Find the Domain square root of -8x+16

$\sqrt{- 8 x + 16}$

Set the radicand in $\sqrt{- 8 x + 16}$ greater than or equal to $0$ to find where the expression is defined.

$- 8 x + 16 \geq 0$

Solve for $x$ .

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Subtract $16$ from both sides of the inequality.

$- 8 x \geq - 16$

Divide each term by $- 8$ and simplify.

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Divide each term in $- 8 x \geq - 16$ by $- 8$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 8 x}{- 8} \leq \frac{- 16}{- 8}$

Cancel the common factor of $- 8$ .

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Cancel the common factor.

$\frac{- 8 x}{- 8} \leq \frac{- 16}{- 8}$

Divide $x$ by $1$ .

$x \leq \frac{- 16}{- 8}$

Divide $- 16$ by $- 8$ .

$x \leq 2$

The solution consists of all of the true intervals.

$x \leq 2$

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

Interval Notation:

$(- \infty, 2]$

Set-Builder Notation:

${x | x \leq 2}$

Find the Domain square root of -8x+16

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