# Show that for any choice of `y_1, y_2, y_3` , there is one and only one parabola `y = ax^2+bx+c` passing by the points `(1, y_1)` , `(2, y_2)` and `(3, y_3)` .

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If parabola `y=ax^2+bx+c` passes through point `(x_1,y_1)` then it must satisfy the following equation

`ax_1^2+bx_1+c=y_1`

That is if we put `x_1` instead of `x` we must get `y_1.`

From three points we have we get system of three equations.

`a+b+c=y_1`

`4a+2b+c=y_2`

`9a+3b+c=y_3`

In matrix form that would be

`[[1,1,1],[4,2,1],[9,3,1]][[a],[b],[c]]=[[y_1],[y_2],[y_3]]`

This system will have unique solution if determinant of its matrix is not equal to zero. Let us then calculate that determinant by using Sarrus rule (you can also use Laplace expansion).

`|[1,1,1],[4,2,1],[9,3,1]|=2+9+12-18-3-4=-2 ne 0`

**Hence our system has exactly one solution, that is our three points define exactly one polynomial of degree less than 3.** Note that our solution will not necessarily be parabola, if all three points lie on a line we will get `a=0` hence a line.