Find the Domain and Range y=-1/3 square root of x+1

$y = - \frac{1}{3} \sqrt{x + 1}$

Step 1

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Step 2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, 0]$

Set-Builder Notation:

${y | y \leq 0}$

Step 5

Determine the domain and range.

Domain: $[- 1, \infty), {x | x \geq - 1}$

Range: $(- \infty, 0], {y | y \leq 0}$

Step 6