1. **State the problem:** We have a cubic function $$f(x) = 2x^3 + px^2 - 11x + q$$ where $$p, q \in \mathbb{Z}$$. We know: - The graph cuts the x-axis at $$x = \frac{1}{2}$$, so $$f\left(\frac{1}{2}\right) = 0$$. - The graph cuts the y-axis at $$(0,6)$$, so $$f(0) = 6$$. 2. **Use the y-intercept to find $$q$$:** Since $$f(0) = q = 6$$, we have $$q = 6$$. 3. **Use the x-intercept to find $$p$$:** Substitute $$x = \frac{1}{2}$$ and $$q = 6$$ into the function: $$f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + p\left(\frac{1}{2}\right)^2 - 11\left(\frac{1}{2}\right) + 6 = 0$$ Calculate each term: $$2 \times \frac{1}{8} = \frac{1}{4}$$ $$p \times \frac{1}{4} = \frac{p}{4}$$ $$-11 \times \frac{1}{2} = -\frac{11}{2}$$ So the equation becomes: $$\frac{1}{4} + \frac{p}{4} - \frac{11}{2} + 6 = 0$$ 4. **Simplify the equation:** Combine constants: $$-\frac{11}{2} + 6 = -\frac{11}{2} + \frac{12}{2} = \frac{1}{2}$$ So: $$\frac{1}{4} + \frac{p}{4} + \frac{1}{2} = 0$$ 5. **Combine all terms with common denominator 4:** $$\frac{1}{4} + \frac{p}{4} + \frac{2}{4} = 0$$ This simplifies to: $$\frac{1 + p + 2}{4} = 0$$ 6. **Multiply both sides by 4 to clear denominator:** $$\cancel{4} \times \frac{1 + p + 2}{\cancel{4}} = 0 \times 4$$ $$1 + p + 2 = 0$$ 7. **Solve for $$p$$:** $$p + 3 = 0$$ $$p = -3$$ **Final answers:** $$p = -3, \quad q = 6$$