g(X)=2x^2+8x

Question

find the minimum value or maximum value and give the coordinates of the vertex

Answer 1

g(x)
= 2x^2 + 8x
= 2(x^2 + 4x)
= 2(x + 2)^2 - 2(4)
= 2(x + 2)^2 - 8

Notice that (x + 2)^2 >= 0 for all value of x due to the square.
Thus, the minimum value will occur when 2(x + 2)^2 = 0.
When 2(x + 2)^2 = 0, g(x) = -8.
When 2(x + 2)^2 = 0, x = -2.

Hence, the minimum value of g(x) is -8, and the vertex is at (-2, -8).