Solve the problem
-In one U.S. city, the quadratic function $f ( x ) = 0.0042 x ^ { 2 } - 0.46 x + 36.44$ models the median, or average, age, $y$, 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) 1953,36 years old
B) 1936,49 years old
C) $1955,23.8$ years old
D) 1955,49 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 $a = 0.0042$ and $b = -0.46$. Plugging these values in, we get $-\frac{-0.46}{2(0.0042)} = 54.76$. Since $x$ represents years after 1900, we add this to 1900 to get 1954.76, which rounds to 1955. To find the average age at that year, we substitute $x = 54.76$ into the original equation to get $f(54.76) = 0.0042(54.76)^2 - 0.46(54.76) + 36.44$, which approximately equals 23.8. Therefore, the correct answer is 1955, with an average age of 23.8 years old.

Solve the Problem -In One U f(x)=0.0042x2−0.46x+36.44f ( x ) = 0.0042 x ^ { 2 } - 0.46 x + 36.44f(x)=0.0042x2−0.46x+36.44

Solve the Problem
-In One U $f ( x ) = 0.0042 x ^ { 2 } - 0.46 x + 36.44$