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

1 Answer
Jun 2, 2016

Answer:

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]}