📚 SQA Higher Mathematics: Case Study Practice Step-by-Step | SQA高级数学：案例分析实战演练
In SQA Higher Mathematics, case study questions bring together differentiation, coordinate geometry, and problem-solving skills. This article walks through a complete worked example involving stationary points, tangents, normals, and area calculation – all typical of examination-style problems. By following each step carefully, you will strengthen your understanding of how to apply theory to a multi-part question.
在 SQA 高级数学中，案例分析题综合考查微分、坐标几何以及问题解决能力。本文将通过一个完整范例，逐步讲解如何求解驻点、切线、法线以及相关面积计算——这些都是考试中的典型题型。跟随每一步详细解析，有助于巩固运用理论知识解决多部分问题的能力。
1. Case Study Problem Statement | 案例问题陈述
A curve C has equation y = x³ − 6x² + 9x + 2. The tasks are: (a) find the stationary points on C and determine their nature; (b) find the equation of the tangent to C at x = 0.5; (c) find the equation of the normal to C at x = 0.5; (d) calculate the area of the triangle formed by the tangent, the normal, and the y‑axis.
已知曲线 C 的方程为 y = x³ − 6x² + 9x + 2。要求完成以下任务：(a) 求曲线 C 上的驻点并判断其性质；(b) 求曲线在 x = 0.5 处的切线方程；(c) 求曲线在 x = 0.5 处的法线方程；(d) 计算由该切线、法线和 y 轴所围成三角形的面积。
2. Differentiation Refresher | 微分知识回顾
To tackle this problem, recall the power rule: if y = xⁿ, then dy/dx = n xⁿ⁻¹. For a cubic y = x³ − 6x² + 9x + 2, differentiate term by term. The derivative is dy/dx = 3x² − 12x + 9. The second derivative, d²y/dx² = 6x − 12, will help classify stationary points.
要解决这个问题，回顾幂函数求导法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。对于三次函数 y = x³ − 6x² + 9x + 2，逐项求导可得 dy/dx = 3x² − 12x + 9。二阶导数 d²y/dx² = 6x − 12 将用于判断驻点性质。
3. Finding Stationary Points | 求驻点
Stationary points occur where dy/dx = 0. Set 3x² − 12x + 9 = 0. Divide by 3 to get x² − 4x + 3 = 0, which factorises as (x − 1)(x − 3) = 0. Thus x = 1 or x = 3. Substitute into y: when x = 1, y = 1 − 6 + 9 + 2 = 6; when x = 3, y = 27 − 54 + 27 + 2 = 2. The stationary points are (1, 6) and (3, 2).
驻点出现在 dy/dx = 0 时。令 3x² − 12x + 9 = 0。两边除以 3 得 x² − 4x + 3 = 0，因式分解为 (x − 1)(x − 3) = 0。因此 x = 1 或 x = 3。代入原函数：当 x = 1 时，y = 1 − 6 + 9 + 2 = 6；当 x = 3 时，y = 27 − 54 + 27 + 2 = 2。驻点为 (1, 6) 和 (3, 2)。
4. Nature of Stationary Points | 驻点的性质判定
Use the second derivative d²y/dx² = 6x − 12. At x = 1, d²y/dx² = 6(1) − 12 = −6 < 0, so (1, 6) is a local maximum. At x = 3, d²y/dx² = 6(3) − 12 = 6 > 0, so (3, 2) is a local minimum. The table below summarises the results.
使用二阶导数 d²y/dx² = 6x − 12 进行判断。在 x = 1 处，d²y/dx² = 6(1) − 12 = −6 < 0，故 (1, 6) 为极大值点。在 x = 3 处，d²y/dx² = 6(3) − 12 = 6 > 0，故 (3, 2) 为极小值点。下表总结了这一结果。
| x | y | d²y/dx² | Nature / 性质 |
|---|---|---|---|
| 1 | 6 | −6 | Maximum / 极大 |
| 3 | 2 | 6 | Minimum / 极小 |
5. Equation of the Tangent | 切线方程
At x = 0.5, the y‑coordinate is y = (0.5)³ − 6(0.5)² + 9(0.5) + 2 = 0.125 − 1.5 + 4.5 + 2 = 5.125. Write this as 41/8 for exact work. The gradient m_t = dy/dx at x = 0.5: 3(0.5)² − 12(0.5) + 9 = 0.75 − 6 + 9 = 3.75 = 15/4. Using y − y₁ = m(x − x₁), tangent equation is y − 41/8 = (15/4)(x − 0.5).
