Question 115443
{{{f(x)=(2x-3)/(x^2+2)}}}


Rational functions have a vertical asymptote where the denominator goes to zero.  But the denominator of this function has no real zeros because there is no real number x such that {{{x^2=-2}}}.  Therefore there are no vertical asymptotes.


To investigate the possibility of a horizontal asymptote, examine the function in terms of what happens when x gets very large.  We can see that the denominator is going to get larger much faster than the numerator in this case, so the function is going to tend to zero.  Since {{{x=3/2}}} is a zero of the function, the function will be positive on the interval ({{{3/2}}},{{{infinity}}}) and will be negative on the interval ({{{-infinity}}},{{{3/2}}})
So we know that as x decreases without bound, the function will tend to zero, but be negative.


What I have given you is an intuitive look at the problem.  To actually determine and prove the horizontal asymptote to this function you need to use the concept of a limit from Calculus.  In this case, we need to use the general rule that:


If g(x) and h(x) are functions such that the degree of g is smaller than the degree of h, then {{{lim(x->infinity,(g(x)/h(x)))=0}}}.  Since the degree of {{{2x-3}}} is 1 and the degree of {{{x^2+2}}} is 2, we can say that {{{lim(x->infinity,((2x-3)/(x^2+2)))=0}}}.  This is sufficient to establish the horizontal asymptote at {{{y=0}}}.  We also know that this is the only horizontal asymptote because rational functions have at most 1 horizonal asymptote.


{{{graph(600,600,-10,10,-10,10,(2x-3)/(x^2+2))}}}