Example 4.3.4: Graph Transformations of a Cubic Function | 例4.3.4：三次函数图像变换

📚 Example 4.3.4: Graph Transformations of a Cubic Function | 例4.3.4：三次函数图像变换

This article offers a complete walkthrough of Example 4.3.4 from the AQA A-Level Mathematics syllabus. We dissect a typical problem involving a cubic curve and apply a sequence of graph transformations. We will show not only how to obtain the final equation, but also why each order of transformation is essential. By the end, you should be confident with translations, reflections, and stretches, and be able to handle similar questions in the Pure Mathematics exam.

本文将对 AQA A-Level 数学大纲中的例 4.3.4 进行完整解析。我们选取一个典型的三次曲线问题，并应用一系列图像变换。我们不仅要展示如何得到最终方程，还要解释为什么每一步变换的顺序至关重要。学完本文后，你应该能够熟练处理平移、反射和伸缩变换，并在纯数考试中应对同类问题。

1. The Example Problem | 1. 例题描述

Consider the function f(x) = x³ − 3x² + 2. The graph of y = f(x) is transformed by applying the equation y = −2 f(x + 1) + 4. Part (a) asks us to write down the equation of the transformed curve in expanded polynomial form. Part (b) asks us to describe the sequence of transformations that map the original graph onto the new graph.

已知函数 f(x) = x³ − 3x² + 2。曲线 y = f(x) 经过变换得到新方程 y = −2 f(x + 1) + 4。第 (a) 问要求我们写出变换后曲线的多项式展开式。第 (b) 问要求我们描述从原图到新图所经历的变换顺序。

2. Understanding the Original Function | 2. 理解原函数

Before transforming, it is useful to know a few basic features of f(x). The function is a cubic polynomial, so its graph has a general ‘S’ shape. We can find the y-intercept by setting x = 0:

在进行变换之前，有必要了解 f(x) 的基本特征。这是一个三次多项式，因此图像大致呈现“S”形。令 x = 0，可以求出 y 轴截距：

f(0) = 0³ − 3(0)² + 2 = 2

So the original curve passes through (0, 2). We also note that as x → ∞, f(x) → ∞, and as x → −∞, f(x) → −∞ because the leading coefficient of x³ is positive.

因此原曲线经过点 (0, 2)。我们还注意到，因为 x³ 的系数为正，所以当 x → ∞ 时 f(x) → ∞；当 x → −∞ 时 f(x) → −∞。

3. Transformation Step 1: Horizontal Shift y = f(x + 1) | 3. 变换第一步：水平平移 y = f(x + 1)

In the expression f(x + 1), the input x is replaced by x + 1. This causes the graph to shift horizontally to the left by 1 unit. It is easy to make a sign error here: x + 1 means left, not right. To verify, consider a point on the original curve such as (0, 2). After the shift, this point moves to (−1, 2).

在表达式 f(x + 1) 中，输入 x 被替换为 x + 1。这会使图像向左平移 1 个单位。这里很容易弄错符号：x + 1 表示向左，而不是向右。我们可以验证：原曲线上有一点 (0, 2)，平移后该点变为 (−1, 2)。

Consequently, we compute:

因此，我们计算：

f(x + 1) = (x + 1)³ − 3(x + 1)² + 2

Expansion gives:

展开得到：

(x + 1)³ = x³ + 3x² + 3x + 1

3(x + 1)² = 3x² + 6x + 3

Thus:

因此：

f(x + 1) = x³ + 3x² + 3x + 1 − 3x² − 6x − 3 + 2 = x³ − 3x

So after the first transformation, the equation is y = x³ − 3x.

所以在第一步变换后，方程为 y = x³ − 3x。

4. Transformation Step 2: Vertical Stretch and Reflection y = −2 f(x + 1) | 4. 变换第二步：竖直伸缩与反射 y = −2 f(x + 1)

Now we multiply f(x + 1) by −2. Multiplication by −2 does two things at once. First, applying the factor 2 vertically stretches the graph by a factor of 2. Second, the negative sign reflects the graph in the x-axis. This means every y-coordinate is multiplied by −2.

