Trigonometry Examples

$y = 2 | - 3 x + 4 | - 3$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = 2 | - 3 x + 4 | - 3$ is $(\frac{4}{3}, - 3)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $- 3 x + 4$ equal to $0$ . In this case, $- 3 x + 4 = 0$ .

$- 3 x + 4 = 0$

Step 1.2

Solve the equation $- 3 x + 4 = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Subtract $4$ from both sides of the equation.

$- 3 x = - 4$

Step 1.2.2

Divide each term in $- 3 x = - 4$ by $- 3$ and simplify.

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Step 1.2.2.1

Divide each term in $- 3 x = - 4$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 4}{- 3}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $- 3$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 4}{- 3}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 3}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{4}{3}$

Step 1.3

Replace the variable $x$ with $\frac{4}{3}$ in the expression.

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

Step 1.4

Simplify $2 | - 3 (\frac{4}{3}) + 4 | - 3$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Simplify each term.

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Step 1.4.1.1.1

Cancel the common factor of $3$ .

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Step 1.4.1.1.1.1

Factor $3$ out of $- 3$ .

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

Step 1.4.1.1.1.2

Cancel the common factor.

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

Step 1.4.1.1.1.3

Rewrite the expression.

$y = 2 | - 1 \cdot 4 + 4 | - 3$

Step 1.4.1.1.2

Multiply $- 1$ by $4$ .

$y = 2 | - 4 + 4 | - 3$

Step 1.4.1.2

Add $- 4$ and $4$ .

$y = 2 | 0 | - 3$

Step 1.4.1.3

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

$y = 2 \cdot 0 - 3$

Step 1.4.1.4

Multiply $2$ by $0$ .

$y = 0 - 3$

Step 1.4.2

Subtract $3$ from $0$ .

$y = - 3$

Step 1.5

The absolute value vertex is $(\frac{4}{3}, - 3)$ .

$(\frac{4}{3}, - 3)$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

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 $- 1$ into $f (x) = 2 | - 3 x + 4 | - 3$ . In this case, the point is $(- 1, 11)$ .

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Step 3.1.1

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

$f (- 1) = 2 | - 3 \cdot - 1 + 4 | - 3$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Multiply $- 3$ by $- 1$ .

$f (- 1) = 2 | 3 + 4 | - 3$

Step 3.1.2.1.2

Add $3$ and $4$ .

$f (- 1) = 2 | 7 | - 3$

Step 3.1.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$f (- 1) = 2 \cdot 7 - 3$

Step 3.1.2.1.4

Multiply $2$ by $7$ .

$f (- 1) = 14 - 3$

Step 3.1.2.2

Subtract $3$ from $14$ .

$f (- 1) = 11$

Step 3.1.2.3

The final answer is $11$ .

$y = 11$

Step 3.2

Substitute the $x$ value $0$ into $f (x) = 2 | - 3 x + 4 | - 3$ . In this case, the point is $(0, 5)$ .

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Step 3.2.1

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

$f (0) = 2 | - 3 \cdot 0 + 4 | - 3$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $- 3$ by $0$ .

$f (0) = 2 | 0 + 4 | - 3$

Step 3.2.2.1.2

Add $0$ and $4$ .

$f (0) = 2 | 4 | - 3$

Step 3.2.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$f (0) = 2 \cdot 4 - 3$

Step 3.2.2.1.4

Multiply $2$ by $4$ .

$f (0) = 8 - 3$

Step 3.2.2.2

Subtract $3$ from $8$ .

$f (0) = 5$

Step 3.2.2.3

The final answer is $5$ .

$y = 5$

Step 3.3

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

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Step 3.3.1

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

$f (1) = 2 | - 3 \cdot 1 + 4 | - 3$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $- 3$ by $1$ .

$f (1) = 2 | - 3 + 4 | - 3$

Step 3.3.2.1.2

Add $- 3$ and $4$ .

$f (1) = 2 | 1 | - 3$

Step 3.3.2.1.3

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

$f (1) = 2 \cdot 1 - 3$

Step 3.3.2.1.4

Multiply $2$ by $1$ .

$f (1) = 2 - 3$

Step 3.3.2.2

Subtract $3$ from $2$ .

$f (1) = - 1$

Step 3.3.2.3

The final answer is $- 1$ .

$y = - 1$

Step 3.4

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

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Step 3.4.1

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

$f (2) = 2 | - 3 \cdot 2 + 4 | - 3$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply $- 3$ by $2$ .

$f (2) = 2 | - 6 + 4 | - 3$

Step 3.4.2.1.2

Add $- 6$ and $4$ .

$f (2) = 2 | - 2 | - 3$

Step 3.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (2) = 2 \cdot 2 - 3$

Step 3.4.2.1.4

Multiply $2$ by $2$ .

$f (2) = 4 - 3$

Step 3.4.2.2

Subtract $3$ from $4$ .

$f (2) = 1$

Step 3.4.2.3

The final answer is $1$ .

$y = 1$

Step 3.5

The absolute value can be graphed using the points around the vertex $(\frac{4}{3}, - 3), (- 1, 11), (0, 5), (1, - 1), (2, 1)$

$\begin{array}{c} x & y \\ - 1 & 11 \\ 0 & 5 \\ 1 & - 1 \\ \frac{4}{3} & - 3 \\ 2 & 1 \end{array}$

Step 4