1. **State the problem:** We need to sketch a possible graph of a polynomial with the following characteristics: - Degree 3 (cubic polynomial) - Negative y-intercept - No relative maximum or minimum points - End behavior of $Q3 \to Q1$ (meaning as $x \to -\infty$, $y \to -\infty$ and as $x \to \infty$, $y \to \infty$) 2. **Recall the end behavior of cubic polynomials:** - For a cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$, if $a > 0$, the end behavior is $Q3 \to Q1$ (left end down, right end up). - If $a < 0$, the end behavior is $Q2 \to Q4$ (left end up, right end down). 3. **No relative max or min means:** - The polynomial is strictly increasing or strictly decreasing. - For a cubic, this happens if the derivative has no real roots (no critical points). 4. **Constructing such a polynomial:** - Let $f(x) = x^3 + d$ (simplest cubic with positive leading coefficient and no quadratic or linear terms). - The derivative is $f'(x) = 3x^2$, which is zero only at $x=0$ (a point of inflection, not max or min). 5. **Check y-intercept:** - $f(0) = d$. - We want negative y-intercept, so choose $d < 0$. 6. **Example polynomial:** $$f(x) = x^3 - 1$$ - End behavior: as $x \to -\infty$, $f(x) \to -\infty$ (Q3), as $x \to \infty$, $f(x) \to \infty$ (Q1). - Y-intercept: $f(0) = -1$ (negative). - Derivative: $f'(x) = 3x^2$, zero only at $x=0$ (inflection point, no max or min). 7. **Summary:** - The polynomial $f(x) = x^3 - 1$ satisfies all conditions. **Final answer:** $$f(x) = x^3 - 1$$