Question

Third-degree, with zeros of -2,-1, and 3, and passes through the point (2, 11).
Construct a polynomial function with the stated properties.

Answer 1

Answer:

$P(x) = -\frac{11}{12}(x+2)(x+1)(x-3)$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$, in which a is the leading coefficient.

Zeros of -2,-1, and 3

This means that:

$P(x) = a(x - (-2))(x - (-1))(x - 3) = a(x+2)(x+1)(x-3)$

Passes through the point (2, 11).

This means that when $x = 2, P(x) = 11$. We use this to find the leading coefficient.

$P(x) = a(x+2)(x+1)(x-3)$

$a(2+2)(2+1)(2-3) = 11$

$-12a = 11$

$a = -\frac{11}{12}$

So the polynomial is:

$P(x) = -\frac{11}{12}(x+2)(x+1)(x-3)$