下图显示了 $0 \\le x < 2a$ 时 $y = f(x)$ 的部分图像草图。垂直渐近线 $x = 2a$ 也是该图像的一条对称轴。

1. Sketch $y = f(x)$ for $0 \le x \le 4a$ using the given line of symmetry.

Since the graph has a line of symmetry at $x = 2a$, the portion for $2a < x \le 4a$ is the reflection of the graph for $0 \le x < 2a$.

• The original $x$-intercept at $x = a$ reflects to a new $x$-intercept at $x = 3a$.
• The original starting point $(0, -2)$ reflects to the endpoint $(4a, -2)$.
• The vertical asymptote remains at $x = 2a$.

2. Sketch the reciprocal graph $y = \frac{1}{f(x)}$ for $0 \le x \le 4a$.

For the graph of $y = \frac{1}{f(x)}$:

• The $x$-intercepts of $f(x)$ at $x = a$ and $x = 3a$ become vertical asymptotes.
• As $x \to 2a$, $f(x) \to +\infty$, so $\frac{1}{f(x)} \to 0$. Since $f(x)$ is positive on both sides of $2a$, $\frac{1}{f(x)}$ touches the $x$-axis at $(2a, 0)$, forming a local minimum.
• The $y$-intercept is $\frac{1}{f(0)} = \frac{1}{-2} = -0.5$.
• The endpoint at $x = 4a$ is $\left(4a, -0.5\right)$.
• For $0 \le x < a$ and $3a < x \le 4a$, $f(x)$ is negative, so $\frac{1}{f(x)}$ is negative and approaches $-\infty$ at the asymptotes.
• For $a < x < 3a$, $f(x)$ is positive, so $\frac{1}{f(x)}$ is positive and approaches $+\infty$ at the asymptotes.

3. Sketch $y = f\left(\frac{x}{2} + a\right)$ for $-2a \le x \le 6a$ by determining its transformations.

Let $X = \frac{x}{2} + a \implies x = 2(X - a)$. This corresponds to a horizontal translation left by $a$, followed by a horizontal stretch by a factor of $2$.

• Vertical Asymptote: $X = 2a \implies x = 2(2a - a) = 2a$.
• $x$-intercepts: $X = a \implies x = 0$ and $X = 3a \implies x = 4a$.
• $y$-intercept: Occurs at $x = 0$, where $y = f(a) = 0$.
• Endpoints:
  When $X = 0$, $x = -2a$ and $y = f(0) = -2$.
  When $X = 4a$, $x = 6a$ and $y = f(4a) = -2$.
  The overall shape is identical to $f(x)$ but stretched horizontally. It passes through $(-2a, -2)$, $(0,0)$, rises to the asymptote at $x=2a$, falls from the asymptote to pass through $(4a, 0)$, and ends at $(6a, -2)$.