The graph of y=tan[(1)/(4)(x-(pi )/(2))]+1 is shown. What is the period of the function? square DONE V

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The graph of y=tan[(1)/(4)(x-(pi )/(2))]+1 is shown. What is the period of the function? square DONE V

Answer 1

To find the period of the function \(y=\tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1\), we need to look at the argument of the tangent function and determine how much \(x\) needs to change for the function to complete one full cycle.The general form of a tangent function is \(y = \tan(bx - c)\), where the period \(P\) is given by \(P = \frac{\pi}{|b|}\).In the given function, \(y=\tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1\), the coefficient \(b\) in front of \(x\) is \(\frac{1}{4}\).Step 1: Identify the coefficient \(b\) that affects the period of the tangent function.The coefficient \(b\) is \(\frac{1}{4}\).Step 2: Use the formula for the period of the tangent function.The period \(P\) of the tangent function is given by \(P = \frac{\pi}{|b|}\).Step 3: Substitute the value of \(b\) into the formula.\(P = \frac{\pi}{|\frac{1}{4}|}\)Step 4: Calculate the period.\(P = \frac{\pi}{\frac{1}{4}}\)\(P = \pi \times 4\)\(P = 4\pi\)Answer: The period of the function \(y=\tan \left[\frac{1}{4}\left(x-\frac{\pi}{2}\right)\right]+1\) is \(4\pi\).

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The graph of y=tan[(1)/(4)(x-(pi )/(2))]+1 is shown. What is the period of the function? square DONE V

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