SOLUTION

The basic shape of a cubic is:

Keeping this in mind helps with the sketching process.

1. x-intercepts:

when x = 0, x = -3 and x = 3

2. y-intercepts:

When x = 0, y = 0.

3. maxima and minima?

dy/dx = 0 when x = -√3 or x = √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:

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:

Now

d²y/dx² = 0 when x = 0

and

d²y/dx² changes sign from negative (concave down) to positive (concave up) as x passes through 0.

So we are ready to sketch the curve: