Verified Answer :
- The function \(y = 2 \sec \left(x - \frac{\pi}{3}\right)\) has vertical asymptotes where \(\cos\left(x - \frac{\pi}{3}\right) = 0\), specifically at \(x = \frac{5\pi}{6} + k\pi\).
- Upward-opening branches have local minima at \(y = 2\), with the first occurring at \(x = \frac{\pi}{3}\).
- Downward-opening branches have local maxima at \(y = -2\), with the first occurring at \(x = \frac{4\pi}{3}\).
- Comparing these key points and asymptotes with the provided options will identify the correct graph of \(\boxed{y = 2 \sec \left(x - \frac{\pi}{3}\right)}\).
Explanation
- Analyzing the Function Features
The function \(y = 2 \sec \left(x - \frac{\pi}{3}\right)\) is a transformation of the basic secant function \(f(x) = \sec(x)\). To understand its graph, we analyze its key features, such as vertical asymptotes, local extrema, and domain. The secant function is defined as the reciprocal of the cosine function: \(\sec(\theta) = \frac{1}{\cos(\theta)}\). Therefore, the function \(y = 2 \sec \left(x - \frac{\pi}{3}\right)\) is undefined whenever \(\cos\left(x - \frac{\pi}{3}\right) = 0\).
- Determining Vertical Asymptotes
Vertical asymptotes occur where the denominator of the secant function is zero. We set the argument of the cosine function equal to the values where it vanishes:
\[x - \frac{\pi}{3} = \frac{\pi}{2} + k\pi, \quad \text{for any integer } k.\]
Solving for \(x\), we get:
\[x = \frac{\pi}{2} + \frac{\pi}{3} + k\pi = \frac{3\pi}{6} + \frac{2\pi}{6} + k\pi = \frac{5\pi}{6} + k\pi.\]
Thus, the vertical asymptotes for this function are located at \(x = \frac{5\pi}{6}, \frac{11\pi}{6}, \dots\) and so on.
- Identifying Local Extrema
Local extrema occur where the absolute value of the cosine function is at its maximum, i.e., where \(\cos\left(x - \frac{\pi}{3}\right) = 1\) or \(-1\).
- For local minima of the upward-opening branches, set \(\cos\left(x - \frac{\pi}{3}\right) = 1\):
\[x - \frac{\pi}{3} = 2k\pi \implies x = \frac{\pi}{3} + 2k\pi.\]
At \(x = \frac{\pi}{3}\), \(y = 2 \sec(0) = 2(1) = 2\). So, there is a local minimum at \(\left(\frac{\pi}{3}, 2\right)\).
- For local maxima of the downward-opening branches, set \(\cos\left(x - \frac{\pi}{3}\right) = -1\):
\[x - \frac{\pi}{3} = \pi + 2k\pi \implies x = \frac{4\pi}{3} + 2k\pi.\]
At \(x = \frac{4\pi}{3}\), \(y = 2 \sec(\pi) = 2(-1) = -2\). So, there is a local maximum at \(\left(\frac{4\pi}{3}, -2\right)\).
- Conclusion and Identifying the Graph
To correctly identify the graph of \(y = 2 \sec \left(x - \frac{\pi}{3}\right)\), look for these specific characteristics:
- The function has vertical asymptotes at \(x = \frac{5\pi}{6} + k\pi\).
- Upward-opening branches have local minima at \(y=2\), the first being at \(\left(\frac{\pi}{3}, 2\right)\).
- Downward-opening branches have local maxima at \(y=-2\), the first being at \(\left(\frac{4\pi}{3}, -2\right)\).
The graph should show a vertical stretch of a factor of 2 compared to the standard secant curve. Given these points and asymptotes, you can identify the correct graph among the provided options.
Examples
In engineering, secant functions appear when modeling certain types of periodic motion or calculating distances in specialized surveying equipment. For example, if a rotating beacon light is placed at a specific distance from a straight wall, the distance of the light beam from a reference point on the wall at a given angle can be modeled using a secant function. Understanding how transformations like \(y = a \sec(x - c)\) affect the graph helps engineers predict the beam's behavior and ensure it illuminates the required area effectively.