📚 Example 4.4.1: Graph Transformations and Curve Sketching | 例 4.4.1：图像变换与曲线草图
In AQA A-Level Mathematics, graph transformations are a core part of the pure mathematics specification. Being able to start from a known function and apply translations, stretches, and reflections quickly and accurately is essential for both the AS and A-Level examinations. This article works through a typical Example 4.4.1 in detail, showing how to sketch a family of related curves and how to check your work.
在 AQA A-Level 数学中，图像变换是纯数学考试大纲的核心内容。能够从已知函数出发，快速、准确地应用平移、伸缩和反射变换，对 AS 和 A-Level 考试都至关重要。本文将详细讲解一个典型的例 4.4.1，展示如何绘制一族相关曲线，并检查你的答案。
1. The Set Up of Example 4.4.1 | 例 4.4.1 的题目设置
the graph of y = f(x), then use transformations to sketch y = f(x) + 2, y = f(2x), and y = |f(x)| tical translation, horizontal stretch, and modulus transformation in one coherent example.
本例中，我们从二次函数 f(x) = x² - 4x + 3 出发。任务是在不同坐标系中分别画出 y = f(x)、y = f(x) + 2、y = f(2x) 和 y = |f(x)| 的图像。这道题将垂直平移、水平伸缩和绝对值变换整合在一个完整的例题中。
Before applying any transformation, we must understand the key features of the parent function f(x). This includes its shape, vertex, axis of symmetry, and intercepts. Once these features are known, each transformation can be applied point by point or through the general mapping rules.
在应用任何变换之前，我们必须先掌握原始函数 f(x) 的关键特征。这些特征包括形状、顶点、对称轴以及截距。一旦掌握了这些特征，就可以逐点或通过一般映射规则来应用每一个变换。
2. Find the Key Features of f(x) = x² - 4x + 3 | 求 f(x) = x² - 4x + 3 的关键特征
atic with a positive leading coefficient, so its graph is a U-shaped parabola. To locate its vertex, we can complete the square.
函数 f(x) = x² - 4x + 3 是一个二次函数，且首项系数为正，因此它的图像是一条开口向上的 U 形抛物线。为了找到顶点，我们可以使用配方法。
r the x-intercepts, solve f(x) = 0. These points give a reliable framework for the sketch.
同时，求出 x 轴截距和 y 轴截距也很有帮助。求 y 轴截距时令 x = 0；求 x 轴截距时解方程 f(x) = 0。这些点为绘制图像提供了可靠的框架。
3. Complete the Square to Locate the Vertex | 配方法求顶点
lete the square. The coefficient of x is -4, half of this is -2, and (-2)² = 4. Therefore:
从 f(x) = x² - 4x + 3 开始，我们进行配方。x 的系数是 -4，它的一半是 -2，而 (-2)² = 4。因此：
f(x) = (x - 2)² - 1
cause the squared term is positive, the parabola opens upwards, so this vertex is a minimum point.
这个完全平方形式表明顶点坐标为 (2, -1)。对称轴是铅直线 x = 2。由于平方项为正，抛物线开口向上，因此该顶点是一个最小值点。
e (1, 0) and (3, 0). The y-intercept is found by substituting x = 0, which gives f(0) = 3.
利用完全平方形式解方程 f(x) = 0，得到 (x - 2)² = 1，所以 x - 2 = ±1。因此 x = 1 或 x = 3。x 轴截距为 (1, 0) 和 (3, 0)。y 轴截距通过代入 x = 0 求得，即 f(0) = 3。
4. Sketch the Parent Graph y = f(x) | 画原图像 y = f(x)
We now have four key points for the parent graph: the vertex (2, -1), the x-intercepts (1, 0) and (3, 0), and the y-intercept (0, 3). The graph is a U-shaped parabola passing through these points, with its lowest point at (2, -1).
