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.
