How do you find the asymptotes for #f(x) = 1 /(x^2-2x-8)#?

1 Answer
May 29, 2016

Answer:

See below.

Explanation:

In this particular case, only two things cause asymptotic behaviour: if the denominator is zero, or as #x# tends to infinity.

Let's study these individually:

If the denominator is zero:
Then #x^2-2x-8=0#
By factorising: #(x-4)(x+2)=0#
Finding solutions: #x=4# or #x=-2#

Therefore, the above lines will cause asymptotic behaviour as the curve approaches. You can substitute values of x such as #x=3.9 or 4.1# to determine whether the curve approaches positive or negative infinity from these points (since limits are usually defined as having a positive and negative side, for example, #lim_(xrarr4^-)f(x)# which would be determined by checking #x=3.9#)

As #x# tends to infinity:
As this occurs, only the term which the largest order will matter (order refers to the power or exponent value). We consider both the top and bottom of the fraction of #f(x)# separately. The highest order of the top is just #1# (#x^0#) and on the bottom this is #x^2#. So as #x# tends to infinity, only the #1/x^2# will affect the value of f(x) significantly. But in this case, the asymptote is #f(x)=0# (or #y=0#) since if #x# tends to infinity then #f(x)# will get increasingly smaller.

So the three asymptote said are #x=4#, #x=-2# and #y=0#.