Tell the maximum number of zeros that the polynomial function may have. Then use Descartes' Rule of Signs to determine how many positive and how many negative real zeros the polynomial function may have. Do not attempt to find the zeros. \[ f(x)=9 x^{7}+x^{3}-x^{2}+6 \] What is the maximum number of zeros that this polynomial function can have? 7 How many positive real zeros can the function have? (Use a comma to separate answers as needed)

Step 1 ：The maximum number of zeros that a polynomial function can have is equal to its degree. In this case, the degree of the polynomial is 7, so the maximum number of zeros is 7.

Step 2 ：Descartes' Rule of Signs is used to determine the possible number of positive and negative real zeros of a polynomial. According to the rule, the number of positive real zeros of a polynomial is equal to the number of sign changes in the polynomial's terms, or less than that by an even number. The number of negative real zeros is found by applying the rule to the polynomial obtained by replacing x by -x.

Step 3 ：To find the number of positive real zeros, we need to count the number of sign changes in the polynomial's terms. The terms of the polynomial are \(9x^7\), \(x^3\), \(-x^2\), and 6. The signs of these terms are positive, positive, negative, and positive, respectively. So, there are 2 sign changes.

Step 4 ：To find the number of negative real zeros, we replace x by -x in the polynomial to get \(-9x^7 + x^3 + x^2 + 6\). The signs of these terms are negative, positive, positive, and positive, respectively. So, there is 1 sign change.

Step 5 ：Final Answer: The maximum number of zeros that this polynomial function can have is \(\boxed{7}\). The function can have \(\boxed{1}\) positive real zero and \(\boxed{0}\) negative real zeros.