$y=ax^3+bx^2+cx+d$ cubic function, goes through these points $(0,1) ,(-1,-2), (1,2), (2,9)$

Question

How can I solve it without using matrix? I tried it to solve by using systems. But I have no idea how deal with "$0$"

Answer 1

Just solve the system$$\left\{\begin{array}{l}d=1\\-a+b-c+d=-2\\a+b+c+d=2\\8a+4b+2c+d=9.\end{array}\right.$$

Answer 2

You can substitute every point in the equation and solve a system in 4 unknown (a,b,c,d) with four equation, one for every point. The system is $$d=1$$ $$-a+b-c+d=-2$$ $$a+b+c+d=2$$ $$8a+4b+2c+d=9$$

so the point with $0$ tells you that $d=1$. From this point you can use every method you want to solve it. (Putting it on a linear system solver I find that $a=\frac{4}{3}$, $b=-1$, $c=\frac{2}{3}$ and $d=1$).

Answer 3

Given equation, $$y = ax^3 + bx^2 + cx + d$$ ,On putting the points $(0,1),(-1,-2),(1,2),(2,9)$ following system of equation are obtained $$\begin{align} d &=1\\ -a+b-c+d &=-2\\a+b+c+d &=2\\8a+4b+2c+d &=9.\end{align}.$$

On adding $(2)$ and $(3)$ and using $(1)$, $ 2b + 2 = 0 \Rightarrow b = -1$

On multiplying $(3)$ by $8$ ,subtracting it from $(4)$, and using value of $b$, gives $c = 2/3$ , and finally placing all this value in $(2)$ gives $a = 4/3$

Hence, values are $ a = 4/3, b = -1 , c = 2/3$ and $d = 1$

Answer 4

Since we have the values of the polynomial at successive integers (-1,0,1 and 2), one way to find the interpolating polynomial is to use a table of finite differences: $$\begin{matrix} -2\\ &3\\ 1 & &-2\\ &1 & &8\\ 2 & &6\\ &7\\ 9 \end{matrix}$$ Each column in the table is derived from the the previous column by applying the forward difference operator $\Delta(x) = f(x+1) - f(x)$. We can then plug the differences on the upper diagonal into Newton's Forward Difference Formula: $$f(s) = -2 + (3) s + (-2)(1/2!) s(s-1) + 8 (1/3!)s(s-1)(s-2)$$ But the Newton formula assumes the series of $s$ values starts at $0$, whereas our data starts at $-1$, so we need to make the substitution $s=x+1$ in order to fit the data given: $$f(x) = -2 + (3) (x+1) + (-2)(1/2!)(x+1)x + 8(1/3!) (x+1)x(x-1)$$

Expanding this polynomial will yield the coefficients $a,b,c$ and $d$.