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Chapter 4: Problem 33

Sketch the graph of the function. (Include two full periods.) $$y=2 \csc (x-\pi)$$

Short Answer

Expert verified
The graph of \(y=2\csc(x-\pi)\) is similar to the graph of standard cosecant \(y=\csc(x)\). It is vertically stretched by a factor of 2 and shifted to the right by \(\pi\) units. Vertical asymptotes occur at \(x=n\pi + \pi\), and significant points are \(x=\frac{3\pi}{2}\) where \(y=2\) and \(x=\frac{\pi}{2}\) where \(y=-2\).

Step by step solution

01

Understanding the Function

Identify the function as a transformed cosecant function of the form \(y=A\csc(Bx-C)\). Here, the amplitude \(A\) is 2, the period-relation constant \(B\) is 1 (implying a period of \(2\pi\)), and the horizontal translation \(C\) is \(\pi\). Thus, the graph will be similar to the graph of \(y=\csc(x)\) but is vertically stretched by a factor of 2 and translated \(\pi\) units to the right.
02

Determine Points of Undefinition

The original cosecant function \(y=\csc(x)\) is undefined at every multiple of \(\pi\). However, due to the translation by \(\pi\), for the function of interest, \(y=2\csc(x-\pi)\) is undefined at \(x=n\pi + \pi\) where \(n\) is an integer.
03

Determine Key Points for Plotting

\(\csc(x)\) equals 1 when \(sin(x)=1\) and \(\csc(x)\) equals -1 when \(sin(x)=-1\). Hence, find the values of \(x\) that result in these function values, also considering the translation by \(\pi\). For the original \(y=\csc(x)\) function, \(x=\frac{\pi}{2}\) and \(x=-\frac{\pi}{2}\) yields values of 1 and -1 respectively. Apply the translation by \(\pi\) to these x values to get key points for the function.
04

Plotting the Graph

Initially, sketch the standard cosecant graph as a flipped version of the sine graph over two periods, \(0< x < 4\pi\), then vertically stretch it by a factor of 2 (to incorporate the amplitude), and then shift it \(\pi\) units to the right (to incorporate the horizontal translation). Mark the key points and points of undefinition from step 2 and 3, alongside vertical asymptotes at the point of undefinition.

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