For the equation y=3x^2+12x-4, what is the minimum value?

(-2, - 16)

Step-by-step explanation:

the minimum is at the vertex which can be found using - b/2a for the x-value of the vertex.

y = 3x^2 + 12x - 4

= ax^2 + bx - c

b = 12, a = 3

-12 / (2 (3))

-12/6

x = - 2

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

y = - 16

(-2, - 16)