 时。由于导数已经因式分解，我们可以令每个括号分别等于零。
3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3
e curve, one at x = 1 and one at x = 3. These are the x-coordinates of the turning points.
因此曲线上有两个驻点，一个在 x = 1，另一个在 x = 3。这些是拐点的 x 坐标。
5. Step 3: Find the y-Coordinates | 步骤3：计算 y 坐标
nction to find the first y-coordinate. You must use the original f(x), not the derivative.
将 x = 1 代入原函数求第一个 y 坐标。必须使用原函数 f(x)，而不是导数。
f(1) = 1³ − 6(1)² + 9(1) + 1 = 1 − 6 + 9 + 1 = 5
Now substitute x = 3 into f(x). The calculation gives the second stationary point.
现在将 x = 3 代入 f(x)。计算可得到第二个驻点。
f(3) = 3³ − 6(3)² + 9(3) + 1 = 27 − 54 + 27 + 1 = 1
The stationary points are therefore (1, 5) and (3, 1). Always write the coordinates as ordered pairs in your final answer.
因此驻点为 (1, 5) 和 (3, 1)。在最终答案中始终要把坐标写成有序数对。
6. Step 4: Use the Second Derivative Test | 步骤4：使用二阶导数检验
Differentiate f '(x) again to obtain the second derivative. This will tell us whether each stationary point is a local maximum or minimum.
再次对 f '(x) 求导，得到二阶导数。这将告诉我们每个驻点是局部极大值还是局部极小值。
f ''(x) = 6x − 12
Substitute x = 1 into f ''(x). Since f ''(1) is negative, the curve is concave down at x = 1, so (1, 5) is a local maximum.
将 x = 1 代入 f ''(x)。由于 f ''(1) 为负，曲线在 x = 1 处向下凹，因此 (1, 5) 是局部极大值。
f ''(1) = 6(1) − 12 = −6 < 0 ⇒ local maximum
Substitute x = 3 into f ''(x). Since f ''(3) is positive, the curve is concave up at x = 3, so (3, 1) is a local minimum.
将 x = 3 代入 f ''(x)。由于 f ''(3) 为正，曲线在 x = 3 处向上凹，因此 (3, 1) 是局部极小值。
f ''(3) = 6(3) − 12 = 6 > 0 ⇒ local minimum
7. Interpreting Maximum and Minimum Points | 解释极大值与极小值点
A local maximum at (1, 5) means the curve reaches a peak there. Moving from left to right, the gradient changes from positive to negative as x passes through 1.
在 (1, 5) 处的局部极大值意味着曲线在这里到达一个峰值。从左到右移动时，斜率在 x = 1 处由正变为负。
At (3, 1), the curve reaches a trough. The gradient changes from negative to positive as x passes through 3. This is the lowest point in the immediate neighbourhood.
在 (3, 1) 处，曲线到达一个谷底。斜率在 x = 3 处由负变为正。这是邻近区域内的最低点。
mum at x = 3, and then rises again. This classic shape is often examined in AQA questions.
因此这条三次曲线上升到 x = 1 处的极大值，然后下降到 x = 3 处的极小值，之后再次上升。这是 AQA 考试中经常考查的经典形状。
8. Extension: Tangent at a Given Point | 扩展：给定点处的切线
ve at x = 2. First find the y-coordinate by substituting x = 2 into the original function.
现在求曲线在 x = 2 处的切线方程。首先将 x = 2 代入原函数求 y 坐标。
f(2) = 2³ − 6(2)² + 9(2) + 1 = 8 − 24 + 18 + 1 = 3
Next find the gradient of the tangent by substituting x = 2 into f '(x).
然后将 x = 2 代入 f '(x) 求切线斜率。
f '(2) = 3(2)² − 12(2) + 9 = 12 − 24 + 9 = −3
the point-gradient formula y − y₁ = m(x − x₁), the tangent equation is y − 3 = −3(x − 2).
利用点斜式 y − y₁ = m(x − x₁)，切线方程为 y − 3 = −3(x − 2)。
y − 3 = −3(x − 2) ⇒ y = −3x + 9
9. Extension: Normal to the Curve | 扩展：曲线的法线
The normal is perpendicular to the tangent. If the tangent gradient is m, then the normal gradient is −1 ÷ m.
法线垂直于切线。如果切线斜率为 m，那么法线斜率为 −1 ÷ m。
Here the tangent gradient is −3, so the normal gradient is −1 ÷ (−3) = 1/3.
这里切线斜率为 −3，因此法线斜率为 −1 ÷ (−3) = 1/3。
Using the same point (2, 3), the normal equation is y − 3 = (1/3)(x − 2).
使用同一个点 (2, 3)，法线方程为 y − 3 = (1/3)(x − 2)。
y − 3 = (1/3)(x − 2) ⇒ y = (1/3)x + 7/3
10. Common Mistakes and Examiner Advice | 常见错误与考官建议
One common mistake is substituting the x-coordinate into f '(x) instead of f(x) to find the y-coordinate. Always use the original function for y-values and the derivative for gradients.
一个常见错误是将 x 坐标代入 f '(x) 而不是 f(x) 来求 y 坐标。求 y 值时始终使用原函数，求斜率时使用导数。
Another error is forgetting the negative sign when applying the second derivative test. Write down f ''(x) clearly and substitute carefully to avoid sign errors.
另一个错误是在应用二阶导数检验时遗漏负号。清楚地写出 f ''(x) 并仔细代入，以避免符号错误。
When finding normals, do not confuse the tangent gradient with the normal gradient. Use the negative reciprocal relationship and check that the two gradients multiply to −1.
求法线时，不要把切线斜率和法线斜率混淆。使用负倒数关系，并检查两个斜率的乘积是否为 −1。
11. Practice Questions | 练习题目
Try these extension questions to consolidate the method from Example 6.10.1.
请尝试以下扩展问题，巩固例题 6.10.1 中的方法。
- Find the stationary points of y = x³ − 3x² − 9x + 2 and determine their nature.
- Find the equation of the tangent and normal to y = 2x³ − 9x² + 12x − 5 at x = 1.
- the curve y = x³ − 6x² + 9x + 1 has no stationary point for which f(x) is greater than 5.
Check your answers by differentiating twice and verifying the signs of f ''(x) at each stationary point. These small checks can prevent many errors in the exam.
通过两次求导并验证每个驻点处 f ''(x) 的符号来检查答案。这些小检查可以避免考试中的许多错误。
12. Summary and Key Takeaways | 总结与关键要点
The complete method can be summarised in four steps: differentiate, solve f '(x) = 0, find y-coordinates, and use f ''(x) to determine the nature of each stationary point.
完整方法可以总结为四个步骤：求导、解 f '(x) = 0、求 y 坐标、使用 f ''(x) 判断每个驻点的性质。
f(x) = x³ − 6x² + 9x + 1 ⇒ f '(x) = 3x² − 12x + 9 ⇒ f ''(x) = 6x − 12
a local minimum. The tangent at x = 2 is y = −3x + 9, and the normal is y = (1/3)x + 7/3.
驻点为局部极大值点 (1, 5) 和局部极小值点 (3, 1)。x = 2 处的切线为 y = −3x + 9，法线为 y = (1/3)x + 7/3。
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