How do you find vertical, horizontal and oblique asymptotes for # (2x+sqrt(4-x^2)) / (3x^2-x-2)#?

Question

1209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-16T11:54:37

Horizontal: #uarr x = 1 and x=-2/3 darr#. Graph is inserted. Look at the ends #x = +-2#.

To make y real, #-2<=x<=2#.

The equation is #y=(2x+sqrt(4-x^2))/((3x+2)(x-1))#

So, #x-1=0 and 3x+2=0# give vertical asymptotes.

As x is bounded, there is no horizontal asymptote, and for the same

reason, there is no slant asymptote.

The illustrative graph is inserted.

graph{y(x-1)(3x+2)-2x-sqrt(4-x^2)=0 [-10, 10, -5, 5]}