Question

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

Answer 1

vertical asymptote at x = 1
horizontal asymptote at y = 1

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : `x-1=0rArrx=1" is the asymptote"`

You might prefer to consider y as.

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

The result is the same.

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as `xto+-oo,yto0/(1-0)+1`

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