Answer
Refer to the graph below.
Work Step by Step
RECALL:
(1) The function $y=\cot{[b(x-d)]}$ 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 $y=\cot{x}$, whose period is $\pi$, are $x=0$ and $x=\pi$.
Write the given funtion in $y=\cot{[b(x-d)]}$ form by factoring out $3$ within the cotangent function to obtain:
$y=\cot{[3(x+\frac{\pi}{12})]}$
Thus, with $b=3$ and $d=-\frac{\pi}{12}$, the given function has:
period = $\frac{\pi}{3}$
phase (horizontal) shift = $\frac{\pi}{12}$ to the left
To find the consecutive vertical asymptotes, equate $(3x+\frac{\pi}{4})$ to $0$ and to $\pi$, then solve each equation to obtain:
\begin{array}{ccc}
3x+\frac{\pi}{4}&=0 &\text{ or } &3x+\frac{\pi}{4}=\pi
\\3x&=-\frac{\pi}{4} &\text{or} &3x=\frac{3\pi}{4}
\\x&=-\frac{\pi}{12} &\text{or} &x=\frac{\pi}{4}
\end{array}
This means that one period of the given function is in the interval $[-\frac{\pi}{12}, \frac{\pi}{4}]$ and the next period is $[\frac{\pi}{4}, \frac{7\pi}{12}]$.
Dividing each interval into four equal parts give the key x-values: $0, \frac{\pi}{12}, \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{\pi}{2}$.
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.