现在我们有了原图像的四个关键点：顶点 (2, -1)、x 轴截距 (1, 0) 和 (3, 0)，以及 y 轴截距 (0, 3)。图像是一条经过这些点的 U 形抛物线，最低点位于 (2, -1)。
hed vertical line through x = 2. The curve should be smooth and symmetric about this line.
画图时，要清楚地标出顶点和截距。用虚线画出经过 x = 2 的对称轴。曲线应光滑且关于这条线对称。
5. Vertical Translation: y = f(x) + 2 | 垂直平移 y = f(x) + 2
its upwards. This means that every point (x, y) on the original graph moves to (x, y + 2).
变换 y = f(x) + 2 表示将 y = f(x) 的图像垂直向上平移 2 个单位。这意味着原图上的每一个点 (x, y) 都移动到 (x, y + 2)。
h has been lifted; instead, the new curve will cross the y-axis at (0, 5) and may not cross the x-axis at all.
因此，顶点 (2, -1) 移动到 (2, 1)。原 x 轴截距不再位于 x = 1 和 x = 3，因为整个图像已被抬高；新曲线将在 (0, 5) 处穿过 y 轴，并且可能根本不穿过 x 轴。
is negative, so there are no real roots. This confirms that the graph does not cross the x-axis.
为了确认变换后图像与 x 轴的交点，解方程 f(x) + 2 = 0。得到 x² - 4x + 5 = 0。判别式为 (-4)² - 4(1)(5) = 16 - 20 = -4，结果为负，因此没有实数根。这证实了图像不与 x 轴相交。
imum value of 1. The curve is the same shape as the original parabola but shifted up by 2 units.
新函数的完全平方形式为 y = (x - 2)² + 1，最小值为 1。曲线形状与原抛物线相同，但向上平移了 2 个单位。
6. Horizontal Stretch: y = f(2x) | 水平伸缩 y = f(2x)
nts a horizontal stretch by a scale factor of 1/2, or equivalently a compression towards the y-axis. In general, replacing x with kx, where k > 1, compresses the graph horizontally by a factor of 1/k.
变换 y = f(2x) 表示水平伸缩，缩放因子为 1/2，等价于向 y 轴方向压缩。一般地，将 x 替换为 kx（k > 1）时，图像水平压缩为原来的 1/k。
3. This is still a quadratic, but its coefficients have changed, so the graph is narrower.
我们可以直接计算新函数：f(2x) = (2x)² - 4(2x) + 3 = 4x² - 8x + 3。这仍然是二次函数，但系数发生了改变，因此图像更窄。
find the x-intercepts, solve 4x² - 8x + 3 = 0. Using the quadratic formula gives x = [8 ± √64 - 48] / 8 = [8 ± √16] / 8 = [8 ± 4] / 8. Therefore x = 12/8 = 3/2 or x = 4/8 = 1/2.
为求 x 轴截距，解方程 4x² - 8x + 3 = 0。使用求根公式得到 x = [8 ± √64 - 48] / 8 = [8 ± √16] / 8 = [8 ± 4] / 8。因此 x = 12/8 = 3/2 或 x = 4/8 = 1/2。
4(x² - 2x) + 3 = 4[(x - 1)² - 1] + 3 = 4(x - 1)² - 1. Therefore the vertex is at (1, -1).
新图像的顶点可以再次通过配方法求得：4x² - 8x + 3 = 4(x² - 2x) + 3 = 4[(x - 1)² - 1] + 3 = 4(x - 1)² - 1。因此顶点位于 (1, -1)。
Notice that the original vertex (2, -1) has moved to (1, -1), which is exactly the horizontal compression by a factor of 1/2: the x-coordinate is halved, while the y-coordinate remains unchanged.
注意，原顶点 (2, -1) 已移动到 (1, -1)，这正是水平压缩为原来的 1/2 的结果：x 坐标减半，而 y 坐标保持不变。
7. Absolute Value Transformation: y = |f(x)| | 绝对值变换 y = |f(x)|
The transformation y = |f(x)| the x-axis unchanged. This is because the absolute value makes all y-values non-negative.
