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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- Feb 7th 2008, 06:05 PMnathan02079Determine 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! - Feb 7th 2008, 06:12 PMJhevon
- Feb 7th 2008, 06:15 PMnathan02079
"i can tell you what i thought about, but it was not really a mathematically rigorous process"

please tell me =) - Feb 7th 2008, 06:25 PMJhevon
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... - Feb 7th 2008, 06:28 PMnathan02079
- Feb 7th 2008, 06:53 PMSoroban
Hello, nathan02079!

I'll do the easier one first . . .

Quote:

$\displaystyle \begin{array}{c|c} x&y \\ \hline \text{-}3&\text{-}13 \\ \text{-}2&\text{-}9 \\ \text{-}1&\text{-}5 \\ 0&\text{-}1 \\ 1&3 \end{array}$

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

The*first*differences are constant.

. . Hence, the function is of the*first*degree (linear).

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

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

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

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

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

Quote:

$\displaystyle \begin{array}{c|c} x&y \\ \hline \text{-}2&8 \\ \text{-}1&5 \\ 0&4 \\1&5 \\

2&8 \end{array}$

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

The*second*differences are constant.

. . Hence, the function is of the*second*degree (quadratic).

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

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

. . $\displaystyle \begin{array}{ccccccc}f(\text{-}1) = 5: & a(\text{-}1)^2+b(\text{-}1)+c \:=\:5 & \Rightarrow & a - b + c \:=\:5 & {\color{blue}[1]}\\

f(0) = 4: & a\!\cdot\!0^2 + b\!\cdot\!0 + c \:=\:4 & \Rightarrow & c = 4 & {\color{blue}[2]}\\

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: .$\displaystyle \begin{array}{cccc}a - b + 4 \:=\:5 & \Rightarrow & a - b \:=\:1 & {\color{blue}[4]}\\ a + b + 4 \:=\:5 & \Rightarrow & a + b \:=\:1 & {\color{blue}[5]}\end{array}$

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

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