Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms.

Question

Third-degree, with zeros of −3, −1, and 2, and passes through the point (3,12).

Answer 1

through (3,12) with zeros -3, -1, and 2

y = a(x+3)(x+1)(x-2) plug in x=3 y=12 to find "a"

12 = a(6)(4)1) = 24a

a = 12/24 = 1/2

y = .5(x+3)(x+1)(x-2)

2y = (x^2 +4x +3)(x-2) = x^3 +2x^2-5x-6

2y = x^3 +2x^2 -5x -6

Answer 2

3rd degree means your leading term will be cubed (if you multiply out the factors). A zero of x = -3 can be found from a factored polynomial with a factor of (x+3). To write a polynomial with the 3 zeroes given, you could have f(x) = (x+3)(x+1)(x-2). To ensure that this polynomial passes through the point, you could multiply or a add a number. To find this number you could rewrite this function as y = a(x+3)(x+1)(x-2) or y = (x+3)(x+1)(x-2) + a. Then, plug in the x and y coordinates of the point to solve for a.

y = a(x+3)(x+1)(x-2) at (3,12):

12 = a(3+3)(3+1)(3-2)

12 = 24a

a = 1/2

y = 1/2(x+1)(x+3)(x-2)

y = (x+3)(x+1)(x-2) + a at (3,12):

12 =(3+3)(3+1)(3-2) + a

a = -12

y = (x+3)(x+1)(x-2) -12