How do you find the vertical, horizontal and slant asymptotes of: #y= (x+2) /( x^2 -4)#?

1 Answer
May 18, 2016

Answer:

vertical asymptote x = 2
horizontal asymptote y = 0

Explanation:

The first step here is to factor and simplify the function.

#rArr(x+2)/(x^2-4)=(x+2)/((x+2)(x-2))#

and cancelling gives

#cancel((x+2))/(cancel((x+2)) (x-2))=1/(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) , y to 0 #

divide terms on numerator/denominator by x

#rArr(1/x)/(x/x-2/x)=(1/x)/(1-2/x)#

as #xto+-oo , y to 0/(1-0)#

#rArry=0" is the asymptote"#

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