What is the equation of the oblique asymptote # f(x) = (x^2-5x+6)/(x+4)#?

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Answer 1 · 2015-07-09T14:29:34

The line #y = x-9# is an asymptote on both the left and the right.

Rewrite the function using either Polynomial Long Division or the method below.

#f(x) = (x^2-5x+6)/(x+4)#

By long division of polynomials (or, in this case synthetic division will also work), we can get:

#f(x) = x-9 + 42/(x+4)#.

So #y = x-9# is an asymptote.

Long division is a but tedious (difficult) to format nicely here, so I'll show another way to think of this.

#x(x+4) = x^2 +4x# Use this to rewrite:

#f(x) = (x^2+4x-9x+6)/(x+4)#

Notice that we replaced the #-5x# by #4x-9x#. This is done so that we can group and reduce.

We have:

#f(x) = (x^2+4x-9x+6)/(x+4) = (x^2+4x)/(x+4) + (-9x+6)/(x+4)#

# = x + (-9x+6)/(x+4)#

Now, we'll use #-9(x+4) = -9x-36# to get:

#f(x) = x + (-9x-36+42)/(x+4)# (Replacing #+6# with #-36+42# to get:

#f(x) = x + (-9x-36)/(x+4) +42/(x+4)#

# = x-9 + 42/(x+4)#

Writing:

#f(x) = x-9 + 42/(x+4)# we can see that the difference is:

#f(x) - (x-9) = 42/(x+4)#

As #xrarroo# and as #xrarr -oo#, this difference goes to #0#, so

the line #y = x-9# is an asymptote for #f(x)# (on both sides).