Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\sec 2 x $$

Step-by-Step Solution

Answer

The period of \( y = \sec(2x) \) is \( \pi \), with vertical asymptotes at \( x = \frac{\pi}{4} + \frac{k\pi}{2} \).

Step 1: Understand the function

The function given is \( y = \sec(2x) \). The secant function, \( y = \sec(x) \), is the reciprocal of cosine, meaning \( \sec(x) = \frac{1}{\cos(x)} \). This implies that \( \sec(x) \) is undefined whenever \( \cos(x) = 0 \).

Step 2: Determine where cosine is zero

The cosine function, \( \cos(x) \), is zero at odd multiples of \( \frac{\pi}{2} \), i.e., \( x = \frac{\pi}{2} + k\pi \) where \( k \) is an integer. For \( \sec(2x) \), this condition becomes \( \cos(2x) = 0 \). Therefore, \( 2x = \frac{\pi}{2} + k\pi \) which implies \( x = \frac{\pi}{4} + \frac{k\pi}{2} \). This results in vertical asymptotes at these points.

Step 3: Calculate the period

The period of the secant function \( \sec(x) \) is \( 2\pi \) because the cosine function has a period of \( 2\pi \). For \( \sec(2x) \), the function is horizontally compressed by a factor of 2, so the period becomes \( \frac{2\pi}{2} = \pi \). Thus, the period of \( \sec(2x) \) is \( \pi \).

Step 4: Identify key points

Within one period \([0, \pi]\), \( \sec(2x) \) is defined except at the points where there are asymptotes. Thus, choose key points in one period to plot: \( x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} \). Evaluate \( \sec(2x) \) at these points. \( \sec(0) = 1, \sec(\frac{\pi}{2}) = -1, \) whereas at \( x = \frac{\pi}{4}, \frac{3\pi}{4} \) the function is undefined because these are vertical asymptotes.

Step 5: Sketch the graph

On a graph, plot one period from \( 0 \) to \( \pi \). Mark vertical asymptotes at \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \). For \( x = 0 \), plot the point \((0, 1)\) and for \( x = \frac{\pi}{2} \), plot the point \( (\frac{\pi}{2}, -1) \). The secant graph repeats this pattern every period \( \pi \).

Key Concepts

Period of a Function

The period of a function is a measure of how often the function repeats its values over a specific interval. For trigonometric functions, this is especially important as they naturally oscillate.

• The original secant function, \( \sec(x) \), has a period of \( 2\pi \), reflecting the cosine function from which it is derived.
• When a function is altered such as \( y = \sec(2x) \), its period changes. In this case, the coefficient \( 2 \) compresses the function horizontally, affecting how quickly it repeats.
• To find the new period, you can use the formula: New Period = \( \frac{2\pi}{B} \), where \( B \) is the coefficient of \( x \). For \( \sec(2x) \), the period is \( \frac{2\pi}{2} = \pi \).

This means that for \( \sec(2x) \), the function will complete one full cycle from \( 0 \) to \( \pi \) before repeating itself. Understanding how to calculate the period helps in sketching the graph accurately.

Secant Function

The secant function, represented as \( \sec(x) \), is the reciprocal of the cosine function. It is important to understand some key properties of this function.

• Since \( \sec(x) = \frac{1}{\cos(x)} \), it is undefined anywhere the cosine function itself is zero.
• This results in the function displaying vertical asymptotes at these undefined points.
• The graph of the secant function consists of repeating upward and downward curves that approach vertical asymptotes.

Visualizing the secant function and its behavior enhances comprehension especially in relation to graph transformations, facilitating easier graphing of functions like \( \sec(2x) \).

Vertical Asymptotes

Vertical asymptotes are lines that a graph of a function approaches but never actually touches. They are crucial in the graph of the secant function.

• For \( \sec(x) \), the location of vertical asymptotes is determined by where \( \cos(x) = 0 \). These occur at odd multiples of \( \frac{\pi}{2} \), such as \( x = \frac{\pi}{2}, \frac{3\pi}{2} \), and so on.
• For \( \sec(2x) \), the locations of these lines change due to the function's transformation. Solving \( \cos(2x) = 0 \) gives \( x = \frac{\pi}{4} + \frac{k\pi}{2} \).
• These vertical asymptotes appear as gaps in the graph where the function shoots to infinity.

Understanding where these asymptotes occur is crucial for accurately sketching the graph of trigonometric functions.

Graphing Trigonometric Functions

Graphing trigonometric functions involves understanding their shape, period, and asymptotes. Here are some key points to consider when graphing functions like \( \sec(2x) \):

• Identify the period: For \( \sec(2x) \), the period is \( \pi \), meaning the pattern repeats every \( \pi \) units.
• Locate the asymptotes by determining where the cosine function, part of \( \sec(2x) \), becomes zero.
• Mark key points within one period that give defined values, such as where \( \sec(2x) \) is explicitly known, like \( x=0 \) produces \( \sec(0) = 1 \).
• Sketch the graph around these known points, depicting the curves and acknowledging the asymptotic behavior.

Graphing effectively requires practice, but by focusing on these main elements, you can accurately sketch even complex transformations of trigonometric functions.