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!
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...
Hello, nathan02079!
I'll do the easier one first . . .
Take the difference of consecutive values of $\displaystyle y.$$\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: .$\displaystyle \boxed{f(n) \:=\:4n-1}$
Take consecutive differences, then the differences of the differences.$\displaystyle \begin{array}{c|c} x&y \\ \hline \text{-}2&8 \\ \text{-}1&5 \\ 0&4 \\1&5 \\
2&8 \end{array}$
. . $\displaystyle \begin{array}{cccccccccc}\text{y-value} &8 & & 5 & & 4 & & 5 & & 8 \\ \text{1st diff.} & & -3 & & -1 & & +1 & & +3 \\ \text{2nd diff.} & & & +2 & & +2 & & +2 \end{array}$
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: .$\displaystyle \boxed{f(n) \:=\:n^2 + 4}$