## 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 $\frac{1}{2}$ within the tangent function to obtain: $y=\tan{[\frac{1}{2}(x+2\pi)]}$ Thus, with $b=\frac{1}{2}$ and $d=-2\pi$, the given function has: period = $\frac{\pi}{\frac{1}{2}}=2\pi$ phase (horizontal) shift = $2\pi$ to the left To find the consecutive vertical asymptotes, equate $(\frac{x}{2}+\pi)$ to $-\frac{\pi}{2}$ and to $\frac{\pi}{2}$, then solve each equation to obtain: \begin{array}{ccc} \frac{x}{2}+\pi&=-\frac{\pi}{2} &\text{ or } &\frac{x}{2}+\pi=\frac{\pi}{2} \\\frac{x}{2}&=-\frac{3\pi}{2} &\text{or} &\frac{x}{2}=-\frac{\pi}{2} \\x&=-3\pi &\text{or} &x=-\pi \end{array} This means that one period of the given function is in the interval $[-3\pi, -\pi]$ and the next period is $[-\pi, \pi]$. Dividing each interval into four equal parts give the key x-values: $-\frac{5\pi}{2}, -2\pi, -\frac{3\pi}{2}, -\frac{\pi}{2}, 0, \frac{\pi}{2}$. 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.