How do you find the asymptotes for #y= -1/2 sec x#?

Question

Lily Anderson · Accepted Answer

x = an odd multiple of #pi/2#.

sec (odd multiple of #pi/2#) is #+-oo#. The straight lines x= odd multiple of #pi/2# are asymptotic to the periodic graph in both positive and negative directions of y-axis.i

Lily Anderson · Answer

undefined To find the asymptotes of the function $ y = -\frac{1}{2} \sec(x) $, we need to consider the properties of the secant function. Secant has vertical asymptotes where its cosine counterpart, cosine, equals zero. Thus, the vertical asymptotes occur at points where $ \cos(x) = 0 $. The cosine function equals zero at odd multiples of $ \frac{\pi}{2} $. Therefore, the vertical asymptotes of $ y = -\frac{1}{2} \sec(x) $ occur at $ x = (2n + 1) \frac{\pi}{2} $, where $ n $ is an integer.

Lincoln Anderson · Answer

undefined To find the asymptotes for the function $ y = -\frac{1}{2} \sec(x) $, follow these steps:

1. Vertical asymptotes:
   - Vertical asymptotes occur where the denominator of the secant function becomes zero.
   - Secant function has vertical asymptotes at odd multiples of $ \frac{\pi}{2} $, i.e., $ x = \frac{\pi}{2} + k\pi $ where $ k $ is an integer.

2. Horizontal asymptotes:
   - Horizontal asymptotes occur when the function approaches a constant value as $ x $ approaches positive or negative infinity.
   - Since secant oscillates between $ -\infty $ and $ +\infty $, there are no horizontal asymptotes.

Therefore, the vertical asymptotes for $ y = -\frac{1}{2} \sec(x) $ occur at $ x = \frac{\pi}{2} + k\pi $ where $ k $ is an integer, and there are no horizontal asymptotes.