For all real numbers \(x,\) find the minimum value of \((x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2.\)
+6248 my intuition tells me that the minimum will occur when x is the average of where the zeros of each occur.
\(x = \dfrac 1 5(-12-7-3+4+8) = -2\\~\\ v = 262\)
This is the correct answer which you can confirm (and should confirm) by expanding the expression out and
recombining it into a single quadratic expression and determining the vertex of the resulting parabola.
