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

Question

Sophie Adams · Accepted Answer

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

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#.

James Allen · Answer

undefined To find the vertical asymptotes, determine where the denominator equals zero and ensure the numerator is not zero at those points.

To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

To find slant asymptotes, divide the numerator by the denominator using polynomial long division. The quotient will be the equation of the slant asymptote.