91Ó°ÊÓ

The period of the secant function is the interval after which it starts repeating its values. For a given secant function in the form \(y = a \text{sec}(bx)\), we determine the period by the coefficient \(b\). This coefficient affects the frequency of the function. Specifically, the period is calculated using the formula: \[T = \frac{2\pi}{b}\]
In our given function \(y = 2 \text{sec}(\frac{x}{4})\), we have \(b = \frac{1}{4}\). Substituting this into the formula, we get:
\[T = \frac{2\pi}{\frac{1}{4}} = 8\pi\]
This means the secant function \(y = 2 \text{sec}(\frac{x}{4})\) will repeat its pattern every \(8\pi \) units along the x-axis.

Vertical asymptotes occur in a function where the function tends to infinity. For the secant function, these asymptotes happen at the points where the cosine function is zero because division by zero is undefined.
For the secant function in the form \(\text{sec}(bx)\), cosine is zero at \(bx = (2k+1)\frac{\pi}{2}\), where \(k\) is any integer. To find the x-values for our function, substitute \(b = \frac{1}{4}\):
\[\frac{1}{4}x = (2k+1)\frac{\pi}{2}\]
Solving for \(x\), we get:
\[x = 4(2k+1)\frac{\pi}{2} = (2k+1)2\pi\]
This indicates the vertical asymptotes for the function \(y = 2\text{sec}(\frac{x}{4})\) occur at \(x = (2k+1)2\pi \).

The range of a function is all the possible output values (y-values) it can take. The secant function, \(\text{sec}(x)\), corresponds to the reciprocal of the cosine function. Since the cosine function oscillates between \([-1, 1]\), its reciprocal (secant) is outside this interval, taking values in \((-\infty, -1] \cup [1, \infty)\).
For the function \(y = 2 \text{sec}(\frac{x}{4})\), we scale the standard secant function by 2. Thus, anything outside \([-1, 1]\) for the secant becomes outside \([-2, 2]\) for our function:
\[\text{sec}(\frac{x}{4}) otin (-1, 1)\] thus:
\[2 \text{sec}(\frac{x}{4}) otin (-2, 2),\]
So, the range of \(y = 2 \text{sec}(\frac{x}{4}) \) is \((-\infty, -2] \cup [2, \infty)\).