Problem 43 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The period is \( \pi \) and the graph is shifted \( \frac{\pi}{2} \) units to the right.

Step 1: Understanding the Function

The given function is \( y = \sec{2\left( x - \frac{\pi}{2} \right)} \). It's a secant function with a horizontal transformation. The term \( x - \frac{\pi}{2} \) represents a phase shift and the coefficient 2 causes a change in the period.

Step 2: Identifying the Coefficient Impact on Period

The general form of a secant function is \( y = \sec(bx - c) \) where the period is \( \frac{2\pi}{|b|} \). In this function, \( b = 2 \), so the period becomes \( \frac{2\pi}{2} = \pi \).

Step 3: Determining the Phase Shift

The phase shift is given by \( \frac{c}{b} \), where \( c = - \pi \) and \( b = 2 \). So the phase shift is \( \frac{-\pi}{2} \). This means the graph is shifted \( \frac{\pi}{2} \) units to the right.

Step 4: Plotting the Function

We now know the period is \( \pi \) and there's a phase shift of \( \frac{\pi}{2} \) units to the right. Start by plotting the basic secant function, \( y = \sec{bx} \), then apply the horizontal shift to each key point. The asymptotes of the graph repeat every period \( \pi \) and are shifted accordingly.

Step 5: Sketching the Graph

The secant function has vertical asymptotes where the corresponding cosine function has zeros. Since \( \sec{x} = \frac{1}{\cos{x}} \), the periodic points will align vertically with where cosine is zero. Mark these lines every \( \pi \) starting from the phase shift \( \frac{\pi}{2} \) and sketch the curves between them as they stretch infinitely along the vertical purpose of secant's nature.

Key Concepts

Secant Function

The secant function, denoted as \( y = \sec{x} \), is a trigonometric function. It's defined as the reciprocal of the cosine function: \( \sec{x} = \frac{1}{\cos{x}} \). This relationship means the secant function is undefined at the points where the cosine function equals zero. These undefined points lead to vertical asymptotes in the graph of the secant function.

The graph of \( y = \sec{x} \) has a characteristic repeating pattern, with its curves located above and below the x-axis. Each "arc" opens upward or downward between the asymptotes.

Key features of the secant function include:

• Vertical asymptotes coinciding with the cosine function's zeros.
• Intervals where the function takes positive values (where \( \cos{x} > 0 \)) and negative values (where \( \cos{x} < 0 \)).
• Repeated every full cycle of the cosine function's period.

Understanding this behavior is crucial when graphing or transforming secant functions.

Period of Trigonometric Functions

The period of a trigonometric function is the length of one complete cycle of the function. For the standard secant function \( y = \sec{x} \), the period is \( 2\pi \). This period reflects the distance over which the function repeats its shape.

When dealing with transformed secant functions like \( y=\sec(bx-c) \), the period is altered by the coefficient \( b \). To find the new period, use the formula \( \frac{2\pi}{|b|} \).

In the given function \( y = \sec{2\left(x-\frac{\pi}{2} \right)} \), the coefficient \( b \) is 2. Thus, we calculate the period as \( \frac{2\pi}{2} = \pi \), meaning the function repeats every \( \pi \) units. This adjusted period allows you to correctly position the graph on a coordinate plane and account for the cyclical nature of trigonometric functions.

Phase Shift

A phase shift refers to the horizontal displacement of a trigonometric function on its graph. It indicates how far the function is shifted left or right from its usual position.

The general form \( y = \sec(bx-c) \) involves the term \( c \), where the phase shift is calculated as \( \frac{c}{b} \). If \( c > 0 \), the shift is to the right; if \( c < 0 \), it's to the left.

In the exercise at hand, the function is \( y = \sec{2\left(x-\frac{\pi}{2}\right)} \). Here, \( b = 2 \) and \( c = -\pi \). Applying the formula gives us a phase shift of \( \frac{-\pi}{2} = -\frac{\pi}{2} \), which, being negative, means the function moves \( \frac{\pi}{2} \) units to the right.

Understanding the phase shift is pivotal in accurately plotting trigonometric graphs and comprehending how various transformations affect the original function's shape and position on the graph.