Question 982454
<pre>
[Edwin McCravy and AnlytcPhil are the same person!]

{{{g(x)}}}{{{""=""}}}{{{(1-x)(2+3x)/(2x^2+1)}}}

We multiply the numerator out:

{{{g(x)}}}{{{""=""}}}{{{(1+x-3x^2)/(2x^2+1)}}}

Place the numerator in descending order of powers of x:

{{{g(x)}}}{{{""=""}}}{{{(red(-3)x^2+x+1)/(red(2)x^2+1)}}}

Since the degree of the numerator is 2 and the degree
of the denominator is also 2, the equation of the
horizontal asymptote is 

{{{y}}}{{{""=""}}}{{{red(-3)/red(2)}}}

{{{y}}}{{{""=""}}}{{{-3/2}}}

{{{graph(400,1200/7,-7,7,-3,3,(1-x)(2+3x)/(2x^2+1),

(-3/2)*sqrt(sin(9x))/sqrt(sin(9x))

) }}}

The green line is the horizontal asymptote.  It's hard to tell but on
the left the graph actually crosses its horizontal asymptote at x= -3.5,
goes a tiny bit below it and then approaches it from below.  On the right
the graph only approaches the horizontal asymptote from above.

Edwin</pre>