Question 171886


{{{y=(3x)/(x^2+4))}}} 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+4}}}, 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+4=0}}} Set the denominator equal to zero



{{{x^2=0-4}}}Subtract 4 from both sides



{{{x^2=-4}}} Combine like terms on the right side



{{{x=0+-sqrt(-4)}}} Take the square root of both sides

           

Since you CANNOT take the square root of a negative number, the answer is not a real number. So in this case, there are no vertical asymptotes
  
           
           
           
           

Notice if we graph {{{y=(3x)/(x^2+4)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(3x)/(x^2+4)),
blue(line(-20,0,20,0))
)}}} Graph of {{{y=(3x)/(x^2+4))}}}  with the horizontal asymptote {{{y=0}}} (blue line)