How do you find intercepts, extrema, points of inflections, asymptotes and graph `y=x/(x^2+1)`?

The domain of the function is the entire `RR`, as the denominator of the rational function is always `>0`.

We have that:

`lim_(x->-oo) x/(x^2+1) = 0`

`lim_(x->+oo) x/(x^2+1) = 0`

We can see that `y(x)` has the line `y=0` as horizontal asymptote on both sides. We can also see that:

`y(x) < 0` for `x<0`
`y(x) > 0` for `x>0`
`y(x) = 0` for `x=0`

So `x=0` is the only intercept.

`y'(x) = frac(1-x^2) ((x^2+1)^2)`

As the denominator of `y'(x)` is always positive the function is differentiable everywhere and:

`y'(x) <0` for `x in (-oo,-1)` and `in in (1,+oo)`
`y'(x) >0` for `x in (-1,1)`
`y'(x) = 0` for `x=+-1`

Therefore `y(x)` starts from `y=0` at `x->-oo` and decreases until `x=-1` where it reaches a local minimum `y(-1)=-1/2`. It then increases until `x=1` (changing sign at `x=0`) where it reaches a local maximum at `y(1) = 1/2`, and then decreases approaching zero indefinitely as `x->+oo`

`y''(x) = frac (2x(x^2-3)) ((x^2+1)^3)`

so inflection points are: `x=0` and `x=+-sqrt(3)` and the concavity is determined by the sign of `y''(x)`:

for `x in (-oo, -sqrt(3)), y''(x) <0, y(x)` is concave down
for `x in (-sqrt(3),0), y''(x) >0, y(x)` is concave up
for `x in (0, sqrt(3)), y''(x) <0, y(x)` is concave down
for `x in (sqrt(3),+oo), y''(x) >0, y(x)` is concave up