Problem 35 - Precalculus with Calculus Previews

Question

Find the period, \(x\) -intercepts, and the vertical asymptotes of the given function. Sketch at least one cycle of the graph. $$ y=-1+\cot \pi x $$

Step-by-Step Solution

Answer

Period: 1, \( x \)-intercepts: \( x = \frac{1}{2} + k \), vertical asymptotes: \( x = k \).

Step 1: Identify the Basic Characteristics of \( y = \cot \pi x \)

The function \( y = \cot \theta \) has a period of \( \pi \). Since our function is \( y = \cot \pi x \), the period is transformed. Using the formula \( \text{Period} = \frac{\pi}{\text{coefficient of } x} \), the period for \( \cot \pi x \) is \( \frac{\pi}{\pi} = 1 \). Thus, the period of \( y = \cot \pi x \) is 1.

Step 2: Determine \( x \)-intercepts

To find the \( x \)-intercepts of \( y = \cot \pi x \), set \( y = 0 \). Thus, \( \cot \pi x = 1 \). It is known that \( \cot \pi x = 0 \) when \( \pi x = \frac{\pi}{2} + k\pi \) for any integer \( k \). Solving gives \( x = \frac{1}{2} + k \) where \( k \) is an integer. These are the \( x \)-intercepts.

Step 3: Locate the Vertical Asymptotes

Vertical asymptotes occur where \( \cot \pi x \) is undefined, meaning where \( \sin \pi x = 0 \). This occurs at \( \pi x = k\pi \), which simplifies to \( x = k \) for any integer \( k \). Thus, vertical asymptotes exist at all integer values of \( x \).

Step 4: Consider the Transformation \( y = -1 + \cot \pi x \)

The function \( y = -1 + \cot \pi x \) is a vertical shift of \( y = \cot \pi x \) by 1 unit downwards. Thus, for every point on the graph of \( y = \cot \pi x \), the \( y \)-value is decreased by 1, moving the graph lower by one unit.

Step 5: Sketch the Graph

To sketch at least one cycle of the graph, consider the \( x \)-intercepts at \( x = \frac{1}{2} + k \) and vertical asymptotes at \( x = k \), for one cycle from \( x = 0 \) to \( x = 1 \). Draw vertical lines at \( x = 0 \) and \( x = 1 \) to represent asymptotes. Plot \( x = \frac{1}{2} \) as the \( x \)-intercept, since the graph passes through \((\frac{1}{2}, -1)\) due to the vertical shift.

Key Concepts

Cosecant and Cotangent Functions

Cosecant and cotangent functions are trigonometric functions that are closely related to sine and cosine functions. The cotangent function, represented as \( \cot \theta \), is the reciprocal of the tangent function. It can be expressed as \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \). This means it is undefined wherever \( \sin \theta = 0 \).
The cosecant function, \( \csc \theta \), is the reciprocal of the sine function, \( \csc \theta = \frac{1}{\sin \theta} \), and is undefined where \( \sin \theta = 0 \).
For the cotangent function, its basic graph has a period of \( \pi \), in contrast to sine and cosine functions, which have periods of \( 2\pi \). This means the pattern of the graph repeats every \( \pi \) units.
When considering transformations, it is crucial to look at the arguments within the function, such as \( \pi x \) in \( \cot \pi x \), as these impact the function's period and other characteristics.

Vertical Asymptotes

Vertical asymptotes are lines where a function approaches infinity or negative infinity, meaning they represent points on the graph that are undefined. For functions like cotangent, vertical asymptotes occur when the denominator of the function becomes zero.
In the case of the cotangent function \( y = \cot \pi x \), vertical asymptotes appear whenever \( \sin \pi x = 0 \). Solving \( \sin \pi x = 0 \) involves finding integer multiples of \( \pi \), which translates to \( x = k \), where \( k \) is an integer.
These asymptotes divide the graph into segments where the function alternates between positive and negative values. Since the cotangent function tends to \( +\infty \) and \( -\infty \) as it approaches these points, it is important to be aware of asymptote positions when sketching or interpreting the graph.

Graph Transformations

Graph transformations are used to modify and analyze the behavior of trigonometric functions. For the function \( y = -1 + \cot \pi x \), there is a vertical transformation involved. Originally, \( \cot \pi x \) undergoes a vertical shift downward by 1 unit due to the \( -1 \) in \( y = -1 + \cot \pi x \).
This transformation shifts every point on the function down by one unit on the y-axis. As a result, features such as the x-intercepts, which originally occurred at \( x = \frac{1}{2} + k \) for \( y = \cot \pi x \), are downshifted to reflect the subtraction of 1 from the y-values.
Understanding and applying these transformations is vital for students to properly graph and interpret shifts in trigonometric functions. This involves analyzing how each component of a transformation directly affects the graph, ensuring a comprehensive understanding of its behavior.