# find 2 symmetrical points to the function (x+6) (x-2)

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To find symmetrical points, we first have to look for the axis of symmetry of the parabola `f(x) = (x+6)(x-2)` .

We first expand it, and express in the standard form:

`f(x) = (x+6)(x-2)`

`f(x) = x^2 +6x -2x - 12`

`f(x) = x^2 +4x - 12`

Then, we complete the square:

`f(x) = [x^2 + 4x]-12`

`f(x) = [x^2 + 4x + 4] - 12 - 4`

`f(x) = (x+2)^2 - 16`

Hence, the axis of symmetry is `x = -2.`

All points that are mirror images of each other along the axis of symmetry are symmetrical points.

If we move two units to the left and right of `x=-2` , we reach `x=0` and `x=-4` . Hence, the points `(0, -12)` and `(-4,-12)` are symmetrical. If we move to five units to the left and right, `x = 3` or `x=-7` . Hence, the points `(3, 9)` and `(-7, 9)` are also symmetrical.