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

1 Answer
Dec 26, 2016

Answer:

#"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]}