Question

For the curve $y=\frac{3x^3+x^2+x-5}{x^2-1},$ which of the following is NOT true?

• A. curve has one oblique asymptote
• B. curve has no horizontal asymptote
• C. curve has two vertical asymptotes
• D. curve has total even number of asymptotes.

Answer 1

The correct option is C curve has two vertical asymptotes

$y=\frac{3x^3+x^2+x-5}{x^2-1}$
$\Rightarrow y=\frac{(x-1)(3x^2+4x+5)}{(x-1)(x+1)}$
$\underset{x\to -1^{+}}{lim}y\left(x\right)=\infty$
$\Rightarrow x=-1$ is a vertical asymptote.
$\underset{x\to 1}{lim}y\left(x\right)=6$
$\Rightarrow x=1$ is not a vertical asymptote.

Also, $\underset{x\to \pm \infty }{lim}\frac{3x^3+x^2+x-5}{x^2-1}=\pm \infty$ (which is not a fixed value)
So, there is no horizontal asymptote.

For oblique asymptotes :
Let $y=mx+c$ be oblique asymptote.
Then $m=\underset{x\to \pm \infty }{lim}\frac{f\left(x\right)}{x}$
$\Rightarrow m=\underset{x\to \pm \infty }{lim}\frac{3x^3+x^2+x-5}{x^2-1}\cdot \frac{1}{x}=3$
and $c=\underset{x\to \pm \infty }{lim}\left(f\right(x)-mx)$
$\Rightarrow c=\underset{x\to \pm \infty }{lim}\frac{3x^3+x^2+x-5}{x^2-1}-3x=1$
So, the only oblique asymptote is $y=3x+1$