Question 1067666
Degree of numerator will be 2 and degree of denominator will be 3.  You would expect a horizontal asymptote if both of these degrees are the same, but your function is not that way.  Your function f DOES approach 0 for extreme x values, so


y=0 is horizontal asymptote.


{{{graph(300,300,-8,8,-8,8,((3-2x)(x+3))/((2+x)(3-x^2)))}}}


{{{graph(300,300,-15,15,-8,8,((3-2x)(x+3))/((2+x)(3-x^2)))}}}