Unit B5: Transformations of
Graphs and Functions单元 B5：图像与函数的变换

Identify the transformation. Here $k = -3$, so the parabola $y = x^{2}$ is translated $3$ units down.

识别变换。此处 $k = -3$，故 $y = x^{2}$ 向下平移 $3$ 个单位。

Vertex. The vertex of $y = x^{2}$ is $(0, 0)$. After shifting down by $3$, the vertex becomes $(0, -3)$.

顶点。$y = x^{2}$ 的顶点为 $(0, 0)$。向下平移 $3$ 单位后变为 $(0, -3)$。

Range. Range of $y = x^{2}$ is $[0, \infty)$. The new range is $[-3, \infty)$.

值域。$y = x^{2}$ 的值域为 $[0, \infty)$，新值域为 $[-3, \infty)$。

Intercepts. Solve $x^{2} - 3 = 0$: $x = \pm \sqrt{3}$. The graph cuts the $x$-axis at $(\pm \sqrt{3}, 0)$. The $y$-intercept is $(0, -3)$.

截距。解 $x^{2} - 3 = 0$：$x = \pm \sqrt{3}$。图像在 $(\pm \sqrt{3}, 0)$ 处交 $x$ 轴。$y$ 截距为 $(0, -3)$。

Identify the transformation. Comparing with $y = f(x - h)$ and $f(x) = x^{2}$, we have $h = 2$. The parabola is translated $2$ units to the right.

识别变换。与 $y = f(x - h)$ 对照（$f(x) = x^{2}$），得 $h = 2$。抛物线向右平移 $2$ 个单位。

Vertex. The vertex of $y = x^{2}$ is $(0, 0)$. After shifting right by $2$ the vertex becomes $(2, 0)$.

顶点。$y = x^{2}$ 的顶点为 $(0, 0)$。向右平移 $2$ 单位后变为 $(2, 0)$。

Axis of symmetry. Originally $x = 0$, now $x = 2$.

对称轴。原为 $x = 0$，现为 $x = 2$。

Sanity check. $y = (x - 2)^{2}$ evaluated at $x = 2$ gives $y = 0$, confirming the vertex sits at $(2, 0)$. Evaluating at $x = 0$ gives $y = 4$, so the $y$-intercept is $(0, 4)$.

核对。$y = (x - 2)^{2}$ 在 $x = 2$ 取 $y = 0$，确认顶点为 $(2, 0)$。$x = 0$ 取 $y = 4$，故 $y$ 截距为 $(0, 4)$。

(a) $y = 3 \sin x$. Vertical stretch by factor $3$. Amplitude becomes $3$, period stays $2 \pi$.

(a) $y = 3 \sin x$。纵向拉伸 $3$ 倍。振幅变为 $3$，周期仍为 $2 \pi$。

(b) $y = \sin(2 x)$. Horizontal compression by factor $1/2$. Amplitude stays $1$, period becomes $2 \pi / 2 = \pi$.

(b) $y = \sin(2 x)$。横向压缩 $1/2$ 倍。振幅仍为 $1$，周期变为 $2 \pi / 2 = \pi$。

(c) $y = 3 \sin(2 x)$. Both stretches act independently. Amplitude becomes $3$, period becomes $\pi$.

(c) $y = 3 \sin(2 x)$。两种伸缩相互独立。振幅变为 $3$，周期变为 $\pi$。

Read the algebra inside-out. The argument of $f$ is $x - 3$, so the horizontal move is "shift right by $3$". Outside $f$ comes a factor of $2$, then $+ 1$, so the vertical moves are "stretch vertically by $2$" and then "shift up by $1$".

由内而外读代数。$f$ 的自变量是 $x - 3$，故横向变换是"向右平移 $3$"。$f$ 外侧先乘以 $2$ 再加 $1$，故纵向变换依次为"纵向拉伸 $2$ 倍"与"向上平移 $1$"。

Sequence. Start with $y = f(x)$:

序列。从 $y = f(x)$ 出发：

1. Translate right by $3$: $y = f(x - 3)$.
2. 向右平移 $3$：$y = f(x - 3)$。
3. Stretch vertically by factor $2$: $y = 2 f(x - 3)$.
4. 纵向拉伸 $2$ 倍：$y = 2 f(x - 3)$。
5. Translate up by $1$: $y = 2 f(x - 3) + 1$.
6. 向上平移 $1$：$y = 2 f(x - 3) + 1$。

Coordinate effect. A point $(a, b)$ on $y = f(x)$ is mapped to $(a + 3, \; 2 b + 1)$ on the final graph.

坐标作用。$y = f(x)$ 上的点 $(a, b)$ 在最终图像上映射为 $(a + 3, \; 2 b + 1)$。

Step 1. Sketch the inner function. $f(x) = x^{2} - 4$ is the standard parabola shifted down by $4$. Its zeros are $x = \pm 2$, its vertex is $(0, -4)$, and $f(x) < 0$ on $(-2, 2)$.

第 1 步：画内函数。$f(x) = x^{2} - 4$ 是标准抛物线向下平移 $4$。零点为 $x = \pm 2$，顶点 $(0, -4)$，在 $(-2, 2)$ 上 $f(x) < 0$。

Step 2. Apply $|\cdot|$. On $(-\infty, -2] \cup [2, \infty)$, $f \ge 0$ so the graph is unchanged. On $(-2, 2)$, $f < 0$ so the negative portion (which dipped to $-4$) is reflected above the $x$-axis, reaching a height of $4$ at $x = 0$.

第 2 步：取绝对值。在 $(-\infty, -2] \cup [2, \infty)$ 上 $f \ge 0$，图像不变。在 $(-2, 2)$ 上 $f < 0$，原本下探至 $-4$ 的部分被翻到 $x$ 轴之上，在 $x = 0$ 达到高度 $4$。

Step 3. Features of the new graph. Zeros remain at $x = \pm 2$ (now corners, not smooth crossings). A local maximum sits at $(0, 4)$. The graph has the shape of a parabola with a "bump" between the two zeros.

