Edexcel IGCSE Mathematics 018: Complete Worked Solutions | Edexcel IGCSE 数学 018：完整解析

📚 Edexcel IGCSE Mathematics 018 | Edexcel IGCSE 数学 018：完整解析

This article provides a full, step-by-step solution to Edexcel IGCSE Mathematics paper question 018, covering the core algebraic and graphical techniques required. We will break down each part of the question, explain the underlying concepts, and show you exactly how to earn full marks.

本文提供 Edexcel IGCSE 数学试卷第 018 题的完整分步解答，涵盖所需的代数与图象核心技巧。我们将拆解题目的每一部分，解释底层概念，并展示如何拿到满分。

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1. Understanding the Question | 理解题目

The question 018 typically involves a quadratic function, its graph, and a tangent line. You are asked to find the equation of the curve, calculate the gradient of a tangent at a given point, and determine the area under the curve using integration. This is a classic exam question testing your understanding of differentiation, integration, and coordinate geometry.

第 018 题通常涉及一个二次函数、其图象及一条切线。题目要求你求出曲线方程、计算某点处切线的斜率，并利用积分求曲线下的面积。这是一道经典考题，考查你对微分、积分和坐标几何的理解。

The full question (based on Edexcel IGCSE 4MA1 style) is as follows:

完整题目（基于 Edexcel IGCSE 4MA1 风格）如下：

y = x² − 4x + 3

• (a) Find the coordinates of the turning point of the curve. | 求曲线顶点坐标。

• (b) Find the gradient of the curve at the point where x = 5. | 求 x = 5 处曲线的斜率。

• (c) Find the equation of the tangent to the curve at x = 5. | 求 x = 5 处切线的方程。

• (d) Find the area enclosed by the curve, the x-axis, and the lines x = 3 and x = 5. | 求曲线、x 轴及直线 x = 3 与 x = 5 所围成的面积。

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2. Part (a) — Finding the Turning Point | 第 (a) 部分 — 求顶点

The turning point of a quadratic curve y = ax² + bx + c occurs at x = −b⁄(2a). Here a = 1, b = −4, so the x-coordinate is:

二次曲线 y = ax² + bx + c 的顶点出现在 x = −b⁄(2a)。这里 a = 1, b = −4，所以横坐标为：

x = −(−4)⁄(2×1) = 4⁄2 = 2

Substitute x = 2 into the equation: y = 2² − 4(2) + 3 = 4 − 8 + 3 = −1.

将 x = 2 代入方程：y = 2² − 4(2) + 3 = 4 − 8 + 3 = −1。

Therefore the turning point is at (2, −1). Since a = 1 > 0, this is a minimum point.

因此顶点坐标为 (2, −1)。由于 a = 1 > 0，这是极小值点。

Answer (a): Turning point = (2, −1), minimum

答案 (a)：顶点 = (2, −1)，极小值

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3. Part (b) — Gradient of the Curve | 第 (b) 部分 — 曲线斜率

To find the gradient at a specific point, we differentiate the function. The derivative of y = x² − 4x + 3 is:

要求某一点处的斜率，我们需要对函数求导。y = x² − 4x + 3 的导数为：

dy⁄dx = 2x − 4

At x = 5, the gradient is:

在 x = 5 处，斜率为：

dy⁄dx = 2(5) − 4 = 10 − 4 = 6

This means at x = 5, the curve is increasing at a rate of 6 units vertically per unit horizontally.

这意味着在 x = 5 处，曲线在水平方向每增加一个单位，垂直方向增加 6 个单位。

Answer (b): Gradient = 6

答案 (b)：斜率 = 6

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4. Part (c) — Equation of the Tangent | 第 (c) 部分 — 切线方程

A tangent line at a point has the same gradient as the curve at that point. At x = 5, the gradient is 6 (from part b). We also need the y-coordinate at x = 5:

切线上某点的斜率与曲线在该点的斜率相同。在 x = 5 处，斜率为 6（来自第 b 部分）。我们还需要 x = 5 处的 y 坐标：

y = 5² − 4(5) + 3 = 25 − 20 + 3 = 8

The tangent passes through the point (5, 8) with gradient m = 6. Using the equation y − y₁ = m(x − x₁):

切线经过点 (5, 8)，斜率 m = 6。利用点斜式方程 y − y₁ = m(x − x₁)：

y − 8 = 6(x − 5)

y − 8 = 6x − 30

y = 6x − 22

Therefore the tangent equation is y = 6x − 22.

因此切线方程为 y = 6x − 22。

Answer (c): y = 6x − 22

答案 (c)：y = 6x − 22

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5. Part (d) — Area Under the Curve | 第 (d) 部分 — 曲线下面积

The area enclosed by the curve y = x² − 4x + 3, the x-axis, x = 3, and x = 5 requires definite integration. However, we must first check whether the curve crosses the x-axis between these limits.

由曲线 y = x² − 4x + 3、x 轴、x = 3 及 x = 5 所围成的面积需要用定积分计算。但我们必须先检查曲线在区间内是否与 x 轴相交。

Factorise: y = (x − 1)(x − 3). The curve crosses the x-axis at x = 1 and x = 3. Since our lower limit is exactly x = 3, the curve is at the x-axis at the lower boundary, and above the x-axis for all x > 3 (as x² − 4x + 3 = (x−3)(x−1) > 0 for x > 3).

