Why does one and only one parabola pass through the three points given by a function?

Question

I am doing integration by Simpson's method and in the question it says that one and only one parabola is associated with the three points given by the function. Can someone please show me how this is so?

Answer 1

If the parabola $y = a + bx + cx^2$ passes through the points $(x_0, y_0)$, $(x_1, y_1)$, and $(x_2, y_2)$, with $x_0 < x_1 < x_2$, then $$ a(1,1,1) + b(x_0, x_1, x_2) + c(x_0^2, x_1^2, x_2^2) = (y_1, y_2, y_3). $$ Now, $\boldsymbol{(1,1,1), (x_0, x_1, x_2),}$ and $\boldsymbol{(x_0^2, x_1^2, x_2^2)}$ are linearly independent,$^{1}$ So there is only one way to write any vector as a linear combination of them. In particular, $a, b,$ and $c$ are determined.

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$^{1}$ To prove this, you can calculate the determinant: $$ \begin{vmatrix} 1 & 1 & 1 \\ x_0 & x_1 & x_2 \\ x_0^2 & x_1^2 & x_2^2 \end{vmatrix}. $$ It should come out to $$ (x_2 - x_1)(x_3 - x_1)(x_3 - x_2), $$ which is nonzero because $x_1 < x_2 < x_3$.

Answer 2

We know that two parabolas can have a maximum of 4 intersection points so if this is true and we consider any three particular points. There could be 2 parabolas passing through 3 points.