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Sketch two periods of the graph of the function [tex]\( p(x) = \tan \left( x - \frac{\pi}{4} \right) \)[/tex].

Identify the following:

1. Stretching factor: [tex]\(\square\)[/tex]
2. Period: [tex]\( P = \square \pi \)[/tex]

Enter the asymptotes of the function on the domain [tex]\([-P, P]\)[/tex].

Asymptotes: [tex]\( x = \square \)[/tex]


Sagot :

Let's analyze the function [tex]\( p(x) = \tan \left(x - \frac{\pi}{4}\right) \)[/tex].

### Stretching Factor
The function [tex]\( \tan(x) \)[/tex] does not have any vertical stretching factor. Since the given function [tex]\( p(x) = \tan \left(x - \frac{\pi}{4}\right) \)[/tex] is just a horizontal translation of the basic tangent function, there is no stretching factor involved.
- Stretching factor = 1

### Period
The period of the tangent function [tex]\(\tan(x)\)[/tex] is [tex]\( \pi \)[/tex]. Horizontal shifts do not affect the period. Thus,
- Period: [tex]\( P = \pi \)[/tex]

### Asymptotes
The vertical asymptotes of the function [tex]\(\tan(x)\)[/tex] occur at the points where [tex]\(\tan(x)\)[/tex] is undefined, which is at [tex]\( x = \frac{\pi}{2} + k\pi \)[/tex] for any integer [tex]\( k \)[/tex].

For [tex]\( p(x) = \tan \left(x - \frac{\pi}{4}\right) \)[/tex], these asymptotes will be shifted horizontally by [tex]\(\frac{\pi}{4}\)[/tex]:
[tex]\[ x - \frac{\pi}{4} = \frac{\pi}{2} + k\pi \][/tex]
[tex]\[ x = \frac{\pi}{2} + k\pi + \frac{\pi}{4} \][/tex]
[tex]\[ x = \frac{3\pi}{4} + k\pi \][/tex]

We need the asymptotes within the domain [tex]\([-P, P]\)[/tex]:
Here, [tex]\( P = \pi \)[/tex]

1. For [tex]\( k = -1 \)[/tex]:
[tex]\[ x = \frac{3\pi}{4} - \pi = -\frac{\pi}{4} \][/tex]

2. For [tex]\( k = 0 \)[/tex]:
[tex]\[ x = \frac{3\pi}{4} \][/tex]

Thus the asymptotes in the domain [tex]\([-P, P]\)[/tex] are at [tex]\( x = -\frac{\pi}{4} \)[/tex] and [tex]\( x = \frac{3\pi}{4} \)[/tex].

To summarize:
- Stretching factor = 1
- Period: [tex]\( P = \pi \)[/tex]
- Asymptotes: [tex]\( x = -\frac{\pi}{4} ; \frac{3\pi}{4} \)[/tex]