Asymptote problems?

Find all the asymptote of the function #y=(sqrt(x^3+3)-x^3-x^2 )/(x^2-x)#

1 Answer
Conor M. · Shwetank Mauria
Apr 16, 2018

Please see below.

Explanation:

You know that #x# cannot equal #0# because the denominator is #x(x-1)#, but before drawing that vertical asymptote you should check if it makes the numerator #0#.

In that case there would be a hole in the graph. By doing this you find that the numerator wouldn't equal #0# at #x=0# and so you can draw the vertical asymptote at #x=0#.

Doing this process with #x=1# and you find that there isn't a vertical asymptote but a hole at #x=1# as numerator too is #0#.

There is a slant asymptote #y=-x-1# as

#f(x)=(sqrt(x^3+3)-x^3-x^2 )/(x^2-x)=(sqrt(1/x+3/x^4)-x-1 )/(1-1/x)#

and as #x->oo# #y->-x-1#.

Note that the domain cannot go lower than the cube root of #-3# i.e. it cannot be less than #-1.4422#

graph{(sqrt(x^3+3)-x^3-x^2 )/(x^2-x) [-18.3, 21.7, -11.44, 8.56]}

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