Answer
Domain of $h$ is $(-∞, ∞)$
Range of $h$ is $(-∞,9]$
Work Step by Step
$h(x)=-x^2-4x+5$
$a=-1$, $b=-4$, $c=5$
Graphing:
The coefficient of $x$ is $-1$, therefore the parabola opens downward.
Get the vertex: $x=\frac{-b}{2a}$
$x=\frac{-(-4)}{2(-1)}$
$x=-2$
$h(-2)=-(-2^2)-4(-2)+5$
$h(-2)=9$
vertex is at $(-2,9)$
Find the $x$-intercepts by solving $h (x) = 0$.
$0=-x^2-4x+5$
Get the factors:
$-(x-1)(x+5)=0$
$-(x-1)=0$ or $x+5=0$
$x=1$ or $x=-5$
The $x$-intercepts are at points $(1,0)$ and $(-5,0)$
Domain: The graph widens and continues to fall at both ends, and thus, include all real numbers
Domain of $h$ is $(-∞, ∞)$
Range: The parabola’s vertex, (-2, 9), is the highest point on the graph. Because the y-coordinate of the vertex is $9$, outputs on the y-axis fall at or below $9$.
Range of $h$ is $(-∞,9]$
