What are the asymptote(s) and hole(s), if any, of # f(x) =3/x-(8x)/(x^2-3x) #?

1 Answer
Apr 8, 2018

Answer:

Asymptotes: #x=3, x=0, y=0#

Explanation:

#f(x)=3/x-(8x)/(x^2-3x)#

#f(x)=(3(x^2-3x)-8x*x)/(x(x^2-3x)#

For the asymptotes, we look at the denominator.
Since the denominator cannot equal to #0#

ie #x(x^2-3x)=0#

#x^2(x-3)=0#

therefore #x!=0,3 #

For the y asymptotes, we use the limit as #x -> 0#

#lim x->0 (3(x^2-3x)-8x*x)/(x(x^2-3x)#

=#lim x->0 (3x^2-9x-8x^2)/(x(x^2-3x))#

=#lim x->0 (-5x^2-9x)/(x^3-3x^2)#

=#lim x->0 ((-5/x-9/x^2))/(1-3/x)#

=#0#

therefore #y!=0#