📚 Example 4.6.1 — Transformations of Quadratic Graphs | 示例4.6.1——二次函数图像的变换
Example 4.6.1 is a classic AQA A-Level Mathematics problem on graph transformations. In this worked solution, we start with the graph of y = x² and apply a sequence of transformations to obtain y = 2(x − 3)² + 1. We will describe each transformation in the correct order, determine the vertex and line of symmetry of the new parabola, and explain the reasoning that you should replicate in an exam.
示例 4.6.1 是 AQA A-Level 数学中关于图像变换的经典例题。在本篇详解中，我们从 y = x² 的图像出发，依次施加一系列变换得到 y = 2(x − 3)² + 1。我们将按正确顺序描述每一步变换，确定新抛物线的顶点与对称轴，并解释你应该在考试中复现的推理过程。
1. The Original Graph: y = x² | 原函数图像：y = x²
Before applying any transformation, we must recall the key features of the graph of y = x². This is the simplest quadratic graph, often called the parent parabola. It has a vertex (turning point) at the origin (0, 0), a line of symmetry along the y-axis (x = 0), and it opens upwards with a positive gradient on both sides away from the vertex.
在施加任何变换之前，我们必须先回顾 y = x² 图像的关键特征。这是最简单的二次函数图像，常被称为抛物线的母函数。它的顶点（拐点）位于原点 (0, 0)，对称轴为 y 轴（即 x = 0），开口向上，且在顶点两侧的梯度均为正。
A useful mental picture is that the graph passes through (0, 0), (1, 1), (−1, 1), (2, 4) and (−2, 4). Every quadratic graph you meet in Chapter 4 can be traced back to this standard curve through translations, reflections and stretches.
一个有用的直观图像是：该抛物线经过 (0, 0)、(1, 1)、(−1, 1)、(2, 4) 和 (−2, 4)。你在第 4 章遇到的所有二次函数图像，都可以通过平移、反射和拉伸追溯到这条标准曲线。
2. The Four Fundamental Graph Transformations | 四种基本图像变换
AQA A-Level Mathematics requires you to be fluent in four types of transformation applied to a function y = f(x). You must know exactly which change to the equation produces which geometric effect on the graph.
AQA A-Level 数学要求你熟练掌握作用于函数 y = f(x) 的四种图像变换。你必须清楚方程中哪一处改变会对应图像上哪种几何效果。
| Transform 变换 | Equation 方程 | Geometric effect 几何效果 |
| Vertical translation 竖直平移 | y = f(x) + a | Moves up by a if a > 0, down by |a| if a < 0 (a > 0 时向上移动 a，a < 0 时向下移动 |a|) |
| Horizontal translation 水平平移 | y = f(x − b) | Moves right by b if b > 0, left by |b| if b < 0 (b > 0 时向右移动 b，b < 0 时向左移动 |b|) |
| Vertical stretch/compression 竖直拉伸/压缩 | y = k·f(x) | Stretch by scale factor k in the y-direction if k > 1; compression if 0 < k < 1 (k > 1 时沿 y 方向拉伸 k 倍；0 < k < 1 时压缩) |
| Horizontal stretch/compression 水平拉伸/压缩 | y = f(cx) | Compression by factor c in the x-direction if c > 1; stretch if 0 < c < 1 (c > 1 时沿 x 方向压缩 c 倍；0 < c < 1 时拉伸) |
A reflection in the x-axis is a special case of a vertical stretch with k = −1, and a reflection in the y-axis is the case y = f(−x). These appear regularly alongside the transformations above.
关于 x 轴的反射是竖直拉伸中 k = −1 的特殊情形，而关于 y 轴的反射则对应 y = f(−x)。这些变换与上述四种基本变换经常同时出现。
3. What Example 4.6.1 Asks | 示例 4.6.1 的题目要求
Example 4.6.1 in the AQA Pure Mathematics textbook is stated as follows: Starting from the graph of y = x², sketch the graph of y = 2(x − 3)² + 1. State clearly the three transformations applied, the coordinates of the new vertex, and the equation of the axis of symmetry.
