# Thread: Determine a rule for each table of values.

1. ## 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!

2. Originally Posted by nathan02079

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

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

4. Originally Posted by nathan02079
"i can tell you what i thought about, but it was not really a mathematically rigorous process"

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...

5. Originally Posted by Jhevon
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

6. Hello, nathan02079!

I'll do the easier one first . . .

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

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

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

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

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

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