1. The problem states that the points $(3,2)$ and $(-3,2)$ are maximum points, and the minimum value is $-6$ occurring at $x=0$. 2. We want to find a function $y=f(x)$ that satisfies these conditions. 3. Since the function has symmetric maxima at $x=3$ and $x=-3$ with the same $y$-value, and a minimum at $x=0$, a good candidate is a quartic polynomial of the form: $$f(x) = ax^4 + bx^2 + c$$ 4. This form is even (symmetric about the y-axis), which fits the symmetry of the maxima. 5. Using the given points: - At $x=0$, $f(0) = c = -6$ (minimum value) - At $x=3$, $f(3) = a(3)^4 + b(3)^2 + c = 81a + 9b - 6 = 2$ 6. Since $(3,2)$ is a maximum, the derivative at $x=3$ must be zero: $$f'(x) = 4ax^3 + 2bx$$ $$f'(3) = 4a(27) + 2b(3) = 108a + 6b = 0$$ 7. Similarly, at $x=0$, the derivative is zero (minimum): $$f'(0) = 0$$ 8. From $f'(3) = 0$, solve for $b$: $$108a + 6b = 0 \Rightarrow b = -18a$$ 9. Substitute $b$ into the equation for $f(3)$: $$81a + 9(-18a) - 6 = 2 \Rightarrow 81a - 162a - 6 = 2 \Rightarrow -81a = 8 \Rightarrow a = -\frac{8}{81}$$ 10. Then, $$b = -18a = -18 \times -\frac{8}{81} = \frac{144}{81} = \frac{16}{9}$$ 11. The function is: $$f(x) = -\frac{8}{81}x^4 + \frac{16}{9}x^2 - 6$$ 12. This function has maxima at $(3,2)$ and $(-3,2)$ and a minimum of $-6$ at $x=0$ as required.