How do you find vertical, horizontal and oblique asymptotes for #(x^3+8) /(x^2+9)#?

1 Answer
May 19, 2016

Answer:

#(x^3+8)/(x^2+9)# has an oblique asymptote #y=x# and no other asymptotes.

Explanation:

For all Real values of #x#, the denominator #x^2+9 >= 9 > 0#.

So there are no vertical asymptotes.

#(x^3+8)/(x^2+9) = (x^3+9x-9x+8)/(x^2+9)= (x(x^2+9)-9x+8)/(x^2+9)=x+(-9x+8)/(x^2+9)#

Then since the denominator has higher degree than the numerator:

#(-9x+8)/(x^2+9)->0# as #x->+-oo#

Hence:

#(x^3+8)/(x^2+9) = x+(-9x+8)/(x^2+9)#

is asymptotic to #x# as #x->+-oo#

So #f(x)# has no vertical or horizontal asymptotes, but has one oblique asymptote:

#y = x#

graph{(y-(x^3+8)/(x^2+9))(y-x) = 0 [-20, 20, -10, 10]}