Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)=(2x-4)/(x^2-4)`?

Answer 1

vertical asymptote x = -2
horizontal asymptote y = 0

Explanation:

The first step here is to factorise and simplify f(x).

`rArrf(x)=(2x-4)/(x^2-4)=(2cancel((x-2)))/(cancel((x-2))(x+2))=2/(x+2)`

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x + 2 = 0 → x = -2 is the asymptote

Horizontal asymptotes occur as `lim_(xto+-oo),f(x)to0`

divide terms on numerator/denominator by x

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

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

`rArry=0" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here (numerator-degree 0 , denominator-degree 1). Hence there are no slant asymptotes.
graph{2/(x+2) [-10, 10, -5, 5]}