Question

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

Answer 1

Answer:

(-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)