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The graph of [tex]$y=\tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1$[/tex] is shown.

What is the period of the function?


Sagot :

To determine the period of the function [tex]\(y=\tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1\)[/tex], we follow several steps.

Let's start by examining the general properties of the tangent function. The standard form of a tangent function is:
[tex]\[ y = \tan(kx + c) \][/tex]
where [tex]\(k\)[/tex] and [tex]\(c\)[/tex] are constants. The period of the standard tangent function [tex]\( y = \tan(x) \)[/tex] is [tex]\( \pi \)[/tex].

Now, given our function:
[tex]\[ y = \tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1 \][/tex]

The addition of 1 outside the tangent function shifts the graph vertically but does not affect the period. We need to focus on the argument inside the tangent function:
[tex]\[ \tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right] \][/tex]

Let's simplify the argument:
[tex]\[ \frac{1}{4}\left(x-\frac{\pi}{2}\right) \][/tex]

We notice that the coefficient of [tex]\( x \)[/tex] is [tex]\( \frac{1}{4} \)[/tex]. For a tangent function [tex]\( y = \tan(kx) \)[/tex], the period is determined by the coefficient [tex]\( k \)[/tex]. Specifically, the period [tex]\( P \)[/tex] is given by:
[tex]\[ P = \frac{\pi}{k} \][/tex]

In our argument, the effective [tex]\( k \)[/tex] is [tex]\( \frac{1}{4} \)[/tex]. Therefore, we substitute [tex]\( k = \frac{1}{4} \)[/tex] into the period formula:
[tex]\[ P = \frac{\pi}{\frac{1}{4}} \][/tex]

To simplify this expression:
[tex]\[ P = \pi \cdot \frac{4}{1} = 4\pi \][/tex]

Therefore, the period of the function [tex]\( y = \tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1 \)[/tex] is:
[tex]\[ 4\pi \][/tex]

Hence, the period of the function is [tex]\( 4\pi \)[/tex].