Graph H(x) = square root of -x-5

$H (x) = \sqrt{- x} - 5$

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

$- x \geq 0$

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 $- 1$ .

$x \leq 0$

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

Interval Notation:

$(- \infty, 0]$

Set-Builder Notation:

${x | x \leq 0}$

Interval Notation:

$(- \infty, 0]$

Set-Builder Notation:

${x | x \leq 0}$

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = \sqrt{- x} - 5$ .

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

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

Simplify the result.

Simplify each term.

Multiply $- 1$ by $0$ .

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

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

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

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

$f (0) = 0 - 5$

Subtract $5$ from $0$ .

$f (0) = - 5$

The final answer is $- 5$ .

$- 5$

The radical expression end point is $(0, - 5)$ .

$(0, - 5)$

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

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

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

Simplify the result.

Multiply $- 1$ by $- 2$ .

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

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

$y = \sqrt{2} - 5$

Substitute the $x$ value $- 1$ into $f (x) = \sqrt{- x} - 5$ . In this case, the point is $(- 1, - 4)$ .

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

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

Simplify the result.

Simplify each term.

Multiply $- 1$ by $- 1$ .

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

Any root of $1$ is $1$ .

$f (- 1) = 1 - 5$

Subtract $5$ from $1$ .

$f (- 1) = - 4$

The final answer is $- 4$ .

$y = - 4$

The square root can be graphed using the points around the vertex $(0, - 5), (- 2, - 3.59), (- 1, - 4)$

$\begin{array}{c} x & y \\ - 2 & - 3.59 \\ - 1 & - 4 \\ 0 & - 5 \end{array}$

Graph H(x) = square root of -x-5

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