Find the Domain and Range y = square root of 8x+1

$y = \sqrt{8 x + 1}$

Step 1

Set the radicand in $\sqrt{8 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$8 x + 1 \geq 0$

Step 2

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$8 x \geq - 1$

Divide each term in $8 x \geq - 1$ by $8$ and simplify.

Divide each term in $8 x \geq - 1$ by $8$ .

$\frac{8 x}{8} \geq \frac{- 1}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{- 1}{8}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{8}$

Step 3

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

Interval Notation:

$[- \frac{1}{8}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{1}{8}}$

Step 4

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Step 5

Determine the domain and range.

Domain: $[- \frac{1}{8}, \infty), {x | x \geq - \frac{1}{8}}$

Range: $[0, \infty), {y | y \geq 0}$

Step 6