How do you find the Vertical, Horizontal, and Oblique Asymptote given #s(t)=(8t)/sin(t)#?

Question

Serenity Jones · Accepted Answer

Vertical: #uarr x = +-kpi darr, k = +-1, +-2, +-3, ...#

As #t to 0, s to 8#.

# s to +-oo#, as # t to kpi, k = +-1, +-2, +-3, ...#, revealing

vertical asymptotes #x=kpi, k = 0, +-1, +-2, +-3, ...#.

The first graph reveals the trends, for tending towards first pair of

asymptotes #x = +-pi#.

The second graph seems to say 'Grand New Year!'.

graph{(8x)/sin x [-128.1, 128.3, -64, 64]}

graph{(8x)/sin x [-500,500, -1000, 1000]}

Elijah Allen · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $ t $. For $ s(t) = \frac{8t}{\sin(t)} $, the denominator $\sin(t)$ equals zero when $ t = k\pi $, where $ k $ is an integer. So, the vertical asymptotes occur at $ t = k\pi $.

Horizontal asymptotes can be determined by observing the behavior of the function as $ t $ approaches positive or negative infinity. For $ s(t) = \frac{8t}{\sin(t)} $, as $ t $ approaches infinity or negative infinity, the function behaves like $ \frac{8t}{1} $ since $ \sin(t) $ oscillates between -1 and 1. Thus, the horizontal asymptote is $ y = 8t $ or simply $ y = 8 $.

To find the oblique asymptote, perform long division or divide the numerator by the denominator. For $ s(t) = \frac{8t}{\sin(t)} $, long division would result in $ 8 + \frac{8t}{\sin(t)} $. Since $ \frac{8t}{\sin(t)} $ has vertical asymptotes at $ t = k\pi $, the oblique asymptote is the line $ y = 8t $.

So, summarizing:
- Vertical asymptotes: $ t = k\pi $ where $ k $ is an integer.
- Horizontal asymptote: $ y = 8 $.
- Oblique asymptote: $ y = 8t $.