Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `(2x-4)/(x^2-4)`?

Answer 1

`"vertical asymptote at "x=-2`
`"horizontal asymptote at "y=0`

Explanation:

`"let "f(x)=(2x-4)/(x^2-4)`

`"factor numerator/denominator"`

`f(x)=(2cancel((x-2)))/(cancel((x-2))(x+2))=2/(x+2)`

`"the removal of the factor "(x-2)" indicates a removable"`
`"discontinuity (hole) at "x=2`

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+2=0rArrx=-2" is the asymptote"`

`"Horizontal asymptotes occur as"`

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

`"divide terms on numerator/denominator by "x`

`f(x)=(2/x)/(x/x+2/x)=(2/x)/(1+2/x)`

`"as "xto+-oo,f(x)to0/(1+0)`

`y=0" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(2x-4)/(x^2-4) [-10, 10, -5, 5]}