Question

How do you find the asymptotes for `y=(3/(x-2))+1`?

Answer 1

vertical asymptote x = 2
horizontal asymptote y = 1

Explanation:

Begin by expressing y as a single rational function.

`rArry=3/(x-2)+1=3/(x-2)+(x-2)/(x-2)=(3+x-2)/(x-2)`

Thus y simplifies to. `y=(x+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),ytoc" (a constant)"`

divide terms on numerator/denominator by x

`(x/x+1/x)/(x/x-2/x)=(1+1/x)/(1-2/x)`

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

`rArry=1" is the asymptote"`
graph{(x+1)/(x-2) [-10, 10, -5, 5]}