In Exercises 29–44, graph two periods of the given cosecant or secant function. y = −1/2 sec πx
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Identify the given function: \(y = -\frac{1}{2} \sec(\pi x)\). This is a secant function with amplitude scaling and reflection.
Recall that the secant function is the reciprocal of the cosine function, so \(\sec(\theta) = \frac{1}{\cos(\theta)}\). The graph of \(y = \sec(\theta)\) has vertical asymptotes where \(\cos(\theta) = 0\).
Determine the period of the function. The standard period of \(\sec(x)\) is \(2\pi\). For \(\sec(bx)\), the period is \(\frac{2\pi}{b}\). Here, \(b = \pi\), so the period is \(\frac{2\pi}{\pi} = 2\).
Since the problem asks for two periods, the interval to graph is from \(x = 0\) to \(x = 4\) (or any interval of length 4). Identify vertical asymptotes by solving \(\cos(\pi x) = 0\), which occurs at \(\pi x = \frac{\pi}{2} + k\pi\), so \(x = \frac{1}{2} + k\) for integers \(k\).
Plot the key points of \(y = -\frac{1}{2} \sec(\pi x)\) by first plotting \(y = \cos(\pi x)\), then taking the reciprocal to get \(\sec(\pi x)\), applying the vertical stretch by \(\frac{1}{2}\) and reflection (negative sign). Mark vertical asymptotes at \(x = \frac{1}{2} + k\) and sketch the graph between these asymptotes over two periods.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Secant Function and Its Properties
The secant function, sec(x), is the reciprocal of the cosine function, defined as sec(x) = 1/cos(x). It has vertical asymptotes where cos(x) = 0, and its graph consists of branches that extend to infinity near these asymptotes. Understanding its periodicity and behavior is essential for graphing.
The coefficient in front of the secant function, here −1/2, affects the vertical stretch and reflection of the graph. A factor of 1/2 compresses the graph vertically, while the negative sign reflects it across the x-axis. This changes the height and orientation of the secant branches.
Period of the Secant Function with Horizontal Scaling
The period of sec(x) is 2π, but when the function is sec(πx), the period changes to 2 because the input is scaled by π. This horizontal scaling compresses or stretches the graph along the x-axis, affecting where the asymptotes and key points occur. Knowing how to find the period is crucial for graphing two full cycles.