Answer
See the graph below:
Work Step by Step
The two consecutive asymptotes occur at $x+\pi =-\frac{\pi }{2}\text{ and }x+\pi =\frac{\pi }{2}$.
Solve $x+\pi =-\frac{\pi }{2}$ to get
$\begin{align}
& x+\pi =-\frac{\pi }{2}\text{ } \\
& x=-\frac{\pi }{2}-\pi \text{ } \\
& x=-\frac{3\pi }{2}
\end{align}$
Again, solving $x+\pi =\frac{\pi }{2}$, we get
$\begin{align}
& x+\pi =\frac{\pi }{2} \\
& x=\frac{\pi }{2}-\pi \\
& x=-\frac{\pi }{2}
\end{align}$
Now, the x-intercept is in between the two consecutive asymptotes.
The x-intercept is given as follows:
$\begin{align}
& x\text{-intercept = }\frac{-\frac{3\pi }{2}-\frac{\pi }{2}}{2} \\
& =\frac{-2\pi }{2} \\
& =-\pi
\end{align}$
Thus, the graph passes through $\left( -\pi ,0 \right)$ and the x-intercept is $-\pi $. As the coefficient of the provided tangent function is $1$, the points on the graph midway between the x-intercept and the asymptotes have y-coordinates of $-1$ and $1$.
We use the two consecutive asymptotes, $x=-\frac{3\pi }{2}$ and $x=-\frac{\pi }{2}$, to graph one full period of $y=\tan \left( x+\pi \right)$ from $-\frac{3\pi }{2}\text{ to }-\frac{\pi }{2}$.
