📚 IGCSE Workbook G-1 Exercise 218: Completing the Square and Quadratic Graphs | IGCSE 练习册 G-1 第 218 题：配方法与二次函数图像
Exercise 218 in Workbook G-1 focuses on rewriting a quadratic expression by completing the square. You will use the completed-square form to solve a quadratic equation and to identify the turning point of its graph. This combination is extremely common in IGCSE papers, especially in questions worth 5 to 7 marks.
G-1 练习册第 218 题的重点是通过配方法改写二次表达式。你将使用完全平方形式解二次方程，并确定函数图像的顶点。这种组合在 IGCSE 试卷中非常常见，通常占 5 到 7 分。
1. What the Question Looks Like | 题型展示
A typical form of Exercise 218 gives you a quadratic expression such as x² + 6x + 2. The three parts usually ask you to write it in completed-square form, solve the related equation, and state the minimum point of the graph.
第 218 题的典型形式是给出一个二次表达式，例如 x² + 6x + 2。题目通常有三部分：将其写成完全平方形式、解相应方程，并写出图像的最低点坐标。
Express x² + 6x + 2 in the form (x + p)² + q
将 x² + 6x + 2 写成 (x + p)² + q 的形式
Part (a) is therefore about completing the square. Part (b) uses your answer to solve x² + 6x + 2 = 0 in exact surd form. Part (c) asks for the turning point of y = x² + 6x + 2.
因此第 (a) 部分考查配方法。第 (b) 部分使用你的结果解方程 x² + 6x + 2 = 0，并要求保留根号精确值。第 (c) 部分要求写出 y = x² + 6x + 2 的顶点坐标。
2. Key Concept: Completing the Square | 关键概念：配方法
Completing the square works because the expansion of (x + a)² is x² + 2ax + a². If the coefficient of x is b, you take half of b, square it, and then adjust the constant term.
配方法之所以有效，是因为 (x + a)² 展开后是 x² + 2ax + a²。如果 x 的系数是 b，你取出 b 的一半，平方后，再调整常数项。
x² + bx + c = (x + b ÷ 2)² + c – (b ÷ 2)²
For x² + 6x + 2, the coefficient of x is 6. Half of 6 is 3, and 3² is 9. This means x² + 6x can be written as (x + 3)² – 9, because (x + 3)² gives x² + 6x + 9, which is 9 too large.
对于 x² + 6x + 2，x 的系数是 6。6 的一半是 3，而 3² 是 9。这意味着 x² + 6x 可以写成 (x + 3)² – 9，因为 (x + 3)² 展开得到 x² + 6x + 9，比原式多出 9。
3. Step-by-Step Rewriting | 逐步改写
Start with x² + 6x + 2. Add and subtract the square of half the x-coefficient. In this case, half of 6 is 3, so you add and subtract 9. This gives (x² + 6x + 9) – 9 + 2, which simplifies to the completed-square form.
从 x² + 6x + 2 开始。加上并减去 x 系数一半的平方。这里 6 的一半是 3，所以加上并减去 9。得到 (x² + 6x + 9) – 9 + 2，化简后就是完全平方形式。
x² + 6x + 2 = (x + 3)² – 7
This is the required form with p = 3 and q = -7. You should always check by expanding: (x + 3)² – 7 equals x² + 6x + 9 – 7, which is x² + 6x + 2. The check confirms that the rewriting is correct.
这就是所需形式，其中 p = 3，q = -7。你应当始终通过展开来检验：(x + 3)² – 7 等于 x² + 6x + 9 – 7，即 x² + 6x + 2。检验确认改写正确。
4. Solving the Equation | 解方程
Using the completed-square form, the equation x² + 6x + 2 = 0 becomes (x + 3)² – 7 = 0. Add 7 to both sides to isolate the squared term, then take the square root of both sides. Remember to include the positive and negative square root.
使用完全平方形式，方程 x² + 6x + 2 = 0 变为 (x + 3)² – 7 = 0。两边加 7 以分离平方项，然后对两边开平方。记住要取正平方根和负平方根。
(x + 3)² = 7
x + 3 = ±√7
x = -3 ± √7
The two exact solutions are x = -3 + √7 and x = -3 – √7. In IGCSE answers, leaving the roots in surd form is expected unless the question asks for decimal approximations.
两个精确解是 x = -3 + √7 和 x = -3 – √7。在 IGCSE 答案中，除非题目要求写成小数近似值，否则应保留根号形式。
5. Finding the Turning Point | 求顶点
For a quadratic written as y = (x + p)² + q, the graph is a parabola whose turning point is at (-p, q). This is because the squared term (x + p)² is always greater than or equal to zero, so the smallest value of y occurs when x + p = 0.
