Solve the problem.
-In one U.S. city, the quadratic function $f ( x ) = 0.0036 x ^ { 2 } - 0.44 x + 36.95$ models the median, or average, age, $\mathrm { y }$, at which 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) $1961,23.5$ years old
B) $1961,50.4$ years old
C) $1936,50.4$ years old
D) 1953,36 years old

Question

Quizplus · Accepted Answer

To find the year when the average age at first marriage was at a minimum, we need to find the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula $-\frac{b}{2a}$, where $a = 0.0036$ and $b = -0.44$. Plugging these values in gives $-\frac{-0.44}{2(0.0036)} = 61.11$. Since $x$ represents years after 1900, we add 61 to 1900, which rounds to 1961. To find the minimum average age, we substitute $x = 61$ into the original equation, yielding $f(61) = 0.0036(61)^2 - 0.44(61) + 36.95 \approx 23.5$. Therefore, the correct answer is 1961, with an average age of 23.5 years old.

Solve the Problem f(x)=0.0036x2−0.44x+36.95f ( x ) = 0.0036 x ^ { 2 } - 0.44 x + 36.95f(x)=0.0036x2−0.44x+36.95

Solve the Problem $f ( x ) = 0.0036 x ^ { 2 } - 0.44 x + 36.95$