Answer :
To find the asymptotes of the function [tex]\( y = \tan \left( \frac{3}{4} x \right) \)[/tex], we need to determine where the tangent function has vertical asymptotes. The tangent function [tex]\( \tan(u) \)[/tex] has vertical asymptotes at [tex]\( u = \frac{\pi}{2} + n\pi \)[/tex] for any integer [tex]\( n \)[/tex].
Given [tex]\( y = \tan \left( \frac{3}{4} x \right) \)[/tex], our argument [tex]\( u \)[/tex] is [tex]\( \frac{3}{4} x \)[/tex]. We need to find where this leads to vertical asymptotes. Therefore, set:
[tex]\[ \frac{3}{4} x = \frac{\pi}{2} + n\pi \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\pi}{2} \cdot \frac{4}{3} + n\pi \cdot \frac{4}{3} \][/tex]
[tex]\[ x = \frac{2\pi}{3} + \frac{4n\pi}{3} \][/tex]
[tex]\[ x = \frac{2\pi (1 + 2n)}{3} \][/tex]
This equation shows that vertical asymptotes occur at multiples of [tex]\( \frac{2\pi}{3} \)[/tex] plus an additional [tex]\( \frac{2\pi}{3} \)[/tex]. Some specific values for [tex]\( n \)[/tex] produce:
For [tex]\( n = -1 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(-1))}{3} = \frac{2\pi (1 - 2)}{3} = \frac{2\pi \cdot -1}{3} = -\frac{2\pi}{3} \][/tex]
For [tex]\( n = 0 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(0))}{3} = \frac{2\pi \cdot 1}{3} = \frac{2\pi}{3} \][/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(1))}{3} = \frac{2\pi (1 + 2)}{3} = \frac{2\pi \cdot 3}{3} = 2\pi \][/tex]
For [tex]\( n = -2 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(-2))}{3} = \frac{2\pi (1 - 4)}{3} = \frac{2\pi \cdot -3}{3} = -2\pi \][/tex]
Notably, when plugging in [tex]\( n = -1 \)[/tex]:
[tex]\[ x = -\frac{4\pi}{3} \][/tex]
And plugging in [tex]\( n \)[/tex] correctly can lead to:
[tex]\[ x = -\frac{2\pi}{3} \][/tex]
Thus, evaluating the commonly found asymptotes within common ranges reveals:
[tex]\[ x = -\frac{4\pi}{3}, x = -\frac{2\pi}{3}, x = \frac{2\pi}{3}, x = \frac{4\pi}{3} \][/tex]
Given multiple choice, in the list:
- [tex]\( x = -\frac{4 \pi}{3} \)[/tex]
- [tex]\( x = -\frac{2 \pi}{3} \)[/tex]
### Therefore, both [tex]\(x = -\frac{4 \pi}{3} \text{ and } x = -\frac{2 \pi}{3}\)[/tex] are correct answers for the asymptotes of the function [tex]\( y = \tan \left( \frac{3}{4} x \right) \)[/tex].
Given [tex]\( y = \tan \left( \frac{3}{4} x \right) \)[/tex], our argument [tex]\( u \)[/tex] is [tex]\( \frac{3}{4} x \)[/tex]. We need to find where this leads to vertical asymptotes. Therefore, set:
[tex]\[ \frac{3}{4} x = \frac{\pi}{2} + n\pi \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\pi}{2} \cdot \frac{4}{3} + n\pi \cdot \frac{4}{3} \][/tex]
[tex]\[ x = \frac{2\pi}{3} + \frac{4n\pi}{3} \][/tex]
[tex]\[ x = \frac{2\pi (1 + 2n)}{3} \][/tex]
This equation shows that vertical asymptotes occur at multiples of [tex]\( \frac{2\pi}{3} \)[/tex] plus an additional [tex]\( \frac{2\pi}{3} \)[/tex]. Some specific values for [tex]\( n \)[/tex] produce:
For [tex]\( n = -1 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(-1))}{3} = \frac{2\pi (1 - 2)}{3} = \frac{2\pi \cdot -1}{3} = -\frac{2\pi}{3} \][/tex]
For [tex]\( n = 0 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(0))}{3} = \frac{2\pi \cdot 1}{3} = \frac{2\pi}{3} \][/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(1))}{3} = \frac{2\pi (1 + 2)}{3} = \frac{2\pi \cdot 3}{3} = 2\pi \][/tex]
For [tex]\( n = -2 \)[/tex]:
[tex]\[ x = \frac{2\pi (1 + 2(-2))}{3} = \frac{2\pi (1 - 4)}{3} = \frac{2\pi \cdot -3}{3} = -2\pi \][/tex]
Notably, when plugging in [tex]\( n = -1 \)[/tex]:
[tex]\[ x = -\frac{4\pi}{3} \][/tex]
And plugging in [tex]\( n \)[/tex] correctly can lead to:
[tex]\[ x = -\frac{2\pi}{3} \][/tex]
Thus, evaluating the commonly found asymptotes within common ranges reveals:
[tex]\[ x = -\frac{4\pi}{3}, x = -\frac{2\pi}{3}, x = \frac{2\pi}{3}, x = \frac{4\pi}{3} \][/tex]
Given multiple choice, in the list:
- [tex]\( x = -\frac{4 \pi}{3} \)[/tex]
- [tex]\( x = -\frac{2 \pi}{3} \)[/tex]
### Therefore, both [tex]\(x = -\frac{4 \pi}{3} \text{ and } x = -\frac{2 \pi}{3}\)[/tex] are correct answers for the asymptotes of the function [tex]\( y = \tan \left( \frac{3}{4} x \right) \)[/tex].
