This could probably be related to calculus and derivatives, but that might be too advanced for your audience.
You could always show the algebra. For simplicity, use x - 1, x, and x + 1 for your three "typical" x-values. Then your three quadratic terms are:
. . . . .\(\displaystyle \mbox{first term: }\, a(x\, -\, 1)^2\, +\, b(x\, -\, 1)\, +\, c\, =\, ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c\)
. . . . .\(\displaystyle \mbox{second term: }\, ax^2\, +\, bx\, +\, c\)
. . . . .\(\displaystyle \mbox{third term: }\, a(x\, +\, 1)^2\, +\, b(x\, +\, 1)\, +\, c\, =\, ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c\)
Then subtract the first from the second:
. . . . .\(\displaystyle \left(ax^2\, +\, bx\, +\, c \right)\, -\, \left(ax^2\, -\, 2ax\, +\, bx\, +\, a\, -\, b\, +\, c \right)\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, -\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, -\, a\, +\, b\)
...and the second from the third:
. . . . .\(\displaystyle \left(ax^2\, +\, 2ax\, +\, bx\, +\, a\, +\, b\, +\, c \right)\, -\, \left(ax^2\, +\, bx\, +\, c \right)\, =\,\)
. . . . .\(\displaystyle =\, ax^2\, -\, ax^2\, +\, bx\, -\, bx\, +\, c\, -\, c\, +\, 2ax\, +\, a\, +\, b\)
. . . . .\(\displaystyle =\, 2ax\, +\, a\, +\, b\)
Then show the subtraction of these results. You'll be left with a constant.
Have fun!