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

1 Answer
May 18, 2016

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]}