How do you identify all asymptotes or holes for #f(x)=(x^2-1)/(-3x^2+12)#?

1 Answer
Aug 23, 2016

Answer:

vertical asymptotes at x = ± 2
horizontal asymptote at #y=-1/3#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #-3x^2+12=0rArr3x^2=12rArrx^2=4rArrx=+-2#

#rArrx=-2" and " x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x that is #x^2#

#f(x)=(x^2/x^2-1/x^2)/((-3x^2)/x^2+12/x^2)=(1-1/x^2)/(-3+12/x^2)#

as #xto+-oo,f(x)to(1-0)/(-3+0)#

#rArry=-1/3" is the asymptote"#

Holes occur when there are duplicate factors on the numerator/denominator. There are no duplicate factors here hence there are no holes.
graph{(x^2-1)/(-3x^2+12) [-10, 10, -5, 5]}