graph the function over a two - period interval.\ny = - 1 - tan x\nchoose the correct graph…

Explanation:

Step1: Recall properties of $y = \tan x$

The period of $y=\tan x$ is $\pi$. The vertical - asymptotes of $y = \tan x$ occur at $x=\frac{\pi}{2}+n\pi$, $n\in\mathbb{Z}$.

Step2: Analyze $y=-1 - \tan x$

The graph of $y=-1 - \tan x$ is a transformation of the graph of $y = \tan x$. The negative sign in front of $\tan x$ reflects the graph of $y=\tan x$ about the $x$ - axis, and the subtraction of 1 shifts the graph of $- \tan x$ down 1 unit. When $x = 0$, $y=-1-\tan(0)=-1$. The vertical asymptotes of $y=-1 - \tan x$ are still at $x=\frac{\pi}{2}+n\pi$, $n\in\mathbb{Z}$.

Answer:

Without seeing the actual graphs A, B, C, D, we can describe the key - features of the correct graph: It has vertical asymptotes at $x=\frac{\pi}{2}+n\pi$, $n\in\mathbb{Z}$, is reflected about the $x$ - axis compared to $y = \tan x$ and is shifted down 1 unit. If you provide the details of the graphs A, B, C, D, we can further determine the correct one.