Comprehensive Transformations of Function Graphs in IB Mathematics | IB数学：函数图像的综合变换应用

📚 Comprehensive Transformations of Function Graphs in IB Mathematics | IB数学：函数图像的综合变换应用

Function graph transformations are a core topic in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI). A deep understanding of how translations, reflections, stretches and absolute value operations affect a graph allows students to sketch complex functions quickly and solve examination questions with confidence.

函数图像变换是IB数学的核心内容，在分析与方法（AA）以及应用与解释（AI）中都会出现。深入理解平移、反射、伸缩和绝对值运算如何影响图像，能帮助学生快速绘制复杂函数图像，并自信地解决考试问题。

1. The Four Basic Transformations | 四种基本变换

Before combining multiple transformations, it is essential to recall the four basic types that act on a function y = f(x):

在组合多种变换之前，必须回顾作用于函数 y = f(x) 的四种基本类型：

Vertical translation: y = f(x) + a shifts the graph up by a units if a > 0, down if a < 0.

垂直平移：y = f(x) + a 当 a > 0 时图像上移 a 个单位，当 a < 0 时下移。
Horizontal translation: y = f(x − b) shifts the graph right by b units if b > 0, left if b < 0.

水平平移：y = f(x − b) 当 b > 0 时图像右移 b 个单位，当 b < 0 时左移。
Vertical stretch/compression: y = kf(x) stretches vertically by factor k when |k| > 1, compresses when 0 < |k| < 1.

垂直伸缩：y = kf(x) 当 |k| > 1 时纵向拉伸 k 倍，当 0 < |k| < 1 时纵向压缩。
Horizontal stretch/compression: y = f(cx) compresses horizontally by factor 1/c when c > 1, stretches when 0 < c < 1.

水平伸缩：y = f(cx) 当 c > 1 时图像横向压缩到原来的 1/c，当 0 < c < 1 时横向拉伸。

2. Reflections: Across Axes and Origin | 反射：关于轴与原点

Reflections are special cases of stretches with a negative scale factor. Understanding them separately avoids sign errors in composite transformations.

反射是负比例因子的特殊伸缩情况。单独理解它们可以避免复合变换中的符号错误。

y = −f(x) is a reflection in the x-axis. Every point (x, y) becomes (x, −y).

y = −f(x) 是关于 x 轴的反射。每个点 (x, y) 变为 (x, −y)。
y = f(−x) is a reflection in the y-axis. Every point (x, y) becomes (−x, y).

y = f(−x) 是关于 y 轴的反射。每个点 (x, y) 变为 (−x, y)。
y = −f(−x) is a reflection through the origin, equivalent to a 180° rotation.

y = −f(−x) 是关于原点的反射，等价于旋转 180°。

3. Order of Transformations | 变换的顺序

When combining multiple transformations, the order matters, especially when both horizontal translations and stretches are present. The safest approach is to rewrite the transformed function in the standard form:

当组合多种变换时，顺序非常重要，尤其是当水平和垂直方向的平移与伸缩同时存在时。最安全的方法是将变换后的函数写成标准形式：

y = a f(b(x − c)) + d

Here, b is applied to x before the shift c; a is the vertical stretch, and d is the final vertical shift. For y = f(2x − 4), factor out the 2 to get f(2(x − 2)). This shows a horizontal compression by factor 1/2 followed by a shift right by 2 units.

这里，b 先作用于 x，然后才进行平移 c；a 是垂直伸缩，d 是最后的垂直平移。对于 y = f(2x − 4)，先提取因子 2 得到 f(2(x − 2))。这表明先水平压缩到原来的 1/2，然后向右平移 2 个单位。

A common error is to read y = f(2x − 4) as a shift right by 4. Always rewrite the argument as a single coefficient times (x − c).

常见错误是把 y = f(2x − 4) 误读为向右平移 4 个单位。始终要将自变量改写为系数乘以 (x − c) 的形式。

4. Transforming Key Points | 变换关键点

Rather than sketching an entire curve, it is often efficient to transform key points such as intercepts, vertices, asymptotes and endpoints. For a point (x, y) on y = f(x), its image under y = a f(b(x − c)) + d is:

与其画出整条曲线，不如高效地变换关键点，例如交点、顶点、渐近线和端点。对于 y = f(x) 上的点 (x, y)，在 y = a f(b(x − c)) + d 下的像为：

(x, y) → (x/b + c, a y + d)

For example, the vertex of f(x) = x², (0, 0), under y = 2(x − 3)² + 1 becomes (0/1 + 3, 2×0 + 1) = (3, 1). This matches the vertex form directly.

例如，f(x) = x² 的顶点 (0, 0) 在 y = 2(x − 3)² + 1 下变为 (0/1 + 3, 2×0 + 1) = (3, 1)。这与顶点式直接吻合。

5. Vertical Transformations Do Not Change the x-coordinates | 垂直变换不改变 x 坐标

Vertical translations, reflections in the x-axis and vertical stretches only affect the y-coordinate. This means intercepts on the x-axis (where y = 0) remain unchanged under vertical transformations, provided the vertical stretch factor is not zero.

