How do you find the asympotes for #f (x)= (5x-1)/(x^2 + 9)#?

1 Answer
Jul 10, 2015

Answer:

This one has only a horizontal asymptote.

Explanation:

The vertical asymptote would happen if the numerator would go to #=0#. Since #x^2# is always non-negative, the numerator will always be at least #9#

The horizontal asymptote can be found by making #x# larger and larger. The #+1# and #-9# then make less and less of a difference and the whole thing will tend to look like:
#(5x)/x^2~~5/x#
As #x# gets larger #5/x# gets smaller, or:

#lim_(x->+-oo) f(x)=0#

In fact this is not a real asymptote, as the value of #f(x)=0# is also reached when #5x-1=0->x=1/5# (see graph)
graph{(5x-1)/(x^2+9) [-10, 10, -5, 5]}