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Chapter 4: Problem 29

Sketch the graph of the function. (Include two full periods.) $$y=2 \sec 3 x$$

Short Answer

Expert verified
The graph for the given function \(y=2\sec(3x)\) starts at a vertical asymptote at \(x=0\), peaks at \(y=2\) at \(x=(1/6)\pi\) follows with another verical asymptote at \(x=(2/3)\pi\), lows at \(y=-2\) at \(x=\pi\), follows with another vertical asymptote at \(x=(4/3)\pi\), peaks at \(y=2\) at \(x=5\pi/6\), and finally ends the cycle with another vertical asymptote at \(x=2\pi\). It repeats the same pattern every period, \(2\pi/3\).

Step by step solution

01

Identify the Period

The period of the secant function is \(2\pi/k\). For the given function, \(y=2\sec(3x)\), \(k=3\). Thus the period is \(2\pi/3\). This means the function will repeat its pattern every \(2\pi/3\) intervals.
02

Identifying Key Points

For y=sec(x), the secant graph has vertical asymptotes and no particular peak values since sec(x)=1/cos(x), so sec(x) becomes infinity whenever cos(x) = 0. That usually happens at \(x=(1/2)\pi + n\pi\) for \(n\) as an integer. After applying the transformation to sec(x) from the given function \(y=2\sec(3x)\), these points become \(x=(1/6)\pi + (n/3)\pi\). Similarly, sec(x)=1 whenever cos(x)=1, which happens at \(x=2n\pi\). After applying the transformation, these points become \(x=(2/3)n\pi\) at which \(y=2*1=2\), and sec(x)=-1 whenever cos(x)=-1, which happens at \(x=\pi+2n\pi\). After applying the transformation, these points become \(x=(1/3)\pi+(2/3)n\pi\) at which \(y=2*-1=-2\)
03

Drawing the Graph

Start by drawing the vertical asymptotes at \(x=(1/6)\pi + (n/3)\pi\). For two periods, n would be from 0 to 3. So, the asymptotes are at \(x=0, (2/3)\pi, (4/3)\pi, and 2\pi\). Then mark on the graph \(y=2\) at points \(x=(2/3)n\pi\) and \(y=-2\) at points \(x=(1/3)\pi+(2/3)n\pi\). Connect the points on the graph in the shape of the peaks of cosine function but take care not to cross the vertical asymptotes. The base of each peak should touch each vertical asymptote and the peak values will be at points \(y=2\) and \(y=-2\).
04

Repeating the pattern

The secant function repeats the same pattern per period. Therefore, repeat the same sequence of peaks and valleys from the completed period for the next period.

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