## Trigonometry (11th Edition) Clone

RECALL: The function $y=\tan{(x-d)}$: (1) has a period of $\pi$; (2) has the consecutive vertical asymptotes $x=-\frac{\pi}{2}+d$ and $x=\frac{\pi}{2}+d$; and (3) involves a phase shift of $|d|$ of the parent function $y=\tan{x}$ (to the left when $d\lt0$, to the right when $d\gt0$) The given function has $d=\frac{\pi}{2}$. Thus, the given function has: period = $\pi$ phase shift = $|\frac{\pi}{2}|=\frac{\pi}{2}$ to the right One period of the parent function $y=\tan{x}$ is in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ The given function has a phase shift of $\frac{\pi}{2}$ to the right therefore: (i) one period would be in the interval $[-\frac{\pi}{2}+\frac{\pi}{2}, \frac{\pi}{2} + \frac{\pi}{2}\pi]=[0, \pi]$; and (ii) the consecutive vertical asymptotes are $x=-\frac{\pi}{2} + \frac{\pi}{2}=0$ and $x=\frac{\pi}{2}+\frac{\pi}{2}=\pi$ Divide the interval $[0, \pi]$ into four equal parts to obtain the key x-values $\frac{\pi}{4}, \frac{\pi}{2} \frac{3\pi}{4}$. To graph the given function, perform the following steps: (1) Create a table of values using the key x-values listed above. (Refer to the table below.) (2) Graph the consecutive vertical asymptotes $x=0$ and $x=\pi$. (3) Plot each point from the table of values then connect them using a smooth curve. (Refer to the graph in the answer part above.)