📚 The Graphs of y = f(ax + b) and y = [f(x)]² | 函数 y = f(ax + b) 与 y = [f(x)]² 的图像
mpose composite changes and analyse how squaring alters a graph. The first form combines horizontal stretching and translation; the second creates a non‑negative image that magnifies large |f(x)| values and maps roots to touching points on the x‑axis. This article unpacks both graph types with step‑by‑step reasoning and examples.
在 IB 数学中，根据已知函数 f(x) 绘制 y = f(ax + b) 与 y = [f(x)]² 的图像，既考查你对复合变换的分解能力，也考查平方运算如何改变图像特性。第一种形式融合了水平伸缩与平移；第二种则生成非负图像，将 |f(x)| 较大的部分放大，同时把根变为与 x 轴的切点。本文将通过分步推理和实例，剖析这两类函数图像。
1. The Composite View: From f(x) to f(ax + b) | 复合视野：从 f(x) 到 f(ax + b)
The expression f(ax + b) acts on the input x before the function rule is applied. The constants a and b can be arranged in two equivalent ways: f(a(x + b/a)) or as a sequence 'translate then stretch'. IB students need to recognise that the order of operations matters unless the expression is rewritten into the standard f(a(x - h)) format.
表达式 f(ax + b) 在函数规则生效前先对自变量 x 进行加工。常数 a 与 b 有两种等价整理方式：写成 f(a(x + b/a))，或者按“先平移后伸缩”的顺序理解。IB 学生必须意识到，除非把表达式改写成标准的 f(a(x − h)) 格式，否则运算顺序会影响最终的图像结论。
- If a > 1, the graph compresses horizontally; if 0 < a < 1, it stretches.
- 若 a > 1，图像水平压缩；若 0 < a < 1，图像水平拉伸。
- Negative a gives a reflection in the y‑axis after the stretch/translation.
- 负 a 则在伸缩/平移之后再对 y 轴反射。
2. Rewriting to Clarify the Transformation | 重写表达式，明晰变换顺序
+ b/a)). The horizontal shift is then −b/a (left if positive, right if negative), and the horizontal scale factor is 1/a. This eliminates the common mistake of applying the translation b before scaling by a without adjusting the shift.
始终把 a 提取出来，得到 y = f(a(x + b/a))。此时水平平移量为 −b/a（正值则左移，负值则右移），水平缩放因子为 1/a。这样做可以避免不经调整就直接把平移 b 放在比例变换 a 之前所导致的常见错误。
y = f(ax + b) ≡ y = f( a (x + b/a) )
Once in this form, the graph of f(x) undergoes a horizontal stretch by factor 1/|a|, a reflection in the y‑axis if a < 0, and finally a horizontal translation by −b/a.
变成这种形式后，f(x) 的图像先经过水平方向缩放因子 1/|a| 的伸缩，若 a < 0 再对 y 轴反射，最后进行 −b/a 的水平平移。
3. Example 1: y = f(2x − 4) Step by Step | 示例 1：逐步绘制 y = f(2x − 4)
Consider the base graph of f(x). Factor out 2: f(2(x − 2)). This tells us two things:
考虑基本函数 f(x) 的图像。提取 2：f(2(x − 2))。这告诉我们：
- Horizontal shrink by factor 1/2 (all x‑coordinates halved).
- 水平方向压缩为原来的 1/2（所有 x 坐标减半）。
- Then shift right by 2 units.
- 然后向右平移 2 个单位。
(2x − 4)² = 4(x − 2)², which is a parabola with vertex at (2,0) and narrower than y = x².
若 f(x) = x²，则 y = (2x − 4)² = 4(x − 2)²，其图像是顶点在 (2,0) 且比 y = x² 更窄的抛物线。
al asymptote y = 0. The horizontal shrink brings all features closer to the new asymptote.
当原图像有渐近线时，例如 f(x) = 1/x，y = 1/(2x − 4) 的垂直渐近线在 x = 2，水平渐近线为 y = 0。水平压缩让所有特征向新的渐近线靠拢。
4. Avoiding the Order Trap: Translate vs Stretch | 避开顺序陷阱：先平移还是先伸缩
ing f(x) → f(x − 4) then f(2x) does NOT give f(2x − 4); it yields f(2(x − 4)) = f(2x − 8).
