## Precalculus (6th Edition) Blitzer

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 $-2\text{ to }2$.