fill in each blank so that the resulting statement is true. an interval containing one period of y = 3 tan(x…

Explanation:

Step1: Recall the period of tangent function

The period of the tangent function $y = A\tan(Bx - C)$ is $\frac{\pi}{|B|}$. For $y = 3\tan(x-\frac{\pi}{5})$, $B = 1$, so the period is $\pi$.

Step2: Find the interval for one - period

The tangent function $y=\tan u$ has a period on the interval $(-\frac{\pi}{2},\frac{\pi}{2})$. Set $u=x - \frac{\pi}{5}$. Then $-\frac{\pi}{2}<x-\frac{\pi}{5}<\frac{\pi}{2}$. Solving for $x$: Add $\frac{\pi}{5}$ to all parts of the inequality: $-\frac{\pi}{2}+\frac{\pi}{5}<x<\frac{\pi}{2}+\frac{\pi}{5}$. $-\frac{5\pi - 2\pi}{10}<x<\frac{5\pi+2\pi}{10}$, so $-\frac{3\pi}{10}<x<\frac{7\pi}{10}$.

Step3: Find the asymptotes

The asymptotes of $y = \tan u$ occur at $u=-\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. For $u=x-\frac{\pi}{5}$, we set $x-\frac{\pi}{5}=-\frac{\pi}{2}$ and $x-\frac{\pi}{5}=\frac{\pi}{2}$. For $x-\frac{\pi}{5}=-\frac{\pi}{2}$, add $\frac{\pi}{5}$ to both sides: $x=-\frac{\pi}{2}+\frac{\pi}{5}=-\frac{3\pi}{10}$. For $x-\frac{\pi}{5}=\frac{\pi}{2}$, add $\frac{\pi}{5}$ to both sides: $x=\frac{\pi}{2}+\frac{\pi}{5}=\frac{7\pi}{10}$.

Answer:

An interval containing one period of $y = 3\tan(x-\frac{\pi}{5})$ is $(-\frac{3\pi}{10},\frac{7\pi}{10})$. Thus, two consecutive asymptotes occur at $x =-\frac{3\pi}{10}$ and $x=\frac{7\pi}{10}$.