图中，方程为 $y = f(x)$ 的图像用实线表示，方程为 $y = g(x)$ 的图像用虚线表示。

(a) 完整描述一组三个变换，使 $y = f(x)$ 的图像变换为 $y = g(x)$ 的图像。

(b) 求 $g(x)$ 的表达式，写成 $af(bx + c)$ 的形式，其中 $a$、$b$、$c$ 为整数。

先比较两条折线的左右顺序，再比较横向和纵向坐标。实线端点 $(0,3)$ 对应虚线端点 $(-1,6)$。

The horizontal order is reversed, and all heights are doubled. The point $(0,3)$ maps to $(-1,6)$, while $(4,2)$ maps to $(-5,4)$. One sequence is:

1. Reflect in the $y$-axis.
2. Translate by $\begin{pmatrix}-1\\0\end{pmatrix}$.
3. Stretch by scale factor $2$ in the $y$-direction.

The resulting point map is $(x,y)\mapsto(-x-1,2y)$, which also sends the horizontal segment from $x=1$ to $x=2$ to the segment from $x=-3$ to $x=-2$.

若先平移再反射，水平位移的方向会被反射反转。也可以用另一条竖直直线作为反射轴。

Translate first by $\begin{pmatrix}1\\0\end{pmatrix}$, then reflect in the $y$-axis, and finally stretch by scale factor $2$ in the $y$-direction. The horizontal coordinate becomes $-(x+1)$, so this gives the same image.

A further sequence is to stretch by scale factor $2$ in the $y$-direction, translate by $\begin{pmatrix}-4\\0\end{pmatrix}$, and reflect in $x=-\frac52$. The last step sends $x-4$ to $-5-(x-4)=-x-1$.

从新横坐标反推原横坐标。若新横坐标是 $x$，则原横坐标是 $-x-1$，原来的高度还要乘 $2$。

The point map from (a) sends $(u,f(u))$ to $(-u-1,2f(u))$. At new coordinate $x$ we therefore have $u=-x-1$, and

Thus $a=2$, $b=-1$ and $c=-1$.