Asymptotes for a function $$f(x)=\frac{x^2-2ax}{x-a}$$ where $a$ is positive.
The question asks to give equations for the 2 asymptotes. The first is obvious: $x=a$. The other asymptote is $y=x-a$ but I am unsure how to derive this.
 A: 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
$$m=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=\lim_{x\rightarrow \infty} \frac{x^2 - 2ax}{x^2 - ax}=1$$
$$n=\lim_{x\rightarrow \infty} (f(x)-mx)=\lim_{x\rightarrow \infty}  \frac{x^2 - 2ax}{x - a}-x=\lim_{x\rightarrow \infty}  \frac{x^2 - 2ax-x^2+ax}{x - a}=-a$$
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 ($y=x-a$).
A: Option 1; performing (polynomial) long division, you get:
$$\frac{x^2 - 2ax}{x - a} = \color{blue}{x-a}-\color{red}{\frac{a^2}{x-a}}$$
with the asymptote in blue and the remainder, which goes to $0$ for $x\to \pm\infty$, in red.
Option 2; use limits to find the equation of the oblique asymptote $y=\color{blue}{m}x+\color{red}{q}$, with:
$$\color{blue}{m} = \lim_{x \to \pm \infty}\frac{f(x)}{x}  \quad \mbox{and} \quad \color{red}{q} = \lim_{x \to \pm \infty}\bigl( f(x)-\color{blue}{m}x\bigr)$$
provided that both limits exist (which is the case here: $1$ and $-a$ respectively).
A: use that $f(x)$ can be written as $$f(x)=x-a-\frac{a^2}{x-a}$$
