已知y = x^2 + 2kx + k的對稱軸是x + 4 = 0,求該函數的極值

Notice that
y = x^2 + 2kx + k = x^2 + 2kx + k^2 + k - k^2
= (x + k)^2 + k - k^2
Since the axis of symmetry is x + 4 = 0, we have k = 4
and thus it has a global minimum value of
4 - 4^2 = -12

thx!