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.

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

$x = \frac{ - b}{2a}$

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.