Question 294585
Vertical asymptotes occur when the denominator goes to zero.
{{{4x^2-1=0}}}
{{{4x^2=1}}}
{{{x^2=1/4}}}
{{{x=1/2}}} and {{{x=-1/2}}}
The vertical asymptotes are shown below in blue. 
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Horizontal asymptotes occur as x goes to infinity.
Divide numerator and denominator by {{{x^2}}}.
{{{f(x)=(2x^2-3x+2)/(4x^2-1)=(2-3/x+2/x^2)/(4-1/x^2)}}}
Now as x goes to infinity,
{{{f(x)=(2-cross(3/x)+cross(2/x^2))/(4-cross(1/x^2))}}}
the crossed out terms go to zero.
{{{lim( x->infinity, f(x) ) 
= 2/4=1/2}}}
The horizontal asymptote would be y=1/2 (shown below in green). 
There are no oblique asymptotes. 
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{{{drawing( 300, 300, -12, 12, -3, 3, blue(line(-0.5,-10,-0.5,10)),blue(line(0.5,-10,0.5,10)),graph( 300, 300, -12, 12, -3, 3, (2x^2-3x+2)/(4x^2-1),1/2) )}}}