Solve the problem.
-The quadratic function $$f ( x ) = 0.0040 x ^ { 2 } - 0.43 x + 36.98$$ models the median, or average, age, y, at which U.S. men 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) 1953, 36 years old
B) 1936, 48.5 years old
C) 1954, 25.4 years old
D) 1954, 48.5 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 the given function, $a = 0.0040$ and $b = -0.43$, so the year of minimum average age is $x = -\frac{-0.43}{2(0.0040)} = 53.75$, which rounds to 1954 when added to 1900. To find the minimum average age, substitute $x = 53.75$ back into the function: $f(53.75) = 0.0040(53.75)^2 - 0.43(53.75) + 36.98 \approx 25.4$.

Solve the Problem f(x)=0.0040x2−0.43x+36.98f ( x ) = 0.0040 x ^ { 2 } - 0.43 x + 36.98f(x)=0.0040x2−0.43x+36.98

Solve the Problem $f ( x ) = 0.0040 x ^ { 2 } - 0.43 x + 36.98$