How do you identify all asymptotes or holes for #f(x)=-4/(x^2+x-2)#?

1 Answer
Jan 13, 2017

Answer:

vertical asymptotes at x = - 2, x = 1
horizontal asymptote at y = 0

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 : #x^2+x-2=0rArr(x+2)(x-1)=0#

#rArrx=-2" and " x=1" 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)=-(4/x^2)/(x^2/x^2+x/x^2-2/x^2)=-(4/x^2)/(1+1/x-2/x^2)#

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

#rArry=0" is the asymptote"#

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.
graph{-4/(x^2+x-2) [-10, 10, -5, 5]}