Question

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

Answer 1

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]}

Answer 2

`"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]}