#### Answer

opens down and wider

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To identify the opening of the given quadratic function, $
f(x)=-\dfrac{1}{3}(x+6)^2+3
,$ compare $a$ with $0$. If $a$ is greater than $0,$ the graph opens up. Otherwise, it opens down. To determine if the graph is wider, narrower, or the same shape as the graph of $f(x)=x^2,$ compare $|a|$ with $1.$ If it is less than $1,$ the graph is wider. If is greater than $1,$ the graph is narrower. If it is equal to $1,$ then the graph has the same shape.
$\bf{\text{Solution Details:}}$
In the given function, the value of $a$ is $a=
-\dfrac{1}{3}
.$ Since $a
\gt0
,$ then the graph opens $\text{
down
.}$
In the given function, the value of $|a|$ is $|a|=
\dfrac{1}{3}
.$ Since $|a|
\lt1
,$ then the graph is $\text{
wider
}$ than the graph of $f(x)=x^2.$
Hence, the given function has a parabola that $\text{
opens down and wider
}$ than $f(x)=x^2.$