How do you find the vertical, horizontal or slant asymptotes for #(3+2x^2+5x^3) /( 3x^3-8x)#?

1 Answer
Feb 27, 2016

Answer:

Vertical asymptotes are #x=0#, #x=-sqrt(8/3)# and #x=sqrt(8/3)# and horizontal asymptote is #y=5/3#.

Explanation:

To find all the asymptotes for function #y=(3+2x^2+5x^3)/(3x^3-8x)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #(3x^3-8x)=0#.

One factor is obviously #x# as #3x^3-8x=3x(x^2-8/3)#.

Other factors are obviously #(x+sqrt(8/3))(x-sqrt(8/3))# and hence

Vertical asymptotes are #x=0#, #x=-sqrt(8/3)# and #x=sqrt(8/3)#.

As the highest degree of numerator #5x^3# and denominator #3x^3# are equal, we have a horizontal asymptote #y=5/3#.