Problem 47 - Precalculus

Question

In Exercises \(47-56,\) graph the functions over at least one period. $$y=3-2 \sec \left(x-\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

Graph is a vertically shifted and stretched secant, period \(2\pi\), shifted right by \(\pi/2\), with vertical asymptotes.

Step 1: Recognize the Function Type

The given function is of the form \( y = a + b \sec(\theta) \), where \( a = 3 \), \( b = -2 \), and \( \theta = x - \frac{\pi}{2} \). It is a transformation of the basic secant function.

Step 2: Identify Key Features

The standard secant function \( \sec(x) \) has a period of \(2\pi\). However, due to transformations, the function \( y = 3 - 2 \sec(x - \frac{\pi}{2}) \) will maintain the same period, which is \( 2\pi \), but it will be shifted horizontally, scaled, and translated.

Step 3: Determine the Horizontal Shift

Since the argument inside the secant is \( x - \frac{\pi}{2} \), this implies a phase shift to the right by \( \frac{\pi}{2} \). Therefore, the function starts its cycle at \( x = \frac{\pi}{2} \).

Step 4: Apply the Vertical Translation and Stretch

The value \( 3 \) causes a vertical shift upward by 3 units. The coefficient \( -2 \) reflects the graph across the horizontal axis and stretches it by a factor of 2. Thus, the basic secant graph opens downwards and is stretched.

Step 5: Draw the Vertical Asymptotes

Secant functions have vertical asymptotes where the cosine (denominator in its reciprocal form) is zero. For \( \cos(x-\frac{\pi}{2}) = 0 \), the asymptotes are at \( x = \frac{\pi}{2} + k\pi \) for integer \( k \).

Step 6: Graph One Period

Starting from \( x = \frac{\pi}{2} \), graph \( y = 3 - 2 \sec(x - \frac{\pi}{2}) \) over one period from \( \frac{\pi}{2} \) to \( \frac{5\pi}{2} \). Plot key points and the asymptotes to guide the shape of the graph.

Key Concepts

Secant Function

The secant function, denoted as \( \sec(x) \), is a trigonometric function that is defined as the reciprocal of the cosine function. In simpler terms, \( \sec(x) = \frac{1}{\cos(x)} \). Consequently, the secant is undefined wherever the cosine function equals zero, leading to characteristic vertical asymptotes in its graph.
The shape of a secant graph resembles a series of repeated 'U' and 'n' shapes, which result from the function's behavior near its asymptotes.

• Its usual period is \( 2\pi \), matching that of the cosine.
• It features repeating vertical asymptotes at the points where cosines would naturally be zero, such as at \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \) and so on.

For the transformed function \( y = 3 - 2 \sec(x - \frac{\pi}{2}) \), the secant function is adjusted in several ways:

• A vertical shift up by 3 units causes the whole graph to move upwards on the y-axis.
• A vertical stretch and reflection, due to the coefficient \(-2\), stretches the graph by a factor of 2 and reflects it across the horizontal axis.

These transformations alter the appearance of the graph while maintaining the core characteristics of the secant function, such as its period and the positions of its vertical asymptotes.

Period and Phase Shift

The period of a function describes how often it repeats its values. For the basic secant function, this period is \( 2\pi \). This means the graph repeats its shape every \( 2\pi \) units along the x-axis.
In the function \( y = 3 - 2 \sec(x - \frac{\pi}{2}) \), the period remains \( 2\pi \), as the transformations applied do not affect it.

• Period: Still \( 2\pi \).
• The repetition of the pattern starts again after every \( 2\pi \) interval.

The phase shift is the horizontal translation of the graph. It occurs when the input of the trigonometric function is adjusted, for instance from \( x \) to \( x - c \). In this scenario, there is a phase shift of \( \frac{\pi}{2} \) to the right.

• For \( \sec(x - \frac{\pi}{2}) \), each repeat of the function begins at \( x = \frac{\pi}{2} \) instead of \( x = 0 \).
• This shift effectively translates the graph so that all its features, including angles and asymptotes, move rightward.“

Understanding period and phase shift is critical when graphing transformations of trigonometric functions. They determine where the cycles start and how often the pattern repeats.

Vertical Asymptotes

Vertical asymptotes are critical when graphing the secant function. They are vertical lines in the graph where the function approaches infinity or negative infinity due to division by zero. In trigonometric terms, these occur wherever the cosine component is zero, as \( \sec(x) = \frac{1}{\cos(x)} \).

• These points for a standard secant function are at \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \).
• In our transformed function \( y = 3 - 2 \sec(x - \frac{\pi}{2}) \), the phase shift influences these locations.

The effect of the \( \frac{\pi}{2} \) shift can be seen in:

• The shifted asymptote locations are now \( x = \frac{\pi}{2} + k\pi \).
• This means a repeating pattern of asymptotes begins at \( x = \frac{\pi}{2} \), and subsequent ones appear at \( \frac{3\pi}{2}, \frac{5\pi}{2}, \) etc.

Visualizing these asymptotes is essential to draw the secant graph accurately. They inform where the graph cannot exist and guide the shape of the curves that are plotted between these vertical lines.