Graph y = square root of x-2-5

$y = \sqrt{x - 2} - 5$

Step 1

Find the domain for $y = \sqrt{x - 2} - 5$ 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 $\sqrt{x - 2}$ greater than or equal to $0$ to find where the expression is defined.

$x - 2 \geq 0$

Add $2$ to 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:

${x | x \geq 2}$

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${x | x \geq 2}$

Step 2

To find the radical expression end point, substitute the $x$ value $2$ , which is the least value in the domain, into $f (x) = \sqrt{x - 2} - 5$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = \sqrt{(2) - 2} - 5$

Simplify the result.

Simplify each term.

Subtract $2$ from $2$ .

$f (2) = \sqrt{0} - 5$

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

$f (2) = \sqrt{0^{2}} - 5$

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

$f (2) = 0 - 5$

Subtract $5$ from $0$ .

$f (2) = - 5$

The final answer is $- 5$ .

$- 5$

Step 3

The radical expression end point is $(2, - 5)$ .

$(2, - 5)$

Step 4

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 $3$ into $f (x) = \sqrt{x - 2} - 5$ . In this case, the point is $(3, - 4)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = \sqrt{(3) - 2} - 5$

Simplify the result.

Simplify each term.

Subtract $2$ from $3$ .

$f (3) = \sqrt{1} - 5$

Any root of $1$ is $1$ .

$f (3) = 1 - 5$

Subtract $5$ from $1$ .

$f (3) = - 4$

The final answer is $- 4$ .

$y = - 4$

Substitute the $x$ value $4$ into $f (x) = \sqrt{x - 2} - 5$ . In this case, the point is $(4, \sqrt{2} - 5)$ .

Replace the variable $x$ with $4$ in the expression.

$f (4) = \sqrt{(4) - 2} - 5$

Simplify the result.

Subtract $2$ from $4$ .

$f (4) = \sqrt{2} - 5$

The final answer is $\sqrt{2} - 5$ .

$y = \sqrt{2} - 5$

The square root can be graphed using the points around the vertex $(2, - 5), (3, - 4), (4, - 3.59)$

$\begin{array}{c} x & y \\ 2 & - 5 \\ 3 & - 4 \\ 4 & - 3.59 \end{array}$

Step 5

Graph y = square root of x-2-5

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