Find the minimum or maximum y-value for f (x) = 3x^2 + 12x + 8.

Finding x:

x = - b / 2a, where b is the coefficient of the x term and a is the coefficient of the x2 term.

x = - 12 / (2 * 3)

x = - 2

Finding y:

To find y, substitute for x in the given function

y = 3 * (-2) 2 + 12 * (-2) + 4

y = 12 - 24 + 4

y = - 16

Vertex:

The vertex is (-2, - 16)

Since the coefficient of the x2 term is positive, we have a minimum.

The minimum is at - 16.

Your answer:

MIN,-16