Find the Domain and Range square root of -x+2

$\sqrt{- x} + 2$

Step 1

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

$- x \geq 0$

Step 2

Divide each term in $- x \geq 0$ by $- 1$ and simplify.

Divide each term in $- x \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{0}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x \leq 0$

Step 3

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

Interval Notation:

$(- \infty, 0]$

Set-Builder Notation:

${x | x \leq 0}$

Step 4

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

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${y | y \geq 2}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 0], {x | x \leq 0}$

Range: $[2, \infty), {y | y \geq 2}$

Step 6

Find the Domain and Range square root of -x+2

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