Solve the problem.
-The quadratic function models the median, or average, age, y, at which U.S. men $$f ( x ) = 0.0039 x ^ { 2 } - 0.47 x + 36.46$$ were first married x years after 1900. In which year was this average age at a minimum? (Round to the nearest year.) What was the average age at first marriage for that year? (Round to the nearest tenth.)
A) 1936, 50.6 years old
B) 1960, 50.6 years old
C) 1953, 36 years old
D) 1960, 22.3 years old

Question

Quizplus · Accepted Answer

The minimum value of a quadratic function $$f(x) = ax^2 + bx + c$$ occurs at $$x = -\frac{b}{2a}$$. For this function, $$a = 0.0039$$ and $$b = -0.47$$, so $$x = -\frac{-0.47}{2(0.0039)} = 60$$. This means the minimum average age occurred 60 years after 1900, or in 1960. To find the minimum average age, substitute x = 60 into the function: $$f(60) = 0.0039(60)^2 - 0.47(60) + 36.46 = 22.3$$. Therefore, the correct answer is 1960, 22.3 years old.

Solve the Problem f(x)=0.0039x2−0.47x+36.46f ( x ) = 0.0039 x ^ { 2 } - 0.47 x + 36.46f(x)=0.0039x2−0.47x+36.46

Solve the Problem $f ( x ) = 0.0039 x ^ { 2 } - 0.47 x + 36.46$