Answer
See the graph below:
Work Step by Step
The two consecutive asymptotes occur at
$\frac{\pi }{4}x=-\frac{\pi }{2}\text{ and }\frac{\pi }{4}x=\frac{\pi }{2}$.
Solve $\frac{\pi }{4}x=-\frac{\pi }{2}$ to get
$\begin{align}
& \frac{\pi }{4}x=-\frac{\pi }{2}\text{ } \\
& x=-\frac{\pi }{2}\cdot \frac{4}{\pi }\text{ } \\
& x=-2\text{ }
\end{align}$
Again solve $\frac{\pi }{4}x=\frac{\pi }{2}$ to get
$\begin{align}
& \text{ }\frac{\pi }{4}x=\frac{\pi }{2} \\
& x=\frac{\pi }{2}\cdot \frac{4}{\pi } \\
& x=2
\end{align}$
Now, the x-intercept are in between the two consecutive asymptotes.
The x-intercept is given as follows:
$\begin{align}
& x\text{-intercept= }\frac{-2+2}{2} \\
& =\frac{0}{2} \\
& =0
\end{align}$
Thus, the graph passes through (0, 0) and the x-intercept is 0. As the coefficient of the provided tangent 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, $x=-2$ and $x=2$, to graph one full period of $y=-2\tan \frac{\pi }{4}x$ from $-2\text{ to }2$.
