I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$.
I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as.
If anyone could provide their working out for this problem it would be extremely enlightening.
Best Answer
Given four points $(x_i,y_i)$ consider the functions $$f_1(x)=\frac {(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$ so that $f_1(x_1)=1$ and $f_1(x_i)=0, i\neq 1$, and similarly $f_2, f_3, f_4$. Note that the $f_i$ are cubic in $x$.
Then $p(x)=y_1f_1(x)+y_2f_2(x)+y_3f_3(x)+y_4f_4(x)$ is at most a cubic polynomial and passes through the four given points.