Question 846777
{{{f(x) =( 8x^2+2x-3)/ (x^2-4x)}}}
{{{f(x)=((2x-1)(4x+3))/(x(x-4))}}}
Vertical asymptotes occur when denominator goes to zero.
{{{x(x-4)=0}}}
Two solutions:
{{{x=0}}}
and
{{{x-4=0}}}
{{{x=4}}}
Horizontal asymptote : Divide numerator and denominator by {{{x^2}}} and take the limit as {{{x->infinity}}}
{{{lim(x->infinity,( (8+2/x-3/x^2))/ ((1-4/x)))=(8+0-0)/(1-0)=8}}}
There is a horizontal asymptote at {{{y=8}}}
{{{drawing(300,300,-20,20,-20,20,grid(1),green(line(4,-20,4,20)),green(line(0,20,0,-20)),green(line(-20,8,20,8)),graph(300,300,-20,20,-20,20,( 8x^2+2x-3)/ (x^2-4x)))}}}