What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3?

1 Answer
Nov 24, 2015

Answer:

#f(x) = 3x^3-18x^2+33x-18#

Explanation:

To set the zeros of the function, we can first write the function as

#f(x) = (x-1)(x-2)(x-3)#

This gives us #f(1) = f(2) = f(3) = 0#, however #f(0) = -1*-2*-3 = -6#

To fix this, we can multiply by a constant.
#-18 = 3*-6# so we multiply by #3# to obtain

#f(x) = 3(x-1)(x-2)(x-3)#

which has the desired properties. Multiplying that out gives us the cubic function

#f(x) = 3x^3-18x^2+33x-18#