#### Answer

Domain of $f$ is (-∞, ∞)
Range of $f$ is $[-4,∞)$

#### Work Step by Step

$f(x)=(x-3)^2-4$
Expand. $x^2-6x+5$
$a=1$, $b=-6$, $c=5$
Graphing:
The coefficient of $x$ is $1$, therefore the parabola that opens upward.
Get the vertex: $x=\frac{-b}{2a}$
$x=\frac{-(-6)}{2(1)}$
$x=3$
$f(3)=(3-3)^2-4$
$f(3)=-4$
vertex is at $(3,-4)$
Find the $x$-intercepts by solving $f (x) = 0$.
$0=(x-3)^2-4$
$(x-3)^2=4$
Apply the square root property.
$x-3=±\sqrt{4}$
$x=2+3$ or $x=-2+3$
$x=5$ or $x=1$
The $x$-intercepts are at points $(5,0)$ and $(1,0)$
Domain: The graph widens and continues to fall at both ends, and thus, include all real numbers
Domain of $f$ is (-∞, ∞).
Range: The parabola’s vertex, (3, -4), is the lowest point on the graph. Because the y-coordinate of the vertex is $-4$, outputs on the y-axis fall at or above $-4$.
Range of $f$ is $[-4,∞)$