Question

How do you find all the asymptotes for `sqrt(x^2-3x) - x`?

Answer 1

No vertical asymptote, Right horizontal asymptote `y = -3/2`, Left oblique asymptote `y=-2x+3/2`.

Explanation:

`f(x) = sqrt(x^2-3x) - x`

Vertical: None
`f(x)` does not increase without bound (go to infinity) as `x` approaches any (finite) number. So there are no vertical asymptotes.

Horizontal Left: None, Right: `y = -3/2`

Left: `lim_(xrarr-oo)(sqrt(x^2-3x) - x) = oo`

So there is no horizontal asymptote on the left.

Right:
`lim_(xrarroo)(sqrt(x^2-3x) - x)`

`= lim_(xrarroo)((sqrt(x^2-3x) - x))/1 * ((sqrt(x^2-3x) + x))/((sqrt(x^2-3x) + x))`

`= lim_(xrarroo)(x^2-3x - x^2) /(sqrt(x^2-3x) + x)`

`= lim_(xrarroo)(-3x) /(sqrt(x^2)sqrt(1-3/x) + x)`

`= lim_(xrarroo)(-3x) /(x(sqrt(1-3/x) + 1)`

`= -3/(sqrt1+1) = -3/2`

So `y = -3/2` is an asymptote on the right.

Oblique Left: `y = -2x+3/2` (I haven't shown that `b=3/2`)

There is an oblique asymptote on the left, with slope `-2`, but I haven't worked out its `y` intercept.

`f'(x) = (2x-3)/(2sqrt(x^2-3x)) -1`

And `lim_(xrarr-oo)f'(x) = -2`, so the graph of `f` is approaching linearity.

There is a left oblique asymptote of the form: `y = -2x+b`

We know that the vertical distance between the graph of `f` and the line tends to `0` as `x` decreases without bound, so we need to find `b` such that:

`lim_(xrarr-oo)(f(x) - (-2x+b)) = 0`

`lim_(xrarr-oo)(f(x) - (-2x+b)) = lim_(xrarr-oo)((sqrt(x^2-3x)-x) - (-2x+b))`

`= lim_(xrarr-oo)(sqrt(x^2-3x)+(x-b))` (has form `oo-oo`)

`= lim_(xrarr-oo)((x^2-3x)-(x-b)^2)/(sqrt(x^2-3x)-(x-b))`

`= lim_(xrarr-oo)((x^2-3x)-(x^2-2bx+b^2))/(sqrt(x^2)sqrt(1-3/x)-(x-b))`

`= lim_(xrarr-oo)(-3x +2bx - b^2)/(-xsqrt(1-3/x)-x-b))`

`= lim_(xrarr-oo)(-3 +2b - b^2/x)/(-sqrt(1-3/x)-1-b/x)`

`= (-3+2b)/(-1-1) = (-3+2b)/-2`

In order for this limit to be `0`, we must have `b = 3/2`

So the oblique asymptote on the left is `y = -2x+3/2`