How do you find the vertical, horizontal or slant asymptotes for #(3x^2 - 4x + 2) /( 2x^3+3)#?

Question

Lincoln Anderson · Accepted Answer

The vertical asymptote is #x=-(3/2)^(1/3#
No slant asymptote.
The horizontal asymptote is #y=0# Let #f(x)=(3x^2-4x+2)/(2x^3+3)# The domain of #f(x)# is #D_f(x) =RR-{-(3/2)^(1/3)} # As you cannot divide by #0#, #x!=-(3/2)^(1/3# So a vertical asymptote is #x=-(3/2)^(1/3# The degree of the numerator #<# the degree of the denominator, there is To calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator. #lim_(x->-oo)f(x)=lim_(x->-oo)(3x^2)/(2x^3)=lim_(x->-oo)3/(2x)=0^(-)# #lim_(x->+oo)f(x)=lim_(x->+oo)(3x^2)/(2x^3)=lim_(x->+oo)3/(2x)=0^(+)# So, the horizontal asymptote is #y=0# graph{(3x^2-4x+2)/(2x^3+3) [-8.89, 8.885, -4.444, 4.44]}

Greyson Brown · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. The vertical asymptotes occur at these values of x.

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

To find the slant asymptote, perform polynomial long division on the function. The slant asymptote is the quotient obtained from the division when the degree of the numerator is exactly one greater than the degree of the denominator.