1. **Problem statement:** Given the function $$f(t) = -\frac{1}{400} t^{2} + \frac{1}{10} t$$ and the parameterized function $$f_a(t) = - \frac{a}{400} t^{2} + \frac{a}{10} t,$$ where $t$ is time in years and $f(t)$ is in meters per year, we are asked to analyze parts a), b), and c) (only part a) will be solved here as per instructions). 2. **Step a) Find the vertex of the parabola $f(t)$:** The function is quadratic in the form $$f(t) = at^2 + bt + c$$ with $$a = -\frac{1}{400}, b = \frac{1}{10}, c = 0.$$ The vertex $t$-coordinate is given by $$t_v = -\frac{b}{2a}.$$ Substitute values: $$t_v = -\frac{\frac{1}{10}}{2 \times -\frac{1}{400}} = -\frac{\frac{1}{10}}{-\frac{1}{200}} = \frac{\frac{1}{10}}{\frac{1}{200}} = \frac{1}{10} \times 200 = 20.$$ 3. **Calculate the maximum value $f(t_v)$:** $$f(20) = -\frac{1}{400} \times 20^2 + \frac{1}{10} \times 20 = -\frac{1}{400} \times 400 + 2 = -1 + 2 = 1.$$ 4. **Interpretation:** The function reaches its maximum value of 1 meter per year at $t=20$ years. 5. **Summary:** - The vertex of $f(t)$ is at $$t=20,$$ - The maximum value of $f(t)$ is $$1.$$