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