对于写成 y = (x + p)² + q 的二次函数，其图像是一条抛物线，顶点坐标是 (-p, q)。这是因为平方项 (x + p)² 始终大于或等于零，所以当 x + p = 0 时 y 取得最小值。
y = (x + 3)² – 7
Turning point = (-3, -7)
顶点 = (-3, -7)
The coefficient of x² is positive, so the parabola opens upwards. This means the turning point is a minimum point. If the coefficient were negative, the parabola would open downwards and the turning point would be a maximum.
x² 的系数为正，所以抛物线开口向上。这意味着该顶点是最低点。如果系数为负，抛物线将开口向下，顶点将是最高点。
6. Sketching the Graph | 画函数图像
To sketch y = x² + 6x + 2, mark the turning point (-3, -7), the y-intercept (0, 2), and the two roots x = -3 + √7 and x = -3 – √7. The approximate roots are -0.35 and -5.65, which help you place the x-axis crossings.
要画出 y = x² + 6x + 2 的图像，标出顶点 (-3, -7)、y 轴截距 (0, 2) 以及两个根 x = -3 + √7 和 x = -3 – √7。根的近似值约为 -0.35 和 -5.65，有助于确定与 x 轴的交点位置。
| x | y = (x + 3)² – 7 | Point |
| -5.65 | 0 | left root |
| -3 | -7 | minimum turning point |
| 0 | 2 | y-intercept |
| -0.35 | 0 | right root |
The graph is symmetric about the vertical line x = -3. This symmetry means that points on either side of the turning point mirror each other at the same height.
图像关于直线 x = -3 对称。这种对称性意味着顶点两侧的点在同一高度上相互镜像。
7. Common Mistakes | 常见错误
Many students lose marks on completing-the-square questions because of sign errors. A very common mistake is to write (x + 3)² + 7 instead of (x + 3)² – 7. Another frequent error is to state the turning point as (3, -7) instead of (-3, -7).
许多学生在配方法题目中因符号错误而丢分。一个非常常见的错误是把结果写成 (x + 3)² + 7，而不是 (x + 3)² – 7。另一个常见错误是把顶点写成 (3, -7)，而不是 (-3, -7)。
- Forgetting to subtract the square: x² + 6x + 2 is not equal to (x + 3)² + 2.
- Sign of the turning point: (x + p)² + q has vertex (-p, q), not (p, q).
- Missing the negative root: when solving (x + 3)² = 7, both x + 3 = √7 and x + 3 = -√7 must be used.
- Using decimals unnecessarily: exact surd form is expected in many IGCSE mark schemes.
- 忘记减去平方：x² + 6x + 2 不等于 (x + 3)² + 2。
- 顶点符号错误：(x + p)² + q 的顶点是 (-p, q)，而不是 (p, q)。
- 遗漏负根：解 (x + 3)² = 7 时，必须同时使用 x + 3 = √7 和 x + 3 = -√7。
- 不必要地使用小数：许多 IGCSE 评分标准要求保留根号精确形式。
8. Exam Technique | 考试技巧
Always show your working in three clear stages: write the completed-square form, set it equal to zero, and then solve by taking square roots. This structure makes it easy for the examiner to award method marks even if a small arithmetic slip appears later.
始终分三个清晰步骤展示解题过程：写出完全平方形式，令其等于零，然后开平方求解。这种结构能使阅卷人更容易给出方法分，即使后面出现小的计算错误。
After finding the turning point, write it as a coordinate pair with brackets. A statement such as “minimum point = (-3, -7)” is clearer than writing “x = -3, y = -7” without concluding.
求出顶点后，要把结果写成带括号的坐标。像 “最低点 = (-3, -7)” 这样的表述比只写 “x = -3, y = -7” 而不下结论更清晰。
9. Practice Variation | 变式练习
Try the same method on a slightly different quadratic. For example, express x² – 10x + 4 in the form (x + p)² + q, solve x² – 10x + 4 = 0, and state the minimum point of y = x² – 10x + 4.
尝试用一个稍有不同的二次式练习相同方法。例如，将 x² – 10x + 4 写成 (x + p)² + q 的形式，解方程 x² – 10x + 4 = 0，并写出 y = x² – 10x + 4 的最低点坐标。
x² – 10x + 4 = (x – 5)² – 21
x = 5 ± √21
Turning point = (5, -21)
Notice that when the middle term is negative, p becomes negative after rewriting as (x + p)². Many IGCSE questions test this sign change, so practise it carefully.
注意当中间项为负时，改写为 (x + p)² 后 p 会变为负数。许多 IGCSE 题目都会考查这种符号变化，因此要仔细练习。
10. Summary | 小结
Exercise 218 brings together three essential skills: completing the square, solving a quadratic equation in surd form, and interpreting the graph. If you master this sequence, you can confidently answer a wide range of IGCSE algebra and graph questions.
第 218 题整合了三种核心技能：配方法、以根号形式解二次方程，以及解读图像。如果你掌握了这一系列方法，就能自信地解答各种 IGCSE 代数与图像题目。
Remember the key relationship: y = (x + p)² + q has a turning point at (-p, q). Use it to move quickly from algebra to graph features without expanding the expression again.
记住关键关系：y = (x + p)² + q 的顶点是 (-p, q)。利用它可以从代数快速转换到图像特征，而无需再次展开表达式。
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