📚 AQA A-Level Maths Question 4: Differentiation and Curve Sketching | AQA A-Level 数学第4题：微分与曲线草图
4 on the curve y = 2x³ - 9x² + 12x - 1, showing every step you need to secure full marks.
在AQA A-Level数学中，试卷1第4题经常考查多项式函数微分、驻点、切线与曲线草图。本文讲解一道AQA风格的第4题，曲线方程为 y = 2x³ - 9x² + 12x - 1，逐步展示获得满分的全部方法。
1. The Question and Key Ideas | 题目与核心考点
point where x = 1. (d) Sketch the curve C, labelling any stationary points and intercepts.
第4题：曲线 C 的方程为 y = 2x³ - 9x² + 12x - 1。(a) 求 dy/dx 和 d²y/dx²。(b) 求 C 的驻点坐标并判断其性质。(c) 求曲线在 x = 1 处的切线方程。(d) 画出曲线 C 的草图，标出驻点和截距。
quations, applying the second derivative test, and the tangent formula y - y₁ = m(x - x₁).
本题综合了AQA纯数学多个主题：多项式微分、解二次方程、应用二阶导数判别法以及切线公式 y - y₁ = m(x - x₁)。
2. Finding the First Derivative | 求一阶导数
- 1, the derivative of 2x³ is 6x², of -9x² is -18x, of 12x is 12, and of constant -1 is 0.
使用逐项求导公式 d/dx (axⁿ) = naxⁿ⁻¹。对 y = 2x³ - 9x² + 12x - 1 求导，2x³ 的导数为 6x²，-9x² 的导数为 -18x，12x 的导数为 12，常数 -1 的导数为 0。
dy/dx = 6x² - 18x + 12
Always write dy/dx clearly, not y' alone, because AQA mark schemes expect the notation dy/dx in this part.
请始终清楚写出 dy/dx，不要只写 y'，因为AQA评分标准在这一部分要求使用 dy/dx 记号。
3. Finding the Second Derivative | 求二阶导数
Differentiate dy/dx again to obtain d²y/dx². The derivative of 6x² is 12x, of -18x is -18, and of 12 is 0.
再次对 dy/dx 求导得到 d²y/dx²。6x² 的导数为 12x，-18x 的导数为 -18，12 的导数为 0。
d²y/dx² = 12x - 18
The second derivative will be used to decide whether each stationary point is a maximum, minimum or point of inflection.
二阶导数将用于判断每个驻点是极大值、极小值还是拐点。
4. Locating Stationary Points | 定位驻点
ur where dy/dx = 0. Set 6x² - 18x + 12 = 0 and divide through by 6 to get x² - 3x + 2 = 0.
驻点出现在 dy/dx = 0 处。令 6x² - 18x + 12 = 0，两边除以 6 得 x² - 3x + 2 = 0。
- 1)(x - 2) = 0, so x = 1 or x = 2. These are the x-coordinates of the stationary points.
对二次式因式分解：(x - 1)(x - 2) = 0，所以 x = 1 或 x = 2。这两个值就是驻点的横坐标。
nto the original equation y = 2x³ - 9x² + 12x - 1 to find the corresponding y-coordinates.
将每个 x 值代入原方程 y = 2x³ - 9x² + 12x - 1，求对应的纵坐标。
1) - 1 = 2 - 9 + 12 - 1 = 4. At x = 2, y = 2(8) - 9(4) + 12(2) - 1 = 16 - 36 + 24 - 1 = 3.
当 x = 1 时，y = 2(1)³ - 9(1)² + 12(1) - 1 = 2 - 9 + 12 - 1 = 4。当 x = 2 时，y = 2(8) - 9(4) + 12(2) - 1 = 16 - 36 + 24 - 1 = 3。
Stationary points: (1, 4) and (2, 3)
5. Determining the Nature of Stationary Points | 判断驻点性质
Use the second derivative test: substitute each x-coordinate into d²y/dx² = 12x - 18.
应用二阶导数判别法：将每个横坐标代入 d²y/dx² = 12x - 18。
x = 1, d²y/dx² = 12(1) - 18 = -6. Since d²y/dx² < 0, the point (1, 4) is a local maximum.
当 x = 1 时，d²y/dx² = 12(1) - 18 = -6。因 d²y/dx² < 0，所以点 (1, 4) 是局部极大值。
t x = 2, d²y/dx² = 12(2) - 18 = 6. Since d²y/dx² > 0, the point (2, 3) is a local minimum.
当 x = 2 时，d²y/dx² = 12(2) - 18 = 6。因 d²y/dx² > 0，所以点 (2, 3) 是局部极小值。
| x | d²y/dx² | Nature |
|---|---|---|
| 1 | -6 | Maximum |
| 2 | 6 | Minimum |
6. Equation of the Tangent | 求切线方程
d gradient m = dy/dx at x = 1. From dy/dx = 6x² - 18x + 12, at x = 1, m = 6 - 18 + 12 = 0.
曲线在 x = 1 处的切线使用点 (1, 4) 和斜率 m = dy/dx 在 x = 1 处的值。由 dy/dx = 6x² - 18x + 12，当 x = 1 时，m = 6 - 18 + 12 = 0。
A gradient of 0 gives a horizontal tangent. The tangent equation is therefore y = 4.
斜率为0意味着切线是水平的，因此切线方程为 y = 4。
angent equation in its simplest exact form: y = 4, not y - 4 = 0(x - 1) unless simplified.
务必将切线方程写成最简精确形式：y = 4，而不要写成未化简的 y - 4 = 0(x - 1)。
7. Sketching the Curve | 绘制曲线草图
here the curve crosses the y-axis by setting x = 0: y = -1. So the y-intercept is (0, -1).
绘制草图时，先令 x = 0 求曲线与 y 轴交点：y = -1，所以 y 截距为 (0, -1)。
act roots may not be required unless the cubic factors nicely. Here it does not factorise neatly, so a labelled intercept at (0, -1) plus the stationary points is sufficient for the sketch.
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