Results 1 to 3 of 3

Thread: Creating Functions from a data set

  1. #1
    New Member
    Join Date
    Nov 2011
    Location
    Canada
    Posts
    49

    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
    High School: Grade 12 Calculus and Advanced Functions

  2. #2
    Elite Member
    Join Date
    Jan 2005
    Location
    Lexington, MA
    Posts
    5,603
    Hello, Relz!

    Find the polynomial equation that matches the data set.

    . . [tex]\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}[/tex]

    [tex]\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}[/tex]

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

    The general quadratic function is:.[tex]f(x) \:=\:ax^2 + bx + c[/tex]


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

    . . [tex]\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}[/tex]

    Solve the system: .[tex]a = -2,\;b = 3,\;c = 6[/tex]


    Therefore, the function is: .[tex]f(x) \:=\:-2x^2 + 3x + 6[/tex]

  3. #3
    New Member
    Join Date
    Nov 2011
    Location
    Canada
    Posts
    49
    Quote Originally Posted by soroban View Post
    Hello, Relz!


    [tex]\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}[/tex]

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

    The general quadratic function is:.[tex]f(x) \:=\:ax^2 + bx + c[/tex]


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

    . . [tex]\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}[/tex]

    Solve the system: .[tex]a = -2,\;b = 3,\;c = 6[/tex]


    Therefore, the function is: .[tex]f(x) \:=\:-2x^2 + 3x + 6[/tex]
    Okay thank you! I must have screwed up my first differences, haven't done them in a while!
    High School: Grade 12 Calculus and Advanced Functions

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •