Question

The asymptotes of the curve $y=\frac{x^2+2x-1}{x}$ are

• A. $x=0$
• B. $y=0$
• C. $y=x+2$
• D. $x=y+2$

Answer 1

Answer: (A), (C)

$y=x+2$

Given, $y=\frac{x^2+2x-1}{x},$ Domain is $R-\left\{0\right\}$
$\underset{x\to \infty }{lim}y=\infty$ and $\underset{x\to -\infty }{lim}y=-\infty$
So, no horizontal asymptote
$\underset{x\to 0^{+}}{lim}y=-\infty$ and $\underset{x\to 0^{-}}{lim}y=\infty$
$\therefore x=0$ is vertical asymptote
Let $y=mx+c$ be an oblique asymptote (if it exists)
$m=\underset{x\to \pm \infty }{lim}\left(\frac{y}{x}\right)=\underset{x\to \pm \infty }{lim}\left(\frac{x^2+2x-1}{x^2}\right)=1$
$c=\underset{x\to \pm \infty }{lim}(\frac{x^2+2x-1}{x}-x)=\underset{x\to \pm \infty }{lim}\left(\frac{2x-1}{x}\right)=2$
$\therefore y=x+2$ is an oblique asymptote.