Create a polynomial function with the following

Question

Create a polynomial function with the following terms:

Third-degree, with zeros of −3, −2, and 1, and passes through the point (2,9).

Answer 1

P(x) = a(x-(-3))*(x-(-2))*(x-1)

So, P(x) = a*(x+3)*(x+2)*(x-1) is the set of all polynomials with the above zeros.

To find the value of a, substitute the values of x and y given in the point (2,9):

9 = a(2+3)(2+2)(2-1) = a*5*4*1 = 20a

So, a = 9/20.

P(x) = (9/20)*(x+3)*(x+2)*(x-1) is the factored form of the polynomial requested.

Answer 2

Recall, the general formula of a polynomial function is:

f(x) = a(x - p)(x - q)(x -r)…

We wrote only 3 sets of (x - _) because there we are told it is a 3rd degree polynomial function. If it was a 4th degree, then we would write 4 sets.

Substitute each of the given zeros into the formula for p, q, and r

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

Simplify:

f(x) = a(x + 3)(x + 2)(x - 1)

Recall, and point is just an x and y value for that function, and f(x) is the same as a y value, so substitute the values given for the point, (2,9) into the formula, so we can find the “a” value.

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

Simplify:

9 = a( 5)( 4)(1)

9 = 20a

9/20=a

thus,

f(x) = (9/20)(x + 3)(x + 2)(x - 1)

remember, if you have a graphing calculator, put this into y=, graph, and then check the zeros and the table for (2, 9) to make sure it’s correct

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