To find a polynomial function that has the given zeros $0, 1,$ and $9$, we use the fact that if a polynomial has zeros $a, b, c$, then the polynomial can be written in factored form as:

$f(x) = a(x - r_1)(x - r_2)(x - r_3)$ where $r_1, r_2, r_3$ are the zeros, and $a$ is the leading coefficient (which we can take as 1 if not otherwise specified).

Given the zeros $0, 1, 9$, the polynomial will be:

$f(x) = (x - 0)(x - 1)(x - 9)$ Simplify the expression:

$f(x) = x(x - 1)(x - 9)$ Now, expand it:

$f(x) = x \left[ (x - 1)(x - 9) \right]$ First, expand $(x - 1)(x - 9)$:

$(x - 1)(x - 9) = x^2 - 9x - x + 9 = x^2 - 10x + 9$

Now multiply by $x$:

$f(x) = x(x^2 - 10x + 9) = x^3 - 10x^2 + 9x$

Thus, the polynomial function is:

$f(x) = x^3 - 10x^2 + 9x$

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would you verify that 0, 1, and 9 are indeed the zeros of the polynomial?
2. What happens to the polynomial if one of the zeros is repeated, for example, $0, 1, 1, 9$?
3. Can you express the polynomial in terms of its factored form if we add a leading coefficient?
4. How would the polynomial change if the zeros were complex numbers, such as $0, 1, i$?
5. What would the graph of this cubic polynomial look like, and how do the zeros affect its shape?

Tip: The degree of the polynomial equals the number of its zeros (including repeated ones), which can help in identifying the type of polynomial you're dealing with.