Precalculus Examples

$f (x) = \sqrt{x} + 2$

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.

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Step 1.1

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

$x \geq 0$

Step 1.2

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 $f (x) = \sqrt{x} + 2$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

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

Step 2.2.2

Simplify each term.

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

$f (0) = 0 + 2$

Step 2.2.3

Add $0$ and $2$ .

$f (0) = 2$

Step 2.2.4

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.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

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

Step 4.1.2.2

Any root of $1$ is $1$ .

$f (1) = 1 + 2$

Step 4.1.2.3

Add $1$ and $2$ .

$f (1) = 3$

Step 4.1.2.4

The final answer is $3$ .

$y = 3$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Remove parentheses.

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

Step 4.2.2.2

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

$y = \sqrt{2} + 2$

Step 4.3

The square root can be graphed using the points around the vertex $(0, 2), (1, 3), (2, 3.41)$

$\begin{array}{c} x & y \\ 0 & 2 \\ 1 & 3 \\ 2 & 3.41 \end{array}$

Step 5