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

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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]}

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