Question

Find the equation of the cubic curve that passes through the points (4, – 43), (7, 83), (9, 327) and (12, 1053).

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} x: & 4 & 7 & 9 & 12 \\ \hline y: & -43 & 83 & 327 & 1053 \\ \hdashline \end{array}$

By Lagrange’s interpolation formula we have:

$f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3= \frac{(x-7)(x-9)(x-12)}{(4-7)(4-9)(4-12)}(-43)+ \frac{(x-4)(x-9)(x-12)}{(7-4)(7-9)(7-12)} 83+ \frac{(x-4)(x-7)(x-12)}{(9-4)(9-7)(9-12)}327+ \frac{(x-4)(x-7)(x-9)}{(12-4)(12-7)(12-9)} 1053=\frac{43}{120}(x^3-28x^2+255x-756) +\frac{83}{30}(x^3-25x^2+192x-432)-\frac{109}{10}(x^3-23x^2+160x-336)+\frac{351}{40}(x^3-20x^2+127x-252)=x^3-4x^2-7x-15$

Answer: $f(x)=x^3-4x^2-7x-15.$