Problem 12 - Precalculus Essentials

Question

Graph two periods of the given tangent function. $$y=\tan \left(x-\frac{\pi}{4}\right)$$

Step-by-Step Solution

Answer

The requested graph of the tangent function \(y=\tan \left(x-\frac{\pi}{4}\right)\) starts at negative infinity at \(x=-\frac{\pi}{4}\), crosses through \(\frac{\pi}{4}\) in y-axis, and goes to positive infinity at \(x=\frac{3\pi}{4}\). This pattern repeats every \(\pi\) radians for two periods.

Step 1: Understand Tangent Function

The tangent function, \(y=\tan(x)\), is periodic with a period of \(\pi\). The function is undefined when \(\cos(x)=0\), where \(x\) is an odd multiple of \(\frac{\pi}{2}\). Between these points, the graph of the tangent function passes the origin and changes from negative to positive or from positive to negative. This general pattern repeats every \(\pi\) radians.

Step 2: Understand Phase Shift

The function \(y=\tan \left(x-\frac{\pi}{4}\right)\) is a tangent function that has been shifted to the right by \(\frac{\pi}{4}\). This means we have to adjust all the x-values by \(\frac{\pi}{4}\). A general form of tangent function would be \(y=\tan(\omega x - \phi)\), here \(\omega = 1\) and \(\phi = \frac{\pi}{4}\). So the graph is translated to the right by \(\frac{\pi}{4}\).

Step 3: Plot the Parent Function

First of all, one can draw the tangent function without any phase shift. The graph will start at negative infinity at \(x=-\frac{\pi}{2}\), cross through the origin, then go to positive infinity at \(x=+\frac{\pi}{2}\). This pattern repeats every \(\pi\) radians.

Step 4: Apply Phase Shift and Sketch the Function

Applying the phase shift of \(\frac{\pi}{4}\) we adjust all the x-values by adding \(\frac{\pi}{4}\) to each x-value on the parent function. So now, the pattern starts at negative infinity at \(x=-\frac{\pi}{4}\), cross through \(\frac{\pi}{4}\) in y-axis, and go to positive infinity at \(x=\frac{3\pi}{4}\). Just like the parent function, this pattern also repeats every \(\pi\) radians for two periods.