Question

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

Answer 1

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]}