Trigonometry Examples

$| \frac{x^{2} - x - 2}{x - 3} |$

Step 1

Find the domain for $y = | \frac{(x - 2) (x + 1)}{x - 3} |$ 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 1.1

Set the denominator in $\frac{(x - 2) (x + 1)}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Step 1.2

Add $3$ to both sides of the equation.

$x = 3$

Step 1.3

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

Interval Notation:

$(- \infty, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3}$

Interval Notation:

$(- \infty, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3}$

Step 2

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 2.1

Substitute the $x$ value $- 2$ into $f (x) = | \frac{(x - 2) (x + 1)}{x - 3} |$ . In this case, the point is $(- 2, \frac{4}{5})$ .

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Step 2.1.1

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

$f (- 2) = | \frac{((- 2) - 2) ((- 2) + 1)}{(- 2) - 3} |$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify the numerator.

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Step 2.1.2.1.1

Subtract $2$ from $- 2$ .

$f (- 2) = | \frac{- 4 (- 2 + 1)}{- 2 - 3} |$

Step 2.1.2.1.2

Add $- 2$ and $1$ .

$f (- 2) = | \frac{- 4 \cdot - 1}{- 2 - 3} |$

Step 2.1.2.2

Simplify the expression.

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Step 2.1.2.2.1

Subtract $3$ from $- 2$ .

$f (- 2) = | \frac{- 4 \cdot - 1}{- 5} |$

Step 2.1.2.2.2

Multiply $- 4$ by $- 1$ .

$f (- 2) = | \frac{4}{- 5} |$

Step 2.1.2.2.3

Move the negative in front of the fraction.

$f (- 2) = | - \frac{4}{5} |$

Step 2.1.2.3

$- \frac{4}{5}$ is approximately $- 0.8$ which is negative so negate $- \frac{4}{5}$ and remove the absolute value

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

Step 2.1.2.4

The final answer is $\frac{4}{5}$ .

$y = \frac{4}{5}$

Step 2.2

Substitute the $x$ value $- 1$ into $f (x) = | \frac{(x - 2) (x + 1)}{x - 3} |$ . In this case, the point is $(- 1, 0)$ .

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Step 2.2.1

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

$f (- 1) = | \frac{((- 1) - 2) ((- 1) + 1)}{(- 1) - 3} |$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify the numerator.

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Step 2.2.2.1.1

Subtract $2$ from $- 1$ .

$f (- 1) = | \frac{- 3 (- 1 + 1)}{- 1 - 3} |$

Step 2.2.2.1.2

Add $- 1$ and $1$ .

$f (- 1) = | \frac{- 3 \cdot 0}{- 1 - 3} |$

Step 2.2.2.2

Simplify the expression.

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Step 2.2.2.2.1

Subtract $3$ from $- 1$ .

$f (- 1) = | \frac{- 3 \cdot 0}{- 4} |$

Step 2.2.2.2.2

Multiply $- 3$ by $0$ .

$f (- 1) = | \frac{0}{- 4} |$

Step 2.2.2.2.3

Divide $0$ by $- 4$ .

$f (- 1) = | 0 |$

Step 2.2.2.3

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

$f (- 1) = 0$

Step 2.2.2.4

The final answer is $0$ .

$y = 0$

Step 2.3

Substitute the $x$ value $0$ into $f (x) = | \frac{(x - 2) (x + 1)}{x - 3} |$ . In this case, the point is $(0, \frac{2}{3})$ .

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Step 2.3.1

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

$f (0) = | \frac{((0) - 2) ((0) + 1)}{(0) - 3} |$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify the numerator.

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Step 2.3.2.1.1

Subtract $2$ from $0$ .

$f (0) = | \frac{- 2 (0 + 1)}{0 - 3} |$

Step 2.3.2.1.2

Add $0$ and $1$ .

$f (0) = | \frac{- 2 \cdot 1}{0 - 3} |$

Step 2.3.2.2

Reduce the expression by cancelling the common factors.

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Step 2.3.2.2.1

Subtract $3$ from $0$ .

$f (0) = | \frac{- 2 \cdot 1}{- 3} |$

Step 2.3.2.2.2

Multiply $- 2$ by $1$ .

$f (0) = | \frac{- 2}{- 3} |$

Step 2.3.2.2.3

Dividing two negative values results in a positive value.

$f (0) = | \frac{2}{3} |$

Step 2.3.2.3

$\frac{2}{3}$ is approximately $0.\overline{6}$ which is positive so remove the absolute value

$f (0) = \frac{2}{3}$

Step 2.3.2.4

The final answer is $\frac{2}{3}$ .

$y = \frac{2}{3}$

Step 2.4

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

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Step 2.4.1

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

$f (1) = | \frac{((1) - 2) ((1) + 1)}{(1) - 3} |$

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify the numerator.

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Step 2.4.2.1.1

Subtract $2$ from $1$ .

$f (1) = | \frac{- 1 (1 + 1)}{1 - 3} |$

Step 2.4.2.1.2

Add $1$ and $1$ .

$f (1) = | \frac{- 1 \cdot 2}{1 - 3} |$

Step 2.4.2.2

Simplify the expression.

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Step 2.4.2.2.1

Subtract $3$ from $1$ .

$f (1) = | \frac{- 1 \cdot 2}{- 2} |$

Step 2.4.2.2.2

Multiply $- 1$ by $2$ .

$f (1) = | \frac{- 2}{- 2} |$

Step 2.4.2.2.3

Divide $- 2$ by $- 2$ .

$f (1) = | 1 |$

Step 2.4.2.3

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

$f (1) = 1$

Step 2.4.2.4

The final answer is $1$ .

$y = 1$

Step 2.5

The absolute value can be graphed using the points around the vertex $(- 2, 0.8), (- 1, 0), (0, 0.67), (1, 1)$

$\begin{array}{c} x & y \\ - 2 & 0.8 \\ - 1 & 0 \\ 0 & 0.67 \\ 1 & 1 \end{array}$

Step 3