How do you identify all asymptotes or holes for #f(x)=(-x+1)/(x+4)#?

1 Answer
Sep 17, 2016

Answer:

vertical asymptote at x = - 4
horizontal asymptote at y = - 1

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: #x+4=0rArrx=-4" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=((-x)/x+1/x)/(x/x+4/x)=(-1+1/x)/(1+4/x)#

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

#rArry=-1" 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{(-x+1)/(x+4) [-20, 20, -10, 10]}