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

1 Answer
Sep 6, 2015

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#