How do you find the asymptotes for y = (3-4x )/( 6x + sqrt(4x^2-3x-5))?

Dec 6, 2016

Horizontal: $y = - \frac{2}{3}$. The graph is real for $\frac{- \sqrt{89} + 3}{8} \le x \le \frac{\sqrt{89} + 3}{8}$.

Explanation:

To make y real, $\frac{- \sqrt{89} + 3}{8} \le x \le \frac{\sqrt{89} + 3}{8}$.

$y = \frac{- 4 + \frac{3}{x}}{6 + \sqrt{4 - \frac{3}{x} - \frac{1}{x} ^ 2}} \to - \frac{2}{3}$, as $x \to \pm \infty$.

Look at the dead ends $x = \frac{3 \pm \sqrt{89}}{8} = 1.554 \mathmr{and} - 0.804$, nearly.

graph{y(6x+sqrt(4x^2-3x-5))+4x-3=0 [-5, 5, -2.5, 2.5]}