Question

How do you find the vertical, horizontal and slant asymptotes of: `y = (x+1)/(x-3)`?

Answer 1

vertical asymptote x = 3
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 - 3 = 0 → x = 3 is the asymptote

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by x

`(x/x+1/x)/(x/x-3/x)=(1+1/x)/(1-3/x)`

as `xto+-oo,yto(1+0)/(1-0)`

`rArry=1" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(x+1)/(x-3) [-10, 10, -5, 5]}