Question

How do you find the asymptote and graph `y=1/(x+3)`?

Answer 1

`y` has a vertical asymptote at `x=-3`
`y` has a horizontal asymptote at `y=0`
Graph below.

Explanation:

`y=1/(x+3)`

`y` is defined `forall x !=-3`

Let's examine `y` as `x` approaches `-3` from above and from below.

`lim_(x->-3^+) 1/(x+3) ->+oo`

`lim_(x->-3^-) 1/(x+3) ->-oo`

Therefore, `y` is discontinuous at `x=-3` and has a vertical asymptote at that point.

Now consider:

`lim_(x->-oo) y = 0 and lim_(x->+oo) y =0`

Hence the `x-`axis forms a horizontal asymptote.

Therefore, `y` has a horizontal asymptote at `y=0`

`y` is a rectangular hyperbola with the graph below.

graph{1/(x+3) [-7.58, 4.904, -3.4, 2.843]}