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