Question 133517
Asymptotes:


{{{y=(x^4-16)/(x^2-1)}}} Start with the given function




Looking at the numerator {{{x^4-16}}}, we can see that the degree is {{{4}}} since the highest exponent of the numerator is {{{4}}}. For the denominator {{{x^2-1}}}, we can see that the degree is {{{2}}} since the highest exponent of the denominator is {{{2}}}.



<b> Oblique Asymptote: </b>


Since the degree of the numerator (which is {{{4}}}) is greater than the degree of the denominator (which is {{{2}}}), there is no horizontal asymptote. In this case, there's an oblique asymptote


To find the oblique asymptote, simply use polynomial long division


<a href="http://s150.photobucket.com/albums/s91/jim_thompson5910/?action=view&current=long_div.gif" target="_blank"><img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/long_div.gif" border="0" alt="Photobucket"></a>



So the oblique asymptote is the quotient {{{x^2+1}}} (ignore the remainder). So the equation of the oblique asymptote curve is {{{y=x^2+1}}}



--------------------------------------------------




<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}}} or {{{x=1}}}




-----------------------------------------------


Zeros:


Now let's find the zeros of the equation



{{{y=(x^4-16)/(x^2-1)}}} Start with the given function



{{{0=(x^4-16)/(x^2-1)}}} Plug in {{{y=0}}}



Since the denominator can never be equal to zero, this means that the numerator is equal to zero



{{{(x-2)*(x+2)*(x^2+4)=0}}} Factor the left side




Now set each factor equal to zero:


{{{x-2=0}}}, {{{x+2=0}}} or {{{x^2+4=0}}}


Now solve for x for each factor:


{{{x=2}}}, {{{x=-2}}}, {{{x=-2i}}} or {{{x=2i}}}


So the zeros of {{{y=(x^4-16)/(x^2-1)}}} are {{{x=2}}}, {{{x=-2}}} {{{x=-2i}}} or {{{x=2i}}}




So let's use this information to graph {{{y=(x^4-16)/(x^2-1)}}}



{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(x^4-16)/(x^2-1),100,x^2+1),
green(line(1,-20,1,20)),
green(line(-1,-20,-1,20))
)}}} Graph of {{{y=(x^4-16)/(x^2-1))}}}  with the oblique asymptote {{{y=x^2+1}}} (blue curve)  and the vertical asymptotes {{{x=-1}}} and {{{x=1}}}  (green lines)