How do you find the asymptotes for #y=3+2csc x#?

1 Answer
Feb 17, 2017

Answer:

Vertical : #x = kpi, k = 0, +-1, +-2, +-3, ...#. See Socratic depiction.

Explanation:

graph{3+2/sinx [-16,16, -5, 11]}

graph{(3+2/sinx-y)(x-.01y)(x-3.14+.01y)(x+3.14-.01y)=0 [-5, 5, 0, 6]}

The asymptotes # x = 0, pi and -pi# are included in this graph.

The curvilinear/slant/horizontal asymptotes of y = Q(x)+R(x)/A(x) are

given by y =Q(x). if A(x) is unbounded

The vertical asymptotes are given by

x = a zero of A(x).

Here, this form is

y =3+2/sinx.

As #A(x) = sinx in [-1, 1], y = Q(x) = 3# is not an asymptote.

#A(x)=sinx = 0# gives the vertical asymptotes

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