Graph f(x)=|-x|^(1/2)

$f (x) = {| - x |}^{\frac{1}{2}}$

Find the absolute value vertex. In this case, the vertex for $y = {| x |}^{\frac{1}{2}}$ is $(0, 0)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x$ equal to $0$ . In this case, $x = 0$ .

$x = 0$

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

$y = {| 0 |}^{\frac{1}{2}}$

Simplify ${| 0 |}^{\frac{1}{2}}$ .

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 0^{\frac{1}{2}}$

Simplify the expression.

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

$y = {(0^{2})}^{\frac{1}{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 0^{2 (\frac{1}{2})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = 0^{2 (\frac{1}{2})}$

Rewrite the expression.

$y = 0$

Evaluate the exponent.

$y = 0$

The absolute value vertex is $(0, 0)$ .

$(0, 0)$

Find the domain for $f (x) = {| x |}^{\frac{1}{2}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the absolute value function.

Convert expressions with fractional exponents to radicals.

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\sqrt{{| x |}^{1}}$

Anything raised to $1$ is the base itself.

$\sqrt{| x |}$

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

$| x | \geq 0$

Solve for $x$ .

Write $| x | \geq 0$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 0$

Write as a piecewise.

${\begin{cases} x \geq 0 & x \geq 0 \\ - x \geq 0 & x < 0 \end{cases}$

Find the intersection of $x \geq 0$ and $x \geq 0$ .

$x \geq 0$

Solve $- x \geq 0$ when $x < 0$ .

Divide each term in $- x \geq 0$ by $- 1$ and simplify.

Divide each term in $- x \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{0}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x \leq 0$

Find the intersection of $x \leq 0$ and $x < 0$ .

$x < 0$

Find the union of the solutions.

All real numbers

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

Substitute the $x$ value $- 2$ into $f (x) = {| x |}^{\frac{1}{2}}$ . In this case, the point is $(- 2, 2^{\frac{1}{2}})$ .

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

$f (- 2) = {| - 2 |}^{\frac{1}{2}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (- 2) = 2^{\frac{1}{2}}$

The final answer is $2^{\frac{1}{2}}$ .

$y = 2^{\frac{1}{2}}$

Substitute the $x$ value $- 1$ into $f (x) = {| x |}^{\frac{1}{2}}$ . In this case, the point is $(- 1, 1)$ .

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

$f (- 1) = {| - 1 |}^{\frac{1}{2}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (- 1) = 1^{\frac{1}{2}}$

One to any power is one.

$f (- 1) = 1$

The final answer is $1$ .

$y = 1$

Substitute the $x$ value $2$ into $f (x) = {| x |}^{\frac{1}{2}}$ . In this case, the point is $(2, 2^{\frac{1}{2}})$ .

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

$f (2) = {| 2 |}^{\frac{1}{2}}$

Simplify the result.

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (2) = 2^{\frac{1}{2}}$

The final answer is $2^{\frac{1}{2}}$ .

$y = 2^{\frac{1}{2}}$

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, 1.41), (- 1, 1), (1, 1), (2, 1.41)$

$\begin{array}{c} x & y \\ - 2 & 1.41 \\ - 1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 1.41 \end{array}$

Graph f(x)=|-x|^(1/2)

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