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 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]}