How do you find vertical, horizontal and oblique asymptotes for #y=(x+1)/(x+2)#?

1 Answer
Apr 16, 2016

Answer:

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]}