Question 6.5.1: Graphing y = a tan b(x - c) or y = a cot b(x - c) OBJECTIVE ...

1.

\begin{aligned}&y= \underset{\overset{\uparrow }{a=3}}{3} \tan \left[ \underset{\overset{\uparrow }{b=2}}{2}\left(x-\underset{\overset{\uparrow }{c=\frac{\pi}{4}}}{\frac{\pi}{4}}\right)\right] \end{aligned}

vertical stretch factor = |a| = |3| = 3

\text { period }=\frac{\pi}{b}=\frac{\pi}{2}                   \text { phase shift }=c=\frac{\pi}{4}

2.

and

\begin {array} {rl|rl} \begin{aligned}2\left(x-\frac{\pi}{4}\right) &=-\frac{\pi}{2} \\x-\frac{\pi}{4} &=-\frac{\pi}{4} \\x &=-\frac{\pi}{4}+\frac{\pi}{4} \\x &=0\end{aligned}&& \begin{aligned}2\left(x-\frac{\pi}{4}\right) &=\frac{\pi}{2} \\x-\frac{\pi}{4} &=\frac{\pi}{4} \\x &=\frac{\pi}{4}+\frac{\pi}{4} \\x &=\frac{\pi}{2}\end{aligned}\end {array}

3. The interval \left(0, \frac{\pi}{2}\right) \text { has length } \frac{\pi}{2} \text {, and } \frac{1}{4}\left(\frac{\pi}{2}\right)=\frac{\pi}{8} \text {. }

The division points of the interval \left(0, \frac{\pi}{2}\right) are 0+\frac{\pi}{8}=\frac{\pi}{8}, 0+2\left(\frac{\pi}{8}\right)=\frac{\pi}{4} \text {, and } 0+3\left(\frac{\pi}{8}\right)=\frac{3 \pi}{8}.

4.

\begin{array}{|c|c|}\hline x & y=3 \tan \left[2\left(x-\frac{\pi}{4}\right)\right] \\\hline \frac{\pi}{8} & -3 \\\hline \frac{\pi}{4} & 0 \\\hline \frac{3 \pi}{8} & 3 \\\hline\end{array}

5.