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

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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 `

``