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

1 Answer
Jun 10, 2016

Answer:

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]}