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

1 Answer
Nov 27, 2016

Answer:

The vertical asymptote is #x=-(3/2)^(1/3#
No slant asymptote.
The horizontal asymptote is #y=0#

Explanation:

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]}