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

Dec 16, 2016

Horizontal: $\uparrow x = 1 \mathmr{and} x = - \frac{2}{3} \downarrow$. Graph is inserted. Look at the ends $x = \pm 2$.

#### Explanation:

To make y real, $- 2 \le x \le 2$.

The equation is $y = \frac{2 x + \sqrt{4 - {x}^{2}}}{\left(3 x + 2\right) \left(x - 1\right)}$

So, $x - 1 = 0 \mathmr{and} 3 x + 2 = 0$ 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]}