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

Jan 29, 2017

Horizontal asymptote :$- - \frac{3}{2}$. The graph has a gap, when $0 < x < \frac{3}{2}$.

Explanation:

$y = \sqrt{{\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4}} - x$, giving $| x - \frac{3}{2} | \ge \frac{3}{2}$..

So, x is not in $\left(0 , \frac{3}{2}\right)$.

Also, y has the form

$x \left({\left(1 - \frac{3}{x}\right)}^{0.5} - 1\right)$

$= x \left(1 - \frac{3}{2} \left(\frac{1}{x}\right) - \frac{9}{8} \left(\frac{1}{x} ^ 2\right) + \ldots - 1\right)$

$= - \frac{3}{2} - \frac{9}{8} \left(\frac{1}{x}\right)$ + higher powers of (1/x))

$\to - \frac{3}{2}$ , as x to +-oo#.

$\to \pm \infty$, as $x \to \pm \infty$.

So, $y = - \frac{3}{2}$ is the asymptote.