Question

Asymptote(s) of the function, $f\left(x\right)=\frac{x^2+3x+1}{4x^2-9}$ is/are

• A. $x=\frac{3}{2}$
• B. $x=-\frac{3}{2}$
• C. $y=\frac{1}{4}$
• D. $y=-\frac{1}{4}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x=\frac{3}{2}$
B $x=-\frac{3}{2}$
C $y=\frac{1}{4}$

Degree of numerator=2
Degree of denominator=2
Hence
Degree of numerator-Degree of denominator=0.
Hence $y=\frac{coefficientof{x}^{2}inthenumerator}{coefficientof{x}^{2}inthenumerator}=\frac{1}{4}$. Therefore $y=\frac{1}{4}$ is the horizontal asymptote. Now for vertical asymptote the denominator has to be 0. Therefore, $4x^2-9=0$ or $x=\frac{\pm 3}{2}$. Hence in total there are 3 asymptotes.