Find the Domain and Range f(x)=3/4 square root of x+12+3

$f (x) = \frac{3}{4} \sqrt{x + 12} + 3$

Step 1

Set the radicand in $\sqrt{x + 12}$ greater than or equal to $0$ to find where the expression is defined.

$x + 12 \geq 0$

Step 2

Subtract $12$ from both sides of the inequality.

$x \geq - 12$

Step 3

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

Interval Notation:

$[- 12, \infty)$

Set-Builder Notation:

${x | x \geq - 12}$

Step 4

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

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${y | y \geq 3}$

Step 5

Determine the domain and range.

Domain: $[- 12, \infty), {x | x \geq - 12}$

Range: $[3, \infty), {y | y \geq 3}$

Step 6