垂直平移、关于 x 轴的反射和垂直伸缩只影响 y 坐标。这意味着在垂直变换下，x 轴上的交点（即 y = 0 处）保持不变，前提是垂直伸缩因子不为零。

For instance, the roots of f(x) = x² − 4 are x = ±2. The graph of y = 3f(x) + 2 still crosses the x-axis at the same points? Let us check: 3(x² − 4) + 2 = 3x² − 10 has roots x = ±√(10/3), not the same. The reason is that the vertical translation by +2 moves the roots. Vertical transformations alone preserve x-coordinates only when the transformation is y = a f(x) with no added constant. Adding a constant changes the intercepts.

例如，f(x) = x² − 4 的根是 x = ±2。那么 y = 3f(x) + 2 的图像是否仍在相同的点穿过 x 轴？检查一下：3(x² − 4) + 2 = 3x² − 10 的根是 x = ±√(10/3)，并不相同。原因是 +2 的垂直平移移动了根。只有形如 y = a f(x) 且没有额外常数项的垂直变换才保持 x 坐标不变。加上常数项会改变交点。

6. Absolute Value of a Function | 函数的绝对值

The transformation y = |f(x)| reflects the parts of the graph that lie below the x-axis upward, leaving the parts above the axis unchanged. This is a combination of “keep positive” and “reflect negative”.

变换 y = |f(x)| 将图像位于 x 轴下方的部分向上反射，而位于轴上方的部分保持不变。这是“保留正数”和“反射负数”的组合。

If f(x) ≥ 0, then |f(x)| = f(x), so the graph is unchanged.

若 f(x) ≥ 0，则 |f(x)| = f(x)，图像不变。
If f(x) < 0, then |f(x)| = −f(x), so the graph is reflected in the x-axis.

若 f(x) < 0，则 |f(x)| = −f(x)，图像关于 x 轴反射。

All x-intercepts of the original function become “corners” on the transformed graph if the original function crosses the axis transversely.

如果原函数横穿 x 轴，那么它的所有 x 轴交点都会成为变换后图像上的“拐角”。

7. Absolute Value of the Input | 输入变量的绝对值

The transformation y = f(|x|) is different: it discards the graph for x < 0 and replaces it with a reflection of the graph for x > 0 across the y-axis. The resulting graph is always even.

变换 y = f(|x|) 则不同：它舍弃 x < 0 部分的图像，并将 x > 0 部分的图像关于 y 轴反射来替换。得到的图像始终是偶函数图像。

For example, if f(x) = x + 2, then f(|x|) = |x| + 2. The original line crosses the y-axis at 2 and the x-axis at −2. After the transformation, the branch for x < 0 becomes the mirror image of the branch for x > 0, so the new graph is V-shaped with vertex at (0, 2).

例如，若 f(x) = x + 2，则 f(|x|) = |x| + 2。原直线在 y 轴截距为 2，在 x 轴截距为 −2。变换后，x < 0 的分支变成了 x > 0 分支的镜像，因此新图像呈 V 形，顶点在 (0, 2)。

8. Composite Transformations: A Worked Example | 复合变换：一个完整例题

Consider the function f(x) = x². We want to sketch y = −2(x − 1)² + 3.

考虑函数 f(x) = x²。我们希望画出 y = −2(x − 1)² + 3 的图像。

Step 1: Start with y = x².

第一步：从 y = x² 开始。

Step 2: Apply the horizontal shift right by 1: y = (x − 1)². Vertex moves to (1, 0).

第二步：向右平移 1 个单位：y = (x − 1)²。顶点移动到 (1, 0)。

Step 3: Apply the vertical stretch by factor 2: y = 2(x − 1)². The parabola becomes narrower.

第三步：纵向拉伸 2 倍：y = 2(x − 1)²。抛物线变得更窄。

Step 4: Reflect in the x-axis: y = −2(x − 1)². The parabola now opens downward.

第四步：关于 x 轴反射：y = −2(x − 1)²。抛物线现在开口向下。

Step 5: Translate upward by 3: y = −2(x − 1)² + 3. The vertex is now (1, 3).

第五步：上移 3 个单位：y = −2(x − 1)² + 3。顶点现在在 (1, 3)。

Note that the vertical stretch and reflection could be combined into a single vertical stretch by factor −2, but the order of horizontal shift and horizontal stretch matters. Here only a horizontal shift appears, so the order is unambiguous.

注意，垂直拉伸和反射可以合并为一次因子为 −2 的垂直伸缩，但水平平移和水平伸缩的顺序很重要。这里只出现水平平移，因此顺序是明确的。

9. Recognizing Transformations from an Equation | 从方程识别变换

Given a transformed equation, it is useful to identify the original parent function and the sequence of transformations. For example, y = √(4 − x) + 2 can be rewritten as y = √(−(x − 4)) + 2. Starting from y = √x, we reflect in the y-axis to get √(−x), then shift right by 4 (because x − 4 means the zero of the argument is at x = 4), then shift up by 2.