现在把 f(x + 1) 乘以 −2。乘以 −2 会同时产生两个效果。首先，因子 2 使图像在竖直方向拉伸为原来的 2 倍。其次，负号使图像关于 x 轴反射。也就是说，每一个 y 坐标都乘以 −2。

Applying this to y = x³ − 3x gives:

将这个操作应用于 y = x³ − 3x，得到：

y = −2(x³ − 3x) = −2x³ + 6x

Therefore after this step the intermediate equation is y = −2x³ + 6x. Notice that the x-intercepts remain the same because zero multiplied by −2 is still zero.

因此在这一步之后，中间方程为 y = −2x³ + 6x。注意 x 轴交点不变，因为零乘以 −2 仍为零。

5. Transformation Step 3: Vertical Translation y = −2 f(x + 1) + 4 | 5. 变换第三步：竖直平移 y = −2 f(x + 1) + 4

The final operation is adding 4 to the whole expression. This is a vertical translation upwards by 4 units. It shifts every point on the current graph 4 units higher. Therefore we take the previous equation and add 4.

最后一步是在整个表达式上加 4。这是一个竖直向上 4 个单位的平移。它使当前图像上的每一个点都上移 4 个单位。因此我们把上一步的方程加上 4。

y = (−2x³ + 6x) + 4 = −2x³ + 6x + 4

Hence the final transformed equation is:

因此最终变换后的方程为：

y = −2x³ + 6x + 4

6. Combining the Transformations in the Correct Order | 6. 正确处理变换顺序

When a function is written in the form y = a f(bx + c) + d, the order matters. For y = −2 f(x + 1) + 4, we identify the transformations as follows:

当函数写成 y = a f(bx + c) + d 的形式时，顺序非常重要。对于 y = −2 f(x + 1) + 4，我们按如下方式识别变换：

Horizontal translation: replace x with x + 1 → move left by 1 unit.

水平平移：将 x 替换为 x + 1 → 向左移动 1 个单位。
Vertical stretch and reflection: multiply by −2 → stretch vertically by factor 2, then reflect in the x-axis.

竖直伸缩与反射：乘以 −2 → 竖直拉伸为 2 倍，再关于 x 轴反射。
Vertical translation: add 4 → move up by 4 units.

竖直平移：加 4 → 向上移动 4 个单位。

If you perform the vertical shift before the stretch, the final graph would be different. In this example it is safest to apply the scaling/reflection before the final addition of 4.

如果先做竖直平移再做伸缩，最终图像会不同。在本例中，最安全的做法是先在最后加 4 之前完成伸缩与反射。

7. Full Algebraic Verification | 7. 完整代数验证

Let us write the transformation chain step by step to ensure no algebra error has been made.

让我们逐步写出变换链，确保没有代数错误。

Start: y = f(x) = x³ − 3x² + 2

Apply the horizontal shift:

应用水平平移：

y = f(x + 1) = (x + 1)³ − 3(x + 1)² + 2

Expand the cube and square:

展开立方与平方：

(x + 1)³ = x³ + 3x² + 3x + 1

(x + 1)² = x² + 2x + 1

So the expression becomes:

所以表达式变为：

f(x + 1) = x³ + 3x² + 3x + 1 − 3x² − 6x − 3 + 2

Simplification gives:

化简得到：

f(x + 1) = x³ − 3x

Now multiply by −2:

现在乘以 −2：

−2 f(x + 1) = −2x³ + 6x

Finally add 4:

最后加 4：

−2 f(x + 1) + 4 = −2x³ + 6x + 4

This confirms the answer to part (a).

这验证了第 (a) 问的答案。

8. Checking with a Sample Point | 8. 用样本点检验

We can check the transformation by tracking a specific point. The original curve has a y-intercept at (0, 2). Let us see where this point lands after the three transformations.

我们可以通过追踪一个具体的点来检验变换。原曲线在 (0, 2) 处有一个 y 截距。让我们看看这个点经过三个变换后落在哪里。

Transformation	Point (x, y)	变换	点 (x, y)
Start	(0, 2)	起始	(0, 2)
x → x + 1	(−1, 2)	水平平移	(−1, 2)
Multiply by −2	(−1, −4)	伸缩与反射	(−1, −4)
Add 4	(−1, 0)	竖直平移	(−1, 0)

Now substitute x = −1 into the final equation:

现在将 x = −1 代入最终方程：

y = −2(−1)³ + 6(−1) + 4 = 2 − 6 + 4 = 0

This matches the tracked point (−1, 0), so our transformation is consistent.