AQA 纯数学教材中的示例 4.6.1 表述如下：从 y = x² 的图像出发，画出 y = 2(x − 3)² + 1 的图像。请明确指出所施加的三个变换、新顶点的坐标以及对称轴的方程。
The expression 2(x − 3)² + 1 is already in completed-square form, so we do not need to do any algebraic manipulation. Our task is purely to interpret the structure of the expression as a chain of transformations.
表达式 2(x − 3)² + 1 已经是配方后的形式，因此我们不需要进行任何代数化简。我们的任务纯粹是把这个表达式的结构理解为一系列变换的组合。
4. Step 1 — Horizontal Translation: y = (x − 3)² | 第一步——水平平移：y = (x − 3)²
The first transformation comes from replacing x with (x − 3) inside the bracket. The equation changes from y = x² to y = (x − 3)². This is a horizontal translation of the graph of y = x² by 3 units to the right.
第一个变换来自将括号内的 x 替换为 (x − 3)。方程从 y = x² 变为 y = (x − 3)²。这是将 y = x² 的图像向右平移 3 个单位。
It is a common source of confusion that x − 3 moves the graph to the right, not to the left. Remember the rule: to find the new x-coordinate, you solve x − 3 = 0, giving x = 3. Thus the vertex moves from (0, 0) to (3, 0).
一个常见的困惑是：x − 3 明明含有减号，为什么图像反而向右移动？记住这个规律：要找到新的 x 坐标，令 x − 3 = 0，解得 x = 3。因此顶点从 (0, 0) 移动到 (3, 0)。
y = (x − 3)² → Vertex: (3, 0), Axis: x = 3
At this intermediate stage the parabola still has the same shape as y = x²; it has simply slid 3 units to the right along the x-axis.
在此中间阶段，抛物线形状与 y = x² 完全相同；它只是沿 x 轴向右滑动了 3 个单位。
5. Step 2 — Vertical Stretch: y = 2(x − 3)² | 第二步——竖直拉伸：y = 2(x − 3)²
The second transformation is to multiply the whole expression by 2. The equation becomes y = 2(x − 3)². This is a vertical stretch with scale factor 2. Every y-coordinate on the graph of y = (x − 3)² is doubled, while every x-coordinate is unchanged.
第二个变换是将整个表达式乘以 2。方程变为 y = 2(x − 3)²。这是竖直方向上倍率为 2 的拉伸。y = (x − 3)² 图像上每一个点的 y 坐标都变为原来的 2 倍，而所有 x 坐标保持不变。
The vertex is unaffected because its y-coordinate is 0, and doubling 0 still gives 0. However, the point (4, 1) on the previous graph, for example, would move to (4, 2). The parabola becomes narrower in appearance because the vertical scale is larger.
顶点不受影响，因为它的 y 坐标为 0，将 0 翻倍仍然是 0。然而，前一个图像上的点 (4, 1) 会移动到 (4, 2)。抛物线的竖直尺度变大，因此看起来更“瘦长”。
y = 2(x − 3)² → Vertex: (3, 0), opening narrower
Note that the axis of symmetry is still x = 3, because the graph remains symmetric about that vertical line after a vertical stretch.
注意对称轴仍然是 x = 3，因为在竖直拉伸之后，图像关于该竖直线的对称性保持不变。
6. Step 3 — Vertical Translation: y = 2(x − 3)² + 1 | 第三步——竖直平移：y = 2(x − 3)² + 1
The final transformation is to add 1 to the whole expression, giving y = 2(x − 3)² + 1. This is a vertical translation of 1 unit upwards. Every y-coordinate on the graph of y = 2(x − 3)² is increased by 1.
