Find the Domain and Range f(x)=6+3 square root of x+2

$f (x) = 6 + 3 \sqrt{x + 2}$

Set the radicand in $\sqrt{x + 2}$ greater than or equal to $0$ to find where the expression is defined.

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

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

Interval Notation:

$[- 2, \infty)$

Set-Builder Notation:

${x | x \geq - 2}$

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

Interval Notation:

$[6, \infty)$

Set-Builder Notation:

${y | y \geq 6}$

Determine the domain and range.

Domain: $[- 2, \infty), {x | x \geq - 2}$

Range: $[6, \infty), {y | y \geq 6}$