College Algebra, Chapter 4, 4.1, Section 4.1, Problem 36

Determine the maximum or minimum value of the function $f(t) = 10t^2 + 40t + 113$.

Using the quadratic equation $ax^2 + bx + c$, where $a = 10$ and $b = 40$. Thus, the maximum or minimum value occurs at

$\displaystyle x = -\frac{b}{2a} = - \frac{40}{2(10)} = -2$

Since $a > 0$, the function has the minimum value

$\displaystyle f (-2) = 10(-2)^2 + 40(-2) + 113 = 73$