## Trigonometry (11th Edition) Clone

RECALL: (1) The function $y=\tan{[b(x-d)]}$ has: period = $\frac{\pi}{b}$ phase (horizontal) shift = $|d|$, (to the right when $d\gt 0$, to the left when $d\lt0$) (2) The consecutive vertical asymptotes of the function $y=\tan{x}$, whose period is $\pi$, are $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$. Write the given funtion in $y=\tan{[b(x-d)]}$ form by factoring out $2$ within the tangent function to obtain: $y=\tan{[2(x-\frac{\pi}{2})]}$ Thus, with $b=2$ and $d=\frac{\pi}{4}$, the given function has: period = $\frac{\pi}{2}$ phase (horizontal) shift = $\frac{\pi}{2}$ to the right To find the consecutive vertical asymptotes, equate $(2x-\pi)$ to $-\frac{\pi}{2}$ and to $\frac{\pi}{2}$, then solve each equation to obtain: \begin{array}{ccc} 2x-\pi&=-\frac{\pi}{2} &\text{ or } &2x-\pi=\frac{\pi}{2} \\2x&=\frac{\pi}{2} &\text{or} &2x=\frac{3\pi}{2} \\x&=\frac{\pi}{4} &\text{or} &x=\frac{3\pi}{4} \end{array} This means that one period of the given function is in the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]$ and the next period is $[\frac{3\pi}{4}, \frac{5\pi}{4}]$. Dividing each interval into four equal parts give the key x-values: $\frac{3\pi}{8}, \frac{\pi}{2}, \frac{5\pi}{8}, \frac{7\pi}{8}, \pi, \frac{9\pi}{8}$. To graph the given function, perform the following steps: (1) Create a table of values for the given function using the key x-values listed above. (Refer to the attached image table below.) (2) Graph the consecutive vertical asymptotes. (3) Plot each point from the table then connect them using a smooth curve, making sure that the curves are asymptotic with the lines in Step (2) above. Refer to the graph in the answer part above.