Problem 27 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

The period is \(2\pi\); asymptotes are at \(x = 2n\pi - \frac{\pi}{2}\).

Step 1: Identify the Transformations

The given function is \( y = -\frac{1}{2} \cot \left(\frac{1}{2} x + \frac{\pi}{4} \right) \). This represents a transformation of the parent function \( \cot(x) \). The negative sign indicates a vertical reflection, and the \( \frac{1}{2} \) coefficient inside the argument suggests a horizontal stretch. The horizontal phase shift is determined by the \( \frac{\pi}{4} \) inside the argument.

Step 2: Find the Period

The period of the parent \( \cot(x) \) function is \( \pi \). The function \( \cot(bx + c) \) has a period given by \( \frac{\pi}{|b|} \). For \( y = -\frac{1}{2} \cot \left(\frac{1}{2} x + \frac{\pi}{4} \right) \), we have \( b = \frac{1}{2} \), so the period is \( \frac{\pi}{\frac{1}{2}} = 2\pi \).

Step 3: Determine Phase Shift

To find the phase shift, solve for \( x \) in the equation \( \frac{1}{2}x + \frac{\pi}{4} = 0 \). Solving gives \( x = -\frac{\pi}{2} \). This means there is a phase shift of \(-\frac{\pi}{2}\) to the left.

Step 4: Locate the Asymptotes

For \( \cot(x) \), asymptotes occur at \( x = n\pi \) for any integer \( n \). Apply the transformations to find the location of asymptotes for \( y = -\frac{1}{2} \cot \left(\frac{1}{2}x + \frac{\pi}{4} \right) \). Set \( \frac{1}{2}x + \frac{\pi}{4} = n\pi \). Solving for \( x \) gives \( x = 2n\pi - \frac{\pi}{2} \), indicating the vertical asymptotes occur at these values.

Step 5: Sketch the Graph

Using the period \( 2\pi \), phase shift \(-\frac{\pi}{2}\), and the vertical asymptotes, plot one cycle of the \( \cot \) function. Reflect over the x-axis due to the \(-\frac{1}{2} \) factor. The graph decreases within each period from \(-\infty\) to \(\infty\), changing direction at the vertical asymptotes. Take note to mark asymptotes at every \( 2n\pi - \frac{\pi}{2} \) to guide the sketch.

Key Concepts

Cotangent Function

The cotangent function, denoted as \( \cot(x) \), is the reciprocal of the tangent function. In terms of sine and cosine, it can be defined as:

• \( \cot(x) = \frac{\cos(x)}{\sin(x)} \)

The parent graph of \( \cot(x) \) is known for its distinct characteristic of repeating every \( \pi \), which is its period. Unlike sine and cosine, the cotangent graph does not have maximum or minimum points within its period. Instead, its defining feature is the presence of vertical asymptotes, where the function value tends to positive or negative infinity. These asymptotes occur at points where \( \sin(x) = 0 \), specifically where \( x = n\pi \) for any integer \( n \).
The graph of \( \cot(x) \) decreases from \( +\infty \) to \( -\infty \), moving from left to right between consecutive asymptotes.

Graphing Transformations

Transformations alter the appearance of the parent trigonometric function graph, and they can be both horizontal and vertical. For a function like \( y = -\frac{1}{2} \cot \left(\frac{1}{2}x + \frac{\pi}{4}\right) \), we observe several transformations:

• Vertical Reflection: The negative sign indicates a reflection over the x-axis, which flips the graph upside down.
• Vertical Stretch/Compression: The \(-\frac{1}{2}\) factor compresses the graph vertically to half of its original height.
• Horizontal Stretch/Compression: The \( \frac{1}{2} \) factor compresses/stretch horizontally by a factor of 2, effectively changing the period of the graph.
• Horizontal Shift (Phase Shift): The \( \frac{\pi}{4} \) inside the function shifts the entire graph to the left by this amount.

Understanding how each of these transformations affects the graph is crucial in accurately sketching the transformed cotangent function.

Period and Phase Shift

The period of a trigonometric function is the horizontal length of one complete cycle. For \( \cot(x) \), the untransformed period is \( \pi \). When we have a function in the form \( \cot(bx + c) \), the period changes and is calculated by \( \frac{\pi}{|b|} \).
For our function, \( y = -\frac{1}{2} \cot \left(\frac{1}{2}x + \frac{\pi}{4}\right) \), the value of \( b \) is \( \frac{1}{2} \). Thus, the new period is:

• %Formula: \( \frac{\pi}{\frac{1}{2}} = 2\pi \)

Alongside the period, we have a phase shift, which horizontally moves the graph left or right. The phase shift is found by setting the inside of the cotangent function, \( \frac{1}{2}x + \frac{\pi}{4} \), equal to zero and solving for \( x \):

• %Formula: \( \frac{1}{2}x + \frac{\pi}{4} = 0 \)
• \( x = -\frac{\pi}{2} \)

This indicates the graph shifts \( \frac{\pi}{2} \) units left.

Vertical Asymptotes

Vertical asymptotes are lines where the function's value approaches infinity, and for \( \cot(x) \), these are particularly critical in defining the graph's structure. In the parent graph, \( \cot(x) \), asymptotes are at:\( x = n\pi \), for any integer \( n \).
To locate vertical asymptotes in a transformed function, like \( y = -\frac{1}{2} \cot \left(\frac{1}{2}x + \frac{\pi}{4}\right) \), solve for \( x \) in the equation \( \frac{1}{2}x + \frac{\pi}{4} = n\pi \):

• Rearrange to: \( \frac{1}{2}x = n\pi - \frac{\pi}{4} \)
• Multiply by 2: \( x = 2n\pi - \frac{\pi}{2} \)

These new asymptotes guide the sketching of the graph, marking key points where the function is undefined and the value tends to infinity.