Question 729476
{{{y=(x+1)/(2x-4)}}}


Asymptotes of a hyperbola are the lines that pass through center of the hyperbola and touch the hyperbola.

{{{y=(x+1)/(2x-4)}}}....since denominator cannot be zero we will find that value of {{{x}}} that makes it equal to zero

{{{2x-4=0}}}...=>...{{{2x=4}}}...=>...{{{x=2}}}...this is {{{x}}} coordinate of the center and asymptote

now find {{{y}}} value for center

{{{y=(x+1)/2(x-2)}}}....=>..{{{y=(1/2)((x+1)/(x-2))}}}

so, {{{y}}}  coordinate is {{{1/2}}}


{{{ drawing(500, 500, -10, 10, -10, 10,   blue(line(2,10,2,-10)),graph( 500, 500, -10, 10, -10, 10,(x+1)/(2x-4),1/2)) }}}