Let f(x)=3x^2 - 4x - x^2 + 7x + 8. Find the constant k such that f(x)=f(k-x) for all real numbers x.
+130071 3x^2 -4x -x^2 + 7x + 8 simplifies to
2x^2 + 3x + 8 = f(k -x)
So
2x^2 + 3x + 8 = 2(k - x)^2 + 3(k -x) + 8
2x^2 + 3x + 8 = 2 ( k^2 - 2kx + x^2) + 3k -3x + 8
2x^2 + 3x = 2k^2 - 4kx + 2x^2 + 3k - 3x
2k^2 -4kx + 3k -6x= 0
2k ( k - 2x) + 3 ( k - 2x) = 0
(2k + 3) ( k - 2x) = 0 the first factor set = to 0 will give us what we need
2k + 3 = 0
2k = -3
k = -3/2
