Let's analyze the given function:
[tex]\[ f(\theta) = \sec \left(\frac{\theta}{6} + \frac{\pi}{6}\right) \][/tex]
### Amplitude
The secant function does not have an amplitude because it is not bounded. Therefore:
[tex]\[ \text{Amplitude: None} \][/tex]
### Period
To find the period of the function, we use the coefficient of [tex]\(\theta\)[/tex] inside the secant function. The period [tex]\(P\)[/tex] of the secant function is given by:
[tex]\[ P = \frac{2\pi}{\text{coefficient of }\theta} \][/tex]
In our function, the coefficient of [tex]\(\theta\)[/tex] is [tex]\(\frac{1}{6}\)[/tex]. Therefore:
[tex]\[ P = \frac{2\pi}{\frac{1}{6}} = 2\pi \cdot 6 = 12\pi \][/tex]
[tex]\[ \text{Period: } 37.69911184307752 \text{ radians} \][/tex]
### Phase Shift
The phase shift [tex]\(\phi\)[/tex] occurs due to the term added inside the secant function. The phase shift formula is:
[tex]\[ \phi = -\frac{\text{term inside with }\theta}{\text{coefficient of }\theta} \][/tex]
Here, our inside term is [tex]\(\frac{\pi}{6}\)[/tex] and the coefficient of [tex]\(\theta\)[/tex] is [tex]\(\frac{1}{6}\)[/tex]:
[tex]\[ \phi = -\frac{\frac{\pi}{6}}{\frac{1}{6}} = -\pi \][/tex]
[tex]\[ \text{Phase shift: } -3.141592653589793 \text{ radians} \][/tex]
### Vertical Shift
There is no term outside the secant function that vertically shifts the graph up or down, so the vertical shift is:
[tex]\[ \text{Vertical shift: } 0 \][/tex]
### Vertical Asymptotes
Vertical asymptotes occur where the argument of the secant function equals an odd multiple of [tex]\(\frac{\pi}{2}\)[/tex]:
[tex]\[ \frac{\theta}{6} + \frac{\pi}{6} = (2n + 1)\frac{\pi}{2} \][/tex]
where [tex]\(n\)[/tex] is an integer. To solve for [tex]\(\theta\)[/tex]:
[tex]\[ \frac{\theta}{6} = (2n + 1)\frac{\pi}{2} - \frac{\pi}{6} \][/tex]
[tex]\[ \theta = 6 \left[(2n + 1)\frac{\pi}{2} - \frac{\pi}{6}\right] = 6\left[\frac{3(2n + 1)\pi - \pi}{6}\right] = 6\left[\frac{(2n + 1 - \frac{1}{3})\pi}{1}\right] \][/tex]
[tex]\[ \theta = 3\pi(2n + \frac{2}{3}) \][/tex]
Therefore, the vertical asymptotes are located at:
[tex]\[ \theta = 3\pi(2n + \frac{2}{3}) \][/tex]
A few vertical asymptotes within one period are:
[tex]\[
-87.96459430051421, -69.11503837897544, -50.26548245743669, -31.41592653589793, -12.566370614359174, 6.283185307179586, 25.132741228718345, 43.982297150257104, 62.83185307179586, 81.68140899333461, 100.53096491487338
\][/tex]
So, summarizing all parts:
[tex]\[
\begin{align*}
\text{Amplitude:} &\ \text{None} \\
\text{Period:} &\ 37.69911184307752\ \text{radians} \\
\text{Phase Shift:} &\ -3.141592653589793\ \text{radians} \\
\text{Vertical Shift:} &\ 0 \\
\text{Vertical Asymptotes:} &\ -87.96459430051421, -69.11503837897544, -50.26548245743669, -31.41592653589793, -12.566370614359174, 6.283185307179586, 25.132741228718345, 43.982297150257104, 62.83185307179586, 81.68140899333461, 100.53096491487338
\end{align*}
\][/tex]
