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