Graph f(x)< = square root of x-6+2

$f (x) \leq \sqrt{x - 6} + 2$

Find the domain for $y = \sqrt{x - 6} + 2$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

Set the radicand in $\sqrt{x - 6}$ greater than or equal to $0$ to find where the expression is defined.

$x - 6 \geq 0$

Add $6$ to both sides of the inequality.

$x \geq 6$

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

Interval Notation:

$[6, \infty)$

Set-Builder Notation:

${x | x \geq 6}$

Interval Notation:

$[6, \infty)$

Set-Builder Notation:

${x | x \geq 6}$

To find the radical expression end point, substitute the $x$ value $6$ , which is the least value in the domain, into $f (x) = \sqrt{x - 6} + 2$ .

Replace the variable $x$ with $6$ in the expression.

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

Simplify the result.

Simplify each term.

Subtract $6$ from $6$ .

$f (6) = \sqrt{0} + 2$

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

$f (6) = \sqrt{0^{2}} + 2$

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

$f (6) = 0 + 2$

Add $0$ and $2$ .

$f (6) = 2$

The final answer is $2$ .

$2$

The radical expression end point is $(6, 2)$ .

$(6, 2)$

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

Substitute the $x$ value $7$ into $f (x) = \sqrt{x - 6} + 2$ . In this case, the point is $(7, 3)$ .

Replace the variable $x$ with $7$ in the expression.

$f (7) = \sqrt{(7) - 6} + 2$

Simplify the result.

Simplify each term.

Subtract $6$ from $7$ .

$f (7) = \sqrt{1} + 2$

Any root of $1$ is $1$ .

$f (7) = 1 + 2$

Add $1$ and $2$ .

$f (7) = 3$

The final answer is $3$ .

$y = 3$

Substitute the $x$ value $8$ into $f (x) = \sqrt{x - 6} + 2$ . In this case, the point is $(8, \sqrt{2} + 2)$ .

Replace the variable $x$ with $8$ in the expression.

$f (8) = \sqrt{(8) - 6} + 2$

Simplify the result.

Subtract $6$ from $8$ .

$f (8) = \sqrt{2} + 2$

The final answer is $\sqrt{2} + 2$ .

$y = \sqrt{2} + 2$

The square root can be graphed using the points around the vertex $(6, 2), (7, 3), (8, 3.41)$

$\begin{array}{c} x & y \\ 6 & 2 \\ 7 & 3 \\ 8 & 3.41 \end{array}$

Graph f(x)< = square root of x-6+2

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