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

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

Step 1

Set the radicand in $\sqrt{4 x - 12}$ greater than or equal to $0$ to find where the expression is defined.

$4 x - 12 \geq 0$

Step 2

Solve for $x$ .

Add $12$ to both sides of the inequality.

$4 x \geq 12$

Divide each term in $4 x \geq 12$ by $4$ and simplify.

Divide each term in $4 x \geq 12$ by $4$ .

$\frac{4 x}{4} \geq \frac{12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{12}{4}$

Divide $x$ by $1$ .

$x \geq \frac{12}{4}$

Simplify the right side.

Divide $12$ by $4$ .

$x \geq 3$

Step 3

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

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${x | x \geq 3}$

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

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

Step 6