Emissions Estimate E60A6C

1. **State the problem:** We have emissions data for years 1980, 1985, 1990, and 1995 with corresponding emissions $18.8$, $22.6$, $26.0$, and $29.0$ million tons respectively. We want to model this data with a quadratic function $f(x) = ax^2 + bx + c$ where $x$ is the year, and estimate emissions in the year $2070$. 2. **Simplify the variable:** To make calculations easier, define a new variable $t = x - 1980$ so that $t=0$ corresponds to year 1980. Then the data points become: - $t=0$, $f(0) = 18.8$ - $t=5$, $f(5) = 22.6$ - $t=10$, $f(10) = 26.0$ - $t=15$, $f(15) = 29.0$ 3. **Set up the system of equations:** Using $f(t) = a t^2 + b t + c$, plug in the points: $$ \begin{cases} f(0) = c = 18.8 \\ f(5) = 25a + 5b + c = 22.6 \\ f(10) = 100a + 10b + c = 26.0 \end{cases} $$ 4. **Substitute $c=18.8$ into the other equations:** $$ \begin{cases} 25a + 5b + 18.8 = 22.6 \\ 100a + 10b + 18.8 = 26.0 \end{cases} $$ Simplify: $$ \begin{cases} 25a + 5b = 22.6 - 18.8 = 3.8 \\ 100a + 10b = 26.0 - 18.8 = 7.2 \end{cases} $$ 5. **Divide the first equation by 5 and the second by 10 to simplify:** $$ \begin{cases} \cancel{5}(5a + b) = 3.8 \Rightarrow 5a + b = \frac{3.8}{5} = 0.76 \\ \cancel{10}(10a + b) = 7.2 \Rightarrow 10a + b = \frac{7.2}{10} = 0.72 \end{cases} $$ 6. **Subtract the first simplified equation from the second:** $$ (10a + b) - (5a + b) = 0.72 - 0.76 \\ 5a = -0.04 \\ a = -0.008 $$ 7. **Find $b$ using $5a + b = 0.76$:** $$ 5(-0.008) + b = 0.76 \\ -0.04 + b = 0.76 \\ b = 0.8 $$ 8. **Write the quadratic model:** $$ f(t) = -0.008 t^2 + 0.8 t + 18.8 $$ 9. **Estimate emissions in year 2070:** Calculate $t$ for 2070: $$ t = 2070 - 1980 = 90 $$ Calculate $f(90)$: $$ f(90) = -0.008 \times 90^2 + 0.8 \times 90 + 18.8 = -0.008 \times 8100 + 72 + 18.8 = -64.8 + 72 + 18.8 = 26.0 $$ 10. **Conclusion:** The estimated emissions in 2070 are $26.0$ million tons, which corresponds to option B.