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

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Answer 1 · 2017-01-29T07:30:18

Horizontal asymptote : $- -3/2$ . The graph has a gap, when $0< x< 3/2$ .

$y=sqrt((x-3/2)^2-9/4)-x$ , giving $|x-3/2|>=3/2$ ..

So, x is not in $(0, 3/2)$ .

Also, y has the form

$x((1-3/x)^0.5-1)$

$=x(1-3/2(1/x)-9/8(1/x^2)+ ... -1)$

$=-3/2-9/8(1/x)$ + higher powers of (1/x))

$to -3/2$ , as x to +-oo#.

$to +-oo$ , as $x to +-oo$ .

So, $y = -3/2$ is the asymptote.

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