Problem 43 - Precalculus: A Concise Course

Question

Use a graphing utility to graph the function. Include two full periods. $$ y=\tan \left(x-\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

The graph of \( y = \tan(x - \frac{\pi}{4}) \) begins at \( x = \frac{\pi}{4} \) and ends at \( x = \frac{5\pi}{4} \) for its first period, and similarly for subsequent periods. It passes through \( 0 \) at \( x = \frac{3\pi}{4} \), with asymptotes at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Step 1: Identify the period and phase shift

In the given equation \( y=\tan \left(x-\frac{\pi}{4}\right) \), we can see that it's a tangent function with phase shift of \( \frac{\pi}{4} \) to the right. The period of the tangent function is \( \pi \).

Step 2: Determine the primary period

Since the function has been shifted right by \( \frac{\pi}{4} \), the primary period is from \( \frac{\pi}{4} \) to \( \frac{\pi}{4} + \pi = \frac{5\pi}{4} \).

Step 3: Draw the graph

First, plot the middle line of the period at \( x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \), at which \( y = 0 \). Then draw asymptotes at the ends of the period \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). Since it's a tangent graph, it goes to negative infinity at \( x = \frac{\pi}{4} \), goes through \( 0 \) at \( x = \frac{3\pi}{4} \), and climbs to positive infinity at \( x = \frac{5\pi}{4} \). Repeat drawing the graph in the same manner to plot another full period for \( y = \tan(x - \frac{\pi}{4}) \).

Key Concepts

Tangent Function Transformation

Understanding the transformation of the tangent function is essential for graphing it accurately. The general form of the tangent function is given by y = a \tan(b(x - c)) + d, where a affects the vertical stretch or compression, b impacts the period, c represents the phase shift, and d is the vertical translation.

In the exercise y = \tan \bigl(x - \frac{\[pi\]}{4} \bigr), there's a horizontal shift but no vertical stretch, compression, or translation, since the coefficients a and d (which are not written) are both 1 and 0 respectively. The transformation involves sliding the graph to the right by \frac{\[pi\]}{4} units without altering its shape or size. This shift affects the starting and ending points of each period of the tangent curve.

Period of a Tangent Function

The period of a function deals with the length of one full cycle of the curve before it starts repeating. For the standard tangent function y = \tan(x), the period is \[pi\]. Transformations can change this period, determined by the 'b' value in the general tangent equation, but in our exercise, there's no coefficient affecting the 'x', which means the period remains \[pi\].

Therefore, the standard period of the function given in the exercise is not altered. It still completes one cycle every \[pi\] units along the x-axis. However, due to phase shifting, the position of the period is moved right by \frac{\[pi\]}{4}, changing the start and end points of the cycle but not its length. To visualize the period lengths, one can draw tick marks on the x-axis at intervals of \frac{\[pi\]}{4} plus multiples of \[pi\].

Phase Shift in Trigonometry

Phase shift refers to a horizontal movement of a trigonometric function along the x-axis. If the function is y = \tan(b(x - c)), the phase shift is given by the parameter c. It can be thought of as the starting point of the function's cycle. Positive values for c shift the graph to the right, while negative values shift it to the left.

In this exercise, the function y = \tan \bigl(x - \frac{\[pi\]}{4} \bigr) has a phase shift of \frac{\[pi\]}{4} to the right. The function's middle line, which hits y=0 (the x-axis), occurs at \frac{3\[pi\]}{4} rather than at \frac{\[pi\]}{2}, which would have been the case without a phase shift. The graph starts from an asymptote at \frac{\[pi\]}{4} and ends at \frac{5\[pi\]}{4} for one full period, showing how the phase shift affects the function's placement on the coordinate plane.