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

##### 1 Answer

#### Explanation:

To make y real,

Accordingly, we reach dead ens at

y-intercept ( x = 0 ) is 1 and x-intercept ( y = 0 ) is

Note that the curve does not reach the x-axis for positive intercept.

Also,

does not happen, as

Now,

The first is OK. but, the second is negated by

Important note:

The convention

troubled me in making my data compatible with that in the graph. I

had to take negative root, breaking the convention.

graph{y(x-1)(3x-2)-2x-sqrt(4-x^2)=0 [-10.17, 10.17, -5.105, 5.065]}

graph{y(x-1)(3x-2)-2x-sqrt(4-x^2)=0 [-20.34, 20.34, -10.19, 10.15]}