Problem 43 - Precalculus

Question

In Exercises \(29-46,\) graph the functions over the indicated intervals. \(y=2 \sec (2 x-\pi),-2 \pi \leq x \leq 2 \pi\) one period one period

Step-by-Step Solution

Answer

Graph the function by plotting secant values between asymptotes \(-\frac{\pi}{4}\) to \(\frac{7\pi}{4}\), following its \(\pi\) period across \(-2\pi\leq x \leq 2\pi\).

Step 1: Understanding the Function

The given function is a secant function, which is the reciprocal of the cosine function. The function we are dealing with is \(y = 2 \sec(2x - \pi)\). This means we need to focus on the behavior of the cosine function \(\cos(2x - \pi)\) as it will influence the secant function.

Step 2: Determine the Period of the Function

The general form for determining the period of a secant function \(y = a \sec(bx - c)\) is \(\frac{2\pi}{b}\). For our function, \(b = 2\). Thus, the period of the function is \(\frac{2\pi}{2} = \pi\).

Step 3: Identify Asymptotes and Key Points

Set \(\cos(2x - \pi) = 0\) to find vertical asymptotes where the secant function is undefined. This occurs when\[ 2x - \pi = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \] solving for \(x\) gives\[ x = \frac{3\pi}{4} + k\frac{\pi}{2} \] for \(k = -1, 0, 1, 2, 3\). These include, but are not limited to, \(-\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}\).

Step 4: Plot the Basic Shape of One Period

Since the period is \(\pi\), begin plotting from \(-\frac{\pi}{4}\) where the function is undefined. Plot key points within \(-2\pi \le x \le 2\pi\) focusing on one period, noting amplitude from values such as \(y=2\) when \(\cos(2x-\pi)=1\) and \(y=-2\) when \(\cos(2x-\pi)=-1\). Notice symmetrical behavior about the vertical asymptotes.

Step 5: Extend the Graph to Fill the Interval

Using a similar process from Step 4, extend the graph to fill the required interval \(-2\pi \) to \(2\pi\). Repeat the key points and asymptotes due to periodicity. Recognize pattern repetitions beyond one period to ensure continuous plotting.

Key Concepts

Secant Function

The secant function, denoted as "sec," is the reciprocal of the cosine function. In trigonometry, this means that for a given angle, the secant is the length of the hypotenuse of a right-angled triangle divided by the length of the adjacent side. This function is defined as:

• \( \sec(x) = \frac{1}{\cos(x)} \)

The secant function is important because it reveals points where the cosine value is zero, as these points become vertical asymptotes in the graph of the secant function.
When graphing the secant function, it's crucial to understand its relationship with the cosine. Instead of looking at the cosine's rise and fall, you observe the secant's spikes and gaps. Whenever the cosine function crosses the x-axis, the secant will have a vertical asymptote. The secant graph oscillates in a repeating pattern of upward and downward branches that open like hyperbolas between the asymptotes.
For example, in the function \( y = 2 \sec(2x - \pi) \), you would analyze the behavior of \( \cos(2x - \pi) \) in order to construct the graph of the secant. Whenever \( \cos(2x - \pi) = 0 \), you will find a vertical asymptote in the graph of the secant.

Period of a Function

The period of a function dictates how often the pattern of the function repeats. For trigonometric functions, this is calculated based on the function's formula and coefficients.
For a function of the form \( y = a \sec(bx - c) \), the period is derived using the formula:

• Period = \( \frac{2\pi}{b} \)

This formula arises because trigonometric functions in their simplest forms repeat every \( 2\pi \) radians. The factor \( b \) compresses or stretches the graph horizontally by changing how quickly these cycles occur.
In the case of our function \( y = 2 \sec(2x - \pi) \), the coefficient \( b \) is 2, meaning the period is \( \pi \). This indicates that every \( \pi \) units along the x-axis, the pattern of the secant function repeats itself. Understanding the period helps in plotting the function accurately by knowing where to expect similar shapes and cycles on the graph.

Vertical Asymptotes

Vertical asymptotes are lines where a function approaches but never actually reaches; they represent values of x where a function is undefined. For the secant function, vertical asymptotes occur where the cosine component is zero because division by zero leads to undefined values.
To determine the vertical asymptotes in a function like \( y = 2 \sec(2x - \pi) \), set the cosine part equal to zero:

• \( \cos(2x - \pi) = 0 \)

Solving this equation finds the x-values where the secant function will have vertical asymptotes:

• \( 2x - \pi = \frac{\pi}{2} + k\pi \), where \( k \in \mathbb{Z} \)
• \( x = \frac{3\pi}{4} + k\frac{\pi}{2} \)

For \( k = -1, 0, 1, 2, 3 \), this produces specific x-values such as \( -\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \), and more. Vertical asymptotes help define the graph's structure, providing clear points where it will approach infinity upwards or downwards, creating the characteristic spikes of the secant function.