Trigonometry Examples

$f (x) = | \log_{3} (x + 5) |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | \log_{3} (x + 5) |$ is $(- 4, 0)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $\log_{3} (x + 5)$ equal to $0$ . In this case, $\log_{3} (x + 5) = 0$ .

$\log_{3} (x + 5) = 0$

Step 1.2

Solve the equation $\log_{3} (x + 5) = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Rewrite $\log_{3} (x + 5) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$3^{0} = x + 5$

Step 1.2.2

Solve for $x$ .

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Step 1.2.2.1

Rewrite the equation as $x + 5 = 3^{0}$ .

$x + 5 = 3^{0}$

Step 1.2.2.2

Anything raised to $0$ is $1$ .

$x + 5 = 1$

Step 1.2.2.3

Move all terms not containing $x$ to the right side of the equation.

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Step 1.2.2.3.1

Subtract $5$ from both sides of the equation.

$x = 1 - 5$

Step 1.2.2.3.2

Subtract $5$ from $1$ .

$x = - 4$

Step 1.3

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

$y = | \log_{3} ((- 4) + 5) |$

Step 1.4

Simplify $| \log_{3} ((- 4) + 5) |$ .

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Step 1.4.1

Add $- 4$ and $5$ .

$y = | \log_{3} (1) |$

Step 1.4.2

Logarithm base $3$ of $1$ is $0$ .

$y = | 0 |$

Step 1.4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 0$

Step 1.5

The absolute value vertex is $(- 4, 0)$ .

$(- 4, 0)$

Step 2

Find the domain for $f (x) = | \log_{3} (x + 5) |$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the absolute value function.

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Step 2.1

Set the argument in $\log_{3} (x + 5)$ greater than $0$ to find where the expression is defined.

$x + 5 > 0$

Step 2.2

Subtract $5$ from both sides of the inequality.

$x > - 5$

Step 2.3

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

Interval Notation:

$(- 5, \infty)$

Set-Builder Notation:

${x | x > - 5}$

Interval Notation:

$(- 5, \infty)$

Set-Builder Notation:

${x | x > - 5}$

Step 3

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

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Step 3.1

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

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Step 3.1.1

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

$f (- 3) = | \log_{3} ((- 3) + 5) |$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Add $- 3$ and $5$ .

$f (- 3) = | \log_{3} (2) |$

Step 3.1.2.2

$\log_{3} (2)$ is approximately $0.63092975$ which is positive so remove the absolute value

$f (- 3) = \log_{3} (2)$

Step 3.1.2.3

The final answer is $\log_{3} (2)$ .

$y = \log_{3} (2)$

Step 3.2

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

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Step 3.2.1

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

$f (- 2) = | \log_{3} ((- 2) + 5) |$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Add $- 2$ and $5$ .

$f (- 2) = | \log_{3} (3) |$

Step 3.2.2.2

Logarithm base $3$ of $3$ is $1$ .

$f (- 2) = | 1 |$

Step 3.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$f (- 2) = 1$

Step 3.2.2.4

The final answer is $1$ .

$y = 1$

Step 3.3

The absolute value can be graphed using the points around the vertex $(- 4, 0), (- 3, 0.63), (- 2, 1)$

$\begin{array}{c} x & y \\ - 4 & 0 \\ - 3 & 0.63 \\ - 2 & 1 \end{array}$

Step 4