Third-degree, with zeros of −3, −2, and 1, and passes through the point (2,7).

Answer 1

A general third-degree function with those zeros can be written in factored form as:

f(x) = a(x+3)(x+2)(x-1), where a is a constant.

The function has to pass through (2, 7), meaning that f(2) = 7.

We can use this information to solve for the constant, as follows:

f(2) = 7 = a(2+3)(2+2)(2-1)

or

7 = 20a

This means that a = 20/7

So the particular function that meets all of the criteria would be f(x) = (20/7)(x+3)(x+2)(x-1)

If you need it in standard form, you'd need to multiply out all the factors.