最后一个变换是给整个表达式加上 1，得到 y = 2(x − 3)² + 1。这是将图像向上竖直平移 1 个单位。y = 2(x − 3)² 图像上每一个点的 y 坐标都增加 1。
Now the vertex moves from (3, 0) to (3, 1). This is the final position of the turning point. Because no horizontal movement is involved in this step, the axis of symmetry remains x = 3.
现在顶点从 (3, 0) 移动到 (3, 1)。这是拐点的最终位置。因为这一步不涉及水平移动，对称轴仍然是 x = 3。
y = 2(x − 3)² + 1 → Final vertex: (3, 1)
At this point we have the complete target equation. The three transformations in order are: horizontal translation 3 units right, vertical stretch factor 2, vertical translation 1 unit up.
至此我们得到了完整的目标方程。按顺序的三个变换是：向右水平平移 3 个单位、竖直拉伸倍率 2、向上竖直平移 1 个单位。
7. Sketching the Final Graph | 画出最终图像
To sketch the graph accurately, we should locate at least one additional point beyond the vertex. The easiest point to find is the y-intercept, which occurs where x = 0.
要准确地画出图像，我们至少需要找到顶点之外的一个点。最容易求的是 y 轴截距，即当 x = 0 时的点。
Substitute x = 0: y = 2(0 − 3)² + 1 = 2 × 9 + 1 = 19
Therefore the graph passes through (0, 19). By symmetry about x = 3, the point (6, 19) is also on the curve. These two points, together with the vertex (3, 1), are enough to produce a reliable freehand sketch.
因此图像经过点 (0, 19)。根据关于 x = 3 的对称性，点 (6, 19) 也在曲线上。这两个点连同顶点 (3, 1)，足以帮助我们徒手画出一条可靠的草图。
Draw a smooth U-shaped curve through (3, 1), rising steeply on both sides, passing through (0, 19) on the left and (6, 19) on the right. Clearly label the vertex, the intercept, and the axis x = 3 on your sketch.
画一条光滑的 U 形曲线，经过 (3, 1)，两侧迅速上升，左边经过 (0, 19)，右边经过 (6, 19)。在草图上清楚标注顶点、截距和对称轴 x = 3。
8. Why the Order of Transformations Matters | 为什么变换顺序很重要
Suppose we applied the vertical translation before the vertical stretch. We would start from y = x², translate up by 1 to get y = x² + 1, then stretch by factor 2 to get y = 2(x² + 1) = 2x² + 2. This is not the same as our target y = 2(x − 3)² + 1. The two results differ, so the order clearly matters.
假设我们先做竖直平移再做竖直拉伸。我们从 y = x² 出发，向上平移 1 得到 y = x² + 1，再拉伸倍率 2 得到 y = 2(x² + 1) = 2x² + 2。这与目标式 y = 2(x − 3)² + 1 完全不同。两者结果不同，因此变换顺序显然很重要。
A correct way to read the expression 2(x − 3)² + 1 is to work inside the bracket first, then multiply, then add. Algebraically, the order of operations in the expression gives the order of transformations: replace x first, then scale the output, then translate vertically.
解读表达式 2(x − 3)² + 1 的正确方法是：先处理括号内，再乘法，最后加法。代数上，表达式中的运算顺序决定了变换顺序：先替换 x，再缩放输出值，最后竖直平移。
In general, you should always apply transformations in the order: horizontal translations and stretches first, followed by vertical stretches, followed by vertical translations. This matches the natural structure of the equation.
一般情况下，应始终按照如下顺序施加变换：先水平平移和水平伸缩，再做竖直伸缩，最后做竖直平移。这正好与方程的自然结构一致。
9. Common Mistakes in Exam Questions | 考试中的常见错误
The first and most frequent mistake is moving y = (x − 3)² to the left instead of the right. Many students incorrectly think that minus signs inside the bracket mean a leftward shift. Always test with the vertex: x − 3 = 0 gives x = 3, so the vertex moves to x = 3.
