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.