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

Dec 6, 2016

$- 2 \le x \le 2$. y-intercept: x-intercept: $- \frac{2}{\sqrt{5}}$. Dead ends $\left(- 2 , - \frac{1}{6}\right) \mathmr{and} \left(2 , 1\right)$. .Vertical asymptote: x = 1. See the explanation and graphs, for clarity.

#### Explanation:

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

Accordingly, we reach dead ens at $\left(- 2 , - \frac{1}{6}\right) \mathmr{and} \left(2 , 1\right)$.

y-intercept ( x = 0 ) is 1 and x-intercept ( y = 0 ) is $- \frac{2}{\sqrt{5}}$.

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

Also, $y = \left(\frac{1}{x}\right) \frac{2 + \sqrt{1 - \frac{2}{x} ^ 2}}{3 - \frac{1}{x} + \frac{2}{x} ^ 2} \to 0$ as $x \to \pm \infty$

does not happen, as $| x | \le 2$,

Now, $y = \frac{2 x + \sqrt{4 - {x}^{2}}}{\left(x - 1\right) \left(3 x + 2\right)} \to \pm \infty$ as x to 1 and x to -2/3#

The first is OK. but, the second is negated by $x \ge - 2$.

Important note:

The convention $\sqrt{4 - {x}^{2}} \ge 0 \left(u n l i k e {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} = \pm \sqrt{4 - {x}^{2}}\right)$

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]}