Write a quadratic function in standard form whose graph passes through the given points : (2,6) (-2,-2) (1,1)Help !

Let us assume that the quadratic function is f(x) such
that:

`f(x)= ax^2 + bx + c`

``
Given the points: (2, 6) ( -2, -2) and (1,1).

Since the
points are on the graph, then they satisfy the
equation.

Then we will substitute with each point in
f(x).

`f(2)= a(2^2) + b*2 + C
`

`==> 4a + 2b + C = 6
.......(1)`

`f(-2) = a(-2^2) + b*-2 + C
`

`==> 4a - 2b + C = -2
............(2)`

`f(1) = a(1^2) + b*1 +
C`

`==> a + b + C = 1
................(3)`

Now we will solve the
system.

First we will subtract (2) from
(1).

==> 4b = 8 ==> b=
2

Now we will subtract (3) from
(2).

==> 3a -3b =
-3

==> a - b= -1

But b=
2

==> a - 2 = -1 ==> a =
1

==> a+ b+ c =
1

==> 1 +2 + C =
2

==> C = -2

Then the
function is:

`f(x)= x^2 + 2x -2
`

``