Pre-Algebra Examples

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

Step 1

Find the domain for $y = \sqrt{2 x + 5}$ 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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Step 1.1

Set the radicand in $\sqrt{2 x + 5}$ greater than or equal to $0$ to find where the expression is defined.

$2 x + 5 \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Subtract $5$ from both sides of the inequality.

$2 x \geq - 5$

Step 1.2.2

Divide each term in $2 x \geq - 5$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 x \geq - 5$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 5}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 5}{2}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{- 5}{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Move the negative in front of the fraction.

$x \geq - \frac{5}{2}$

Step 1.3

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

Interval Notation:

$[- \frac{5}{2}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{5}{2}}$

Interval Notation:

$[- \frac{5}{2}, \infty)$

Set-Builder Notation:

${x | x \geq - \frac{5}{2}}$

Step 2

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

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Step 2.1

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

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

Step 2.2

Simplify the result.

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Step 2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

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

Step 2.2.1.2

Cancel the common factor.

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

Step 2.2.1.3

Rewrite the expression.

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

Step 2.2.2

Simplify the expression.

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Step 2.2.2.1

Add $- 5$ and $5$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

The final answer is $0$ .

$0$

Step 3

The radical expression end point is $(- \frac{5}{2}, 0)$ .

$(- \frac{5}{2}, 0)$

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.

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Step 4.1

Substitute the $x$ value $- 2$ into $f (x) = \sqrt{2 x + 5}$ . In this case, the point is $(- 2, 1)$ .

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Multiply $2$ by $- 2$ .

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

Step 4.1.2.2

Add $- 4$ and $5$ .

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

Step 4.1.2.3

Any root of $1$ is $1$ .

$f (- 2) = 1$

Step 4.1.2.4

The final answer is $1$ .

$y = 1$

Step 4.2

Substitute the $x$ value $- 1$ into $f (x) = \sqrt{2 x + 5}$ . In this case, the point is $(- 1, \sqrt{3})$ .

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Step 4.2.1

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

$f (- 1) = \sqrt{2 (- 1) + 5}$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Multiply $2$ by $- 1$ .

$f (- 1) = \sqrt{- 2 + 5}$

Step 4.2.2.2

Add $- 2$ and $5$ .

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

Step 4.2.2.3

The final answer is $\sqrt{3}$ .

$y = \sqrt{3}$

Step 4.3

Substitute the $x$ value $0$ into $f (x) = \sqrt{2 x + 5}$ . In this case, the point is $(0, \sqrt{5})$ .

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Step 4.3.1

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

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

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Multiply $2$ by $0$ .

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

Step 4.3.2.2

Add $0$ and $5$ .

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

Step 4.3.2.3

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

$y = \sqrt{5}$

Step 4.4

The square root can be graphed using the points around the vertex $(- 2.5, 0), (- 2, 1), (- 1, 1.73), (0, 2.24)$

$\begin{array}{c} x & y \\ - 2.5 & 0 \\ - 2 & 1 \\ - 1 & 1.73 \\ 0 & 2.24 \end{array}$

Step 5