Question

How do you identify all asymptotes or holes for `f(x)=1/(-2x-2)`?

Answer 1

`"vertical asymptote at " 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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: `-2x-2=0rArrx=-1" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`f(x)=(1/x)/((2x)/x-2/x)=(1/x)/(2-2/x)`

as `xto+-oo,f(x)to0/(2-0)`

`rArry=0" is the asymptote"`

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{1/(-2x-2) [-10, 10, -5, 5]}