Question

y = 3x² + 12x + 8
xis of symmetry:
vertex: ( , )
max or min value:
x | -1 | 0 | 1 | 2 | 3
y | | | | |

Explanation:

Step1: Verify axis of symmetry

For quadratic $y=ax^2+bx+c$, axis is $x=-\frac{b}{2a}$.
$x=-\frac{12}{2\times3}=-2$

Step2: Verify vertex y-value

Substitute $x=-2$ into the equation:
$y=3(-2)^2+12(-2)+8=12-24+8=-4$

Step3: Identify max/min

Since $a=3>0$, parabola opens up, so it has a Min value (not Max).

Step4: Calculate y-values for table

Sub $x=-1$:

$y=3(-1)^2+12(-1)+8=3-12+8=-1$

Sub $x=0$:

$y=3(0)^2+12(0)+8=8$

Sub $x=1$:

$y=3(1)^2+12(1)+8=3+12+8=23$

Sub $x=2$:

$y=3(2)^2+12(2)+8=12+24+8=44$

Sub $x=3$:

$y=3(3)^2+12(3)+8=27+36+8=71$

Answer:

Axis of Symmetry: $\boldsymbol{-2}$ (correct as given)
Vertex: $\boldsymbol{(-2, -4)}$ (correct as given)
Max or Min Value: $\boldsymbol{Min}$ (corrected from Max)
Table y-values:

$x$	$-1$	$0$	$1$	$2$	$3$