Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\sec \left(x-\frac{3 \pi}{4}\right) $$

Step-by-Step Solution

Answer

Period is \(2\pi\). Vertical asymptotes at \(x = \frac{5\pi}{4} + k\pi\).

Step 1: Understand the Secant Function

The secant function, \(y = \sec(x)\), is the reciprocal of the cosine function, \(y = \frac{1}{\cos(x)}\). Its period is the same as that of the cosine function, which is \(2\pi\). However, secant has vertical asymptotes where the cosine function equals zero.

Step 2: Determine the Period of the Given Function

Since the transformation does not involve any changes to the frequency of \(x\) (there's no coefficient multiplying \(x\)), the period remains \(2\pi\). Therefore, the period of \(y = \sec\left(x - \frac{3\pi}{4}\right)\) is \(2\pi\).

Step 3: Identify the Phase Shift

The function \(y = \sec\left(x - \frac{3\pi}{4}\right)\) is shifted to the right by \(\frac{3\pi}{4}\). This shift affects the location of the asymptotes and the key points on the graph.

Step 4: Find the Asymptotes

The asymptotes of the secant function occur where the cosine function is zero. In this transformed function, \(\cos\left(x - \frac{3\pi}{4}\right) = 0\) gives the asymptotes. Solve for \(x\):\[x - \frac{3\pi}{4} = \frac{\pi}{2} + k\pi\quad\text{for any integer } k.\]Thus, the asymptotes are at \[x = \frac{3\pi}{4} + \frac{\pi}{2} + k\pi = \frac{5\pi}{4} + k\pi.\]

Step 5: Sketch the Graph

To sketch the graph:1. Mark the phase shift by moving the graph of \(y = \sec(x)\) to the right by \(\frac{3\pi}{4}\).2. Draw vertical asymptotes at \(x = \frac{5\pi}{4} + k\pi\).3. Since the period is \(2\pi\), repeat the pattern every \(2\pi\).4. The graph will have vertical stretches where the original cosine function is at its maximum and minimum.

Key Concepts

Secant Function

The secant function, often expressed as \(y = \sec(x)\), is a fundamental trigonometric function derived from the cosine function. Specifically, it is the reciprocal of the cosine, or \(y = \frac{1}{\cos(x)}\). Unlike the continuous wave of the cosine function, the secant has distinct characteristics due to its reciprocal nature.

• The secant function has undefined points, or vertical asymptotes, where the cosine value equals zero. This results in noticeable breaks in the secant graph.
• Because it is based on the cosine wave, the secant's general pattern follows the cosine's periodic nature but with these breaks at each cosine zero.

Understanding the secant function involves recognizing both its periodic waves and its unique undefined regions, which divide the graph into separate loops. Analyzing these properties is crucial for graphing and solving equations involving secant.

Period of Trigonometric Functions

The period of a function refers to the interval length it takes for the function to complete a full cycle and start repeating. For the secant function, like the cosine, this period is \(2\pi\).

• This means that, within each interval of length \(2\pi\), the secant function exhibits a repeated pattern of behavior.
• Each segment of the graph within this period will have the same general shape and any shifts or changes to the function must be accounted for within this consistent repeat length.

Since the secant's period matches the cosine's, any transformations applied to the graph, such as shifts or stretches, will likewise adhere to this \(2\pi\) repetition rule. Such a consistent periodic feature helps in predicting the behavior of the graph over broader intervals.

Phase Shift

A phase shift in trigonometric functions refers to the horizontal movement of the graph along the x-axis. For the function \(y = \sec\left(x - \frac{3\pi}{4}\right)\), there is a phase shift to the right by \(\frac{3\pi}{4}\).

• This shift affects the starting point of each cycle of the secant function. Essentially, it drags the entire graph \(\frac{3\pi}{4}\) units to the right.
• Key points such as the peaks, valleys, and vertical asymptotes are all influenced by this shift, necessitating recalibration of the graph's layout.

Accurately recognizing and plotting the phase shift is essential for depicting the true nature of the function and ensuring all features line up appropriately during the cycle. This adjustment helps to maintain the integrity of the graph's periodic repetition and symmetry.

Vertical Asymptotes

Vertical asymptotes are lines where a function approaches but does not touch or cross. They represent values of x where the function becomes undefined, mostly due to division by zero in trigonometric contexts. In the secant function, these occur where \(\cos(x) = 0\).

• For the given function \(y = \sec\left(x - \frac{3\pi}{4}\right)\), the asymptotes are shifted to determine where \(\cos\left(x - \frac{3\pi}{4}\right) = 0\).
• Solving \(x - \frac{3\pi}{4} = \frac{\pi}{2} + k\pi\), where \(k\) is any integer, gives these asymptotes.

Thus, the vertical asymptotes are positioned at \(x = \frac{5\pi}{4} + k\pi\). These lines divide the graph at regular intervals and are an essential feature of the overall shape and structure of the graph. Understanding these boundaries helps in accurately sketching the functions and predicting their behavior.