描述一个能将 $y = e^{x}$ 的图像映射到 $y = -e^{x} + 2$ 图像上的包含两个变换的变换序列。 方程为 $y = x + 6$ 的直线依次经历以下变换： A：沿平行于 x 轴的方向拉伸（缩放），比例因子为 $\tfrac{1}{3}$， B：沿 x 轴正方向平移 2 个单位， C：沿平行于 y 轴的方向拉伸（缩放），比例因子为 $\tfrac{1}{3}$。 求所得直线的方程。

1. Identify the transformation that introduces the negative sign to the function.

$y = e^x \xrightarrow{\text{Reflection in the } x\text{-axis}} y = -e^x$

2. Identify the transformation that adds 2 to the overall function.

$y = -e^x \xrightarrow{\text{Translation of } 2 \text{ units in the positive } y\text{-direction}} y = -e^x + 2$

1. Apply the first transformation (scaling parallel to the $x$-axis).

Scaling by a scale factor of $\frac{1}{3}$ parallel to the $x$-axis means we replace $x$ with $3x$ in the equation.

$y = f(3x) = (3x) + 6$
$y = 3x + 6$

2. Apply the second transformation (translation parallel to the $x$-axis).

Translation of 2 units in the positive $x$-direction means we replace $x$ with $(x - 2)$.

$y = 3(x - 2) + 6$
$y = 3x - 6 + 6$
$y = 3x$

3. Apply the third transformation (scaling parallel to the $y$-axis) to find the final equation.

Scaling by a scale factor of $\frac{1}{3}$ parallel to the $y$-axis means we multiply the entire right-hand side of the function by $\frac{1}{3}$.

$y = \frac{1}{3}(3x)$
$y = x$