📚 AQA A-Level Maths Question 13: Differentiation and Curve Sketching | AQA A-Level 数学第13题：微分与曲线作图
estion 13 using the curve y = x³ − 6x² + 9x + 4, showing the exact steps examiners expect.
AQA A-Level 数学试卷一第13题通常将微分、驻点、切线和曲线作图整合在一个多小问的综合题中。本指南以曲线 y = x³ − 6x² + 9x + 4 为例，逐步展示阅卷官期望的完整解题过程。
1. What Question 13 Tests | 第13题考查什么
A typical AQA Question 13 asks you to analyse a cubic function in stages: differentiate, locate stationary points, classify them, find a tangent and describe where the curve is increasing or decreasing.
典型的 AQA 第13题要求你分阶段分析一个三次函数：求导、确定驻点、判断驻点性质、求切线，并描述曲线在哪些区间递增或递减。
Examiners want clear method marks, so you must show each derivative and each equation before substituting values. Missing a line of working can cost marks even if your final coordinates are correct.
阅卷官看重清晰的方法分，因此你必须先写出每一步导数和方程，再代入数值。即使最终坐标正确，缺少必要的解题步骤也会丢分。
2. Step 1: Differentiate Using the Power Rule | 第一步：用幂法则求导
+ 4, differentiate term by term. The power rule states that if y = xⁿ, then dy/dx = nxⁿ⁻¹.
对于曲线 y = x³ − 6x² + 9x + 4，要逐项求导。幂法则指出，如果 y = xⁿ，那么 dy/dx = nxⁿ⁻¹。
dy/dx = 3x² − 12x + 9
Always simplify the derivative by taking out a common factor where possible. Here all three coefficients are divisible by 3:
始终尽可能提取公因式化简导数。这里三个系数都能被 3 整除：
dy/dx = 3(x² − 4x + 3) = 3(x − 1)(x − 3)
orised form is essential because it makes solving dy/dx = 0 and testing signs much faster.
因式分解后的形式非常重要，因为它能让解 dy/dx = 0 和判断符号变得更加快捷。
3. Step 2: Solve dy/dx = 0 to Locate Stationary Points | 第二步：解 dy/dx = 0 确定驻点
Stationary points occur where the gradient is zero, so set dy/dx = 0 and solve for x.
驻点出现在梯度为零的位置，因此令 dy/dx = 0 并求解 x。
3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3
ation to find the y coordinates: y(1) = 1 − 6 + 9 + 4 = 8 and y(3) = 27 − 54 + 27 + 4 = 4.
将每个 x 值代回原方程求 y 坐标：y(1) = 1 − 6 + 9 + 4 = 8，y(3) = 27 − 54 + 27 + 4 = 4。
So the stationary points are (1, 8) and (3, 4). Always write them as coordinate pairs, not just x values.
因此驻点为 (1, 8) 和 (3, 4)。始终以坐标对的形式写出，而不仅仅是 x 值。
4. Step 3: Use the Second Derivative to Determine Nature | 第三步：用二阶导判断驻点性质
Differentiate dy/dx again to find the second derivative. This tells you whether each stationary point is a local maximum or local minimum.
再次对 dy/dx 求导得到二阶导。二阶导可以告诉你每个驻点是局部最大值还是局部最小值。
d²y/dx² = 6x − 12
he second derivative is negative, the curve is concave down and (1, 8) is a local maximum.
在 x = 1 处，d²y/dx² = 6(1) − 12 = −6。由于二阶导为负，曲线在该点下凹，因此 (1, 8) 是局部最大值点。
the second derivative is positive, the curve is concave up and (3, 4) is a local minimum.
在 x = 3 处，d²y/dx² = 6(3) − 12 = 6。由于二阶导为正，曲线在该点上凹，因此 (3, 4) 是局部最小值点。
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