How do you find vertical, horizontal and oblique asymptotes for #y=(x+1)/(x+2)#?
1 Answer
Apr 16, 2016
vertical asymptote x = -2
horizontal asymptote y = 1
Explanation:
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) to 0 # divide terms on numerator/denominator by x
#(x/x + 1/x)/(x/x + 2/x) = (1 + 1/x)/(1 + 2/x) # as
# xto+-oo , 1/x" and " 2/x to 0 #
#rArr y = 1/1 = 1 " is the asymptote " # Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here , hence there are no oblique asymptotes.
graph{(x+1)/(x+2) [-10, 10, -5, 5]}