Graph g(x) = square root of x-2

$g (x) = \sqrt{x} - 2$

Step 1

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

$x \geq 0$

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 2

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = \sqrt{x} - 2$ .

Replace the variable $x$ with $0$ in the expression.

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

Simplify the result.

Remove parentheses.

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

Simplify each term.

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

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

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

$f (0) = 0 - 2$

Subtract $2$ from $0$ .

$f (0) = - 2$

The final answer is $- 2$ .

$- 2$

Step 3

The radical expression end point is $(0, - 2)$ .

$(0, - 2)$

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 $1$ into $f (x) = \sqrt{x} - 2$ . In this case, the point is $(1, - 1)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = \sqrt{1} - 2$

Simplify the result.

Remove parentheses.

$f (1) = \sqrt{1} - 2$

Any root of $1$ is $1$ .

$f (1) = 1 - 2$

Subtract $2$ from $1$ .

$f (1) = - 1$

The final answer is $- 1$ .

$y = - 1$

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

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

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

Simplify the result.

Remove parentheses.

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

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

$y = \sqrt{2} - 2$

The square root can be graphed using the points around the vertex $(0, - 2), (1, - 1), (2, - 0.59)$

$\begin{array}{c} x & y \\ 0 & - 2 \\ 1 & - 1 \\ 2 & - 0.59 \end{array}$

Step 5

Graph g(x) = square root of x-2

Related Questions

Support LCMGCF.com by disabling your adblocker.