变换 y = |f(x)| 将 y = f(x) 图像中位于 x 轴下方的部分向上反射，而保持 x 轴上或上方的部分不变。这是因为绝对值使所有 y 值都变为非负。
, -1), so the portion between the x-intercepts (1, 0) and (3, 0) lies below the x-axis. Under the modulus transformation, this part is reflected upwards, creating a W-like shape with a local maximum at (2, 1).
对于 f(x) = x² - 4x + 3，其顶点位于 (2, -1)，因此在 x 轴截距 (1, 0) 与 (3, 0) 之间的部分位于 x 轴下方。在绝对值变换下，这一部分被向上反射，形成一个类似 W 的形状，并在 (2, 1) 处形成局部最大值。
The transformed graph can be written as y = |(x - 2)² - 1| < x < 3, the original function is negative, so the graph is reflected to y = 1 - (x - 2)².
变换后的图像可以写为 y = |(x - 2)² - 1|。当 x ≤ 1 或 x ≥ 3 时，原函数为非负，因此图像保持不变。当 1 < x < 3 时，原函数为负，因此图像被反射为 y = 1 - (x - 2)²。
) and (3, 0) and reaches a local maximum of 1 at x = 2. The y-intercept remains at (0, 3).
这样就形成了一条光滑曲线，在 (1, 0) 和 (3, 0) 处与 x 轴相切，并在 x = 2 处达到局部最大值 1。y 轴截距仍为 (0, 3)。
8. Combined Transformations and Order | 组合变换与顺序
o or more transformations in sequence. For example, a question might ask you to sketch y = |f(2x)| or y = f(x) + 2 after a stretch. Understanding the correct order is crucial.
在 AQA 考试题中，你可能会被要求按顺序应用两个或更多变换。例如，题目可能要求你画出 y = |f(2x)| 或先伸缩后平移的图像。理解正确的顺序至关重要。
For horizontal transformations, the replacement of x by kx or x + a is applied inside the function, so the order can affect the result. However, for the transformations in this example, each graph is sketched separately from the parent f(x), so no combination error arises.
对于水平变换，将 x 替换为 kx 或 x + a 是在函数内部进行的，因此顺序会影响结果。不过，在本例的各变换中，每张图都是分别从原始函数 f(x) 直接画出的，因此不会产生组合错误。
If you were asked to sketch y = |f(2x)| nations of horizontal changes and modulus should always be considered from the inside out.
如果你被要求画出 y = |f(2x)|，你应该先将 y = f(x) 水平压缩为原来的 1/2，然后再将负值部分向上反射。水平变化与绝对值组合时，始终应从内到外考虑顺序。
9. Key Comparisons in a Table | 表格关键对比
The table below summarises the key features of each graph in Example 4.4.1. This side-by-side comparison helps you see exactly how each transformation affects the shape and position.
下表总结了例 4.4.1 中各图像的关键特征。这种并排比较有助于你清楚地看到每个变换如何影响图像的形状和位置。
| Function | 函数 | Vertex | 顶点 | x-intercepts | x轴截距 | Shape change | 形状变化 |
|---|---|---|---|
| y = f(x) | (2, -1) | (1, 0), (3, 0) | Original U-shaped parabola | 原 U 形抛物线 |
| y = f(x) + 2 | (2, 1) | None | 无 | Shifted up 2 units | 向上平移 2 个单位 |
| y = f(2x) | (1, -1) | (1/2, 0), (3/2, 0) | Horizontally compressed by factor 1/2 | 水平压缩为 1/2 |
| y = |f(x)| | Local max at (2, 1) | 局部最大值在 (2, 1) | (1, 0), (3, 0) | Negative part reflected upwards | 负值部分向上反射 |
y = f(x) is (0, 3), for y = f(x) + 2 it is (0, 5), for y = f(2x) it is (0, 3), and for y = |f(x)| it is also (0, 3). These checks are useful in exam conditions.
