How to graph cubics?

Question

()

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

2601 views around the world

You can reuse this answer
Creative Commons License

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#