Problem 51 - Algebra and Trigonometry

Question

Find the period and graph the function. $$ y=3 \sec \pi\left(x+\frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The period of the function is 2; it has vertical asymptotes at integer values of \( x \).

Step 1: Identify the Standard Form of the Secant Function

The standard form of the secant function is given by \( y = a \sec(bx + c) + d \). Our function is \( y = 3 \sec\left(\pi \left(x + \frac{1}{2}\right)\right) \). Here, \( a = 3 \), \( b = \pi \), \( c = \pi \times \frac{1}{2} \), and \( d = 0 \).

Step 2: Determine the Period of the Secant Function

The period of the secant function \( y = \sec(bx + c) \) is \( \frac{2\pi}{b} \). In this case, \( b = \pi \), so the period is \( \frac{2\pi}{\pi} = 2 \). Thus, the period of the function is 2.

Step 3: Identify Key Points for Graphing

The secant function is undefined when its corresponding cosine function has zeros (since \( \sec(x) = \frac{1}{\cos(x)} \)). The key points occur at the centers of vertical asymptotes and maximum and minimum values. For the function \( y = 3 \sec\left(\pi \left(x + \frac{1}{2}\right)\right) \), the vertical asymptotes occur when the angle \( \pi(x + \frac{1}{2}) = (2n + 1) \frac{\pi}{2} \), where \( n \) is an integer.

Step 4: Calculate Locations of Vertical Asymptotes

Setting \( \pi(x + \frac{1}{2}) = (2n + 1)\frac{\pi}{2} \), we solve for \( x \): \( \pi x + \frac{\pi}{2} = (2n + 1)\frac{\pi}{2} \) \( \pi x = (2n + 1)\frac{\pi}{2} - \frac{\pi}{2} \) \( \pi x = n\pi \) \( x = n \). So, vertical asymptotes occur at \( x = n \), where \( n \) is an integer.

Step 5: Plotting the Secant Function

Now that we know the period is 2 and vertical asymptotes occur at integer values of \( x \), we can graph the function. Between each consecutive pair of vertical asymptotes, sketch the secant curve which opens upwards between the asymptotes, since the amplitude is positive. Repeat this behavior in each period of the function.

Key Concepts

Secant Function

The secant function, denoted as \( \sec(x) \), is one of the six fundamental trigonometric functions. It is defined as the reciprocal of the cosine function, which means \( \sec(x) = \frac{1}{\cos(x)} \). This implies that the secant function is undefined wherever the cosine function equals zero (since we can't divide by zero).
The secant function is periodic, like all trigonometric functions. This means it repeats its pattern at regular intervals, called periods. In its standard form, the secant function is indicated as \( y = a \sec(bx + c) + d \), where:

• \( a \) is the amplitude, determining how much the function stretches vertically.
• \( b \) modifies the period of the function.
• \( c \) causes a horizontal shift.
• \( d \) moves the function up or down.

Understanding these parameters helps in graphing secant functions effectively.

Graphing Functions

Graphing trigonometric functions like the secant function involves understanding its behavior and recurring components. Key elements to consider when graphing are the vertical asymptotes, turning points (maximums and minimums), and the periodic pattern. Let’s break this down:- **Vertical Asymptotes**: These occur at the values of \( x \) where the cosine function is zero, causing \( \sec(x) \) to be undefined. To find them, solve for when the angle yields a zero cosine result, often described as \( (2n + 1) \frac{\pi}{2} \).- **Maximum and Minimum Values**: Between these asymptotes, the secant function curves will reach their maximum or minimum values. These values occur midpoint between successive asymptotes.The specific function \( y = 3 \sec(\pi(x + \frac{1}{2})) \) alters the standard parameters. Here:

• Amplitude \( a = 3 \): This multiplier scales the graph vertically.
• Horizontal shift brought by \( +\frac{1}{2} \): Shift left by \( \frac{1}{2} \) unit.

When graphing, these adjustments impact where peaks and troughs of the secant function appear.

Period of Functions

The period of a trigonometric function defines the length over which it completes one full cycle. For the standard secant function \( y = \sec(bx + c) \), the period is calculated from the parameter \( b \), using the formula:\[ \text{Period} = \frac{2\pi}{b} \]In our example function \( y = 3 \sec(\pi(x + \frac{1}{2})) \), \( b = \pi \), so:\[ \text{Period} = \frac{2\pi}{\pi} = 2 \]This tells us that the secant pattern repeats every 2 units along the x-axis. Understanding the period is crucial for accurately sketching graphs and predicting function behavior. The repetitive nature of trigonometric functions allows us to anticipate how the function behaves after each full cycle, making calculations and graphing more manageable. Adjusting \( b \) changes how "stretched" or "compressed" the function graph appears, a vital part of trigonometric graph manipulation.