Question #19618

1 Answer
Nov 2, 2016

Many answers possible, but one such answer is #y = -3x^3 + 3/2x^2 + 5/2x + 2#.

Explanation:

A cubic polynomial is of the form #y = Ax^3 + Bx^2 + Cx + D #. Knowing the input/output of the function, we can write a system of equations in #4# variables.

#A + B + C + D = 3#

#D = 2#

#-A + B - C + D = 4#

We instantly know that #D = 2#.

So, #A + B + C = 1# and #-A + B - C = 2#. By elimination, we have that :

#2B = 3#

#B = 3/2#

Resubstitute:

#A + C = -1/2#

#-A + -C = 1/2#

#-(A + C) = 1/2#

#A + C = -1/2#

So, all values of A and C that add to #-1/2# will work. Let's take #A = -3# and #C = 5/2#.

So, the function is #y = -3x^3 + 3/2x^2 + 5/2x + 2#.

Checking, you will find that the function passes through the points labelled above.

Hopefully this helps!