第一个也是最常见的错误，是把 y = (x − 3)² 向左边移动而不是向右。许多学生错误地认为括号内的减号意味着向左平移。始终用顶点来检验：x − 3 = 0 得到 x = 3，因此顶点移动到 x = 3。
The second common mistake is forgetting that the vertical stretch must be applied to the whole expression 2(x − 3)² before adding 1. If you instead write 2(x − 3)² + 1 as a vertical stretch of y = (x − 3)² + 1, you will get y = 2(x − 3)² + 2, which is wrong.
第二个常见错误是忘记竖直拉伸必须先作用于整个表达式 2(x − 3)²，然后才能加 1。如果你错误地把 2(x − 3)² + 1 理解为对 y = (x − 3)² + 1 做竖直拉伸，得到 y = 2(x − 3)² + 2，那就是错的。
The third mistake is stating the axis of symmetry incorrectly. Because the horizontal translation is 3 units right, the axis is x = 3, not x = −3. Double-check the sign of the coordinate of the vertex before writing the final answer.
第三个错误是写错对称轴方程。因为水平平移是向右 3 个单位，对称轴是 x = 3，而不是 x = −3。在写出最终答案之前，务必核对顶点坐标的符号。
10. A General Method for Any Similar Example | 处理同类问题的一般方法
You can apply the same reasoning to any graph of the form y = a(x − h)² + k. The number h gives the horizontal shift (right if h > 0, left if h < 0), the number a gives the vertical stretch factor with a > 0, and k gives the vertical shift. The vertex is (h, k) and the axis of symmetry is x = h.
对于任何形如 y = a(x − h)² + k 的图像，你都可以套用同样的思路。数字 h 决定水平平移（h > 0 向右，h < 0 向左），数字 a 决定竖直拉伸倍率（a > 0），k 决定竖直平移。顶点为 (h, k)，对称轴为 x = h。
If a is negative, the graph is reflected in the x-axis as well as stretched. For example, y = −2(x + 4)² − 5 would have vertex (−4, −5), open downwards, and a vertical stretch factor of 2 after reflection.
如果 a 为负值，则图像在拉伸之外还要关于 x 轴反射。例如 y = −2(x + 4)² − 5 的顶点为 (−4, −5)，开口向下，且在反射之后竖直拉伸倍率为 2。
For non-quadratic functions, the same rules apply. If you know the key points of f(x), you can transform them one by one. Keep a checklist: which points are affected by each transformation, and which are invariant.
对于非二次函数，同样的规则仍然适用。只要你知道 f(x) 的关键点，就可以逐一变换这些点。请使用这个检查清单：每一步变换影响哪些点，哪些点保持不变。
11. Practice Problem | 配套练习
Try this question on your own: The graph of y = x² is transformed to y = 3(x + 2)² − 4. Write down the coordinates of the vertex, the equation of the axis of symmetry, and the y-intercept. Sketch the graph.
请独立尝试这道题：y = x² 的图像经过变换得到 y = 3(x + 2)² − 4。写出顶点坐标、对称轴方程和 y 截距，并画出草图。
Solution check: The horizontal translation is 2 units left, so the vertex is (−2, −4). The vertical stretch factor is 3. At x = 0, y = 3(2)² − 4 = 8, so the y-intercept is 8. The axis of symmetry is x = −2.
答案核对：水平平移是向左 2 个单位，所以顶点为 (−2, −4)。竖直拉伸倍率为 3。当 x = 0 时，y = 3(2)² − 4 = 8，所以 y 截距为 8。对称轴为 x = −2。
12. Conclusion | 总结
Example 4.6.1 is a perfect illustration of the power of transformational thinking in AQA A-Level Mathematics. By breaking y = 2(x − 3)² + 1 into three separate transformations, we can sketch the graph quickly and reliably without plotting many points.
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