identify the period and equation of one asymptote of the function f(x)=3 sec(2/3 x)+3. period: equation of…

Explanation:

Step1: Recall period formula for secant

The general form of a secant - type function is $y = A\sec(Bx - C)+D$, and its period is given by $T=\frac{2\pi}{|B|}$. For the function $f(x)=3\sec(\frac{2}{3}x)+3$, $B = \frac{2}{3}$. $T=\frac{2\pi}{\left|\frac{2}{3}\right|}$

Step2: Calculate the period

$T=\frac{2\pi}{\frac{2}{3}}=2\pi\times\frac{3}{2}=3\pi$.

Step3: Recall asymptote formula for secant

The secant function $y = \sec x$ has asymptotes at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$. For the function $y = \sec(Bx - C)$, the asymptotes are given by $Bx - C=(2n + 1)\frac{\pi}{2}$. In the function $f(x)=3\sec(\frac{2}{3}x)+3$, setting $\frac{2}{3}x=(2n + 1)\frac{\pi}{2}$. Let $n = 0$, then $\frac{2}{3}x=\frac{\pi}{2}$, and $x=\frac{3\pi}{4}$.

Answer:

Period: $3\pi$ Equation of one asymptote: $x=\frac{3\pi}{4}$