Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the function’s domain and range.

\\( y = -\frac{1}{2}sqrt3{x - 4} \\)

\\( x \\)	\\( x - 4 \\)	\\( sqrt3{x - 4} \\)	\\( y = \\)
\\( 3 \\)
\\( 4 \\)
\\( 5 \\)
\\( 12 \\)

Explanation:

Step1: Calculate $x-4$ for each $x$

For $x=-4$: $-4 - 4 = -8$
For $x=3$: $3 - 4 = -1$
For $x=4$: $4 - 4 = 0$
For $x=5$: $5 - 4 = 1$
For $x=12$: $12 - 4 = 8$

Step2: Compute cube roots of results

For $x-4=-8$: $\sqrt[3]{-8} = -2$
For $x-4=-1$: $\sqrt[3]{-1} = -1$
For $x-4=0$: $\sqrt[3]{0} = 0$
For $x-4=1$: $\sqrt[3]{1} = 1$
For $x-4=8$: $\sqrt[3]{8} = 2$

Step3: Calculate $y$ using $y=-\frac{1}{2}\sqrt[3]{x-4}$

For $\sqrt[3]{x-4}=-2$: $y=-\frac{1}{2}(-2)=1$
For $\sqrt[3]{x-4}=-1$: $y=-\frac{1}{2}(-1)=\frac{1}{2}$
For $\sqrt[3]{x-4}=0$: $y=-\frac{1}{2}(0)=0$
For $\sqrt[3]{x-4}=1$: $y=-\frac{1}{2}(1)=-\frac{1}{2}$
For $\sqrt[3]{x-4}=2$: $y=-\frac{1}{2}(2)=-1$

Step4: Identify domain and range

Cube roots accept all real $x$, so domain is all real numbers. The cube root outputs all real numbers, scaled by $-\frac{1}{2}$, so range is all real numbers.

Completed Table:

$x$	$x-4$	$\sqrt[3]{x-4}$	$y$
$3$	$-1$	$-1$	$\frac{1}{2}$
$4$	$0$	$0$	$0$
$5$	$1$	$1$	$-\frac{1}{2}$
$12$	$8$	$2$	$-1$

Answer:

Domain: All real numbers, or $(-\infty, \infty)$
Range: All real numbers, or $(-\infty, \infty)$