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

1 Answer
Jun 25, 2016

Vertical Asymptotes at #x=2 and x=-2#
Horizontal Asymptote at #y=1#;

Explanation:

Vertical asymptote is found by solving the denominator equal to zero. i.e #x^2-4=0 or x^2=4 or x= +- 2#
Horizontal asymptote: Here the degree of numerator and denominator are same. Hence horizontal asymptote #y= 1/1=1# (numerator's leading co efficient/denominator's leading co efficient)
#f(x)= ((x-3)(x+4))/((x+2)(x-2))#Since there is no cancellation , there is no hole.[Ans}