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

Feb 17, 2017

Vertical : $x = k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$. 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 $x = 0 , \pi \mathmr{and} - \pi$ 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 $A \left(x\right) = \sin x \in \left[- 1 , 1\right] , y = Q \left(x\right) = 3$ is not an asymptote.

$A \left(x\right) = \sin x = 0$ gives the vertical asymptotes

$x = k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$