Problem 38 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=-3 \sec \left(\frac{1}{3} x+\frac{\pi}{3}\right) $$

Step-by-Step Solution

Answer

Period is \(6\pi\); asymptotes at \(x = \pi + 3n\pi\).

Step 1: Identify the General Form

The general form of a secant function is given by \( y = a \sec(bx + c) + d \). In our equation, \( a = -3 \), \( b = \frac{1}{3} \), \( c = \frac{\pi}{3} \), and \( d = 0 \). This means the graph is vertically stretched by a factor of 3, flipped over the x-axis, and shifted horizontally.

Step 2: Calculate the Period

The period of a secant function is determined by the coefficient \( b \) in front of \( x \). The period \( T \) is calculated as \( T = \frac{2\pi}{|b|} \). For this equation, \( b = \frac{1}{3} \). Thus, the period \( T \) is \( \frac{2\pi}{\frac{1}{3}} = 6\pi \).

Step 3: Determine Asymptote Locations

Secant functions have vertical asymptotes where the cosine function is zero. The cosine function \( \cos\left(\frac{1}{3}x + \frac{\pi}{3}\right) = 0 \) determines the locations of the asymptotes. This occurs when \( \frac{1}{3}x + \frac{\pi}{3} = \frac{\pi}{2} + n\pi \) for any integer \( n \). Solving for \( x \), \( \frac{1}{3}x = \frac{\pi}{2} + n\pi - \frac{\pi}{3} \). Multiply through by 3 to isolate \( x \): \( x = 3\left(\frac{\pi}{2} + n\pi - \frac{\pi}{3}\right) \). Simplify to find \( x = \pi + 3n\pi \). Thus, asymptotes occur at \( x = \pi + 3n\pi \).

Step 4: Sketch the Graph

Begin by plotting the asymptotes at \( x = \pi + 3n\pi \) for a few values of \( n \). For the secant function, plot \( y = -3 \sec\left(\frac{1}{3}x + \frac{\pi}{3}\right) \), which consists of downward-facing arcs between each pair of asymptotes due to the negative coefficient. The function will repeat every \( 6\pi \). The function will reach a minimum value of \(-3\) at the midpoints between asymptotes.

Key Concepts

Graphing Trigonometric Functions

Graphing trigonometric functions like the secant function requires understanding its general behavior and characteristics. The secant function is closely related to the cosine function. For each section between the vertical asymptotes, the graph of the secant function resembles a series of arcs. These upward or downward-facing arcs depend on the coefficient in front of the secant function.

• The secant function, denoted as \( \sec(x) \), is the reciprocal of the cosine function.
• The graph of \( y = a \sec(bx + c) + d \) typically appears as a sequence of repeating arcs.
• Each arc is located between the vertical asymptotes.
• For the given function \( y = -3 \sec\left(\frac{1}{3}x + \frac{\pi}{3}\right) \), the arcs will face downward.

Plotting secant functions on a graph requires identifying key points, especially the asymptotes and the midpoints of each arc. Understanding these key locations helps in constructing an accurate graph.

Period of a Trigonometric Function

The period of a trigonometric function is the length of one complete cycle of its graph. For the secant function, this is determined by the coefficient \( b \) in front of the \( x \) term. Specifically, the formula for the period \( T \) is \( T = \frac{2\pi}{|b|} \).

• The period tells us how frequently the pattern of the graph repeats itself along the x-axis.
• For the secant function \( y = -3 \sec\left(\frac{1}{3}x + \frac{\pi}{3}\right) \), \( b = \frac{1}{3} \), leading to a period of \( 6\pi \).
• This period means that every \( 6\pi \) units along the x-axis, the secant function's arcs start repeating.

Understanding the period is crucial because it gives insight into the frequency of the arcs along the graph. Larger periods mean more spaced out arcs, while smaller periods result in more frequent arcs.

Asymptotes in Trigonometric Functions

Asymptotes are lines that the graph of a function approaches but never actually touches. In trigonometric functions like secant, vertical asymptotes occur wherever the corresponding cosine function equals zero.

• For \( y = \sec(x) \), vertical asymptotes occur at points where \( \cos(x) = 0 \), because division by zero is undefined.
• In the transformed secant function \( y = -3 \sec\left(\frac{1}{3}x + \frac{\pi}{3}\right) \), these occur where \( \cos\left(\frac{1}{3}x + \frac{\pi}{3}\right) = 0 \).
• This results in vertical asymptotes at \( x = \pi + 3n\pi \) for any integer \( n \).

Vertical asymptotes help define the boundaries of the arcs in a secant function graph. They are crucial for accurately sketching the function, as they indicate where the function rises infinitely and where it "turns around" to form another arc.