How do you find the vertical, horizontal and slant asymptotes of: #y= (x+2) /( x^2 -4)#?
1 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]}