Question 172180


{{{y=(3x^2+4x-12)/(x^2-16))}}} Start with the given function




Looking at the numerator {{{3x^2+4x-12}}}, we can see that the degree is {{{2}}} since the highest exponent of the numerator is {{{2}}}. For the denominator {{{x^2-16}}}, 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 and the denominator are the same, we can find the horizontal asymptote using this procedure:


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


Looking at the numerator {{{3x^2+4x-12}}}, the leading coefficient is {{{3}}}


Looking at the denominator {{{x^2-16}}}, the leading coefficient is {{{1}}}


So the horizontal asymptote is the ratio of the leading coefficients. In other words, simply divide {{{3}}} by {{{1}}} to get {{{(3)/(1)=3}}}



So the horizontal asymptote is {{{y=3}}}






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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^2-16=0}}} Set the denominator equal to zero



{{{x^2=0+16}}}Add 16 to both sides



{{{x^2=16}}} Combine like terms on the right side



{{{x=0+-sqrt(16)}}} Take the square root of both sides



{{{x=sqrt(16)}}} or {{{x=-sqrt(16)}}}  Break up the "plus/minus" to get two equations



{{{x=4}}} or {{{x=-4}}}  Take the square root of 16 to get 4




So the vertical asymptotes are the equations {{{x=4}}} and {{{x=-4}}}

           

Notice if we graph {{{y=(3x^2+4x-12)/(x^2-16)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(3x^2+4x-12)/(x^2-16)),
blue(line(-20,3,20,3)),
green(line(4,-20,4,20)),
green(line(-4,-20,-4,20))
)}}} Graph of {{{y=(3x^2+4x-12)/(x^2-16))}}}  with the horizontal asymptote {{{y=3}}} (blue line)  and the vertical asymptotes {{{x=4}}} and {{{x=-4}}}  (green lines)