Problem 48 - Precalculus: A Concise Course

Question

Use a graphing utility to graph the function. Include two full periods. $$ y=\frac{1}{3} \sec \left(\frac{\pi x}{2}+\frac{\pi}{2}\right) $$

Step-by-Step Solution

Answer

The graph of the function \(y=\frac{1}{3} \sec \left(\frac{\pi x}{2}+\frac{\pi}{2}\right)\) has a lengthwise compression by a factor of 1/3, a period of 4, and it is shifted 1 unit to the left from the typical sec function graph.

Step 1: Identify key transformations

The given function is \(y=\frac{1}{3} \sec \left(\frac{\pi x}{2}+\frac{\pi}{2}\right)\). There are three identifiable transformations from the basic secant function \(y = sec(x)\) to the one given. They are: expansion by a factor of 1/3 along the y-axis, a period change noted by the coefficient of x in the argument of the secant function, and a phase shift to the left by 1 unit since \(\frac{\pi}{2}\) is added in the argument.

Step 2: Calculate the period

The period of the function is given by the formula \(T = \frac{2\pi}{|B|}\), where B is the coefficient of x in the argument. In this case, B equals \(\frac{\pi}{2}\), hence the period T = \(4\).

Step 3: Graph the function

After identifying the key transformations and the period, graph the function using a graphing utility. The y-axis should be stretched by a factor of 1/3. The standard secant graph should be stretched to reflect the period of 4 instead of the standard 2 π, and shifted to the left by 1 unit. Lastly, it's important to remember that secant function has vertical asymptotes where the cosine function equals zero, and it's undefined at these points.

Step 4: Include two full periods

Show at least two complete cycles of the function on your graph to confirm your calculation of the period. In this case, since we discovered our period to be 4, we should ensure our graph displays the function from x = 0 to x = 8.

Key Concepts

Secant Function

The secant function, denoted as \(y = \sec(x)\), is a trigonometric function that is the reciprocal of the cosine function. In simpler terms, \( \sec(x) = \frac{1}{\cos(x)} \). This function is undefined where the cosine of \(x\) is equal to zero because division by zero is impossible.
This leads to vertical asymptotes at points such as \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \text{etc.} \) where the cosine value reaches zero. Graphically, the secant function displays periodic vertical asymptotes and looks like repeating U-shaped or upside-down U-shaped curves between these asymptotes.

Transformations in Trigonometry

Transformation of trigonometric functions involves adjusting the basic equations to reflect changes in amplitude, period, phase shift, and vertical shift. In our equation, \( y=\frac{1}{3} \sec \left(\frac{\pi x}{2}+\frac{\pi}{2}\right) \), we observe several transformations:

• **Amplitude Change:** The factor \( \frac{1}{3} \) compresses the graph vertically, meaning all values of the secant function are reduced to one-third of their original size.
• **Phase Shift:** The addition of \( \frac{\pi}{2} \) within the secant argument results in a phase shift to the left. Here, every point on the graph of the secant function moves leftward by one unit.

Bringing these transformations together gives us the unique form and position seen in the posed exercise.

Period of Trigonometric Functions

When exploring the period of trigonometric functions, we focus on how much an interval needs to be completed before the function starts repeating its pattern. The period of the basic secant function \( y = \sec(x) \) is \(2\pi\). However, when modifications occur, as seen in \( B \) of \( \left(\frac{\pi x}{2}+\frac{\pi}{2}\right) \), the period changes.
To compute the new period, you use the formula \( T = \frac{2\pi}{|B|} \). With \( B \) located in our problem as \( \frac{\pi}{2} \), the resulting period for this function becomes \(4\). This alteration means that the secant function, now transformed, will complete a full cycle every 4 units on the \(x\)-axis.

Vertical Asymptotes

Vertical asymptotes are lines where the value of the function heads towards infinity. They occur in situations where the denominator of the function equals zero. In this case, for the secant function, they will appear wherever the cosine portion of the function is zero.
Reviewing our transformed function, \( y=\frac{1}{3} \sec \left(\frac{\pi x}{2}+\frac{\pi}{2}\right) \), we find that vertical asymptotes appear at values of \(x\) where \( \sin \left(\frac{\pi x}{2}+\frac{\pi}{2}\right) \) equals zero. These will create the boundaries between the repeating U-shaped arcs seen in the graph of the secant function. This makes them an essential part of understanding how and where the secant function behaves abruptly on the plane.