How do you identify the horizontal asymptotes for the following function? #f(x)=(x-8)/(x+3)#. #y=#?

1 Answer
1s2s2p
Jun 12, 2018

#y=1#

Explanation:

When it comes to horizontal asymptotes, there is an easy way to find them is you are given #(f(x))/(g(x))# where #f(x)# and #g(x)# are both in the form #A+Bx+Cx^2+Dx^3+cdots#

First, to determine horizontal asymptotes, look at the degree with the highest power in #f(x)# and #g(x)#.

So, in #3x^5-2x^2+3x#, #x_"max"=3x^5#

In #-2x^2+4x-1#, #x_"max"=-2x^2#

  1. #f(x)_text(max)>g(x)_text(max), "no horizontal asymptotes"#

  2. #f(x)_ "max">g(x)_"max", y=(f(x)_text(max))/(g(x)_text(max))#

  3. #f(x)_text(max) < g(x)_text(max), y=0#

Here:
#f(x)=x-8# and #f(x)_text(max)=x#

#g(x)=x+3# and #g(x)_"max"=x#

Since both powers are the same (#x^1#) we use the 2nd rule:
#x/x=1# so the horizontal asymptote will be #y=1#

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