Question

In one U.S.​ city, the quadratic function f (x )equals 0.0039 x squared minus 0.42 x plus 36.79 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)?

Answer 1

Answer:

The average age was minimum at 1954 and the average age is 25.5.

Step-by-step explanation:

The given quadratic function is

$f(x)=0.0039x^2-0.42x+36.79$

It models the​ median, or​ average, age,​ y, at which men were first married x years after 1900.

In the above equation leading coefficient is positive, so it is an upward parabola and vertex of an upward parabola, is point of minima.

We need to find the year in which the average age was at a minimum​.

If a quadratic polynomial is $f(x)=ax^2+bx+c$, then vertex is

$Vertex=(-\dfrac{b}{2a},f(-\dfrac{b}{2a}))$

$-\dfrac{b}{2a}=-\dfrac{(-0.42)}{2(0.0039)}=53.846153\approx 54$

54 years after 1900 is

$1900+54=1954$

Substitute x=54 in the given function.

$f(54)=0.0039(54)^2-0.42(54)+36.79=25.4824\approx 25.5$

Therefore, the average age was minimum at 1954 and the average age is 25.5.