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

1 Answer
Dec 20, 2016

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]}