# Oblique asymptote position

Let $f(x)=\frac{x^2-5x+7}{x-2}$, I have calculated the oblique asymptote and it is: y=x-3.

So, I study the position with the function. For this:

$f(100) = 97'...$ and $f(-100)=-103'...$

This result is not logic, I think that $f(100)$ would be $100'$ of this form, the function would be over the oblique asymptote.

Someone can explain it? Thank you

## 1 Answer

Your division to find the oblique asymptote gives that $\displaystyle f(x)=x-3+\frac{1}{x-2}$, so

you are correct that the graph of the function is above the asymptote when $x>2$, and it is below the asymptote when $x<2$.

For example, $f(100)\approx 97.0102>97=100-3$.