Question

practice assignment 8.2 graphs of the other trigonometric functions
score: 5/9 answered: 8/9
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question 9
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write an equation in the form ( y = c + a sec ( b ( x - d ) ) ) or ( y = c + a csc ( b ( x - d ) ) ) for the function graphed above.
question help: video

Explanation:

Step1: Determine the type of function

The graph has vertical asymptotes and U - shaped curves. The general form of the secant function is \(y = c + a\sec(b(x - d))\). The standard secant function \(y=\sec(x)\) has a period of \(2\pi\).

Step2: Find the period

Looking at the graph, the distance between two consecutive vertical asymptotes (or the distance between two consecutive "U - shapes") is \(2\). The formula for the period of \(y = a\sec(b(x - d))+c\) is \(T=\frac{2\pi}{|b|}\). Since \(T = 2\), we have \(\frac{2\pi}{|b|}=2\), so \(b=\pi\)

Step3: Find the vertical shift \(c\)

The mid - line of the graph (the line halfway between the minimum and maximum of the U - shaped curves) is \(y = 0\). So \(c = 0\)

Step4: Find the amplitude \(a\)

The minimum value of the function (the bottom of the U - shape) is \(y=- 5\) and the maximum value (the top of the U - shape) is \(y = 5\). For the secant function \(y=a\sec(b(x - d))+c\), the amplitude \(|a|\) is related to the distance from the mid - line. Since \(c = 0\), \(a = 5\)

Step5: Find the phase shift \(d\)

The graph of \(y = 5\sec(\pi x)\) has a vertical asymptote at \(x=\frac{1}{2}+n\) for \(n\in\mathbb{Z}\). The given graph has a vertical asymptote at \(x = 0\). The general form \(y=a\sec(b(x - d))+c\). If we set \(b(x - d)=\frac{\pi}{2}+n\pi\). When \(n = 0\) and \(b=\pi\), \(\pi(x - d)=\frac{\pi}{2}\), \(x=d+\frac{1}{2}\). Since there is a vertical asymptote at \(x = 0\), \(d=-\frac{1}{2}\)

Answer:

\(y = 5\sec(\pi(x+\frac{1}{2}))\)