Question

Number of asymptotes of the function, $f\left(x\right)=\frac{x+3}{x^2+9}$ is

• A. $1$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is A $1$

$y=(\frac{x+3}{x^2+9})$

for vertical asymptote, find the point where denominator equal to $0$

here $x^2+9$ is always greater than $0$, hence there is no vertical asymptote

$horizontal$ $asymptote$

line $y=L$ is a horizontal asymptote of fuction if either ${}_{x\to \infty }^{lim}f\left(x\right)=L$ and ${}_{x\to -\infty }^{lim}f\left(x\right)=L$

where $L$ is finite

${}_{x{\to }_{-}^{+}\infty }^{lim}(\frac{x+3}{x^2+9})=0$

hence horizonatal asymptote is $y=0$

$slant$ $asymptote$

since the degree of the numerator is less than the denominator, hence there is no slant asymptote

no. of asymptote $=1$