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Chapter 4: Problem 34

Sketch one full period of the graph of each function. $$y=\frac{3}{2} \csc 3 x$$

Short Answer

Expert verified
The graph of the function \( y=\frac{3}{2} \csc (3 x) \) has an amplitude of \( \frac{3}{2} \), a period of \( \frac{2\pi}{3} \), a phase shift of 0 and a vertical shift of 0. Asymptotes will appear at points where \( \sin(x)=0 \).

Step by step solution

01

Identify the Amplitude

The amplitude of the function \( y=A \csc(Bx-C) \) is given by the absolute value of \( A \). In this case, \( A=\frac{3}{2} \) hence, the amplitude is \( |\frac{3}{2}|=\frac{3}{2} \).
02

Identify the Period

The period \( T \) of the function \( y=A \csc(Bx-C) \) is given by \( \frac{2\pi}{|B|} \). In our case, \( B=3 \). Hence, the period of the function is \( T=\frac{2\pi}{3} \).
03

Identify the Phase Shift

The phase shift of the function \( y=A \csc(Bx-C) \) is given by \( C/B \). The given function does not contain any 'C' and hence, the phase shift is \( 0 \).
04

Identify the Vertical Shift

The vertical shift of the function \( y=A \csc(Bx-C)+D \) is given by \( D \). As no 'D' value is given in the original function equation, the vertical shift is \( 0 \).
05

Graphing the Function

Now plot the graph using the earlier calculated values. The graph will have vertical asymptotes at each \( x \) for which \( \sin(x)=0 \), since the cosecant function \( \csc(x) \) is defined as \( \frac{1}{\sin(x)} \). The peaks of the graph will happen where the \( \sin \) function is at its minimum and maximum. Sketch the object based on these.

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