Problem 22 - Precalculus: Functions and Graphs

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$y=\cot \frac{1}{2} x$$

Step-by-Step Solution

Answer

The period of the function is \( 2\pi \). Asymptotes occur at \( x = 0, \pm 2\pi, \pm 4\pi, \ldots \).

Step 1: Identify the Base Function

The given function is \( y = \cot\left(\frac{1}{2} x\right) \). The base function is \( \cot(x) \), whose period is \( \pi \). To find the period of \( y = \cot\left(\frac{1}{2} x\right) \), we need to modify this base period by the function's coefficient.

Step 2: Calculate the New Period

To find the period of \( y = \cot\left(\frac{1}{2} x\right) \), divide the base period \( \pi \) by the coefficient of \( x \) inside the cotangent function, which is \( \frac{1}{2} \). The period is calculated as follows: \[ \text{New Period} = \frac{\pi}{\frac{1}{2}} = 2\pi \] This means the function repeats every \( 2\pi \) units.

Step 3: Determine the Asymptotes

For the cotangent function, vertical asymptotes occur where the function is undefined. The asymptotes for \( \cot(x) \) are located at \( x = k\pi \), where \( k \) is an integer. For \( y = \cot\left(\frac{1}{2} x\right) \), these asymptotes occur where \[ \frac{1}{2} x = k\pi \rightarrow x = 2k\pi \] So, the asymptotes will be at \( x = 0, \pm 2\pi, \pm 4\pi, \ldots \).

Step 4: Sketch the Graph

To sketch \( y = \cot\left(\frac{1}{2} x\right) \), begin by plotting the vertical asymptotes at \( x = 0, \pm 2\pi, \pm 4\pi \). Between each pair of consecutive asymptotes (for instance, between \( 0 \) and \( 2\pi \)), the graph of the cotangent function will descend from positive infinity to negative infinity, crossing the x-axis at \( x = \pi \), due to the nature of the cotangent curve.

Key Concepts

Period of Trigonometric Functions

Understanding the period of trigonometric functions is crucial for graphing them correctly. The period is the horizontal length over which the function completes one full cycle before repeating. For a standard cotangent function, \( \cot(x) \), the period is \( \pi \). In the modified cotangent function \( y = \cot\left(\frac{1}{2} x\right) \), we consider the coefficient of \( x \) to define the new period.
To find this, take the period of the base cotangent function (\( \pi \)) and divide it by the coefficient \( \frac{1}{2} \). The calculation looks like this:

• Base period of \( \cot(x) \): \( \pi \)
• Adjustment by the coefficient: \( \text{New Period} = \frac{\pi}{\frac{1}{2}} = 2\pi \)

As a result, our function \( y = \cot\left(\frac{1}{2} x\right) \) repeats every \( 2\pi \) units. This means it only starts to repeat its pattern after covering a longer span on the x-axis compared to \( \cot(x) \).

Trigonometric Asymptotes

Where the function is undefined, vertical asymptotes are found. For the cotangent function \( \cot(x) \), these occur at \( x = k\pi \) (where \( k \) is any integer), creating breaks in the graph. Asymptotes help determine the start and end points of a repeating section and show where the graph will shoot off to infinity.
For the function \( y = \cot\left(\frac{1}{2} x\right) \), the process involves finding where the inner function equals \( k\pi \). Solve \[ \frac{1}{2}x = k\pi \] This leads to:

• \( x = 2k\pi \)

The vertical asymptotes for \( y = \cot\left(\frac{1}{2} x\right) \) are therefore at \( x = 0, \pm 2\pi, \pm 4\pi, \ldots \). Recognizing these points is vital for correctly sketching the graph because they represent the places where the graph is not connected.

Graphing Trigonometric Functions

Graphing trigonometric functions involves accurately mapping out their peaks, troughs, and repeating cycles on a coordinate plane. For functions like the cotangent, where asymptotes play an important role, aligning your graph with these lines is essential. Start by plotting the asymptotes, which act as boundaries.
Next, determine key points between asymptotes:

• Identify where the function crosses the x-axis
• For \( \cot\left(\frac{1}{2} x\right) \), this happens halfway between asymptotes, i.e., at \( x = \pi, \pm 3\pi, \pm 5\pi, \ldots \)
• The graph descends from positive to negative infinity between each pair of asymptotes

Each cycle of the graph will look the same, slowly transitioning from top to bottom, defining the characteristic shape of the cotangent curve. Consistency in these repeating cycles is a hallmark of trigonometric graphs, allowing prediction and interpretation of their behavior over each period.

Cotangent Graph Sketching

Cotangent graphs, such as \( y = \cot\left(\frac{1}{2} x\right) \), have a visually distinctive descending shape commonly beginning near positive infinity and ending near negative infinity. Their smooth, unbroken descent spans the space between asymptotes.To start sketching:

• Mark the vertical asymptotes: At \( x = 0, \pm 2\pi, \pm 4\pi, \ldots \)
• Identify x-intercepts: Midway between asymptotes—e.g., \( x = \pi \)
• Fill in the descending curve using these guide points, ensuring each part of the graph smoothly transitions

Keep in mind that the spacing of these features is determined by the calculated period \( 2\pi \). This extension compared to the standard period \( \pi \) means making sure the completed graph covers more of the horizontal axis. Practice and familiarity with these elements make mastering cotangent sketching much simpler.