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

##### 2 Answers

The vertical asymptote is

#### Explanation:

The denominator must be

Therefore, the vertical asymptote is

As, the degree of the numerator

Also,

The horizontal asymptote is

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

#### 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]}