The problem asks us to analyze the graph of the function $y = \tan(\theta - \frac{\pi}{2})$. We are to determine its characteristics based on its graph.

1. Problem Description

The problem asks us to analyze the graph of the function $y = \tan(\theta - \frac{\pi}{2})$. We are to determine its characteristics based on its graph.

2. Solution Steps

The general form of the tangent function is $y = a \tan(b(x-c)) + d$, where:

- $|a|$ is the vertical stretch factor.

- The period is $\frac{\pi}{|b|}$.

- $c$ is the horizontal shift.

- $d$ is the vertical shift.

For the given function $y = \tan(\theta - \frac{\pi}{2})$, we have $a = 1$, $b = 1$, $c = \frac{\pi}{2}$, and $d = 0$.

Therefore, the period of the function is $\frac{\pi}{|1|} = \pi$, and the graph is shifted horizontally by $\frac{\pi}{2}$ to the right.

The standard tangent function $y = \tan(\theta)$ has vertical asymptotes at $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

For $y = \tan(\theta - \frac{\pi}{2})$, the vertical asymptotes occur when $\theta - \frac{\pi}{2} = \frac{\pi}{2} + n\pi$, which simplifies to $\theta = \pi + n\pi = (n+1)\pi$.

So, the vertical asymptotes are at $\theta = \pi, 2\pi, 3\pi, \dots$ and $\theta = 0, -\pi, -2\pi, \dots$

From the graph, we can see that the period is indeed $\pi$.

The standard $\tan(x)$ function has asymptotes at $x = \pm\frac{\pi}{2}$.

The given function $y = \tan(\theta - \frac{\pi}{2})$ has asymptotes shifted by $\frac{\pi}{2}$ to the right. Thus, asymptotes are at $\frac{\pi}{2} \pm \frac{\pi}{2} + n\pi$ where n is an integer. When $n = 0$ and we choose the positive option, we get $\frac{\pi}{2} + \frac{\pi}{2} = \pi$ and when we choose the negative option we get zero.

3. Final Answer

The period of the function $y = \tan(\theta - \frac{\pi}{2})$ is $\pi$. The graph has vertical asymptotes at integer multiples of $\pi$.