Pre-Algebra Examples

$- 15 \sin (x + 10) - 3$

Step 1

Use the form $a \sin (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

$a = - 15$

$b = 1$

$c = - 10$

$d = - 3$

Step 2

Find the amplitude $| a |$ .

Amplitude: $15$

Step 3

Find the period using the formula $\frac{2 π}{| b |}$ .

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Step 3.1

Find the period of $- 15 \sin (x + 10)$ .

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Step 3.1.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.1.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.1.3

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

$\frac{2 π}{1}$

Step 3.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2

Find the period of $- 3$ .

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Step 3.2.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.3

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

$\frac{2 π}{1}$

Step 3.2.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$2 π$

Step 4

Find the phase shift using the formula $\frac{c}{b}$ .

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Step 4.1

The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Step 4.2

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{- 10}{1}$

Step 4.3

Divide $- 10$ by $1$ .

Phase Shift: $- 10$

Step 5

List the properties of the trigonometric function.

Amplitude: $15$

Period: $2 π$

Phase Shift: $- 10$ ( $10$ to the left)

Vertical Shift: $- 3$

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = - 10$ .

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Step 6.1.1

Replace the variable $x$ with $- 10$ in the expression.

$f (- 10) = - 15 \sin ((- 10) + 10) - 3$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Simplify each term.

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Step 6.1.2.1.1

Add $- 10$ and $10$ .

$f (- 10) = - 15 \sin (0) - 3$

Step 6.1.2.1.2

The exact value of $\sin (0)$ is $0$ .

$f (- 10) = - 15 \cdot 0 - 3$

Step 6.1.2.1.3

Multiply $- 15$ by $0$ .

$f (- 10) = 0 - 3$

Step 6.1.2.2

Subtract $3$ from $0$ .

$f (- 10) = - 3$

Step 6.1.2.3

The final answer is $- 3$ .

$- 3$

Step 6.2

Find the point at $x = \frac{π}{2} - 10$ .

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Step 6.2.1

Replace the variable $x$ with $\frac{π}{2} - 10$ in the expression.

$f (\frac{π}{2} - 10) = - 15 \sin ((\frac{π}{2} - 10) + 10) - 3$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Add $- 10$ and $10$ .

$f (\frac{π}{2} - 10) = - 15 \sin (\frac{π}{2} + 0) - 3$

Step 6.2.2.1.2

Add $\frac{π}{2}$ and $0$ .

$f (\frac{π}{2} - 10) = - 15 \sin (\frac{π}{2}) - 3$

Step 6.2.2.1.3

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2} - 10) = - 15 \cdot 1 - 3$

Step 6.2.2.1.4

Multiply $- 15$ by $1$ .

$f (\frac{π}{2} - 10) = - 15 - 3$

Step 6.2.2.2

Subtract $3$ from $- 15$ .

$f (\frac{π}{2} - 10) = - 18$

Step 6.2.2.3

The final answer is $- 18$ .

$- 18$

Step 6.3

Find the point at $x = π - 10$ .

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Step 6.3.1

Replace the variable $x$ with $π - 10$ in the expression.

$f (π - 10) = - 15 \sin ((π - 10) + 10) - 3$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Simplify each term.

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Step 6.3.2.1.1

Add $- 10$ and $10$ .

$f (π - 10) = - 15 \sin (π + 0) - 3$

Step 6.3.2.1.2

Add $π$ and $0$ .

$f (π - 10) = - 15 \sin (π) - 3$

Step 6.3.2.1.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (π - 10) = - 15 \sin (0) - 3$

Step 6.3.2.1.4

The exact value of $\sin (0)$ is $0$ .

$f (π - 10) = - 15 \cdot 0 - 3$

Step 6.3.2.1.5

Multiply $- 15$ by $0$ .

$f (π - 10) = 0 - 3$

Step 6.3.2.2

Subtract $3$ from $0$ .

$f (π - 10) = - 3$

Step 6.3.2.3

The final answer is $- 3$ .

$- 3$

Step 6.4

Find the point at $x = \frac{3 π}{2} - 10$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{3 π}{2} - 10$ in the expression.

$f (\frac{3 π}{2} - 10) = - 15 \sin ((\frac{3 π}{2} - 10) + 10) - 3$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

Add $- 10$ and $10$ .

$f (\frac{3 π}{2} - 10) = - 15 \sin (\frac{3 π}{2} + 0) - 3$

Step 6.4.2.1.2

Add $\frac{3 π}{2}$ and $0$ .

$f (\frac{3 π}{2} - 10) = - 15 \sin (\frac{3 π}{2}) - 3$

Step 6.4.2.1.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (\frac{3 π}{2} - 10) = - 15 (- \sin (\frac{π}{2})) - 3$

Step 6.4.2.1.4

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{3 π}{2} - 10) = - 15 (- 1 \cdot 1) - 3$

Step 6.4.2.1.5

Multiply $- 15 (- 1 \cdot 1)$ .

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Step 6.4.2.1.5.1

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2} - 10) = - 15 \cdot - 1 - 3$

Step 6.4.2.1.5.2

Multiply $- 15$ by $- 1$ .

$f (\frac{3 π}{2} - 10) = 15 - 3$

Step 6.4.2.2

Subtract $3$ from $15$ .

$f (\frac{3 π}{2} - 10) = 12$

Step 6.4.2.3

The final answer is $12$ .

$12$

Step 6.5

Find the point at $x = 2 π - 10$ .

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Step 6.5.1

Replace the variable $x$ with $2 π - 10$ in the expression.

$f (2 π - 10) = - 15 \sin ((2 π - 10) + 10) - 3$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Simplify each term.

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Step 6.5.2.1.1

Add $- 10$ and $10$ .

$f (2 π - 10) = - 15 \sin (2 π + 0) - 3$

Step 6.5.2.1.2

Add $2 π$ and $0$ .

$f (2 π - 10) = - 15 \sin (2 π) - 3$

Step 6.5.2.1.3

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (2 π - 10) = - 15 \sin (0) - 3$

Step 6.5.2.1.4

The exact value of $\sin (0)$ is $0$ .

$f (2 π - 10) = - 15 \cdot 0 - 3$

Step 6.5.2.1.5

Multiply $- 15$ by $0$ .

$f (2 π - 10) = 0 - 3$

Step 6.5.2.2

Subtract $3$ from $0$ .

$f (2 π - 10) = - 3$

Step 6.5.2.3

The final answer is $- 3$ .

$- 3$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 10 & - 3 \\ \frac{π}{2} - 10 & - 18 \\ π - 10 & - 3 \\ \frac{3 π}{2} - 10 & 12 \\ 2 π - 10 & - 3 \end{array}$

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Amplitude: $15$

Period: $2 π$

Phase Shift: $- 10$ ( $10$ to the left)

Vertical Shift: $- 3$

$\begin{array}{c} x & f (x) \\ - 10 & - 3 \\ \frac{π}{2} - 10 & - 18 \\ π - 10 & - 3 \\ \frac{3 π}{2} - 10 & 12 \\ 2 π - 10 & - 3 \end{array}$

Step 8