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Math Help - Determine a rule for each table of values.

  1. #1
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    Smile Determine a rule for each table of values.

    x,y
    1, -2
    2,-1
    3,0
    4,1
    5,2

    rule: y=x-3

    x,y
    -2,8
    -1,5
    0,4
    1,5
    2,8

    rule: _____

    x,y
    -3,-13
    -2,-9
    -1,-5
    0,-1
    1,3

    rule: _____

    Help? Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nathan02079 View Post

    x,y
    -2,8
    -1,5
    0,4
    1,5
    2,8

    rule: _____
    by inspection, y = x^2 + 4


    x,y
    -3,-13
    -2,-9
    -1,-5
    0,-1
    1,3

    rule: _____
    by inspection, y = 4x - 1

    i can tell you what i thought about, but it was not really a mathematically rigorous process
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  3. #3
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    "i can tell you what i thought about, but it was not really a mathematically rigorous process"

    please tell me =)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nathan02079 View Post
    "i can tell you what i thought about, but it was not really a mathematically rigorous process"

    please tell me =)
    as i said, this is not sure to get you the answer with things like this, i just lucked out. it is better to think of things this way before i went into the real mathematical process, which can get complicated computation-wise.

    i looked at the 0's first. in the first problem, 0 --> 4. so i assume y = (some manipulation of x) + 4.

    so then i subtracted 4 from all the other y-values, and tried to figure out what i would do to x to get them. i thought about multiplying by a negative number to make it positive, since all the negative numbers went to positive, but it did not work for all. so then i thought of squaring. and it worked!

    same for the second. i saw 0 --> -1, so i thought y = f(x) - 1. after playing around a bit, 4x was the obvious choice

    sorry i can't explain more, but i'm in a rush...
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    as i said, this is not sure to get you the answer with things like this, i just lucked out. it is better to think of things this way before i went into the real mathematical process, which can get complicated computation-wise.

    i looked at the 0's first. in the first problem, 0 --> 4. so i assume y = (some manipulation of x) + 4.

    so then i subtracted 4 from all the other y-values, and tried to figure out what i would do to x to get them. i thought about multiplying by a negative number to make it positive, since all the negative numbers went to positive, but it did not work for all. so then i thought of squaring. and it worked!

    same for the second. i saw 0 --> -1, so i thought y = f(x) - 1. after playing around a bit, 4x was the obvious choice

    sorry i can't explain more, but i'm in a rush...
    hmmm....never thought about working it backwards before...lol i knew it was something to do with the 0. Thanks
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  6. #6
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    Hello, nathan02079!

    I'll do the easier one first . . .


    \begin{array}{c|c} x&y \\ \hline \text{-}3&\text{-}13 \\ \text{-}2&\text{-}9 \\ \text{-}1&\text{-}5 \\ 0&\text{-}1 \\ 1&3 \end{array}
    Take the difference of consecutive values of y.

    \begin{array}{cccccccccc}\text{y-value} &-13 & & -9 & & -5 & & -1 & & 3 \\ \text{1st diff.} && +4 & & +4 & & +4 & & +4\end{array}

    The first differences are constant.
    . . Hence, the function is of the first degree (linear).

    The general linear function is: . f(n) \:=\:an + b


    Use any two values from the table to set up a system of equations.

    . . \begin{array}{ccccccc}f(0) = \text{-}1: &a(0) + b \:=\:\text{-}1 & \Rightarrow & b \:=\:\text{-}1 \\ f(1) = 3: & a(1) + b \:=\:3 & \Rightarrow & a + b \:=\:3 \end{array}

    Hence: . a \,=\,4,\;b\,=\,-1


    Therefore: . \boxed{f(n) \:=\:4n-1}




    \begin{array}{c|c} x&y \\ \hline \text{-}2&8 \\ \text{-}1&5 \\ 0&4 \\1&5 \\<br />
2&8 \end{array}
    Take consecutive differences, then the differences of the differences.

    . . \begin{array}{cccccccccc}\text{y-value} &8 & & 5 & & 4 & & 5 & & 8 \\ \text{1st diff.} & & -3 & & -1 & & +1 & & +3 \\ \text{2nd diff.} &  & & +2 & & +2 & & +2 \end{array}

    The second differences are constant.
    . . Hence, the function is of the second degree (quadratic).

    The general quadratic function is: . f(n) \:=\:an^2 + bn + c


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

    . . \begin{array}{ccccccc}f(\text{-}1) = 5: & a(\text{-}1)^2+b(\text{-}1)+c \:=\:5 & \Rightarrow & a - b + c \:=\:5 & {\color{blue}[1]}\\<br />
f(0) = 4: & a\!\cdot\!0^2 + b\!\cdot\!0 + c \:=\:4 & \Rightarrow & c = 4 & {\color{blue}[2]}\\<br />
f(1) = 5: & a\!\cdot\!1^2 + b\!\cdot\!1 + c \:=\:5 & \Rightarrow & a + b + c \:=\:5 & {\color{blue}[3]}\end{array}

    Then [1] and [3] become: . \begin{array}{cccc}a - b + 4 \:=\:5 & \Rightarrow & a - b \:=\:1 & {\color{blue}[4]}\\ a + b + 4 \:=\:5 & \Rightarrow & a + b \:=\:1 & {\color{blue}[5]}\end{array}

    Add [4] and [5]: . 2a \:=\:2\quad\Rightarrow\quad a \:=\:1 \quad\Rightarrow\quad b \:=\:0


    Therefore: . \boxed{f(n) \:=\:n^2 + 4}

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