1. **Problem Statement:** Write a profit function for producing and selling $x$ thousand notebook computers given the cost function $C(x) = 4000 + 500x$ thousand dollars and the price-demand function $p(x) = 2000 - 60x$ for $1 \leq x \leq 25$. 2. **Formula and Explanation:** Profit $P(x)$ is revenue minus cost. Revenue $R(x) = x \times p(x)$ where $x$ is in thousands and $p(x)$ is price per thousand units. 3. **Write Revenue Function:** $$R(x) = x \times (2000 - 60x) = 2000x - 60x^2$$ 4. **Write Profit Function:** $$P(x) = R(x) - C(x) = (2000x - 60x^2) - (4000 + 500x)$$ 5. **Simplify Profit Function:** $$P(x) = 2000x - 60x^2 - 4000 - 500x = -60x^2 + (2000x - 500x) - 4000 = -60x^2 + 1500x - 4000$$ 6. **Domain:** Since $x$ represents thousands of notebooks sold, domain is $1 \leq x \leq 25$. 7. **Complete Table 13 (Profit values):** Calculate $P(x)$ for $x = 1,5,10,15,20,25$: - $P(1) = -60(1)^2 + 1500(1) - 4000 = -60 + 1500 - 4000 = -2560$ (given) - $P(5) = -60(25) + 1500(5) - 4000 = -1500 + 7500 - 4000 = 2000$ - $P(10) = -60(100) + 1500(10) - 4000 = -6000 + 15000 - 4000 = 5000$ - $P(15) = -60(225) + 1500(15) - 4000 = -13500 + 22500 - 4000 = 5000$ - $P(20) = -60(400) + 1500(20) - 4000 = -24000 + 30000 - 4000 = 1000$ - $P(25) = -60(625) + 1500(25) - 4000 = -37500 + 37500 - 4000 = -4000$ 8. **Summary Table 13:** | x (thousands) | P(x) (thousand $) | |---------------|-------------------| | 1 | -2560 | | 5 | 2000 | | 10 | 5000 | | 15 | 5000 | | 20 | 1000 | | 25 | -4000 | 9. **Graphing:** Plot points $(1,-2560), (5,2000), (10,5000), (15,5000), (20,1000), (25,-4000)$ on coordinate plane with $x$-axis from 1 to 25 and $y$-axis in thousands. 10. **Extra function:** For $y = x^2$ with $x$ from 1 to 5, points are $(1,1), (2,4), (3,9), (4,16), (5,25)$. **Final profit function:** $$P(x) = -60x^2 + 1500x - 4000, \quad 1 \leq x \leq 25$$