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.

$y = -\frac{1}{2}sqrt3{x + 1}$

column by column

$x$	$x + 1$	$sqrt3{x + 1}$	$y =$
$-2$
$-1$
$0$
$7$

Explanation:

Step1: Calculate $x+1$ for each $x$

For $x=-9$: $-9+1=-8$
For $x=-2$: $-2+1=-1$
For $x=-1$: $-1+1=0$
For $x=0$: $0+1=1$
For $x=7$: $7+1=8$

Step2: Compute $\sqrt[3]{x+1}$

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

Step3: Find $y=-\frac{1}{2}\sqrt[3]{x+1}$

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

Step4: Identify domain

Cube roots accept all real numbers, so $x$ can be any real number.

Step5: Identify range

Outputs of cube roots are all real numbers, scaling by $-\frac{1}{2}$ preserves all real numbers as outputs.

Answer:

Completed Table:

$x$	$x+1$	$\sqrt[3]{x+1}$	$y=-\frac{1}{2}\sqrt[3]{x+1}$
$-2$	$-1$	$-1$	$\frac{1}{2}$
$-1$	$0$	$0$	$0$
$0$	$1$	$1$	$-\frac{1}{2}$
$7$	$8$	$2$	$-1$

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

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