Question 145523


{{{y=(x+2)/(x^2-5x+6))}}} Start with the given function




Looking at the numerator {{{x+2}}}, we can see that the degree is {{{1}}} since the highest exponent of the numerator is {{{1}}}. For the denominator {{{x^2-5x+6}}}, 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^2-5x+6=0}}} Set the denominator equal to zero



      Now let's use the quadratic formula to solve for x. If you need help with the quadratic formula, check out this <a href="http://www.algebra.com/algebra/homework/quadratic/quadratic-formula.solver">solver</a>.

      

      After using the quadratic formula, we get the solutions

      {{{x=3}}} or {{{x=2}}}


      So this means the vertical asymptotes are {{{x=3}}} or {{{x=2}}}
      

Notice if we graph {{{y=(x+2)/(x^2-5x+6)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(x+2)/(x^2-5x+6)),
blue(line(-20,0,20,0)),
green(line(3,-20,3,20)),
green(line(2,-20,2,20))
)}}} Graph of {{{y=(x+2)/(x^2-5x+6))}}}  with the horizontal asymptote {{{y=0}}} (blue line)  and the vertical asymptotes {{{x=3}}} and {{{x=2}}} (green lines)