一个常见错误是先把图像水平移动 b 个单位，再按 1/a 缩放。只有当使用 f(a(x + b/a)) 并将平移视为 −b/a 时，这种顺序才正确。若直接对 f(x) 先做 f(x − 4) 再做 f(2x)，得到的并非 f(2x − 4)，而是 f(2(x − 4)) = f(2x − 8)。
To keep the intended graph, always factor first. Then execute transformations on x from the inside out, mirroring the order of operations.
为得到正确的图像，务必先提取 a。然后按照运算顺序从内向外对 x 实施变换。
| Correct: f(x) → f(x + b/a) → f(a(x + b/a)) | 正确：f(x) → f(x + b/a) → f(a(x + b/a)) |
| Incorrect: f(x) → f(x + b) → f(ax + b) when a ≠ 1 | 错误：当 a ≠ 1 时，f(x) → f(x + b) → f(ax + b) |
5. Effect on Domain, Range, and Key Points | 对定义域、值域及关键点的影响
Under y = f(ax + b), the x‑ e range is unchanged unless combined with a vertical transformation, which is absent here.
在变换 y = f(ax + b) 下，原图像上每个点 (x₀, y₀) 的 x 坐标变为 (x₀ − b)/a。因此，已知的根 x = p 会变成 x = (p − b)/a。由于没有伴随竖直变换，值域保持不变。
- Domain boundaries shift accordingly: limits or intervals transform via the same linear mapping.
- 定义域的边界也相应平移：区间端点通过相同的线性映射变换。
- Turning points retain their y‑values but have new x‑coordinates (x′ = (x − b)/a).
- 极值点纵坐标不变，新横坐标为 x′ = (x − b)/a。
6. Entering the Squared Graph: y = [f(x)]² | 走进平方图像：y = [f(x)]²
The graph of y = [f(x)]² is obtained by squaring every y‑coordinate of y = f(x). This operation preserves the x‑intercepts (roots) but makes them 'bounce' points, because slopes on either side become non‑negative. All parts of the original graph that lie below the x‑axis are reflected upwards.
y = [f(x)]² 的图像可通过将 y = f(x) 的每个点的纵坐标平方得到。这一操作保留了 x 轴截距（根），但使这些点成为“反弹”点，因为两侧的斜率经平方后均变为非负。原图像位于 x 轴下方的部分全部向上翻折。
- )]² = 0. Roots become roots, but with horizontal tangent if the original crossed the axis.
- 当 f(x) = 0，[f(x)]² = 0。根依然是根，但如果原图穿过 x 轴，该处会出现水平切线。
- Regions where |f(x)| > 1 are magnified; where |f(x)| < 1 they are flattened closer to the x‑axis.
- |f(x)| > 1 的区域被放大；|f(x)| < 1 的区域则被压平，更贴近 x 轴。
7. Non‑negativity and Its Consequences | 非负性及其后果
Since a square is never negative, the image of y = [f(x)]² lies entirely on or above the x‑ merge with a previously negative asymptote becoming positive, e.g., y = −2 becomes y = 4.
因为平方始终非负，y = [f(x)]² 的图像完全位于 x 轴或其上方。任何一条水平渐近线 y = c 会变为 y = c²；原先负的水平渐近线会变成正数，如 y = −2 变成 y = 4。
ptotes remain unchanged in position (x = k) because squaring does not move the undefined x‑values. However, the behaviour on either side may now both approach +∞.
垂直渐近线的位置（x = k）保持不变，因为平方并不改变无意义的 x 值。不过渐近线两侧的趋势可能均趋向 +∞。
8. Stationary Points and Multiplication of Gradients | 驻点与斜率的乘积效应
Differentiating y = [f(x)]² with the chain rule gives dy/dx = 2 f(x) f'(x). This reveals that stationary points occur either where f(x) = 0 or where f'(x) = 0. Thus every turning point or stationary point of f(x) survives, but its y‑alue is squared, which may change a maximum into a minimum if the original y was negative.
利用链式法则对 y = [f(x)]² 求导，得 dy/dx = 2 f(x) f'(x)。由此可见，驻点出现在 f(x) = 0 或 f'(x) = 0 处。所以 f(x) 的所有极值点或驻点保留下来，但纵坐标被平方；如果原纵坐标为负，原本的极大值点可能变成极小值点。
- If f(x) has a local maximum at (−1, −3), the squared graph shows a local minimum at (−1, 9).
- 若 f(x) 在 (−1, −3) 有局部极大值，平方图像则在 (−1, 9) 呈现局部极小值。
- Points where f(x) = 0 are local minima of [f(x)]², provided f'(x) ≠ 0, since the gradient changes sign through zero.
