Quadratic function word problem

[alyren]

The quadratic function f(x) = 0.0041x^2 - 0.44x +36.08 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 nearest year). What was the average age at first marriage for that year? ( round to nearest tenth.)

So far i can only plug in the number and my equation is y=0.0041(1900)^2 - 0.44(1900) +36.08. i don't fully understand what to do. i don't know if my equation were setup right.

 

[masters]

The quadratic function f(x) = 0.0041x^2 - 0.44x +36.08 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 nearest year). What was the average age at first marriage for that year? ( round to nearest tenth.)

So far i can only plug in the number and my equation is y=0.0041(1900)^2 - 0.44(1900) +36.08. i don't fully understand what to do. i don't know if my equation were setup right.

Hi alyren,

I think I understand what you're asking. If this sounds like jibberish, maybe someone else could weigh in.

What was the average age at first marriage for that year? ( round to nearest tenth.)

I think, for this, you need to find f(0) if you're looking for the age at 1900.

f(0) = 36.08 or approximately 36.1 years.

One year after 1900, you would find f(1), etc.

In which year was this average age at a minimum?( round to nearest year).

As far as the minimum is concerned you need to find the y-coordinate of the vertex of the parabola modeled here.

You can get the x-coordinate of the vertex by using the formula \(\displaystyle x=-\frac{b}{2a}\), and using the standard form quadratic equation:

\(\displaystyle ax^2+bx+c=0\).

So, in your case: \(\displaystyle f(x) = 0.0041x^2 - 0.44x +36.08 \longrightarrow 0.0041x^2 - 0.44x +36.08 = 0\)

and, \(\displaystyle x=-\frac{-.44}{.0082} \longrightarrow \frac{2200}{41}\)

This comes out to about 53.7 years after 1900 that the minimum occurred. Rounding up to the nearest year would be 54 years from 1900 or the year 1954.

Substitute 54 back into your original quadratic and solve for y. This will become your minimum age eight and a half months into 1953. Rounded to the nearest year would be 1954.

Looks like the average minimum age for first marriage that year was about 24.3 years old.

 

[EddyTeddy]

How did you get 2200/41

 

[Dr.Peterson]

How did you get 2200/41

Why are you asking a question of someone who was here 10 years ago??

It appears that they simplified the fraction shown before it. There was no need to do that; what they actually used on the next line was the decimal form, which you can get by just dividing 0.44 by 0.0082.

 

[EddyTeddy]

Why are you asking a question of someone who was here 10 years ago??

It appears that they simplified the fraction shown before it. There was no need to do that; what they actually used on the next line was the decimal form, which you can get by just dividing 0.44 by 0.0082.

Karen I didnt know its against the law to ask question at any time regardles of when the person was here last

 

[Steven G]

Karen I didnt know its against the law to ask question at any time regardles of when the person was here last

You can ask a question anytime you want, even if the post is 10 years old. Dr Peterson was just asking why you addressed your question to Masters 10 years after their post. Masters already received her PhD in Physics and is thinking about retiring soon.

 

[Steven G]

[MATH]x=\dfrac{-b}{2a} = \dfrac{.44}{.0082} = \dfrac{4400}{82} = \dfrac{2200}{41}[/MATH]
If you are unclear how I did this then please ask.

 

[EddyTeddy]

I asked cause i had a midterm and had similar question so i was trying to plug in my numbers instead of the ines plugged here and i was confused as to how it went from .44 to 4400 thats all i know the formula -b over 2a but i couldnt figure out avarage age

 

[Steven G]

I asked cause i had a midterm and had similar question so i was trying to plug in my numbers instead of the ines plugged here and i was confused as to how it went from .44 to 4400 thats all i know the formula -b over 2a but i couldnt figure out avarage age

I moved the decimals 4 places to the right in the numerator and denominators as that got me two numbers without any decimals.