Question 369295
<pre>
You're a little mixed up about how to find horizontal asymptotes.
You don't set the numerator equal to 0 to find the horizontal
asymptote.  Apparently that's what you were thinking.  You need to
review how to find horizontal asymptotes.

{{{y=(x-1)/(2x^2+5x-3)}}}

You found vertical asymptotes OK:

Set denominator = 0

{{{2x^2+5x-3 = 0}}}
Factor:
{{{(2x-1)(x+3)=0}}}

{{{2x-1=0}}}
{{{2x=1}}}
{{{x=1/2}}}  That's the equation of one vertical asymptote

{{{x+3=0}}}
{{{x=-3}}}  That's the equation of the other asymptote

Here they are:

{{{drawing(400,400,-5,5,-5,5, green(line(1/2,-11,1/2,11),line(-3,-11,-3,11)),
graph(400,400,-5,5,-5,5)  )}}} 

Since the largest exponent in the top is smaller than the exponent in the
bottom, the x-axis, whose equation is y=0, is the horizontal asymptote:

{{{drawing(400,400,-5,5,-5,5, green(line(1/2,-11,1/2,11),line(-3,-11,-3,11),

line(-11,0,11,0)),
graph(400,400,-5,5,-5,5,(x-1)/(2x^2+5x-3))  )}}} 

It's hard to tell what the graph does on the right. It crosses the
x-axis at 1, it goes up just a tiny bit (less than one tenth),
then it goes back down and approaches the x-axis on the right.

Edwin</pre>