## Algebra 2 (1st Edition)

Let's compare $f(x)=\frac{1}{9000}x^2-\frac{7}{15}x+500$ to $f(x)=ax^2+bx+c$ We can see that $a=\frac{1}{9000}, b=-\frac{7}{15}, c=0$. a>0, hence the graph opens up, and its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-\frac{7}{15}}{2\cdot \frac{1}{9000}}=2100.$ Hence the minimum value is $f(2100)=\frac{1}{9000}x(2100)^2-\frac{7}{15}(2100)+500=10.$