# 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]}