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.

$y = 2^{x+1}$

$x$	$x + 1$	$y = 2^{x+1}$
$-2$
$-1$
$0$
$1$
$2$

Explanation:

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

For $x=-3$: $-3 + 1 = -2$
For $x=-2$: $-2 + 1 = -1$
For $x=-1$: $-1 + 1 = 0$
For $x=0$: $0 + 1 = 1$
For $x=1$: $1 + 1 = 2$
For $x=2$: $2 + 1 = 3$

Step2: Calculate $y=2^{x+1}$ for each value

For $x=-3$: $2^{-2} = \frac{1}{2^2} = \frac{1}{4}$
For $x=-2$: $2^{-1} = \frac{1}{2^1} = \frac{1}{2}$
For $x=-1$: $2^{0} = 1$
For $x=0$: $2^{1} = 2$
For $x=1$: $2^{2} = 4$
For $x=2$: $2^{3} = 8$

Step3: Identify domain of the function

The function $y=2^{x+1}$ is an exponential function, which accepts all real numbers as input for $x$.

Step4: Identify range of the function

Exponential functions of the form $a^{kx+c}$ (where $a>1$) produce only positive real numbers, and approach 0 but never reach it.

Answer:

Completed Table:

$x$	$x+1$	$y=2^{x+1}$
$-2$	$-1$	$\frac{1}{2}$
$-1$	$0$	$1$
$0$	$1$	$2$
$1$	$2$	$4$
$2$	$3$	$8$

Domain and Range:

• Domain: All real numbers, or $(-\infty, \infty)$
• Range: All positive real numbers, or $(0, \infty)$