📚 Example 1.3.1: Completing the Square | 示例1.3.1：配方法
Welcome to this worked example from the AQA A-Level Mathematics syllabus. In this revision article, we will solve Example 1.3.1 step by step, explaining the method of completing the square and its application to finding the minimum point of a quadratic function.
欢迎阅读AQA A-Level数学课程中的这道示例题。在这篇复习文章中，我们将逐步解答示例1.3.1，阐释配方法及其在求二次函数最小点中的应用。
1. Problem Statement | 问题陈述
in the completed-square form a(x + b)² + c, and hence state the coordinates of its minimum point.
我们给定二次函数 f(x) = 2x² − 4x + 5。要求将其写成完全平方形式 a(x + b)² + c，并由此指出其最小点坐标。
2. Identifying Coefficients | 系数识别
The general quadratic form is ax² + bx + c. For f(x), we have a = 2, b = −4, and c = 5.
一般二次形式为 ax² + bx + c。对 f(x)，我们得到 a = 2，b = −4，c = 5。
3. Step 1: Factor Out the Leading Coefficient | 第一步：提取二次项系数
st two terms while keeping c separate. This prepares the bracket for completing the square.
因为 a = 2，我们从前两项中提出因子2，同时保持 c 单独。这为括号内配方法做好准备。
f(x) = 2(x² − 2x) + 5
Notice that we have not changed the function; the expression is equivalent to the original.
注意我们并没有改变函数；该表达式与原式等价。
4. Step 2: Complete the Square Inside the Brackets | 第二步：在括号内完成平方
Inside the parentheses we have x² − 2x. The coefficient of x is −2. Half of this is −1, and its square is 1. We add and subtract 1 inside the bracket:
括号内是 x² − 2x。x 的系数是 −2。它的一半是 −1，其平方是 1。我们在括号内加上并减去 1：
f(x) = 2[(x − 1)² − 1] + 5
This technique is valid because adding and subtracting the same value preserves equality.
该技巧有效是因为加上和减去同一个值不改变等式。
5. Step 3: Distribute and Simplify | 第三步：分配并化简
Now multiply the bracket
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