graph two periods of the given secant function.\n\n$y = - \\frac{1}{5} \\sec 3 \\pi x$\n\nchoose the correct…

Explanation:

Step1: Find the period of the function

The general form of the secant function is (y = A\sec(Bx - C)+D). The period of the secant function (y=\sec(Bx)) is given by (T=\frac{2\pi}{|B|}). For the function (y =-\frac{1}{5}\sec(3\pi x)), we have (B = 3\pi). So the period (T=\frac{2\pi}{3\pi}=\frac{2}{3}).

Step2: Analyze the amplitude and reflection

The amplitude of (y = A\sec(Bx)) is (|A|). Here (A=-\frac{1}{5}), so (|A|=\frac{1}{5}). The negative sign indicates a reflection about the (x) - axis.

Since the period is (\frac{2}{3}), two - periods will span an (x) - interval of length (2\times\frac{2}{3}=\frac{4}{3}).

We know that the vertical asymptotes of (y = \sec(x)) occur at (x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}). For (y=\sec(3\pi x)), the vertical asymptotes occur when (3\pi x=(2n + 1)\frac{\pi}{2}), or (x=\frac{2n + 1}{6},n\in\mathbb{Z})

When (x = 0), (y=-\frac{1}{5}\sec(0)=-\frac{1}{5}\times1=-\frac{1}{5})

We can also consider the key points of the cosine function (since (\sec x=\frac{1}{\cos x})). The function (y =-\frac{1}{5}\sec(3\pi x)) has the same period as (y=\cos(3\pi x)) but is the reciprocal (with a reflection and vertical shrink).

The standard secant function (y = \sec x) has a "U - shape" in each period. Since (A=-\frac{1}{5}), the graph of (y =-\frac{1}{5}\sec(3\pi x)) is a vertically shrunken ((y) values range from (-\frac{1}{5}) to (\frac{1}{5}) in terms of the extremum of the "U - shape") and reflected (because of the negative sign) version of (y=\sec(3\pi x))

Answer:

B.