Verified Answer :
- The function is a transformation of the secant function with vertical shift \(D=2\) and vertical stretch \(A=3\).
- The period is \(\pi\) due to the frequency factor \(B=2\), and the midline is the line \(y=2\).
- The local minima are at \(y=5\) and local maxima are at \(y=-1\).
- Vertical asymptotes occur at \(x = \frac{3\pi}{4} + k\frac{\pi}{2}\) for any integer \(k\).
Explanation
- Analysis of the Function Components
The function \(y = 3 \sec \left[2\left(x - \frac{\pi}{2}\right)\right] + 2\) is a transformed version of the standard secant function \(y = \sec(x)\). The general form for such a transformation is \(y = A \sec[B(x - C)] + D\). By comparing the given function to this general form, we identify:
- The vertical stretch factor \(A = 3\).
- The frequency factor \(B = 2\), which modifies the period of the function.
- The phase shift \(C = \frac{\pi}{2}\), indicating a horizontal shift to the right.
- The vertical shift \(D = 2\), which establishes the midline around which the secant curves oscillate.
- Determining Period and Midline
The period \(T\) of the secant function is determined by the formula \(T = \frac{2\pi}{|B|}\). With \(B = 2\), the period is \(T = \frac{2\pi}{2} = \pi\). The vertical shift \(D = 2\) defines the midline of the function, meaning the graph oscillates relative to the horizontal line \(y = 2\). This midline acts as the central reference for the local minima and maxima of the function.
- Identifying Local Extrema and Range
The vertical range of the local extrema is determined by the vertical stretch \(A\) and vertical shift \(D\). The secant function's local minima (relative to its own branches) occur at \(y = D + A = 2 + 3 = 5\), while local maxima (relative to its own branches) occur at \(y = D - A = 2 - 3 = -1\). Consequently, the graph will have U-shaped branches with local minima at \(y = 5\) and inverted U-shaped branches with local maxima at \(y = -1\).
- Locating Vertical Asymptotes
Vertical asymptotes for the secant function occur where the cosine component in the denominator is zero. Setting the argument \(2(x - \frac{\pi}{2}) = \frac{\pi}{2} + k\pi\) for any integer \(k\), we find:
\(x - \frac{\pi}{2} = \frac{\pi}{4} + \frac{k\pi}{2}\)
\(x = \frac{3\pi}{4} + \frac{k\pi}{2}\)
This results in vertical asymptotes at values like \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4},\dots\). Between these asymptotes, the function alternates between values greater than or equal to 5 and less than or equal to -1. For instance, at \(x = \frac{\pi}{2}\), the function value is \(y = 3 \sec(0) + 2 = 5\), which is a local minimum.
Examples
Understanding the graph of a secant function like \(y = 3 \sec \left[2\left(x - \frac{\pi}{2}\right)\right] + 2\) is essential in fields such as electrical engineering and physics. For example, when analyzing the voltage or current in a circuit containing specific resonance properties, trigonometric functions often describe the waveform. Knowing how transformations like shifts and stretches affect these graphs allows engineers to predict the behavior of signals over time and identify critical points like maximum amplitudes (extrema) and points of discontinuity (asymptotes) where the system might behave unpredictably.