Answer
Vertex: $(-2, 4)$
Opens upward
x-intercept: none
y-intercept: 16
Work Step by Step
$f(x)=3x^2+12x+16$
$f(x)=3(x^2+4x)+16$
$f(x)=3(x^2+4x+(4/2)^2)+16-3*(4/2)^2)$
$f(x)=3(x^2+4x+(2)^2)+16-3*(2)^2)$
$f(x)=3(x^2+4x+4)+16-3*(4)$
$f(x)=3(x+2)^2+16-12$
$f(x)=3(x+2)^2+4$
Vertex: $(-2, 4)$
Opens upward
$x=0$
$f(x)=3x^2+12x+16$
$f(0)=3*0^2+12*0+16$
$f(0)=3*0+0+16$
$f(0)=0+16$
$f(0)=16$
Since the vertex of the graph has a minimum y-value of 4 and opens up, there are no lower points. Thus, there are no x-intercepts.
y-intercept: 16
