Question

graph two periods of the given secant function.

$y = - \frac{1}{5} \sec 3 \pi x$

choose the correct graph below.

a.
b.
c.

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.