Question : Is there a cubic polynomial that takes these values
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline
y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}$
You substitute the x with the values your given.
Then you get an equation with each of the first 4 values, which gives you a system of 4 equations with 4 unknowns (a,b,c,d).
Then see if the remaining values are such that the equations are satisfied.
I suggest you first consider 0,1,-1 because they're nice values to deal with
Hello, zorro!
You don't understand a thing we're saying, do you?
Is there a cubic polynomial that takes these values?
. . $\displaystyle \begin{array}{|c||c|c|c|c|c|c|}x & \text{-}2 & \text{-}1 & 0 & 1 & 3 & 7\\ \hline
y & \text{-}56 & 16 & \text{-}2 & \text{-}2 & 4 & 376 \end{array}$
The general cubic polynomial is: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx + d $
. . and we must determine the four coefficients: $\displaystyle a,b,c,d.$
We can use the first four values of the function and create a system of equations.
. . $\displaystyle \begin{array}{ccccccccc}
f(\text{-}2) = 56: & \text{-}8a + 4b - 2c + d &=& \text{-}56 \\
f(\text{-}1) = 16: & \text{-}a + b - c + d &=& 16 \\
f(0) = 0: & 0 + 0 + 0 + d &=& \text{-}2 \\
f(1) = \text{-}2: & a + b + c + d &=& \text{-}2 \end{array}$
Solve the system of equations: .$\displaystyle a = 18,\;b = 9,\;c = -27,\;d = -2$
Hence, the cubic is: .$\displaystyle f(x) \;=\;18x^3 + 9x^2 - 27x - 2$
. . which goes through the first four points.
But: .$\displaystyle f(3) \:=\:18(3^3) + 9(3^2) - 27(3) - 2 \;=\;322$
. . It does not go through the fifth point.
Therefore, there is no cubic polynomial through the given six points.
we only need four points to constuct a cubic polynomial so
I choose those points which are located at $\displaystyle x = -2 , -1 , 0 , 1 $
To find the cubic polynomial which passes through those points
we need to find the first , second and the third difference ,
$\displaystyle -56 , 16 , -2 , -2 $
$\displaystyle \delta = 16 - (-56) , -2 - 16 , -2 - (-2) = 72 , -18 , 0$
$\displaystyle \delta^2 = - 90 , 18 $
$\displaystyle \delta^3 = 108 $
Therefore , by using Newton's Interpolation Formula ,
the cubic polynomial is :
$\displaystyle -56 + 72 \binom{x+2}{1} - 90\binom{x+2}{2} + 108\binom{x+2}{3} $
To check whether there is a cubic polynomial passing through the other points either , we just sub. $\displaystyle x = 3 , x = 7 $
and see if the values the polynomial gives us equal to $\displaystyle 4 $ and $\displaystyle 376 $ respectively .