Because the degree of the numerator is exactly #1# more than the degree of the denominator, there is an oblique asymptote. To find it, divide # (x^2 - 5x + 6)/(x-4)#
I don't have a great format for long division for Socratic, but this may help:
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
What do we need to multiply #x# by, to get #x^2#? We need to multiply by #x#
#" " " " " " "x#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
Now multiply #x# times the divisor, #x-4#, to get #x^2-4x# and write that under the dividend.
#" " " " " " "# #x#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
#" " " " " "# #x^2-4x#
#" " " "##-----#
Now we need to subtract #x^2-4x# from the dividend. (Many find it simpler to change the signs and add.)
#" " " " " " "# #x#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
#" " " " " "# #x^2-4x#
#" " " "##-----#
#" " " " " "" "# #-x# #+6#
Now, what do we need to multiply #x# (the first term of the divisor) by to get #-x#? We need to multiply by #-1#
#" " " " " " "# #x# #-1#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
#" " " " " "# #x^2-4x#
#" " " "##-----#
#" " " " " "" "# #-x# #+6#
Do the multiplication: #-1xx(x-4)# and write the result underneath:
#" " " " " " "# #x# #-1#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
#" " " " " "# #x^2-4x#
#" " " "##-----#
#" " " " " "" "# #-x# #+6#
#" " " " " "" "# #-x# #+4#
Now subtract (change the signs and add), to get:
#" " " " " " "# #x# #-1#
#" " " "##-----#
#x-4 |# #x^2# #-5x# #+6#
#" " " " " "# #x^2-4x#
#" " " "##-----#
#" " " " " "" "# #-x# #+6#
#" " " " " "" "# #-x# #+4#
#" " " "##-----#
#" " " " " "" "" " " "# #+2#
Now we see that:
#(x^2-5x+6)/(x-4) = x-1+2/(x-4)# The difference (subtraction) between #y=f(x)# and the line #y=x-1# is the remainder term: #2/(x-4)#.
As #x# gets very very large, whether positive or negative, this difference gets closer and closer to #0#. So the graph of #f(x)# gets closer and closer to the line #y=x-1#
The line #y=x-1# is an oblique aymptote for the graph of #f#.