注意 y = f(x) 的 y 轴截距为 (0, 3)，y = f(x) + 2 的 y 轴截距为 (0, 5)，y = f(2x) 的 y 轴截距为 (0, 3)，而 y = |f(x)| 的 y 轴截距也为 (0, 3)。这些检查在考试中非常有用。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
ph horizontally by a factor of 1/2, not 2. The coefficient inside the function has the opposite effect to what many students first expect.
一个常见错误是混淆垂直伸缩和水平伸缩。请记住，y = f(2x) 是将图像水平压缩为原来的 1/2，而不是 2 倍。函数内部的系数所产生的效果与许多学生最初的预期相反。
e mapping rule, for example (x, y) → (x, y + 2) for y = f(x) + 2, prevents this confusion.
另一个常见错误是忘记垂直平移只影响 y 坐标，而水平变换只影响 x 坐标。写出映射规则，例如对于 y = f(x) + 2 使用 (x, y) → (x, y + 2)，可以避免这种混淆。
re y ≥ 0 exactly where they are, and reflect points where y < 0 upwards across the x-axis.
对于绝对值图像，学生有时会反射整条曲线，而不是只反射负值部分。规则是：保持所有 y ≥ 0 的点不动，将 y < 0 的点关于 x 轴向上反射。
In the exam, always label key points on your sketch. Even if your drawing is not perfectly to scale, clear labels for the vertex, intercepts, and turning points will earn marks.
在考试中，一定要在草图上标出关键点。即使你的绘图不完全按比例，清楚地标出顶点、截距和转折点也能得分。
11. Check Your Understanding | 自测练习
Test yourself by working through these quick questions based on Example 4.4.1. Try to sketch each graph before scrolling back to check the answers.
通过完成以下基于例 4.4.1 的快速练习来检验自己的理解。在回看答案之前，先尝试画出每个图像。
- Find the vertex of the graph y = f(x + 3), where f(x) = x² - 4x + 3. | 求 y = f(x + 3) 的顶点，其中 f(x) = x² - 4x + 3。
- Write down the equation of the graph obtained when y = f(x) is reflected in the y-axis. | 写出将 y = f(x) 关于 y 轴反射后所得图像的方程。
- Sketch y = 2f(x) and describe how it differs from y = f(x). | 画出 y = 2f(x) 并描述它与 y = f(x) 有何不同。
he third, y = 2f(x) is a vertical stretch by a factor of 2, so the vertex becomes (2, -2).
对于第一个问题，将 x 替换为 x + 3 会将图像向左平移 3 个单位，因此顶点从 (2, -1) 移到 (-1, -1)。对于第二个问题，关于 y 轴反射得到 y = f(-x) = x² + 4x + 3。对于第三个问题，y = 2f(x) 是垂直拉伸为原来的 2 倍，因此顶点变为 (2, -2)。
12. Summary and Final Reminders | 总结与最终提醒
Example 4.4.1 brings together three essential AQA transformations: a vertical translation, a horizontal stretch, and a modulus transformation. The key is to first identify the vertex and intercepts of the original function, then apply each transformation systematically using coordinate mapping.
例 4.4.1 汇集了三种重要的 AQA 变换：垂直平移、水平伸缩和绝对值变换。关键是先确定原函数的顶点和截距，然后利用坐标映射系统地应用每个变换。
Always write the transformed function in its simplified form when possible, as this allows you to find the exact coordinates of key points. Completing the square is a powerful tool for quadratics because it reveals the vertex and the range of the function immediately.
在可能的情况下，始终写出简化后的变换函数，这样你就能找到关键点的精确坐标。配方法是处理二次函数的有力工具，因为它能立即显示顶点和函数的值域。
Finally, practise sketch graphs regularly. The speed and accuracy you develop through examples like this one will make a significant difference in the AQA A-Level Mathematics examination.
最后，要经常练习画图。通过这类例题所培养的速度和准确性，将在 AQA A-Level 数学考试中带来显著的提升。
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