这与追踪得到的点 (−1, 0) 一致，因此我们的变换是正确的。

9. Common Pitfalls and How to Avoid Them | 9. 常见陷阱与应对方法

Students frequently make errors in transformation questions. Below are some common pitfalls connected with this example.

学生在变换题中经常出错。下面是与本例相关的常见陷阱。

Incorrect horizontal direction. Remember that the shift x + 1 moves the graph left, not right. Use a test point if you are unsure.

水平方向搞错。记住 x + 1 使图像向左移动，而不是向右。如果不确定，可以用一个测试点验证。
Forgetting to expand fully. When a cubic expression like (x + 1)³ is expanded, skipping terms can lead to an incorrect polynomial. Always write out the expansion.

忘记完全展开。像 (x + 1)³ 这样的三次式展开时，漏写项会导致错误的多项式。一定要写出完整展开。
Mixing the order of vertical stretch and translation. In y = a f(x) + k, the stretch by a must be applied before adding k. If you add k first, your graph will be vertically stretched by different amounts.

混淆竖直伸缩和平移的顺序。在 y = a f(x) + k 中，必须先进行系数 a 的伸缩，再添加 k。如果先加 k，图像的竖直伸缩程度就会不同。
Making sign errors when substituting. For x = −1, (−1)³ = −1 and (−1)² = 1. Double-check negative powers.

代入时出现符号错误。对 x = −1，(−1)³ = −1，(−1)² = 1。要反复检查负数次方。

10. Practice Example | 10. 练习例题

Try this similar problem on your own: Let g(x) = 2x³ + x² − x. Describe the transformations needed to sketch y = 3 g(x − 2) − 5. Then expand the resulting polynomial.

请自己尝试一个类似的题目：设 g(x) = 2x³ + x² − x。描述画出 y = 3 g(x − 2) − 5 所需要的变换。然后展开得到的多项式。

Solution outline:

解答思路：

Horizontal shift right by 2: g(x − 2).

水平向右平移 2 个单位：g(x − 2)。
Vertical stretch by factor 3: 3 g(x − 2).

竖直拉伸为 3 倍：3 g(x − 2)。
Vertical translation down by 5: 3 g(x − 2) − 5.

竖直向下平移 5 个单位：3 g(x − 2) − 5。

Expanded, the polynomial becomes 3[2(x − 2)³ + (x − 2)² − (x − 2)] − 5, which simplifies to 6x³ − 35x² + 64x − 43.

展开后，多项式变为 3[2(x − 2)³ + (x − 2)² − (x − 2)] − 5，化简为 6x³ − 35x² + 64x − 43。

11. General Rules for Graph Transformations | 11. 图像变换的一般规则

The following table summarises the transformations that appear in this example. You should memorise them for the AQA examination.

下表总结了本例中出现的变换。你应该在 AQA 考试中牢记这些规则。

Algebraic Expression	Transformation	代数表达	变换效果
f(x) + a	Vertical translation up by a (a > 0)	f(x) + a	竖直向上平移 a
f(x + a)	Horizontal shift left by a (a > 0)	f(x + a)	水平向左平移 a
− f(x)	Reflection in the x-axis	− f(x)	关于 x 轴反射
a f(x)	Vertical stretch by factor a (a > 0)	a f(x)	竖直拉伸为 a 倍

In Example 4.3.4, the coefficient −2 combines reflection in the x-axis and a vertical stretch by factor 2.

在例 4.3.4 中，系数 −2 结合了 x 轴反射和竖直拉伸 2 倍。

12. Conclusion | 12. 小结

We have completed Example 4.3.4 from start to finish. The original curve y = x³ − 3x² + 2 was transformed horizontally, vertically, and by reflection to give the final polynomial y = −2x³ + 6x + 4. We also checked the result using a sample point and examined common mistakes. Remember to apply transformations in the correct order, expand carefully, and always test your final equation when possible.

我们已经完整完成了例 4.3.4。原曲线 y = x³ − 3x² + 2 经过水平、竖直和反射变换，得到最终多项式 y = −2x³ + 6x + 4。我们还使用了样本点进行验证，并检查了常见错误。请记住按正确顺序进行变换，仔细展开，并在可能的情况下始终检验你的最终方程。

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Example 4.3.4: Graph Transformations of a Cubic Function | 例4.3.4：三次函数图像变换