Question

Find the minimum or maximum y-value for f(x)=3x^2+12x+4.

Answer 1

The vertex is (-2,-8)
So the y-value would be -8

Answer 2

`\bf \huge \red{ANSWER}`

For the function f(x) =
`{3x}^(2) + 12x + 4`

, we need to find the vertex. The vertex is found by first finding
`x`, and then substituting for
`x` in the function.

`\bf \huge \blue{Finding \: x}`

`x = ( - b)/(2a)`

where b is the coefficient of the
`x` term and a is the coefficient of the
`{x}^(2)` term.

`x = \frac{ - 12}{ {2}^(3) }`

`x = - 2`

`\bf \huge \green{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`

`\bf \huge \purple{vertex}`

The vertex is
`( - 2 - 16)`

Since the coefficient of the
`{x}^(2)`

term is positive, we have a minimum.

The minimum is at -16.