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)

Now y=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:

(-1.7,10.4)
(0,0)
(1.7,-10.4)

Graph of y=x^3-9x.

The following points are indicated with dots:

Local maximum (-1.7,10.4)

Point of inflection (0,0)

Local minimum (1.7,-10.4)

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