# What are the asymptote(s) and hole(s) of: f(x)=(x^2+x-12)/(x^2-4)?

Vertical Asymptotes at $x = 2 \mathmr{and} x = - 2$
Horizontal Asymptote at $y = 1$;
Vertical asymptote is found by solving the denominator equal to zero. i.e ${x}^{2} - 4 = 0 \mathmr{and} {x}^{2} = 4 \mathmr{and} x = \pm 2$
Horizontal asymptote: Here the degree of numerator and denominator are same. Hence horizontal asymptote $y = \frac{1}{1} = 1$ (numerator's leading co efficient/denominator's leading co efficient)
$f \left(x\right) = \frac{\left(x - 3\right) \left(x + 4\right)}{\left(x + 2\right) \left(x - 2\right)}$Since there is no cancellation , there is no hole.[Ans}