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]}