What are the asymptotes for #f(x) = (x+3)/(x-3)#?

2 Answers
Jul 3, 2018

Answer:

The vertical asymptote is #x=3# and the horizontal asymptote is #y=1#

Explanation:

The denominator must be #!=0#

#x-3!=0#

#x!=3#

Therefore, the vertical asymptote is #x=3#

As, the degree of the numerator #=# to the degree of the denominator, There is no slant asymptote.

Also,

#lim_(x->+oo)(x+3)/(x-3)=lim_(x->+oo)(x/x)=1#

#lim_(x->-oo)(x+3)/(x-3)=lim_(x->-oo)(x/x)=1#

The horizontal asymptote is #y=1#

graph{(y-(x+3)/(x-3))(y-1)=0 [-18.02, 18.03, -9.01, 9.01]}

Jul 3, 2018

Answer:

#"vertical asymptote at "x=3"#
#"horizontal asymptote at "y=1#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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-3=0rArrx=3" is the asymptote"#

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" (a constant)"#

#"divide terms on numerator/denominator by "x#

#f(x)=(x/x+3/x)/(x/x-3/x)=(1+3/x)/(1-3/x)#

#"as "xto+-oo,f(x)to(1+0)/(1-0)#

#y=1" is the asymptote"#
graph{(x+3)/(x-3) [-20, 20, -10, 10]}