Thread: Creating Functions from a data set

1. Creating Functions from a data set

Okay so I have three questions like this but I will only post the one! I have to find the polynomial equation that matches the data set. In class we are learning how to determine if the equation is linear, quadratic, cubic, quartic, etc. using the differences so I think that may be where I need to go but I tried finding the first, second and third differences to see if it was linear, quadratic or cubic but the last point doesn't fit. Should I graph this and go from there or am I missing something?

Given:

X Y
-3 -21
-2 -8
-1 1
0 6
1 7
2 4
3 -3

2. Hello, Relz!

Find the polynomial equation that matches the data set.

. . $\begin{array}{cc}x & y \\ \hline \text{-}3 & \text{-}21 \\ \text{-}2 & \text{-}8 \\ \text{-}1 & 1 \\ 0 & 6 \\ 1 & 7 \\ 2 & 4 \\ 3 & \text{-}3 \end{arry}$

$\begin{array}{c|ccccccccccccc} x & \text{-}3 && \text{-}2 && \text{-}1 && 0 && 1 && 2 && 3 \\ \hline y & \text{-}21 && \text{-}8 && 1 && 6 && 7 && 4 && \text{-}3 \\ \hline \text{1st diff} && 13 && 9 && 5 && 1 && \text{-}3 && \text{-}7 \\ \hline \text{2nd diff} &&& \text{-}4 && \text{-}4 && \text{-}4 && \text{-}4 && \text{-}4 \\ \hine \end{array}$

The second differences are constant.
. . Hence, the polynomial fuction is of the second degree.

The general quadratic function is:.$f(x) \:=\:ax^2 + bx + c$

Use three values from the table and set up a system of equations.

. . $\begin{array}{cccccccccc}f(0) = 6: & a(0^2) + b(0) + c &=& 6 \\ f(1) = 7: & a(1^2) + b(1) + c &=& 7 \\ f(2) = 4: & a(2^2) + b(2) + c &=& 4 \end{array}$

Solve the system: .$a = -2,\;b = 3,\;c = 6$

Therefore, the function is: .$f(x) \:=\:-2x^2 + 3x + 6$

3. Originally Posted by soroban
Hello, Relz!

$\begin{array}{c|ccccccccccccc} x & \text{-}3 && \text{-}2 && \text{-}1 && 0 && 1 && 2 && 3 \\ \hline y & \text{-}21 && \text{-}8 && 1 && 6 && 7 && 4 && \text{-}3 \\ \hline \text{1st diff} && 13 && 9 && 5 && 1 && \text{-}3 && \text{-}7 \\ \hline \text{2nd diff} &&& \text{-}4 && \text{-}4 && \text{-}4 && \text{-}4 && \text{-}4 \\ \hine \end{array}$

The second differences are constant.
. . Hence, the polynomial fuction is of the second degree.

The general quadratic function is:.$f(x) \:=\:ax^2 + bx + c$

Use three values from the table and set up a system of equations.

. . $\begin{array}{cccccccccc}f(0) = 6: & a(0^2) + b(0) + c &=& 6 \\ f(1) = 7: & a(1^2) + b(1) + c &=& 7 \\ f(2) = 4: & a(2^2) + b(2) + c &=& 4 \end{array}$

Solve the system: .$a = -2,\;b = 3,\;c = 6$

Therefore, the function is: .$f(x) \:=\:-2x^2 + 3x + 6$
Okay thank you! I must have screwed up my first differences, haven't done them in a while!

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