/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Sketch one full period of the gr... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch one full period of the graph of each function. $$y=\frac{3}{4} \sec x$$

Short Answer

Expert verified
The graph of \(y=\frac{3}{4} \sec(x)\) is a vertically compressed version of the standard secant function graph. It spans a period of \(2\pi\) radians, starting and ending at \(y= \frac{3}{4}\) for \(x=0\) and \(x=2\pi\), respectively and has vertical asymptotes at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). The curve rises from \(y= \frac{3}{4}\) up to the vertical asymptote at \(x=\frac{\pi}{2}\), skips to negative infinity at the same x-value, rises back up to \(y= -\frac{3}{4}\) at \(x=\pi\), goes down again to the asymptote at \(x=\frac{3\pi}{2}\), and finally skips back up to \(y= \frac{3}{4}\) at \(x=2\pi\).

Step by step solution

01

Understand the base function

Secant (sec) is the reciprocal of the cosine (cos) function, i.e., \(sec(x) = \frac{1}{cos(x)}\), for all \(x\) where \(cos(x)\neq0\). We know that the cosine function repeats itself every \(2\pi\) radians and there are critical points at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\), where the cosine function equals 0 and secant is undefined.
02

Apply the coefficient

The function in this exercise is \(y=\frac{3}{4} \sec x\). So, the specific function is a vertically compressed version of the standard secant curve. Every \(y\) value of the secant function is multiplied by \(\frac{3}{4}\) here. Hence, the highest and lowest points of our curve will be \(\frac{3}{4}\) times the usual secant function's value at these points.
03

Plot the function

At \(x=0\), \(cos(x)=1\) and hence \(sec(x) = \frac{1}{cos(x)} = 1\). Multiply this by the coefficient \(\frac{3}{4}\) to get \(y= \frac{3}{4}\). Similarly calculate the y values for critical points \(x=\frac{\pi}{2}\), \(x=\pi\), \(x=\frac{3\pi}{2}\) and \(x=2\pi\). You will get y values as undefined for \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). Connect the points to sketch the graph of \(y=\frac{3}{4} \sec(x)\), remembering that at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\), secant function has vertical asymptotes.

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