How do you identify all asymptotes or holes for #f(x)=(x^2+2x-8)/(-4x)#?

1 Answer
Nov 26, 2016

Answer:

vertical asymptote at x = 0
slant asymptote #y=-1/4x-1/2#

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

Slant asymptotes occur when the degree of the numerator > degree of the numerator. This is the case here ( numerator-degree 2, denominator-degree 1 ) Hence there is a slant asymptote.

#f(x)=(x^2+2x-8)/(-4x)=x^2/(-4x)+(2x)/(-4x)-8/(-4x)#

#rArrf(x)=-1/4x-1/2+2/x#

as #xto+-oo,f(x)to-1/4x-1/2+0#

#rArry=-1/4x-1/2" 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{(x^2+2x-8)/(-4x) [-10, 10, -5, 5]}