Pre-Algebra Examples

$f (x) = 3 - 2 \sin (π x)$

Step 1

Rewrite the expression as $- 2 \sin (π x) + 3$ .

$- 2 \sin (π x) + 3$

Step 2

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

$a = - 2$

$b = π$

$c = 0$

$d = 3$

Step 3

Find the amplitude $| a |$ .

Amplitude: $2$

Step 4

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

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Step 4.1

Find the period of $- 2 \sin (π x)$ .

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Step 4.1.1

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

$\frac{2 π}{| b |}$

Step 4.1.2

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

$\frac{2 π}{| π |}$

Step 4.1.3

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 4.1.4

Cancel the common factor of $π$ .

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Step 4.1.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 4.1.4.2

Divide $2$ by $1$ .

$2$

Step 4.2

Find the period of $3$ .

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Step 4.2.1

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

$\frac{2 π}{| b |}$

Step 4.2.2

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

$\frac{2 π}{| π |}$

Step 4.2.3

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 4.2.4

Cancel the common factor of $π$ .

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Step 4.2.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 4.2.4.2

Divide $2$ by $1$ .

$2$

Step 4.3

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

$2$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

Phase Shift: $\frac{0}{π}$

Step 5.3

Divide $0$ by $π$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: $2$

Period: $2$

Phase Shift: None

Vertical Shift: $3$

Step 7

Select a few points to graph.

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Step 7.1

Find the point at $x = 0$ .

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Step 7.1.1

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

$f (0) = 3 - 2 \sin (π (0))$

Step 7.1.2

Simplify the result.

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Step 7.1.2.1

Simplify each term.

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Step 7.1.2.1.1

Multiply $π$ by $0$ .

$f (0) = 3 - 2 \sin (0)$

Step 7.1.2.1.2

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

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

Step 7.1.2.1.3

Multiply $- 2$ by $0$ .

$f (0) = 3 + 0$

Step 7.1.2.2

Add $3$ and $0$ .

$f (0) = 3$

Step 7.1.2.3

The final answer is $3$ .

$3$

Step 7.2

Find the point at $x = \frac{1}{2}$ .

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Step 7.2.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = 3 - 2 \sin (π (\frac{1}{2}))$

Step 7.2.2

Simplify the result.

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Step 7.2.2.1

Simplify each term.

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Step 7.2.2.1.1

Combine $π$ and $\frac{1}{2}$ .

$f (\frac{1}{2}) = 3 - 2 \sin (\frac{π}{2})$

Step 7.2.2.1.2

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

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

Step 7.2.2.1.3

Multiply $- 2$ by $1$ .

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

Step 7.2.2.2

Subtract $2$ from $3$ .

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

Step 7.2.2.3

The final answer is $1$ .

$1$

Step 7.3

Find the point at $x = 1$ .

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Step 7.3.1

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

$f (1) = 3 - 2 \sin (π (1))$

Step 7.3.2

Simplify the result.

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Step 7.3.2.1

Simplify each term.

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Step 7.3.2.1.1

Multiply $π$ by $1$ .

$f (1) = 3 - 2 \sin (π)$

Step 7.3.2.1.2

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

$f (1) = 3 - 2 \sin (0)$

Step 7.3.2.1.3

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

$f (1) = 3 - 2 \cdot 0$

Step 7.3.2.1.4

Multiply $- 2$ by $0$ .

$f (1) = 3 + 0$

Step 7.3.2.2

Add $3$ and $0$ .

$f (1) = 3$

Step 7.3.2.3

The final answer is $3$ .

$3$

Step 7.4

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

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Step 7.4.1

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

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

Step 7.4.2

Simplify the result.

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Step 7.4.2.1

Simplify each term.

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Step 7.4.2.1.1

Combine $π$ and $\frac{3}{2}$ .

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

Step 7.4.2.1.2

Move $3$ to the left of $π$ .

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

Step 7.4.2.1.3

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.

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

Step 7.4.2.1.4

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

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

Step 7.4.2.1.5

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 7.4.2.1.5.1

Multiply $- 1$ by $1$ .

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

Step 7.4.2.1.5.2

Multiply $- 2$ by $- 1$ .

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

Step 7.4.2.2

Add $3$ and $2$ .

$f (\frac{3}{2}) = 5$

Step 7.4.2.3

The final answer is $5$ .

$5$

Step 7.5

Find the point at $x = 2$ .

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Step 7.5.1

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

$f (2) = 3 - 2 \sin (π (2))$

Step 7.5.2

Simplify the result.

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Step 7.5.2.1

Simplify each term.

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Step 7.5.2.1.1

Move $2$ to the left of $π$ .

$f (2) = 3 - 2 \sin (2 \cdot π)$

Step 7.5.2.1.2

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

$f (2) = 3 - 2 \sin (0)$

Step 7.5.2.1.3

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

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

Step 7.5.2.1.4

Multiply $- 2$ by $0$ .

$f (2) = 3 + 0$

Step 7.5.2.2

Add $3$ and $0$ .

$f (2) = 3$

Step 7.5.2.3

The final answer is $3$ .

$3$

Step 7.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 3 \\ \frac{1}{2} & 1 \\ 1 & 3 \\ \frac{3}{2} & 5 \\ 2 & 3 \end{array}$

Step 8

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

Amplitude: $2$

Period: $2$

Phase Shift: None

Vertical Shift: $3$

$\begin{array}{c} x & f (x) \\ 0 & 3 \\ \frac{1}{2} & 1 \\ 1 & 3 \\ \frac{3}{2} & 5 \\ 2 & 3 \end{array}$

Step 9