Precalculus (6th Edition) Blitzer

The two consecutive asymptotes occur at $x-\frac{\pi }{4}=-\frac{\pi }{2}\text{ and }x-\frac{\pi }{4}=\frac{\pi }{2}$. We solve $x-\frac{\pi }{4}=-\frac{\pi }{2}$ to get \begin{align} & x-\frac{\pi }{4}=-\frac{\pi }{2}\text{ } \\ & x=-\frac{\pi }{2}+\frac{\pi }{4} \\ & x=-\frac{\pi }{4} \end{align} Again, solve $x-\frac{\pi }{4}=\frac{\pi }{2}$ to get \begin{align} & \text{ }x-\frac{\pi }{4}=\frac{\pi }{2} \\ & x=\frac{\pi }{2}+\frac{\pi }{4} \\ & \text{ }x=\frac{3\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 }{4}+\frac{3\pi }{4} \right)}{2} \\ & =\frac{\left( \frac{\pi }{2} \right)}{2} \\ & =\frac{\pi }{4} \end{align} Thus, the graph passes through $\left( \frac{\pi }{4},0 \right)$ and the x-intercept is $\frac{\pi }{4}$. 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, $x=-\frac{\pi }{4}$ and $x=\frac{3\pi }{4}$, to graph one full period of $y=-\tan \left( x-\frac{\pi }{4} \right)$ from $-\frac{\pi }{4}\text{ to }\frac{3\pi }{4}$.