1 2 Graph the function y sec 2x.
To draw the graph, plot three consecutive asymptotes and two points, one on each side of the second asymptote. Then click on the graph icon
The Correct Answer and Explanation is :
To graph the function ( y = \sec(2x) ), we need to understand the characteristics of the secant function, which is the reciprocal of the cosine function. The graph of ( \sec(2x) ) behaves similarly to ( \sec(x) ), but with a few key changes due to the transformation involving ( 2x ).
Steps to Graph ( y = \sec(2x) ):
- Determine the asymptotes:
The secant function has vertical asymptotes where the cosine function equals zero, because ( \sec(x) = \frac{1}{\cos(x)} ). Therefore, we first need to find where ( \cos(2x) = 0 ). The cosine function equals zero at ( \frac{\pi}{2} + n\pi ), where ( n ) is an integer. Since we have ( 2x ) inside the cosine, we solve for ( x ):
[
2x = \frac{\pi}{2} + n\pi
]
[
x = \frac{\pi}{4} + \frac{n\pi}{2}
]
Thus, the vertical asymptotes for ( y = \sec(2x) ) occur at ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots ). - Plot two points:
After finding the asymptotes, we select a few values of ( x ) to plot points. Typically, we choose points that are near the asymptotes, since secant curves are undefined at the asymptotes and approach infinity. Choose ( x = 0 ) and ( x = \frac{\pi}{8} ) for illustration. For ( x = 0 ),
[
y = \sec(2(0)) = \sec(0) = 1
]
So, the point is ( (0, 1) ). For ( x = \frac{\pi}{8} ),
[
y = \sec\left(2 \times \frac{\pi}{8}\right) = \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}
]
So, the point is ( \left(\frac{\pi}{8}, \sqrt{2}\right) ). - Sketch the graph:
Now that we know the vertical asymptotes and two points on the graph, sketch the curve. Between each pair of asymptotes, the graph should form a U-shaped curve that approaches infinity near the asymptotes and curves toward the horizontal line ( y = 1 ) or ( y = -1 ).
Graphing Example:
The graph of ( y = \sec(2x) ) will show alternating “hills” and “valleys” between vertical asymptotes, with the curve approaching positive and negative infinity at the asymptotes.
For the full graph, look at the provided image. The key elements are the locations of the vertical asymptotes, which are ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots ), and the general shape of the curve as described.