8.4 graphing tangent\n5) sketch one period of y = tan(π/4 θ) - 2\nperiod 1/9 ×\nvertical asymptotes 1/8 ×\ny…

Explanation:

Step1: Find the period

The general form of the tangent - function is $y = A\tan(B\theta - C)+D$. For the function $y=\tan(\frac{\pi}{4}\theta)-2$, where $B = \frac{\pi}{4}$. The period of the tangent function $y = \tan(B\theta)$ is given by $T=\frac{\pi}{|B|}$. So, $T=\frac{\pi}{\frac{\pi}{4}}=4$.

Step2: Find the vertical asymptotes

The vertical asymptotes of the tangent function $y = \tan(B\theta)$ occur at $B\theta=\frac{\pi}{2}+k\pi$, $k\in\mathbb{Z}$. For $y=\tan(\frac{\pi}{4}\theta)$, we set $\frac{\pi}{4}\theta=\frac{\pi}{2}+k\pi$. Solving for $\theta$ gives $\theta = 2 + 4k$, $k\in\mathbb{Z}$. For one - period, when $k = 0$, $\theta = 2$ and when $k=- 1$, $\theta=-2$.

Step3: Find the y - intercept

To find the y - intercept, we set $\theta = 0$ in the function $y=\tan(\frac{\pi}{4}\theta)-2$. Then $y=\tan(0)-2=0 - 2=-2$.

To sketch the graph:

1. Mark the vertical asymptotes at $\theta=-2$ and $\theta = 2$ for one period.
2. Plot the y - intercept at the point $(0,-2)$.
3. Recall the shape of the tangent function, which has a vertical stretch and shift as per the given function, and draw the curve between the vertical asymptotes.

Answer:

Period: 4; Vertical Asymptotes: $\theta=-2,\theta = 2$; Y - Intercept: - 2. The graph has vertical asymptotes at $\theta=-2$ and $\theta = 2$, passes through the point $(0,-2)$ and has the characteristic shape of the tangent function within the period $(-2,2)$.