#### Answer

Domain of $g$ is $(-∞, ∞)$
Range of $g$ is $(-∞,5]$

#### Work Step by Step

$g(x)=5-(x+2)^2$
Expand.
$5-(x^2+4x+4)$
$-x^2-4x+1$
$a=-1$, $b=-4$, $c=1$
Graphing:
The coefficient of $x$ is $-1$, therefore the parabola opens downward.
Get the vertex: $x=\frac{-b}{2a}$
$x=\frac{-(-4)}{2(-1)}$
$x=-2$
$g(-2)=5-(-2+2)^2$
$g(-2)=5$
vertex is at $(-2,5)$
Find the $x$-intercepts by solving $g (x) = 0$.
$0=5-(x+2)^2$
$(x+2)^2=5$
Apply the square root property.
$x+2=±\sqrt{5}$
$x=\sqrt{5}-2$ or $x=-\sqrt{5}-2$
The $x$-intercepts are at points $(\sqrt{5}-2,0)$ and $(-\sqrt{5}-2,0)$
Domain: The graph widens and continues to fall at both ends, and thus, include all real numbers
Domain of $g$ is $(-∞, ∞)$
Range: The parabola’s vertex, (-2, 5), is the highest point on the graph. Because the y-coordinate of the vertex is $5$, outputs on the y-axis fall at or below $5$.
Range of $g$ is $(-∞,5]$