`y = x^3-3x^2-9x`

When x = 0 then `y = (x^3-3x^2-9x)_(x=0) = 0`

*So the graph goes through (0,0).*

When y = 0 then;

`x^3-3x^2-9x = 0`

`x(x^2-3x-9) = 0`

`x^2-3x-9 = 0`

Solving x^2-3x-9 = 0 yields;

`x = (-(-3)+-sqrt((-3)^2-4xx1xx9(-9)))/(2xx1)`

x = 4.85 and x = -1.85

** So we have y = 0 or the graph cuts x-axis at `x = 4.85` and **`x = -1.85`

`y = x^3-3x^2-9x`

`dy/dx = 3x^2-6x-9`

At minimums and maximums of the graph `dy/dx = 0`

`3x^2-6x-9 = 0`

`x^2-2x-3 = 0`

`x^2-3x+x-3 = 0`

`x(x-3)+1(x-3) = 0`

`(x-3)(x+1) = 0`

This will happen when;

x=3 and x = -1

Now we know we have maximum/minimum at x = 3 and x = -1.

We have to figure out what exactly is at each point whether maximum or minimum.

If `(d^y)/dx^2 > 0` in the changing point in that point we have a minimum.

If `(d^y)/dx^2 < 0` in the changing point in that point we have a maximum.

`dy/dx = 3x^2-6x-9`

`(d^y)/dx^2 = 6x-6 = 6(x-1)`

Our changing points are x = 3 and x = -1

` ((d^y)/dx^2)_(x=3) = 12 > 0`

So at x = 3 we have a minimum point in the graph.

`((d^y)/dx^2)_(x=-1) = -12 > 0`

So at x = -1 we have a maximum point in the graph.

When x = 3 then ` y = 3^3-3xx3^2-9xx3 = -27`

When x = -1 then y = 5

*So the maximum is at (-1,5) and minimum is at (3,-27)*

`y = x^3-3x^2-9x`

`y = x^3(1-3/x-9/x^2)`

When `xrarr -oo` ;

`y = lim_(xrarroo)x^3(1-3/x-9/x^2) = -oo(1-3/(-oo)-9/(-oo)`

`y = -oo(1-0-0) = -oo`

*When `xrarr -oo` then `y = -oo` *

*Similarly we can show that when `x rarr oo` then `y = oo` *

No using the above facts you can easily graph the function. Below I have shown the look of it.

You can see everything that we have obtained above, in the graph.