## Intermediate Algebra for College Students (7th Edition)

Domain of $g$ is $(-∞, ∞)$ Range of $g$ is $(-∞,5]$
$g(x)=5-(x+2)^2$ 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 $(-∞,5]$