1. **State the problem:** We are given a graph of a polynomial function with x-intercepts near $x \approx -9, -6, 4, 8$ and local extrema at $(-8,-5)$ (minimum), $(-2,9)$ (maximum), $(4,0)$ (minimum), and $(7,3)$ (maximum). We need to find: (A) The minimum degree of the polynomial. (B) Whether the leading coefficient is positive or negative. 2. **Minimum degree of polynomial:** - The minimum degree of a polynomial is at least the number of distinct x-intercepts if all are simple roots. - Here, there are 4 distinct x-intercepts: $-9, -6, 4, 8$. - The polynomial also has 3 turning points (local maxima and minima), which means the degree must be at least one more than the number of turning points. - Number of turning points = 3, so minimum degree $\geq 3 + 1 = 4$. - Since the number of roots is 4 and minimum degree must be at least 4, the minimum degree is 4. 3. **Leading coefficient sign:** - The graph falls sharply below $-10$ on the far right, indicating the polynomial tends to $-\infty$ as $x \to +\infty$. - For even degree polynomials, if leading coefficient is positive, $f(x) \to +\infty$ as $x \to \pm \infty$. - If leading coefficient is negative, $f(x) \to -\infty$ as $x \to \pm \infty$. - Since the right end goes down to $-\infty$, the leading coefficient is negative. **Final answers:** (A) Minimum degree = 4 (B) Leading coefficient is negative