Question

The asymptote(s) of the curve $y=3x+\frac{2}{x}$ is/are:

• A. $x=0$
• B. $y=0$
• C. $y=3x+1$
• D. $y=3x$

Answer 1

Answer: (A), (D)

$y=3x$

Given, $y=3x+\frac{2}{x}$, Domain is $R-\left\{0\right\}$
$i)x\to 0^{+}\Rightarrow y\to \infty$
$x\to 0^{-}\Rightarrow y\to -\infty$
$\therefore x=0$ is a vertical asymptote
$ii)x\to \infty \Rightarrow y\to \infty$
$x\to -\infty \Rightarrow y\to -\infty$
So, it may have an oblique asymptote.
Let oblique asymptote be $y=mx+c$
$m=\underset{x\to \pm \infty }{lim}\left(\frac{y}{x}\right)=\underset{x\to \pm \infty }{lim}(3+\frac{2}{x^2})=3$
$c=\underset{x\to \pm \infty }{lim}(y-mx)=\underset{x\to \pm \infty }{lim}(3x+\frac{2}{x}-3x)=0$
$\therefore$ Oblique asymptote is $y=3x$