- 若 f(x) = 0 且 f'(x) ≠ 0，则该点是 [f(x)]² 的局部极小值点，因为导数通过零点时符号改变。
9. Roots Become Touching Points: The 'Bounce' | 根变为切点：“反弹”效应
At a simple root where f(x) changes sign, the function [f(x)]² touches the x‑axis and turns back, producing a local minimum of zero. The derivative there is zero, making it a stationary point. If the original had a double root, the squared graph may have a higher‑order flatness, resembling a quartic touch.
在 f(x) 变号的单根处，[f(x)]² 会触碰 x 轴后反弹，形成一个值为零的局部极小值。该处导数为零，因此是一个驻点。如果原函数有重根，那么平方图像在该点可能具有更高阶的平坦性，类似四次函数触碰到 x 轴。
ing at x = 2. f(x) = (x − 1)² gives [f(x)]² = (x − 1)⁴, touching with a wider flat bottom.
示例：f(x) = x − 2 得到 [f(x)]² = (x − 2)²，是一条在 x = 2 触底的抛物线。f(x) = (x − 1)² 得到 [f(x)]² = (x − 1)⁴，触底更平坦。
10. Symmetry Inherited and Acquired | 继承与获得的对称性
[f(x)]², making y = [f(x)]² an even function. This introduces reflective symmetry in the y‑axis even if f(x) lacked it.
若 f(x) 是偶函数、奇函数或具有线对称，平方运算可能改变或强化对称性。奇函数 f(x) 满足 f(−x) = −f(x)，平方后得 [f(−x)]² = [−f(x)]² = [f(x)]²，因而 y = [f(x)]² 变为偶函数。这导致图像即使原本没有 y 轴对称，也会在平方后获得 y 轴反射对称。
- Vertical asymptotes preserve their symmetry; a square graph can appear mirrored if the original was odd.
- 垂直渐近线保持对称；如果原函数是奇函数，平方后的图像可能表现出镜像对称。
- Even functions remain even when squared, reinforcing symmetry.
- 偶函数平方后仍是偶函数，对称性增强。
11. Sketching Strategy Combined: Transform then Square | 综合绘图策略：先变换再平方
t plot y = f(ax + b) using the horizontal transformation rules. Then square all obtained y‑coordinates. This two‑ d stationary points on the transformed graph, then apply the square mapping to the values.
当要求绘制 y = [f(ax + b)]² 时，先利用水平变换规则画出 y = f(ax + b) 的图像，再将所有纵坐标平方。这种两步顺序可以避免混淆：在变换后的图像上标出新截距、渐近线和驻点，然后对这些点的纵坐标施以平方映射。
For example, starting from f(x) = √x, consider y = [√2x + 6]². The inside function g(x) = √2x + 6 estriction remains, so the final graph is the ray y = 2x + 6, x ≥ −3, not the entire line.
例如，从 f(x) = √x 出发，考虑 y = [√2x + 6]²。内层函数 g(x) = √2x + 6 的定义域为 x ≥ −3，值域 y ≥ 0。平方后得到 2x + 6，但定义域限制保留，因此最终图像是射线 y = 2x + 6, x ≥ −3，而不是整条直线。
12. Summary and IB Exam Tips | 总结与 IB 考试技巧
t roots bounce, negative regions reflect, and stationary points are squared in value. In exams, label all transformed coordinates clearly and mark the behaviour near asymptotes and roots to secure method marks.
掌握 y = f(ax + b) 的要诀是始终提取 a，从内向外进行水平变换。对于 y = [f(x)]²，记住根会反弹、负值区域向上翻折、驻点的纵坐标被平方。考试时，清晰标注所有变换后的坐标，并明确渐近线和根附近的变化趋势，这样能够确保拿到方法分。
- Rewrite f(ax + b) as f(a(x + b/a)) before describing transformations.
- 在描述变换前，先将 f(ax + b) 改写为 f(a(x + b/a))。
- For [f(x)]², sketch f(x) lightly, then square the heights; erase or adjust below the x‑axis.
- 对 [f(x)]²，先用虚线画出 f(x)，再将纵坐标平方；擦除或调整 x 轴下方的部分。
- Check for spurious solutions when simplifying algebraically; the graph of [f(x)]² may hide domain restrictions (e.g., squares of square roots).
- 代数简化时注意增根；[f(x)]² 的图像可能隐藏定义域限制（例如根号平方的情况）。
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