Question

How do you sketch the graph `y=x^3-4x^2` using the first and second derivatives?

Answer 1

`y=x^3-4x^2`

1. General Observations

A cubic function and the cubic coefficient is +ve, so it will have the classic cubic shape with a maximum to the left of a minimum,

2. Roots

`y=0 => x^3 - 4x^2 = 0`
`:. x^2(x-4) = 0`
`:. x=0 "(double root"), x=4`

So we can already deduce that there must be a local maximum at `x=0` that touches the axis (because of the double root) and a local minimum in between `x=0` and `x=4` (or else it could not be a cubic with a +ve cubic coefficient).

3. Turning Points

`y=x^3-4x^2 => y' = 3x^2-8x`
At min/max `y'(0)=0 => 3x^2-8x = 0`
`:. x(3x-8) = 0`
`:. x=0 "(as expected)", x=8/3 (~~2.7)`

When `x=0 => y=0`
When `x=8/3 => y= 512/27-256/9=-245/2 (~~-9.5)`

4. Nature of Turning Points

`y' = 3x^2-8x => y''=6x-8`

When `x=0 => y''<0 => "maximum (as expected)"`
When `x=8/3 => y''>0 => "minimum (as expected)"`

5. The Graph

There is now information to plot the graph, here I will use the actual graph:

graph{x^3-4x^2 [-10, 10, -15, 10]}