1. **Problem statement:** We need to sketch a quartic polynomial function $y = f(x)$ such that $f(x) > 0$ when $x < -5$, $2 < x < 3$, and $x > 4$. We also need to write an inequality statement for this function. 2. **Understanding the problem:** A quartic polynomial is degree 4, so it can have up to 4 real roots. The function is positive in three intervals: $(-\infty, -5)$, $(2, 3)$, and $(4, \infty)$. This means the function crosses the x-axis at points that separate these intervals where the sign changes. 3. **Determining roots:** The roots must be at $x = -5$, $x = 2$, $x = 3$, and $x = 4$ to create the intervals where $f(x)$ changes sign. 4. **Sign analysis:** Since $f(x) > 0$ for $x < -5$, the polynomial is positive to the left of $-5$. The sign changes at each root, so the intervals between roots alternate signs. 5. **Constructing the polynomial:** The roots are $-5$, $2$, $3$, and $4$. The polynomial can be written as: $$f(x) = a(x + 5)(x - 2)(x - 3)(x - 4)$$ 6. **Determining leading coefficient $a$:** Since $f(x) > 0$ for $x < -5$, and the degree is even, the ends of the graph go to $+\infty$ if $a > 0$. So choose $a > 0$. 7. **Inequality statement:** The function is positive in the intervals $x < -5$, $2 < x < 3$, and $x > 4$. So the inequality is: $$f(x) > 0 \text{ for } x < -5, \quad 2 < x < 3, \quad \text{and} \quad x > 4$$ 8. **Summary:** The quartic polynomial is $$f(x) = a(x + 5)(x - 2)(x - 3)(x - 4), \quad a > 0$$ and the inequality describing where $f(x)$ is positive is $$f(x) > 0 \text{ for } x < -5, \quad 2 < x < 3, \quad x > 4.$$