Find Three Ordered Pair Solutions f(x) = square root of x+4

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

Step 1

Write $f (x) = \sqrt{x} + 4$ as an equation.

$y = \sqrt{x} + 4$

Step 2

Choose any value for $x$ that is in the domain to plug into the equation.

Step 3

Choose $0$ to substitute in for $x$ to find the ordered pair.

Simplify $\sqrt{0} + 4$ .

Remove parentheses.

$y = \sqrt{0} + 4$

Simplify each term.

Rewrite $0$ as $0^{2}$ .

$y = \sqrt{0^{2}} + 4$

Pull terms out from under the radical, assuming positive real numbers.

$y = 0 + 4$

Add $0$ and $4$ .

$y = 4$

Use the $x$ and $y$ values to form the ordered pair.

$(0, 4)$

Step 4

Choose $1$ to substitute in for $x$ to find the ordered pair.

Simplify $\sqrt{1} + 4$ .

Remove parentheses.

$y = \sqrt{1} + 4$

Any root of $1$ is $1$ .

$y = 1 + 4$

Add $1$ and $4$ .

$y = 5$

Use the $x$ and $y$ values to form the ordered pair.

$(1, 5)$

Step 5

Choose $2$ to substitute in for $x$ to find the ordered pair.

Remove parentheses.

$y = \sqrt{2} + 4$

Use the $x$ and $y$ values to form the ordered pair.

$(2, \sqrt{2} + 4)$

Step 6

These are three possible solutions to the equation.

$(0, 4), (1, 5), (2, \sqrt{2} + 4)$

Step 7