Question

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

Answer 1

`"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]}