I puruse this site often to help my kids with math and find it interesting. I have a problem of my own that seems easy but I can't seem to figure out. I'm trying to figure out a formula so that if x=100 then y=1 and if x=50 then y=50 and if x=1 then y=100.
Inverse relationship
- Thread starter MrCrum
- Start date
Hello, MrCrum!
This is neither a simple problem nor a pleasant one.
We want a function that contains these points: .\(\displaystyle (100,1),\
50,50),\1,100).\)
Since they are not collinear, the "simplest" function would be a quadratic.
The general quadratic function is: .\(\displaystyle f(x) \:=\:ax^2 + bx + c\)
Substitute the three points and form a system of equations:
. . \(\displaystyle \begin{array}{ccccccccc}(100,1) & 10,000a + 100b + c &=& 1 \\
(50,50) & 2500a + 50b + c &=& 50 \\
(1,100) & a + b + c & = & 100 \end{array}\)
After a lot of algebra, we get: .\(\displaystyle a = \frac{1}{2450},\;b = \text{-}\frac{2551}{2450},\;c = \frac{4951}{49}\)
Therefore, the function is: .\(\displaystyle f(x) \;=\;\dfrac{1}{2450}x^2 - \dfrac{2551}{2450}x + \dfrac{4951}{49}\)
This is neither a simple problem nor a pleasant one.
We want a function that contains these points: .\(\displaystyle (100,1),\
50,50),\1,100).\)Since they are not collinear, the "simplest" function would be a quadratic.
The general quadratic function is: .\(\displaystyle f(x) \:=\:ax^2 + bx + c\)
Substitute the three points and form a system of equations:
. . \(\displaystyle \begin{array}{ccccccccc}(100,1) & 10,000a + 100b + c &=& 1 \\
(50,50) & 2500a + 50b + c &=& 50 \\
(1,100) & a + b + c & = & 100 \end{array}\)
After a lot of algebra, we get: .\(\displaystyle a = \frac{1}{2450},\;b = \text{-}\frac{2551}{2450},\;c = \frac{4951}{49}\)
Therefore, the function is: .\(\displaystyle f(x) \;=\;\dfrac{1}{2450}x^2 - \dfrac{2551}{2450}x + \dfrac{4951}{49}\)
Wow, and thanks for the quick response. I thought it was just the cobwebs in my head but it took me a while just to figure out how you did what you did to figure it out. You've been a great help since I've been diligently working on it since I posted it and was still getting nowhere. Thanks again and Good Luck to you!!