How do you sketch the graph of #f(x)=x^3-3x^2#?

Calculate derivative :

#f'(x) = 3 x^2 - 6 x = 3x (x-2)#

So you can study the sign of #f'(x)# :

• #f'(x) < 0 iff x in ]0,2[#
• #f'(x) > 0 iff x in ]-oo, 0[ cup ]2,+oo[#.

You get now the variations of #f# :
- #f# is decreasing on #[0,2]#
- #f# is creasing on #[-oo, 0]# and on #[2,+oo]#.

Remark that #f# has :
- local maximum in 0 with #f(0)=0#
- local minimum in 2 with #f(2) =-4#.
There are two horizontal tangents at 0 and at 2.

• For the limits, apply the rule :
  #lim_{x->-oo} x^3-3x^2 = lim_{x->-oo} x^3 = -oo#
  #lim_{x->+oo} x^3-3x^2 = lim_{x->+oo} x^3 = +oo#

Finally :

graph{x^3 - 3x^2 [-8.42, 13.78, -6.62, 4.48]}

You can start by setting #x=0# that gives you #y=f(0)=0# so your curve passes through the origin.
Setting #y=0# you get #x^3-3x^2=0# that gives #x=0# and #x=3#
When #x->+oo# #f(x)->+oo# as well while when #x->-oo# then #f(x)->-oo#.
Points of maximum or minimum are found by setting the first derivative equal to zero:
#f'(x)=3x^2-6x#
and
#3x^2-6x=0# gives: #3x(x-2)=0#
and #x=0# and #x=2#
When #x=0# #y=0# and when #x=2# #y=-4#
Setting the second derivative equal to zero will give us inflection point(s):
#f''(x)=6x-6#
and
#6x-6=0#
so that #x=1# and #y=-2#
And finally: