Question

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

• A. $x=0$
• B. $y=0$
• C. Both
• D. None

Answer 1

Answer: (B)

$y=0$

Degree of numerator=1
Degree of denominator=2
Hence
Degree of numerator $-$ Degree of denominator $=-1.$
Hence $y=0$. Therefore $y=0$ is the horizontal asymptote. Now for vertical asymptote the denominator has to be 0. Therefore, $x^2+9=0$. Hence for no real values of x is $x^2+9=0$. Hence there is only one asymptote and that is $y=0$.