因式分解：y = (x − 1)(x − 3)。曲线在 x = 1 和 x = 3 处与 x 轴相交。由于积分的下限恰好是 x = 3，曲线在下边界处位于 x 轴上，且在 x > 3 时曲线位于 x 轴上方（因为 x > 3 时 x² − 4x + 3 = (x−3)(x−1) > 0）。

Thus, the area is given by the definite integral:

因此面积为定积分：

Area = ∫₃⁵ (x² − 4x + 3) dx

Find the antiderivative:

求不定积分：

∫ (x² − 4x + 3) dx = x³⁄3 − 2x² + 3x + C

Evaluate from 3 to 5:

从 3 到 5 求定积分值：

F(5) = 5³⁄3 − 2(5²) + 3(5) = 125⁄3 − 50 + 15 = 125⁄3 − 35 = 125⁄3 − 105⁄3 = 20⁄3

F(3) = 3³⁄3 − 2(3²) + 3(3) = 27⁄3 − 18 + 9 = 9 − 18 + 9 = 0

Therefore:

因此：

Area = F(5) − F(3) = 20⁄3 − 0 = 20⁄3

The area is 20⁄3 square units (approximately 6.67).

面积为 20⁄3 平方单位（约等于 6.67）。

Answer (d): Area = 20⁄3 square units

答案 (d)：面积 = 20⁄3 平方单位

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6. Common Mistakes to Avoid | 常见错误提醒

Many students lose marks on this type of question. Here are the most common pitfalls and how to avoid them:

许多学生在此类题目上失分。以下是最常见的陷阱及避免方法：

• Wrong turning point formula: Do not confuse −b⁄(2a) with the quadratic formula. The turning point x-coordinate uses just −b⁄(2a), not the ±√(b²−4ac)

• 错误的顶点公式：不要将 −b⁄(2a) 与求根公式混淆。顶点横坐标仅使用 −b⁄(2a)，不带 ±√(b²−4ac)。

• Forgetting to substitute y-coordinate: The turning point is a coordinate pair, not just the x-value.

• 忘记代回求 y 坐标：顶点是一个坐标对，而不只是 x 值。

• Tangent equation errors: Use y − y₁ = m(x − x₁) consistently. A common error is using y = mx + c and then miscalculating c.

• 切线方程错误：始终使用点斜式 y − y₁ = m(x − x₁)。常见错误是使用斜截式 y = mx + c 时算错 c 值。

• Integration limits: Always evaluate the antiderivative at the upper limit first and subtract the lower limit. Do not reverse the order.

• 积分上下限：务必先用上限代入不定积分，再减去下限代入的结果。不要颠倒顺序。

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7. Exam Technique Tips | 应试技巧

To maximise your score on this question, follow these strategies:

要在本题中获得最高分，请遵循以下策略：

• Show all working: In IGCSE mathematics, method marks are awarded even if the final answer is incorrect. Write out each step clearly.

• 展示完整过程：在 IGCSE 数学考试中，即使最终答案错误，方法分仍然会被授予。清楚地写出每一步。

• Check your tangent equation: Substitute x = 5 into your tangent equation. It should give y = 8. This is a quick check.

• 验证切线方程：将 x = 5 代入你的切线方程，应得到 y = 8。这是一个快速验证方法。

• Sketch the graph: A quick sketch helps you visualise the area you are finding and confirm whether integration is appropriate or if you need to split regions.

• 画草图：快速画图有助于你直观理解所求面积，并确认是否可以直接积分，或是否需要分割区域。

• Units and form: Give area in square units, and if a question says “give your answer in the form a⁄b,” make sure your answer is simplified.

• 单位与形式：面积使用平方单位；若题目要求“以 a⁄b 形式作答”，请确保答案已化简。

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8. Alternative Method for Part (a) | 第 (a) 部分的替代方法

Another approach to find the turning point is completing the square. For y = x² − 4x + 3:

另一种求顶点的方法是配方法。对于 y = x² − 4x + 3：

y = (x − 2)² − 4 + 3 = (x − 2)² − 1

In the form y = (x − p)² + q, the vertex is (p, q), giving (2, −1). This method is often faster and less error-prone than using the formula.

在 y = (x − p)² + q 的形式中，顶点为 (p, q)，即 (2, −1)。这种方法通常更快，也不容易出错。

For part (b), you could also use the general rule that the derivative of xⁿ is nxⁿ⁻¹, applied term by term — this is exactly what we did. Ensure this rule is memorised perfectly.

对于第 (b) 部分，你也可以利用幂函数求导法则：xⁿ 的导数为 nxⁿ⁻¹，逐项求导——这正是我们之前所做的。确保牢记这一法则。

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9. Practice Question | 练习题

Try this similar question on your own to reinforce your understanding:

请尝试以下类似题目以巩固理解：

y = x² − 6x + 5

• (a) Find the turning point. | 求顶点。

• (b) Find the gradient at x = 4. | 求 x = 4 处的斜率。

• (c) Find the equation of the tangent at x = 4. | 求 x = 4 处的切线方程。

• (d) Find the area enclosed by the curve, the x-axis, x = 2 and x = 4. | 求曲线、x 轴及 x = 2、x = 4 所围成的面积。

Answers: (a) (3, −4); (b) 2; (c) y = 2x − 11; (d) 4⁄3 square units.

答案：(a) (3, −4)；(b) 2；(c) y = 2x − 11；(d) 4⁄3 平方单位。

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10. Summary | 总结

In this worked solution for Edexcel IGCSE Mathematics question 018, we covered the key steps: identifying the turning point using x = −b⁄(2a), differentiating to find the gradient, constructing the tangent equation using the point-slope form, and calculating the area under the curve through definite integration.

在 Edexcel IGCSE 数学第 018 题的完整解析中，我们涵盖了关键步骤：使用 x = −b⁄(2a) 求顶点、求导得斜率、用点斜式构造切线方程，以及通过定积分计算曲线下面积。

Remember to practise these techniques thoroughly, show clear working in the exam, and always double-check your results with quick mental arithmetic where possible.

记住要彻底练习这些技巧，考试中清晰展示步骤，并尽可能用快速心算复查结果。

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