`y = (2x)/ (x^2+2x-15)`

By factoring `x^2+2x-15` you will get `x^2+2x-15 = (x+5)(x-3)`

`y = (2x)/((x+5)(x-3))`

Now you can see that at x = -5 and x = 3 the graph goes to `oo` .

Also we can see that at x = 0 then y = 0. This means the graph goes through origin.

`y = (2x)/((x+5)(x-3))`

`y = (2x)/(x^2(1+5/x)(1-3/x))`

`y = 2/(x(1+5/x)(1-3/x))`

Now it is clear that when `x rarr+-oo` then `y rarr0` .

`lim_(xrarr+-oo)y = 0`

The stationary point of the graph is given by the first derivative.

`y = (2x)/ (x^2+2x-15)`

`dy/dx = ((x^2+2x-15)xx2-2x(2x+2))/(x^2+2x-15)^2`

`dy/dx = (2x^2+4x-30-4x^2-4x)/(x^2+2x-15)^2`

`dy/dx = (-2x^2-30)/(x^2+2x-15)^2`

`dy/dx = -(x^2+15)/(x^2+2x-15)^2`

Here `(x^2+15)>0` always. So the graph has no real stationary points.

*Using the data we can plot the graph.*