1. The problem asks to write an equation in factored form for the 6th degree polynomial shown in the top-right graph. 2. From the description, the polynomial is degree 6 and has zeros near $x=-4$, $x=-1$, $x=3$, and passes through $(0,6)$. 3. The curve crosses the x-axis near $x=-4$ (implying a root with odd multiplicity), touches/turns at $x=-1$ (implying even multiplicity), and touches/turns again at $x=3$ (implying even multiplicity). 4. Since the polynomial is degree 6, the sum of multiplicities of roots must be 6. 5. Assign multiplicities: - Root at $x=-4$: multiplicity 1 (crosses) - Root at $x=-1$: multiplicity 2 (touches/turns) - Root at $x=3$: multiplicity 2 (touches/turns) 6. That accounts for $1 + 2 + 2 = 5$ multiplicities. The remaining multiplicity is 1, which must be at another root or the root at $x=2$ (local max near 2) is not a root. 7. Since no root is mentioned at $x=2$, assume the last root is at $x=0$ or the polynomial has a leading factor. 8. The polynomial passes through $(0,6)$, so we can use this to find the leading coefficient $a$. 9. The factored form is: $$f(x) = a(x+4)(x+1)^2(x-3)^2$$ 10. Substitute $x=0$ and $f(0)=6$: $$6 = a(0+4)(0+1)^2(0-3)^2 = a \times 4 \times 1 \times 9 = 36a$$ 11. Solve for $a$: $$a = \frac{6}{36} = \frac{1}{6}$$ 12. Final factored form: $$f(x) = \frac{1}{6}(x+4)(x+1)^2(x-3)^2$$ This matches the degree 6 polynomial with the given roots and behavior.