Question

Find a third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3. Using complete sentences, explain how you found it.

Answer 1

since the zeros are given, then you know what the polynomial is in factored form... can you give the factors?

Answer 2

no

Answer 3

a zeros is where the function crosses the x-axis...
and since the zeros of the polynomial is given as x=1, 2, 3, then your polynomial will hold true for

0 = a(x-1)(x-2)(x-3)

Answer 4

can you multiply out (x-1)(x-2)(x-3) for me?

Answer 5

x^3 - 6x^2 + 11x -6

Answer 6

ok..
so the equation will look like this: \(\large y=a(x^3 - 6x^2 + 11x - 6) \)

now you have to meet the condition f(0) = -12

Answer 7

distribute the "a" so you'll have \(\large y=ax^3-6ax^2+11ax-6a \)
when x=0, y=-12.
you must figure out what "a" is for this to be true.