Problem 48 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

Period is 1. Graph has vertical asymptotes at \( x = \frac{3}{4} + \frac{n}{2} \) and oscillates between \( \frac{1}{2} \) and \( -\frac{1}{2} \).

Step 1: Identify the Period of the Base Function

The base function is \( \sec(x) \), which has the same period as \( \cos(x) \), specifically \( 2\pi \). For \( \sec(kx) \), the period is adjusted and becomes \( \frac{2\pi}{|k|} \). Here, the coefficient of \( x \) inside the secant function is \( 2\pi \), so the period will be \( \frac{2\pi}{2\pi} = 1 \).

Step 2: Understand the Horizontal Shift

Rewrite the expression inside the secant function: \( 2\pi x - \pi \) can be factored to \( 2\pi(x - \frac{1}{2}) \). This indicates a horizontal shift to the right by \( \frac{1}{2} \).

Step 3: Calculate Key Points for the Graph

Determine the key points for one period of \( \sec(x) \), which involves identifying the intervals where \( \sec(x) \) is undefined (where \( \cos(x) = 0 \)). For \( \sec(2\pi x - \pi) \), set \( 2\pi x - \pi = \frac{\pi}{2} + n\pi \) for integers \( n \). Solving gives \( x = \frac{3}{4} + \frac{n}{2} \), representing vertical asymptotes at these values.

Step 4: Graph the Function

Plot the vertical asymptotes at \( x = \frac{3}{4} + \frac{n}{2} \). Between each pair of asymptotes, the function \( y = \frac{1}{2}\sec(2\pi x - \pi) \) will oscillate, with peaks at \( \pm\frac{1}{2} \). Mark the key point at \( x = \frac{1}{2} \) where the function has a minimum (due to the horizontal shift) at \( y = \frac{1}{2} \) and at \( x = 1 \), where the function also has a maximum at \( y = -\frac{1}{2} \).

Step 5: Interpret the Full Graph

With the graph plotted between \( x = \frac{1}{4} \) and \( x = \frac{3}{4} \), etc., it repeats with a period of \( 1 \). Ensure to express the vertical stretch/compression factor of \( \frac{1}{2} \), as the typical \( 1 \) and \( -1 \) peak values of \( \sec(x) \) become \( \frac{1}{2} \) and \( -\frac{1}{2} \).

Key Concepts

Graphing Trigonometric Functions

Graphing trigonometric functions involves plotting these functions on a coordinate plane to visualize their behavior. The function y = \( \frac{1}{2} \sec(2\pi x-\pi) \) is based on the secant function, which is derived from the cosine function. To graph any trigonometric function, consider the following steps:

• **Determine the period**: The period determines how often the function repeats its pattern.
• **Identify key points:** These include the maximum and minimum points, as well as points where the function is undefined.
• **Mark asymptotes:** For secant functions, vertical asymptotes occur where the cosine value is zero.

By having these elements laid out, you can successfully plot the graph of a secant function like the one in the exercise. Always remember that for \( y = a \sec(bx - c) \), the amplitude, marked as \( |a| \), affects the height of peaks in the graph.
Given that the function here changes the normal secant pattern with a coefficient of \( \frac{1}{2} \), the graph of \( y = \pm \frac{1}{2} \) stretches or compresses the secant peaks and valleys accordingly.

Trigonometric Function Period

The period of a trigonometric function is the length of one complete cycle before it repeats. For the function \( y = \frac{1}{2} \sec(2\pi x-\pi) \), it is essential to determine the function's period to graph it accurately. The base period of the \( \sec(x) \) function, like \( \cos(x) \), is \( 2\pi \). However, transformations require adjusting this period.

• **Coefficients Impact:** When the argument of the secant is altered, like \( \sec(kx) \), the period adjusts to \( \frac{2\pi}{|k|} \).
• **Example from the function:** Here, \( k \) is equal to \( 2\pi \), resulting in a new period of \( \frac{2\pi}{2\pi} = 1 \).

This means that instead of repeating every \( 2\pi \) units, the function completes a cycle every 1 unit on the x-axis. Knowing this helps in marking accurately spaced intervals when graphed. Just as with any other trigonometric function, the period plays a critical role when dealing with horizontal stretches and compressions.

Horizontal Shift in Trigonometry

In trigonometry, horizontal shifts, also known as phase shifts, move the graph of a function left or right along the x-axis. For a function like \( y = \frac{1}{2} \sec(2\pi x-\pi) \), recognizing and calculating the horizontal shift allows us to precisely position key points or features of the graph.
Factoring the expression \( 2\pi x-\pi \) produces \( 2\pi(x-\frac{1}{2}) \), indicating a shift to the right by \( \frac{1}{2} \). Understanding this step is crucial for graphing because:

• The secant function's behavior, in this case, will start at \( \frac{1}{2} \) instead of zero.
• This influences where vertical asymptotes and crossings appear.

By knowing each shift, you can adjust the positioning of the graph relative to the x-axis effectively. This helps make the graph aligned with its intended mathematical representation, ensuring no errors in plotting and interpretation.