当 x = 0.5 时，y 坐标为 y = (0.5)³ − 6(0.5)² + 9(0.5) + 2 = 0.125 − 1.5 + 4.5 + 2 = 5.125。为精确计算，写成分数 41/8。切线斜率 m_t 为 x = 0.5 处的 dy/dx：3(0.5)² − 12(0.5) + 9 = 0.75 − 6 + 9 = 3.75 = 15/4。利用点斜式 y − y₁ = m(x − x₁)，得切线方程为 y − 41/8 = (15/4)(x − 0.5)。
6. Equation of the Normal | 法线方程
The normal is perpendicular to the tangent, so its gradient m_n = −1 / m_t = −4/15. Using the same point (0.5, 41/8), the normal equation is y − 41/8 = (−4/15)(x − 0.5). Both equations are now ready for intersection work.
法线与切线垂直，因此其斜率 m_n = −1 / m_t = −4/15。使用同一点 (0.5, 41/8)，得法线方程为 y − 41/8 = (−4/15)(x − 0.5)。切线方程与法线方程均已得出，可供后续求交点用。
7. Intercepts on the y‑axis | 求 y 轴截距
To find where the tangent meets the y‑axis, set x = 0: y − 41/8 = (15/4)(−0.5) = −15/8. So y = 41/8 − 15/8 = 26/8 = 13/4 = 3.25. For the normal, set x = 0: y − 41/8 = (−4/15)(−0.5) = 2/15. Hence y = 41/8 + 2/15 = (615 + 16)/120 = 631/120. The intercepts are A(0, 13/4) and B(0, 631/120).
求切线与 y 轴的交点，令 x = 0：y − 41/8 = (15/4)(−0.5) = −15/8，故 y = 41/8 − 15/8 = 26/8 = 13/4 = 3.25。对于法线，令 x = 0：y − 41/8 = (−4/15)(−0.5) = 2/15，因此 y = 41/8 + 2/15 = (615 + 16)/120 = 631/120。得到交点 A(0, 13/4) 和 B(0, 631/120)。
8. Triangle Area Calculation | 三角形面积计算
The triangle is formed by the points P(0.5, 41/8), A(0, 13/4), and B(0, 631/120). The base AB lies on the y‑axis with length |631/120 − 13/4| = |631/120 − 390/120| = 241/120. The perpendicular height is the horizontal distance from P to the y‑axis, which is 0.5 = 1/2. Area = ½ × base × height = ½ × (241/120) × (1/2) = 241/480 square units.
三角形由点 P(0.5, 41/8)、A(0, 13/4) 和 B(0, 631/120) 构成。底边 AB 位于 y 轴上，其长度为 |631/120 − 13/4| = |631/120 − 390/120| = 241/120。垂直高度为点 P 到 y 轴的水平距离，即 0.5 = 1/2。面积 = ½ × 底 × 高 = ½ × (241/120) × (1/2) = 241/480 平方单位。
9. Verification and Checking | 验证与检查
Use decimal approximations to check reasonableness. Tangent intercept ≈ 3.25, normal intercept ≈ 5.2583, difference ≈ 2.0083. Half‑base ≈ 1.00415; height = 0.5; area ≈ 0.5021. The exact area 241/480 = 0.502083… matches. Also verify the stationary points by sketching or with a sign diagram of dy/dx.
可用小数近似值检验合理性。切线截距约为 3.25，法线截距约为 5.2583，差约为 2.0083。半底边长约为 1.00415，高为 0.5，面积约为 0.5021。精确面积 241/480 = 0.502083… 吻合。也可通过草图或 dy/dx 的符号表验证驻点性质。
10. Common Mistakes and Summary | 常见错误与总结
Typical errors include forgetting to substitute x into the original function for the y‑coordinate, confusing gradient of tangent and normal, and misapplying the area formula. Always keep exact fractions until the final answer to avoid rounding issues. This case study demonstrates how differentiation links to coordinate geometry and applied problem solving – a vital skill in SQA Higher Mathematics.
常见错误包括：忘记将 x 代入原函数求 y 坐标、混淆切线与法线斜率、错误使用面积公式。务必保留精确分数直至最后答案，以避免舍入误差。本案例分析展示了微分如何与坐标几何和应用性问题紧密结合——这是 SQA 高级数学中的一项核心技能。
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