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

1 Answer
Jan 1, 2017

Answer:

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]}