How do you sketch the graph y=x^3-4x^2 using the first and second derivatives?
1 Answer
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 betweenx=0 andx=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/maxy'(0)=0 => 3x^2-8x = 0
:. x(3x-8) = 0
:. x=0 "(as expected)", x=8/3 (~~2.7) When
x=0 => y=0
Whenx=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)"
Whenx=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]}