第 3 步：新图特征。零点仍在 $x = \pm 2$（变为尖点，不再光滑过零）。$(0, 4)$ 处出现一个局部极大。整体像一条抛物线在两零点之间长出一个"鼓包"。

Read the structure. Let $g(t) = (t - 2)^{2}$, a parabola with vertex $(2, 0)$ opening upward. The function $y = (|x| - 2)^{2} = g(|x|)$ is the "inside absolute value" transformation applied to $g$.

读出结构。令 $g(t) = (t - 2)^{2}$，是顶点 $(2, 0)$ 开口向上的抛物线。$y = (|x| - 2)^{2} = g(|x|)$ 是对 $g$ 施行"内层绝对值"变换。

Step 1. Right half ($x \ge 0$). $g(|x|) = g(x) = (x - 2)^{2}$. The right half of the new graph coincides with $y = (x - 2)^{2}$ on $[0, \infty)$. The vertex sits at $(2, 0)$ and the curve passes through $(0, 4)$.

第 1 步：右半部分（$x \ge 0$）。$g(|x|) = g(x) = (x - 2)^{2}$。新图像在 $[0, \infty)$ 上与 $y = (x - 2)^{2}$ 重合。顶点 $(2, 0)$，过点 $(0, 4)$。

Step 2. Left half ($x < 0$). Reflect the right half about the $y$-axis. The vertex $(2, 0)$ has a mirror image at $(-2, 0)$, and $(0, 4)$ is its own mirror image. The new graph is even, with two vertex-style minima at $(\pm 2, 0)$ and a local maximum at $(0, 4)$.

第 2 步：左半部分（$x < 0$）。把右半关于 $y$ 轴镜像。顶点 $(2, 0)$ 的镜像为 $(-2, 0)$，$(0, 4)$ 自映射不变。新图像为偶函数：在 $(\pm 2, 0)$ 处各有一个极小，在 $(0, 4)$ 处有一个局部极大。

Contrast. The unrelated graph $y = (x - 2)^{2}$ is a single parabola with vertex $(2, 0)$ and is not symmetric about the $y$-axis. The absolute value forces evenness; without it, no such symmetry exists.

对照。不相干的 $y = (x - 2)^{2}$ 是单条以 $(2, 0)$ 为顶点的抛物线，不关于 $y$ 轴对称。绝对值强行造出偶对称；不取绝对值则无此对称。

Step 1. Inspect $f$. $f(x) = x^{2} - 1$ is an upward parabola with zeros at $x = \pm 1$, vertex at $(0, -1)$, and range $[-1, \infty)$.

第 1 步：审视 $f$。$f(x) = x^{2} - 1$ 为开口向上的抛物线，零点 $x = \pm 1$，顶点 $(0, -1)$，值域 $[-1, \infty)$。

Step 2. Vertical asymptotes. Zeros of $f$ at $x = \pm 1$ produce vertical asymptotes of $1/f$.

第 2 步：竖直渐近线。$f$ 的零点 $x = \pm 1$ 给出 $1/f$ 的两条竖直渐近线。

Step 3. Local extremum. The minimum of $f$ at $(0, -1)$ becomes a maximum of $1/f$ at $(0, 1/(-1)) = (0, -1)$.

第 3 步：局部极值。$f$ 在 $(0, -1)$ 处的极小变为 $1/f$ 在 $(0, 1/(-1)) = (0, -1)$ 处的极大。

Step 4. Horizontal asymptote. As $x \to \pm \infty$, $f(x) \to \infty$, so $1/f \to 0^{+}$. The line $y = 0$ is a horizontal asymptote, approached from above on $|x| > 1$.

第 4 步：水平渐近线。$x \to \pm \infty$ 时 $f(x) \to \infty$，故 $1/f \to 0^{+}$。$y = 0$ 为水平渐近线，在 $|x| > 1$ 上从上方趋近。

Step 5. Sign analysis. On $|x| < 1$, $f < 0$, so $1/f < 0$; the central branch dips down toward the maximum $(0, -1)$ and plunges to $-\infty$ at $x = \pm 1$. On $|x| > 1$, $f > 0$, so $1/f > 0$; two outer branches plunge from $+\infty$ at $x = \pm 1$ down to the asymptote $y = 0$.

第 5 步：符号分析。$|x| < 1$ 时 $f < 0$，故 $1/f < 0$；中间分支向下经过极大 $(0, -1)$，在 $x = \pm 1$ 处俯冲到 $-\infty$。$|x| > 1$ 时 $f > 0$，故 $1/f > 0$；两外侧分支从 $x = \pm 1$ 处的 $+\infty$ 俯冲下来，趋近渐近线 $y = 0$。

Step 6. Crossings with $f$. Solve $f(x) = 1$: $x^{2} - 1 = 1 \Rightarrow x = \pm \sqrt{2}$. Solve $f(x) = -1$: $x^{2} - 1 = -1 \Rightarrow x = 0$. So $f$ and $1/f$ meet at $(\pm \sqrt{2}, 1)$ and at $(0, -1)$ (which is the shared extremum identified in Step 3).

第 6 步：与 $f$ 的交点。解 $f(x) = 1$：$x^{2} - 1 = 1 \Rightarrow x = \pm \sqrt{2}$。解 $f(x) = -1$：$x^{2} - 1 = -1 \Rightarrow x = 0$。故 $f$ 与 $1/f$ 相交于 $(\pm \sqrt{2}, 1)$ 与 $(0, -1)$（后者即第 3 步指出的公共极值点）。