How do you sketch one cycle of #y=sec(2x)#?

Question

Oliver Atkinson · Accepted Answer

Graph data: Period : #pi; x in[-pi/2, pi/2]. y in [-pi/2, pi/2]#;
the asymptotes in this single period: #uarr x=+-pi/4 darr#.

#|y|=|sec(2x)|>=1# Period is# (2pi)/2=pi=3.1416#, nearly, #y to +-oo#, as #x to (2k+1)pi/2, k = 0, +-1, +-2, +-3, ...#The graph{(y-1/cos(2x))=0 [-1.57, 1.57, -1.57, 1.57]}

Josiah Anderson · Answer

undefined To sketch one cycle of the function $ y = \sec(2x) $, you can follow these steps:

1. Identify the key points and asymptotes: 
   - The secant function has vertical asymptotes where the cosine function is zero. In this case, vertical asymptotes occur when $ \cos(2x) = 0 $, which happens when $ 2x = \frac{\pi}{2} + n\pi $, where $ n $ is an integer.
   - The secant function has peaks and valleys where the cosine function reaches its maximum and minimum values. Since the range of the secant function is all real numbers except where it is undefined, there are no restrictions on the maximum and minimum values.

2. Determine the critical points:
   - Find the critical points by setting $ \cos(2x) = 1 $ and $ \cos(2x) = -1 $. These points correspond to the peaks and valleys of the secant function.

3. Sketch the graph:
   - Start by plotting the vertical asymptotes at $ x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots $.
   - Plot the critical points where $ \cos(2x) = 1 $ and $ \cos(2x) = -1 $.
   - Connect the critical points and asymptotes with smooth curves to complete the sketch of one cycle of the secant function.

Remember that the secant function is undefined at its vertical asymptotes, so be sure to leave gaps in the graph at those points. Additionally, since the secant function oscillates between positive and negative infinity, the graph will approach the asymptotes but will never touch them.