Question

Vertical asymptotes of the function, $y=\frac{x^2-x-6}{x^2-9}$ is/are

• A. $x=-2$
• B. $x=3$
• C. $x=-3$
• D. $x=\pm 3$

Answer 1

The correct option is C $x=-3$

Degree of numerator=2
Degree of denominator=2
Hence
Degree of numerator $-$ Degree of denominator $=0$.
Hence $y=y=\frac{coefficientof{x}^{2}inthenumerator}{coefficientof{x}^{2}inthenumerator}=1$. Therefore $y=1$ is the horizontal asymptote. Now for vertical asymptote $\frac{x^2-x-6}{x^2-9}=\frac{(x-3)(x+2)}{(x-3)(x+3)}=\frac{x+2}{x+3}$. Now we get a vertical asymptote when the denominator is 0. Hence the vertical asymptote is $x+3=0$ or $x=-3$.