question identify the period and equation of one asymptote of the function f(x)=2 sec(π/4 x)-1. period…

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 $f(x)=2\sec(\frac{\pi}{4}x)-1$, where $B = \frac{\pi}{4}$. $T=\frac{2\pi}{\frac{\pi}{4}}$

Step2: Calculate the period

Simplify the expression for the period: $\frac{2\pi}{\frac{\pi}{4}}=2\pi\times\frac{4}{\pi}=8$.

Step3: Recall the 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 $y = 2\sec(\frac{\pi}{4}x)-1$, we set $\frac{\pi}{4}x=(2n + 1)\frac{\pi}{2}$. Solve for $x$: $\frac{\pi}{4}x=(2n + 1)\frac{\pi}{2}$ $x=(2n + 1)\frac{\pi}{2}\times\frac{4}{\pi}=4n + 2$. When $n = 0$, one asymptote is $x = 2$.

Answer:

Period: $8$ Equation of one asymptote: $x = 2$