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

1 Answer
Dec 4, 2016

Answer:

The oblique asymptote is #y=x#
No vertical and horizontal asymptotes

Explanation:

The domain of #f(x)# is #D_f(x)=RR#

The denominator #x^2+9>0#, #AAx in RR#

So there are no vertical asymptotes.

As the degree of the numerator is #># degree of the denominator, there is an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##x^3+8##color(white)(aaaa)##∣##x^2+9#

#color(white)(aaaa)##x^3+9x##color(white)(aaa)##∣##x#

#color(white)(aaaa)##0-9x+8#

So,

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

So, the oblique asymptote is #y=x#

To calculate the limits as #x-+-oo#, we take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)f(x)=lim_(x->+-oo)x^3/x^2=lim_(x->+-oo)x=+-oo#

graph{(y-(x^3+8)/(x^2+9))(y-x)=0 [-5.546, 5.55, -2.773, 2.774]}