The **basic shape** of a cubic is:

Keeping this in mind helps with the sketching process.

**1. x-intercepts:**

`{: (y,=x^3-9x),(,=x(x^2-9)),(,=x(x+3)(x-3)),(,=0) :}`

when

`x = 0`, `x = -3` and `x = 3`

**2. y-intercepts:**

When `x = 0`, `y = 0`.

**3. maxima and minima?**

`{: (dy/dx,=3x^2-9),(,=3(x^2-3)),(,=3(x+sqrt(3))(x-sqrt(3))),(,=0) :}`

when

`x=-sqrt(3)` or `x=sqrt(3)`

So we have max or min at approximately `(-1.7,10.4)` and `(1.7,-10.4)`.

[We could check which is which by trying some points near `-1.7` and `+1.7` to determine what the sign changes are. But we need to find the second derivative anyway for points of inflection, so we use that to determine max or min.]

**4. Second derivative:**

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

If `x = -1.7`, `y^('') < 0`, so MAX at `(-1.7,10.4)`

If `x = +1.7`, `y^('') > 0`, so MIN at `(1.7,-10.4)`

**5. Point of inflection:**

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

Now

`(d^2y)/(dx^2)=0` when `x = 0`

and

`(d^2y)/(dx^2)`

changes sign from negative (concave down) to positive
(concave up) as *x* passes through `0`.

So we are ready to sketch the curve:

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