3 years ago

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

Mathematics

1 answer:

Neko [114]3 years ago

4 0

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)

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