Graph y=- square root of x-3

$y = - \sqrt{x - 3}$

Find the domain for $y = - \sqrt{x - 3}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Set the radicand in $\sqrt{x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

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

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${x | x \geq 3}$

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${x | x \geq 3}$

To find the radical expression end point, substitute the $x$ value $3$ , which is the least value in the domain, into $f (x) = - \sqrt{x - 3}$ .

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

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

Simplify the result.

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Subtract $3$ from $3$ .

$f (3) = - \sqrt{0}$

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

$f (3) = - \sqrt{0^{2}}$

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

$f (3) = - 0$

Multiply $- 1$ by $0$ .

$f (3) = 0$

The final answer is $0$ .

$0$

The radical expression end point is $(3, 0)$ .

$(3, 0)$

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.

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Substitute the $x$ value $4$ into $f (x) = - \sqrt{x - 3}$ . In this case, the point is $(4, - 1)$ .

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

$f (4) = - \sqrt{(4) - 3}$

Simplify the result.

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Subtract $3$ from $4$ .

$f (4) = - \sqrt{1}$

Any root of $1$ is $1$ .

$f (4) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (4) = - 1$

The final answer is $- 1$ .

$y = - 1$

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

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Replace the variable $x$ with $5$ in the expression.

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

Simplify the result.

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Subtract $3$ from $5$ .

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

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

$y = - \sqrt{2}$

The square root can be graphed using the points around the vertex $(3, 0), (4, - 1), (5, - 1.41)$

$\begin{array}{c} x & y \\ 3 & 0 \\ 4 & - 1 \\ 5 & - 1.41 \end{array}$

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