Find the Domain y = square root of 9+8x

$y = \sqrt{9 + 8 x}$

Step 1

Set the radicand in $\sqrt{9 + 8 x}$ greater than or equal to $0$ to find where the expression is defined.

$9 + 8 x \geq 0$

Step 2

Solve for $x$ .

Subtract $9$ from both sides of the inequality.

$8 x \geq - 9$

Divide each term in $8 x \geq - 9$ by $8$ and simplify.

Divide each term in $8 x \geq - 9$ by $8$ .

$\frac{8 x}{8} \geq \frac{- 9}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{- 9}{8}$

Divide $x$ by $1$ .

$x \geq \frac{- 9}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{9}{8}$

Step 3

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

Interval Notation:

$[- \frac{9}{8}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{9}{8}}$

Step 4