📚 Example 4.3.5: Transformations of Graphs | 例4.3.5：图形变换
This article works through Example 4.3.5 from the AQA A-Level Mathematics Pure Year 1 curriculum (Chapter 4: Graphs and Transformations). This example focuses on applying a combined translation to a cubic function and determining the equation of the transformed curve. We will explore the underlying theory, step-by-step reasoning, common pitfalls, and verification methods.
本文解析AQA A-Level数学纯数学第1册课程（第4章：图形与变换）中的例4.3.5。该例题着重于对三次函数施以复合平移，并求出变换后曲线的方程。我们将深入探讨其理论背景、逐步推理过程、常见陷阱及验证方法。
Graph transformations are a cornerstone of A-Level mathematics. They appear throughout the pure mathematics papers, in mechanics problems involving displacement–time graphs, and even in statistics when analysing density curves. A firm grasp of how translations affect a function’s equation is therefore not only examinable in its own right but also a prerequisite for later topics such as curve sketching, numerical methods and mathematical modelling.
图形变换是A-Level数学的基石。它不仅出现在纯数学试卷中，还出现在力学中涉及位移-时间图的问题里，甚至在分析密度曲线的统计学中也有应用。因此，牢固掌握平移对函数方程的影响，不仅是考试的直接考点，也是后续主题（如曲线草图、数值方法和数学建模）的必备基础。
1. The Problem Statement | 题目陈述
The curve with equation y = x³ is translated by vector (2, −3). Determine the equation of the new curve and describe the geometrical effect of this transformation.
已知曲线 y = x³ 沿向量 (2, −3) 平移。求新曲线的方程，并描述该变换的几何效果。
This type of question appears regularly in AQA Paper 1 of the Pure Mathematics examination. It tests three key skills: interpreting a translation vector correctly, substituting the appropriate expression into the original function, and understanding the relationship between an algebraic equation and its geometric representation.
这类问题经常出现在AQA纯数学试卷1中。它考查三项关键技能：正确解读平移向量、将相应表达式代入原函数，以及理解代数方程与几何表示之间的关系。
Because this example appears in Section 4.3, students are expected to solve it without a calculator and to be able to sketch the resulting graph accurately on a set of axes. The problem looks simple, but it encapsulates several concepts that underpin all later transformation work.
由于该例题出现在第4.3节，学生应在不使用计算器的情况下解答，并能在坐标系中准确绘制出结果图形。问题看似简单，却蕴含着支撑后续所有变换工作的多个核心概念。
2. Key Concept: Translations | 关键概念：平移
A translation is a transformation that slides a graph horizontally and/or vertically without rotating, reflecting or resizing it. Because the shape of the graph is preserved, every single point on the curve moves by the same displacement vector. This distinguishes a translation from a reflection (which flips the graph) or a stretch (which changes its scale).
平移是指将图像沿水平方向和/或垂直方向滑动，而不旋转、反射或改变其大小。由于图像的形状保持不变，曲线上的每一个点都以相同的位移向量移动。这使平移区别于反射（翻转图像）或伸缩（改变其比例）。
For a function y = f(x), the following translation rules apply. This table is essential and should be memorised:
对于函数 y = f(x)，以下平移规则适用。此表至关重要，应牢记：
| Translation vector | New equation | Geometric effect |
| (a, 0) | y = f(x − a) | Shift a units right (if a > 0) or left (if a < 0) |
| (0, b) | y = f(x) + b | Shift b units up (if b > 0) or down (if b < 0) |
| (a, b) | y = f(x − a) + b | Shift by vector (a, b) |
| 平移向量 | 新方程 | 几何效果 |
| (a, 0) | y = f(x − a) | 向右移动a个单位（a > 0）或向左移动（a < 0） |
| (0, b) | y = f(x) + b | 向上移动b个单位（b > 0）或向下移动（b < 0） |
| (a, b) | y = f(x − a) + b | 沿向量 (a, b) 移动 |
Notice the crucial sign convention: a positive value of a in the vector results in a minus sign inside the function bracket, as in x − a. This is the single most common source of sign errors in graph transformation questions.
请注意关键的正负号约定：向量中a为正数时，函数括号内出现减号，如 x − a。这是图形变换问题中正负号错误最常见的来源。
Similarly, the second component b is added outside the function, directly to the entire expression. The horizontal component always acts inside the function argument, whereas the vertical component always acts outside — this distinction is fundamental.
类似地，第二个分量b在函数外部加上，直接作用于整个表达式。水平分量始终作用于函数自变量内部，而垂直分量始终作用在外部——这一区别是根本性的。
3. Reading the Vector | 解读向量
The given translation vector is (2, −3). We identify a = 2 and b = −3. This means every point on the curve moves 2 units to the right (positive x-direction) and 3 units downwards (negative y-direction).
给定的平移向量为 (2, −3)。我们识别出 a = 2，b = −3。这意味着曲线上的每个点向右移动2个单位（x正方向），向下移动3个单位（y负方向）。
It is helpful to visualise the movement as a diagonal displacement. If you were standing at any point on the original curve, you would take two steps to the right and three steps downward to land on the corresponding point of the transformed curve. The horizontal distance moved is 2 units; the vertical distance moved is 3 units — and because the movement is downward, b is negative.
将移动视为对角位移有助于理解。如果你站在原曲线上的任意一点，你需要向右走两步、向下走三步，才能到达变换后曲线上对应的点。水平移动距离为2个单位；垂直移动距离为3个单位——因为移动方向向下，所以b为负。
Once we have correctly identified the components, we can proceed with the substitution. The order in which we apply the two components does not affect the final answer, but for pedagogical clarity we will first handle the horizontal component, then the vertical component.
一旦正确识别了分量，我们就可以进行代换。两个分量应用的先后顺序不影响最终答案，但为了教学的清晰性，我们首先处理水平分量，然后处理垂直分量。
4. Applying the Horizontal Translation | 应用水平平移
We begin with the horizontal component. Since a = 2, the instruction is to replace every occurrence of x in the original function with (x − 2).
我们从水平分量开始。由于 a = 2，操作指令是将原函数中的每一个 x 替换为 (x − 2)。
The original function is f(x) = x³. Performing the substitution x → (x − 2) gives:
原函数为 f(x) = x³。执行代换 x → (x − 2) 后得到：
y = (x − 2)³
Let us reason about why this works. The original curve y = x³ passes through the point (0, 0). When the curve is translated 2 units to the right, this point moves to (2, 0). For the new equation to pass through (2, 0), we need the value of x³ to be zero when x = 2; the expression (x − 2)³ achieves precisely this because (2 − 2)³ = 0.
让我们思考为什么这样做有效。原曲线 y = x³ 经过点 (0,
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