给定一个变换后的方程，识别原始母函数和变换序列非常有用。例如，y = √(4 − x) + 2 可以改写为 y = √(−(x − 4)) + 2。从 y = √x 出发，先关于 y 轴反射得到 √(−x)，然后向右平移 4 个单位（因为 x − 4 意味着参数的零点在 x = 4），再上移 2 个单位。

Always factor out coefficients in the argument before reading shifts. For instance, y = (2x + 6)³ should be written as y = (2(x + 3))³, indicating a horizontal compression by 1/2 and a shift left by 3.

在读取平移量之前，始终要把参数中的系数提取出来。例如，y = (2x + 6)³ 应写成 y = (2(x + 3))³，表示水平压缩 1/2 并向左平移 3 个单位。

10. Inverse Functions and Reflections | 反函数与反射

The graph of the inverse function y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. This is a fundamental transformation that connects a function to its inverse. When combined with other transformations, it is useful to remember that the roles of x and y are swapped.

反函数 y = f⁻¹(x) 的图像是 y = f(x) 关于直线 y = x 的反射。这是一个基本变换，将函数与其反函数联系起来。当与其他变换结合时，记住 x 和 y 的角色互换非常有用。

For example, if f(x) = 2x + 1, then f⁻¹(x) = (x − 1)/2. The point (1, 3) on f corresponds to (3, 1) on the inverse, and both lie on the line y = x as mirror images.

例如，若 f(x) = 2x + 1，则 f⁻¹(x) = (x − 1)/2。f 上的点 (1, 3) 对应反函数上的 (3, 1)，两者关于直线 y = x 互为镜像。

11. Common Pitfalls and Exam Tips | 常见误区与考试技巧

Students often lose marks on transformation problems due to a few repeated mistakes. Being aware of them can significantly improve accuracy.

学生在变换题目中常因几个重复出现的错误而失分。意识到这些错误可以显著提高准确性。

Mistake 1: Confusing horizontal shift direction. y = f(x + 2) shifts left, not right, because the zero of x + 2 is at x = −2.

错误一：混淆水平平移方向。y = f(x + 2) 向左平移，而非向右，因为 x + 2 的零点在 x = −2。
Mistake 2: Applying vertical stretch before vertical translation incorrectly. In y = 2f(x) + 3, the stretch is applied first, then the translation. The order is determined by the standard order of operations in the expression.

错误二：错误地先垂直平移再垂直伸缩。在 y = 2f(x) + 3 中，先伸缩后平移。顺序由表达式中的运算先后决定。
Mistake 3: Forgetting that horizontal transformations are “opposite” in intuition: y = f(2x) compresses rather than stretches.

错误三：忘记水平变换与直觉相反：y = f(2x) 是压缩而非拉伸。
Exam tip: Always test a known point, such as (0,0) if it lies on the original graph, through the transformed equation to verify your sketch.

考试技巧：始终选取一个已知点（如原图上的 (0,0)）代入变换后的方程验证你的草图。

12. Putting It All Together | 综合应用

Consider the function g(x) = |−2(x + 1)³ + 4|. We can analyze it step by step.

考虑函数 g(x) = |−2(x + 1)³ + 4|。我们可以逐步分析。

Start with f(x) = x³. Shift left by 1 to get (x + 1)³. Stretch vertically by factor 2 and reflect in the x-axis to get −2(x + 1)³. Translate upward by 4 to get h(x) = −2(x + 1)³ + 4. Finally, take the absolute value: g(x) = |h(x)|. This reflects the negative parts of h above the x-axis.

从 f(x) = x³ 开始。向左平移 1 得到 (x + 1)³。纵向拉伸 2 倍并关于 x 轴反射得到 −2(x + 1)³。上移 4 得到 h(x) = −2(x + 1)³ + 4。最后取绝对值：g(x) = |h(x)|。这将 h 的负值部分反射到 x 轴上方。

The point where h(x) = 0 is found by solving −2(x + 1)³ + 4 = 0, giving (x + 1)³ = 2, so x = ∛2 − 1 ≈ 0.26. At this point, the graph of g touches the x-axis but does not cross it because the cubic crosses the axis and the absolute value reflects one side.

h(x) = 0 的点通过解 −2(x + 1)³ + 4 = 0 得到，即 (x + 1)³ = 2，所以 x = ∛2 − 1 ≈ 0.26。在这个点，g 的图像接触 x 轴但不穿过，因为三次函数穿过轴，而绝对值将一侧反射。

In an examination, you would not need to plot every point, but you should be able to describe the sequence of transformations and identify key features such as the vertex-like “corner” at the zero of h, the intercepts and the general shape.

在考试中，你不需要画出每个点，但你应该能够描述变换序列并识别关键特征，例如在 h 的零点处的“拐角”、截距和大致形状。

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Comprehensive Transformations of Function Graphs in IB Mathematics | IB数学：函数图像的综合变换应用