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

1 Answer
May 20, 2017

Answer:

#"vertical asymptote at " x=0#
#"oblique 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"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote.

#"dividing numerator by denominator gives"#

#f(x)=-1/4x+1/2-8/(-4x)=-1/4x+1/2+2/x#

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

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