Trigonometry Examples

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

Step 1

Find the domain for $y = \sqrt{x - 1 - 2}$ 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 $\sqrt{x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$x - 3 \geq 0$

Step 1.2

Add $3$ to both sides of the inequality.

$x \geq 3$

Step 1.3

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:

${x | x \geq 3}$

Step 2

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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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Subtract $3$ from $3$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

$f (3) = 0$

Step 2.2.4

The final answer is $0$ .

$0$

Step 3

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

$(3, 0)$

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Subtract $3$ from $4$ .

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

Step 4.1.2.2

Any root of $1$ is $1$ .

$f (4) = 1$

Step 4.1.2.3

The final answer is $1$ .

$y = 1$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Subtract $3$ from $5$ .

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

Step 4.2.2.2

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

$y = \sqrt{2}$

Step 4.3

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

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

Step 5