How to graph cubics?

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For question a, can somebody please tell me how to get to the answer in the photo?

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Answer 1 · 2017-04-09T05:24:03

#y=x^3-3x^2+3x-1#

The general equation of a cubic polynomial is: #y= ax^3+bx^2+cx+d#

From looking at the graph in the question labelled '(a)', it appears that:

# y=-1# at #x=0# (1)

#y=0# at #x=1# (2)

Slope of #y=0# at #x=1# (3)

#y# has an inflection point at #x=1# (4)

Assuming (1) through (4) above are true:

#(1) -> d=-1# {A]

#(2) -> a+b+c+d =0# [B]

#y' = 3ax^2+2bx+c#

Hence #(3) -> 3a+2b+c =0# [C]

#y'' = 6ax+2b#

Hence #(4) -> 6a+2b=0#

#:. b=-3a# [D}

[D] in [C} #-> 3a-6a+c=0#

#:. c=3a# [E]

[A], [D] and [E] in [B] #-> a -3a+3a -1 =0#

Hence: #a=1# #-> b=-3# and #c=3#

#:.# our cubic is: #y=x^3-3x^2+3x-1#