Asymptotes for a function

Remember that an oblique asymptote is defined by

• $y=mx+n$ with $m,n\in \mathbb{R}$.

with

$$m=\lim_{x\rightarrow \infty}\frac{f(x)}{x}$$

and

$$n=\lim_{x\rightarrow \infty} (f(x)-mx)$$

when both limit exist (at $+\infty$ and/or $-\infty$).

In your case it is easy verify that

$$m=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=\lim_{x\rightarrow \infty} \frac{x^2 - 2ax}{x^2 - ax}=1$$

$$n=\lim_{x\rightarrow \infty} (f(x)-mx)=\lim_{x\rightarrow \infty} \frac{x^2 - 2ax}{x - a}-x=\lim_{x\rightarrow \infty} \frac{x^2 - 2ax-x^2+ax}{x - a}=-a$$

Rewrite the function: $$f(x)=\frac{x^2-2ax}{x-a}=\frac{x^2-2ax+a^2-a^2}{x-a}=\frac{(x-a)^2-a^2}{x-a}=x-a-\frac{a^2}{x-a}$$ Now both the stated asymptotes are obvious: a vertical one ($x=a$) and an oblique one ($y=x-a$).