Parabola

Interpolate parabola \(y=ax^2+bx+c\) through points:

Variant 1
\((-1,-9)\), \((1,-3)\) and \((2{,}3)\)
Resolution
For the equation \(y=ax^2+bx+c\) we seek coefficients \(a\), \(b\) and \(c\), s.t. for \(x=-1\) they satisfy \(a-b+c=-9\). Similarly, for \(x=1\) and \(x=2\) we get equations \(a+b+c=-3\) and \(4a+2b+c=3\).

From the first pair we deduce that \(b=3\). Then we get \(a=1\) from the second and the third. Finally we obtain that \(c=-7\).
Result
The parabola equation is \(y=x^2+3x-7\).
Variant 2
\((-1{,}10)\), \((1{,}4)\) and \((4{,}25)\)
Result
\(y=2x^2-3x+5\)
Variant 3
\((-1{,}0)\), \((1{,}4)\) and \((2,-6)\)
Result
\(y=-4x^2+2x+6\)
Variant 4
\((-2,-7)\), \((3,-17)\) and \((4,-37)\)
Result
\(y=-3x^2+x+7\)
Variant 5
\((-3,-33)\), \((2,-8)\) and \((4,-68)\)
Result
\(y=-5x^2+12\)
Variant 6
\((-2,-27)\), \((4{,}15)\) and \((7{,}36)\)
Result
\(y=7x-13\), (it is a line, i.e. a "degnerate parabola")
Variant 7
\((5{,}20)\), \((6{,}30)\) and \((11{,}110)\)
Result
\(y=x^2-x\)