Answer :
To determine which x-values are vertical asymptotes of the function [tex]\( y = \sec(x) \)[/tex], we need to understand where the secant function becomes undefined. The secant function is defined as:
[tex]\[ y = \sec(x) = \frac{1}{\cos(x)} \][/tex]
The secant function becomes undefined wherever [tex]\(\cos(x) = 0\)[/tex], because division by zero is undefined. Therefore, we need to find the x-values where [tex]\(\cos(x) = 0\)[/tex].
The cosine function, [tex]\(\cos(x)\)[/tex], equals zero at:
[tex]\[ x = \frac{(2n+1)\pi}{2} \][/tex]
where [tex]\( n \)[/tex] is any integer. This is because the cosine function has zeros at every odd multiple of [tex]\(\frac{\pi}{2}\)[/tex].
Let’s check each of the given options to see if they correspond to one of these points.
1. [tex]\( x = -2\pi \)[/tex]
- For [tex]\( x = -2\pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-2\pi = \frac{(2 \cdot -2 + 1)\pi}{2} = \frac{-3\pi}{2}\)[/tex], which is not correct because [tex]\(-3\pi/2\)[/tex] is not equivalent to [tex]\(-2\pi\)[/tex].
- So, [tex]\( x = -2\pi \)[/tex] is not an asymptote.
2. [tex]\( x = -\frac{\pi}{6} \)[/tex]
- For [tex]\( x = -\frac{\pi}{6} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-\frac{\pi}{6}\)[/tex] cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = -\frac{\pi}{6} \)[/tex] is not an asymptote.
3. [tex]\( x = \pi \)[/tex]
- For [tex]\( x = \pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\pi = \frac{2\pi}{2}\)[/tex], which cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = \pi \)[/tex] is not an asymptote.
4. [tex]\( x = \frac{3\pi}{2} \)[/tex]
- For [tex]\( x = \frac{3\pi}{2} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\frac{3\pi}{2} = \frac{(2 \cdot 1 + 1)\pi}{2}\)[/tex], which matches the form [tex]\(\frac{(2n+1)\pi}{2}\)[/tex] for [tex]\( n = 1 \)[/tex].
- Therefore, [tex]\( x = \frac{3\pi}{2} \)[/tex] is an asymptote.
Therefore, the correct answer is:
[tex]\[ x = \frac{3\pi}{2} \][/tex]
The option that corresponds to this is:
[tex]\[ 4 \][/tex]
[tex]\[ y = \sec(x) = \frac{1}{\cos(x)} \][/tex]
The secant function becomes undefined wherever [tex]\(\cos(x) = 0\)[/tex], because division by zero is undefined. Therefore, we need to find the x-values where [tex]\(\cos(x) = 0\)[/tex].
The cosine function, [tex]\(\cos(x)\)[/tex], equals zero at:
[tex]\[ x = \frac{(2n+1)\pi}{2} \][/tex]
where [tex]\( n \)[/tex] is any integer. This is because the cosine function has zeros at every odd multiple of [tex]\(\frac{\pi}{2}\)[/tex].
Let’s check each of the given options to see if they correspond to one of these points.
1. [tex]\( x = -2\pi \)[/tex]
- For [tex]\( x = -2\pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-2\pi = \frac{(2 \cdot -2 + 1)\pi}{2} = \frac{-3\pi}{2}\)[/tex], which is not correct because [tex]\(-3\pi/2\)[/tex] is not equivalent to [tex]\(-2\pi\)[/tex].
- So, [tex]\( x = -2\pi \)[/tex] is not an asymptote.
2. [tex]\( x = -\frac{\pi}{6} \)[/tex]
- For [tex]\( x = -\frac{\pi}{6} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-\frac{\pi}{6}\)[/tex] cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = -\frac{\pi}{6} \)[/tex] is not an asymptote.
3. [tex]\( x = \pi \)[/tex]
- For [tex]\( x = \pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\pi = \frac{2\pi}{2}\)[/tex], which cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = \pi \)[/tex] is not an asymptote.
4. [tex]\( x = \frac{3\pi}{2} \)[/tex]
- For [tex]\( x = \frac{3\pi}{2} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\frac{3\pi}{2} = \frac{(2 \cdot 1 + 1)\pi}{2}\)[/tex], which matches the form [tex]\(\frac{(2n+1)\pi}{2}\)[/tex] for [tex]\( n = 1 \)[/tex].
- Therefore, [tex]\( x = \frac{3\pi}{2} \)[/tex] is an asymptote.
Therefore, the correct answer is:
[tex]\[ x = \frac{3\pi}{2} \][/tex]
The option that corresponds to this is:
[tex]\[ 4 \][/tex]
