Example 4.5.2: Transformations of Quadratic Graphs | 例题4.5.2：二次函数图像的变换

📚 Example 4.5.2: Transformations of Quadratic Graphs | 例题4.5.2：二次函数图像的变换

Welcome to this focused revision article, centred on Example 4.5.2 from the AQA A-Level Mathematics syllabus. In this example, we take a quadratic function and apply a sequence of translations and reflections to its graph. By the end, you will understand exactly how each algebraic change to the equation corresponds to a geometric change on the coordinate plane.

欢迎阅读本篇专题复习文章，核心围绕 AQA A-Level 数学考纲中的例题 4.5.2。在本例中，我们取一个二次函数，并对其图像依次施加平移与反射变换。学完本文后，你将清楚理解方程上的每一步代数变化，如何对应到坐标平面上的几何变化。

1. The Quadratic Function f(x) = x² − 4x + 7 | 二次函数 f(x) = x² − 4x + 7

The function we work with in Example 4.5.2 is \(f(x) = x^2 – 4x + 7\), where \(x \in \mathbb{R}\). This is a quadratic function, so its graph is a parabola. Before performing any transformations, we should understand its basic shape and the position of its vertex (turning point).

例题 4.5.2 中我们使用的函数是 f(x) = x² − 4x + 7，其中 x ∈ ℝ。这是一个二次函数，其图像为抛物线。在进行任何变换之前，我们应先了解它的基本形状以及顶点（极值点）的位置。

To locate the vertex, we complete the square. This is an essential skill in AQA Pure Mathematics, as it converts a quadratic into a form that reveals translations directly.

为确定顶点位置，我们需要配方。这是 AQA 纯数学中的必备技能，因为它能将二次函数化为可直接揭示平移量的形式。

f(x) = x² − 4x + 7 = (x − 2)² + 3

Therefore, the vertex is at (2, 3), and since the coefficient of x² is positive, the parabola opens upward. The graph of y = f(x) is exactly the graph of y = x² translated by the vector (2, 3).

因此，顶点坐标为 (2, 3)，且由于 x² 的系数为正，抛物线开口向上。y = f(x) 的图像正是 y = x² 的图像沿向量 (2, 3) 平移后的结果。

2. Completed Square Form and Its Meaning | 配方法形式及其含义

The completed square form f(x) = (x − 2)² + 3 packs all the information about the parabola’s position into one expression. The term (x − 2)² tells us that the graph of x² has been shifted 2 units to the right; the +3 tells us it has also been shifted 3 units upward.

配方法形式 f(x) = (x − 2)² + 3 将所有关于抛物线位置的信息浓缩在一个表达式之中。(x − 2)² 这一项告诉我们 x² 的图像向右平移了 2 个单位；+3 告诉我们它同时向上平移了 3 个单位。

This viewpoint is crucial for Example 4.5.2. When we later apply further translations or reflections, we can either track the vertex geometrically or manipulate the algebraic expression — and both approaches must agree.

这一视角对例题 4.5.2 至关重要。当我们后续施加额外的平移或反射时，既可以从几何上追踪顶点的位置，也可以在代数上操作表达式——两种方法得到的结果必须一致。

For the original function, the vertex is a reference point. After any sequence of transformations, the new vertex tells us the final position of the parabola, and the coefficient of x² tells us whether it opens upward (positive) or downward (negative).

对原函数而言，顶点是一个参考点。在任意一系列变换之后，新的顶点告诉我们抛物线的最终位置，而 x² 的系数告诉我们其开口方向是朝上（正）还是朝下（负）。

3. Core Transformation Rules | 核心变换规则

AQA A-Level Mathematics expects you to know the following graph-transformation rules fluently. Applied to y = f(x), each rule changes the equation in a specific way.

AQA A-Level 数学要求你熟练掌握以下图像变换规则。对 y = f(x) 应用每条规则时，方程都按特定方式发生变化。

Notice the crucial sign convention: a translation of (a, 0) is written inside the function as f(x − a), not f(x + a). This counter-intuitive rule is a common source of lost marks.

请注意关键的符号约定：沿向量 (a, 0) 平移在函数内部写作 f(x − a)，而不是 f(x + a)。这个与直觉相反的规则是常见失分点。

Similarly, a translation by (a, b) combined is written as y = f(x − a) + b. We will apply this in part (a) of the example.

类似地，沿向量 (a, b) 的组合平移写作 y = f(x − a) + b。我们将在例题第（a）问中应用此规则。

4. Translation Vectors and Order of Operations | 平移向量与运算顺序

When two translations are applied one after another, the combined movement is simply the vector sum of the two translations. For instance, a translation by (3, 0) followed by a translation by (0, −2) is equivalent to a single translation by (3, 0) + (0, −2) = (3, −2).

当两个平移依次施加时，总位移就是两个平移向量的和。例如，先沿 (3, 0) 平移，再沿 (0, −2) 平移，等价于一次沿 (3, 0) + (0, −2) = (3, −2) 的平移。

This is true for translations, but it does not generally apply to reflections. The order of a reflection and a translation matters: reflecting first then translating usually produces a different result from translating first then reflecting.

这一点对平移成立，但对反射一般不成立。反射与平移的顺序至关重要：先反射再平移，与先平移再反射，通常得到不同的结果。

In Example 4.5.2 part (a), we only have translations. Therefore, we can fold the two steps into one. In part (b), we have a reflection followed by a translation, so we must apply the operations in the stated order.

在例题 4.5.2 的（a）问中，只有

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