Problem 38 - Precalculus Essentials

Question

Graph two periods of the given cosecant or secant function. $$y=-\frac{1}{2} \csc \pi x$$

Step-by-Step Solution

Answer

The graph of the function \(y=-\frac{1}{2} \csc \pi x\) contains two periods. It features 'U' shaped figures between asymptotes, which occur at every \(x = k . 2\), where \(k\) is an integer. Due to the negative amplitude, it's reflected over the x-axis.

Step 1: Identify the Period

The period of the basic cosecant function is \(2\pi\). However, the period might change because of the coefficient of \(x\), \(\pi\) in this case. The period of the transformed function is calculated as the basic period divided by the absolute value of this coefficient. Therefore, the period of the given function is \(2\pi / |\pi| = 2\)

Step 2: Identify the Amplitude

The amplitude is the absolute value of the coefficient of the cosecant function. In this case, it is \(|-1/2| = 1/2\). Since the coefficient is negative, the basic graph needs to be reflected over the x-axis.

Step 3: Identify the X-intercepts and Asymptotes

Since the cosecant function is undefined for \(x = 0\), the graph of this function has asymptotes at these values. For the given function with a period of 2, these values will fall at \(x = k . 2\), where \(k\) is an integer. Additionally, since the graph for cosecant is undefined wherever sine is equal to zero, there are no x-intercepts for this function.

Step 4: Sketch the Graph

Using the details from above steps, begin graphing. Since the period is 2, start by graphing the period between \(x = 0\) and \(x = 2\). Reflect these points over the x-axis to account for the negative amplitude. Extend this pattern for two periods. Add in the asymptotes at \(x = k . 2\), where \(k\) is an integer. Since the graph moves up and down between the asymptotes, the final graph will show a series of 'U' shapes, inverted due to the negative coefficient.