Question

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

Answer 1

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]}