Question 125252
a)



{{{y=(2x+3)/(x+2))}}} Start with the given function




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



<b> Horizontal Asymptote: </b>

Since the degree of the numerator and the denominator are the same, we can find the horizontal asymptote using this procedure:


To find the horizontal aysmptote, first we need to find the leading coefficients of the numerator and the denominator.


Looking at the numerator {{{2x+3}}}, the leading coefficient is {{{2}}}


Looking at the denominator {{{x+2}}}, the leading coefficient is {{{1}}}


So the horizontal aysmptote is the ratio of the leading coefficients. In other words, simply divide {{{2}}} by {{{1}}} to get {{{(2)/(1)=2}}}



So the horizontal asymptote is {{{y=2}}}






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<b> Vertical Asymptote: </b>

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


{{{x+2=0}}} Set the denominator equal to zero



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



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



So the vertical asymptote is {{{x=-2}}}



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


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




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b)




{{{y=(5x)/(x^2+1))}}} Start with the given function




Looking at the numerator {{{5x}}}, 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 aysmptote, 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}}}Subtract 1 from both sides



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



{{{x=0+-sqrt(-1)}}} 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=(5x)/(x^2+1)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(5x)/(x^2+1)),
blue(line(-20,0,20,0))
)}}} Graph of {{{y=(5x)/(x^2+1))}}}  with the horizontal asymptote {{{y=0}}} (blue line) . Notice how there are no vertical asymptotes