1. **State the problem:** We have the population model for an Ontario city given by the quadratic equation $$P = 14t^2 + 820t + 42000,$$ where $t$ is the number of years since 2000. We need to solve the following: a) Find the population in 1991. b) Find the year(s) when the population was 30000. c) Determine the year when the population was the smallest. 2. **Recall the formula and rules:** - The quadratic formula is $$P = at^2 + bt + c$$ with $a=14$, $b=820$, and $c=42000$. - To find population at a specific year, substitute $t$. - To find $t$ for a given population, solve the quadratic equation. - The vertex of the parabola gives the minimum or maximum population depending on $a$. 3. **Solve part (a): Population in 1991** - Since $t=0$ corresponds to 2000, 1991 is $t = 1991 - 2000 = -9$. - Calculate: $$P(-9) = 14(-9)^2 + 820(-9) + 42000 = 14 \times 81 - 7380 + 42000 = 1134 - 7380 + 42000$$ $$= (1134 - 7380) + 42000 = -6246 + 42000 = 35754$$ 4. **Solve part (b): Year(s) when population was 30000** - Set $P = 30000$: $$30000 = 14t^2 + 820t + 42000$$ - Rearrange: $$14t^2 + 820t + 42000 - 30000 = 0$$ $$14t^2 + 820t + 12000 = 0$$ - Divide entire equation by 2 to simplify: $$\cancel{14}^7 t^2 + \cancel{820}^{410} t + \cancel{12000}^{6000} = 0$$ - Use quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-410 \pm \sqrt{410^2 - 4 \times 7 \times 6000}}{2 \times 7}$$ - Calculate discriminant: $$410^2 = 168100$$ $$4 \times 7 \times 6000 = 168000$$ $$\sqrt{168100 - 168000} = \sqrt{100} = 10$$ - Calculate roots: $$t = \frac{-410 \pm 10}{14}$$ - First root: $$t = \frac{-410 + 10}{14} = \frac{-400}{14} = -\frac{200}{7} \approx -28.57$$ - Second root: $$t = \frac{-410 - 10}{14} = \frac{-420}{14} = -30$$ - Convert $t$ to years: $$\text{Year} = 2000 + t$$ - So years are approximately: $$2000 - 28.57 \approx 1971.43$$ $$2000 - 30 = 1970$$ 5. **Solve part (c): Year when population was smallest** - Since $a=14 > 0$, parabola opens upward, so vertex is minimum. - Vertex $t$ coordinate: $$t = -\frac{b}{2a} = -\frac{820}{2 \times 14} = -\frac{820}{28} = -29.29$$ - Year: $$2000 - 29.29 = 1970.71$$ - Minimum population: $$P(-29.29) = 14(-29.29)^2 + 820(-29.29) + 42000$$ Calculate: $$(-29.29)^2 = 857.9$$ $$14 \times 857.9 = 12010.6$$ $$820 \times (-29.29) = -24017.8$$ $$P = 12010.6 - 24017.8 + 42000 = 29992.8$$ **Final answers:** a) Population in 1991 is approximately 35754. b) Population was 30000 around years 1970 and 1971. c) The fewest people lived around year 1971 with population about 29993.