Problem 37 - Precalculus: Functions and Graphs

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$y=-\frac{1}{3} \sec \left(\frac{1}{2} x+\frac{\pi}{4}\right)$$

Step-by-Step Solution

Answer

The period of the function is \(4\pi\), with vertical asymptotes at \(x=\pi+2k\pi\).

Step 1: Understand the components of the secant function

The function given is of the form \( y = a \cdot \sec(bx + c) \). In this exercise, \( a = -\frac{1}{3} \), \( b = \frac{1}{2} \), and \( c = \frac{\pi}{4} \). The presence of the negative \( a \) value implies a reflection across the x-axis.

Step 2: Determine the period of the secant function

The standard period of the secant function \( \sec(x) \) is \( 2\pi \). For a transformed function \( \sec(bx) \), the period becomes \( \frac{2\pi}{|b|} \). For this function, \( b = \frac{1}{2} \), so the period is \( \frac{2\pi}{\frac{1}{2}} = 4\pi \).

Step 3: Find the phase shift

The expression \( \frac{1}{2}x + \frac{\pi}{4} = 0 \) gives the starting point of the first cycle. Solving for \( x \), we have \( \frac{1}{2}x = -\frac{\pi}{4} \), thus \( x = -\frac{\pi}{2} \). This is the phase shift to the left by \( \frac{\pi}{2} \).

Step 4: Determine the vertical asymptotes

Secant functions have vertical asymptotes where the corresponding cosine function is zero. Solve \( \cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) = 0 \). This occurs at \( \frac{1}{2}x + \frac{\pi}{4} = \frac{\pi}{2} + k\pi \) where \( k \) is an integer. Solving for \( x \), \( x = 2\left(\frac{\pi}{4}\right) + 2k\pi = \pi + 2k\pi \). So, the asymptotes are at \( x = \pi + 2k\pi \).

Step 5: Sketch the graph

Start by drawing the vertical asymptotes on the graph at \( x = \pi + 2k\pi \). Then plot the secant function, which oscillates between these lines, considering the phase shift and the reflection across the x-axis due to the negative amplitude. The graph's top and bottom curves will be at \( y = -\frac{1}{3} \) and \( y = \frac{1}{3} \) respectively between each pair of asymptotes.

Key Concepts

Secant Function

The secant function is an essential trigonometric function denoted as \( \sec(x) \), which is the reciprocal of the cosine function. It is expressed as \( \sec(x) = \frac{1}{\cos(x)} \). The secant function is undefined whenever the cosine function is zero, leading to its distinctive feature of having vertical asymptotes. Because of its nature, the secant graph consists of repeating U-shaped and upside-down U-shaped curves.

• The graph is interrupted by vertical asymptotes, which correspond to the points where the cosine graph crosses its x-axis.
• The secant function's graph overlaps with the cosines wherever the cosine graph reaches its peak or trough.

Particularly, in the graph of the secant function provided in the exercise, the negative value of \( a \) indicates a reflection across the x-axis. This is evident in the function \( y = -\frac{1}{3} \sec\left(\frac{1}{2} x + \frac{\pi}{4}\right) \), which flips the secant's typical curve direction.

Period of Trigonometric Functions

The period of a trigonometric function is the interval over which the function's graph repeats itself. For the basic secant function \( \sec(x) \), this is \( 2\pi \) because it completes a full cycle as the cosine function—which it is based on—completes its cycle. In functions such as \( y = a \sec(bx + c) \), the period is adjusted by the factor \( b \). The formula to calculate the new period is \( \frac{2\pi}{|b|} \). This is key when graphing transformations of the secant function, as it helps in determining how often the full secant curve will repeat along the x-axis.

• In our example \( y = -\frac{1}{3} \sec\left(\frac{1}{2} x + \frac{\pi}{4}\right) \), \( b = \frac{1}{2} \), leading to a period of \( 4\pi \).
• This increase in period creates a wider span between each repeating section of the graph.

Phase Shift

Phase shift refers to the horizontal movement of a trigonometric graph along the x-axis. It is determined via the expression inside the trigonometric function. If the function is \( y = a \sec(bx + c) \), the phase shift is calculated by solving the equation \( bx + c = 0 \) for \( x \). This gives the starting point of the function's cycle.

• In the exercise, \( bx + c \) is \( \frac{1}{2}x + \frac{\pi}{4} \), leading to a phase shift of \( \frac{-\pi}{2} \).
• This shift moves the entire graph to the left by \( \frac{\pi}{2} \).

Understanding the phase shift helps in accurately placing the graph on a coordinate plane, ensuring the function's behavior is correctly represented from its adjusted start point.

Vertical Asymptotes

Vertical asymptotes are key features of the secant graph. They occur where the cosine function, which the secant is a reciprocal of, equals zero. The secant graph will approach infinity in the positive or negative direction at these points, causing undefined values along these vertical lines.

• To find vertical asymptotes, solve: \( \cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) = 0 \). In our example graph, this occurs at \( x = \pi + 2k\pi \) for integer values of \( k \).
• The graph is drawn around these asymptotes, never touching them, but illustrating the values approaching infinity.

Drawing these asymptotes correctly on the graph is crucial for depicting the intervals over which the secant function alternates between positive and negative infinite values.