Question 140335
{{{y=(3x)/(x^2-1))}}} Start with the given function




Looking at the numerator {{{3x}}}, we can see that the degree is {{{1}}} since the highest exponent of the numerator is {{{1}}}. For the denominator {{{x^2-1}}}, we can see that the degree is {{{2}}} since the highest exponent of the denominator is {{{2}}}.



<b> Horizontal Asymptote: </b>


Since the degree of the numerator (which is {{{1}}}) is less than the degree of the denominator (which is {{{2}}}), the horizontal asymptote is always {{{y=0}}}


So the horizontal asymptote is {{{y=0}}}




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<b> Vertical Asymptote: </b>

To find the vertical asymptote, just set the denominator equal to zero and solve for x


{{{x^2-1=0}}} Set the denominator equal to zero



{{{x^2=0+1}}}Add 1 to both sides



{{{x^2=1}}} Combine like terms on the right side



{{{x=0+-sqrt(1)}}} Take the square root of both sides



{{{x=1}}} or {{{x=-1}}} Simplify


So the vertical asymptotes are {{{x=1}}} and {{{x=-1}}} 

           

Notice if we graph {{{y=(3x)/(x^2-1)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(3x)/(x^2-1)),
blue(line(-20,0,20,0)),
green(line(1,-20,1,20)),
green(line(-1,-20,-1,20))
)}}} Graph of {{{y=(3x)/(x^2-1))}}}  with the horizontal asymptote {{{y=0}}} (blue line)  and the vertical asymptotes {{{x=1}}} and {{{x=-1}}}   (green lines)