Question

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

Answer 1

`-2<=x<=2`. y-intercept: x-intercept: `-2/sqrt5`. Dead ends `(-2, -1/6) and (2, 1)`. .Vertical asymptote: x = 1. See the explanation and graphs, for clarity.

Explanation:

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

Accordingly, we reach dead ens at `(-2, -1/6) and (2, 1)`.

y-intercept ( x = 0 ) is 1 and x-intercept ( y = 0 ) is `-2/sqrt5`.

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

Also, `y = (1/x)(2+sqrt(1-2/x^2))/(3-1/x+2/x^2) to 0` as `x to +-oo`

does not happen, as `|x|<=2`,

Now, `y =(2x+sqrt(4-x^2))/((x-1)(3x+2)) to +-oo` as x to 1 and x to -2/3#

The first is OK. but, the second is negated by `x >=-2`.

Important note:

The convention `sqrt(4-x^2)>=0 ( unlike (4-x^2)^(1/2)=+-sqrt(4-x^2))`

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]}