Algebra Examples

$y = \frac{1}{2} \sin (x - π) + 1$

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.

$a = \frac{1}{2}$

$b = 1$

$c = π$

$d = 1$

Step 2

Find the amplitude $| a |$ .

Amplitude: $\frac{1}{2}$

Step 3

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

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Step 3.1

Find the period of $\frac{\sin (x - π)}{2}$ .

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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 $1$ .

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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: $\frac{π}{1}$

Step 4.3

Divide $π$ by $1$ .

Phase Shift: $π$

Step 5

List the properties of the trigonometric function.

Amplitude: $\frac{1}{2}$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $1$

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = π$ .

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Step 6.1.1

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

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

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Subtract $π$ from $π$ .

$f (π) = \frac{\sin (0)}{2} + 1$

Step 6.1.2.2

Simplify each term.

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Step 6.1.2.2.1

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

$f (π) = \frac{0}{2} + 1$

Step 6.1.2.2.2

Divide $0$ by $2$ .

$f (π) = 0 + 1$

Step 6.1.2.3

Add $0$ and $1$ .

$f (π) = 1$

Step 6.1.2.4

The final answer is $1$ .

$1$

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.2.2.2

Combine fractions.

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Step 6.2.2.2.1

Combine $- π$ and $\frac{2}{2}$ .

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

Step 6.2.2.2.2

Combine the numerators over the common denominator.

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

Step 6.2.2.3

Simplify each term.

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Step 6.2.2.3.1

Simplify the numerator.

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Step 6.2.2.3.1.1

Multiply $2$ by $- 1$ .

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

Step 6.2.2.3.1.2

Subtract $2 π$ from $3 π$ .

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

Step 6.2.2.3.2

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

$f (\frac{3 π}{2}) = \frac{1}{2} + 1$

Step 6.2.2.4

Simplify the expression.

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Step 6.2.2.4.1

Write $1$ as a fraction with a common denominator.

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

Step 6.2.2.4.2

Combine the numerators over the common denominator.

$f (\frac{3 π}{2}) = \frac{1 + 2}{2}$

Step 6.2.2.4.3

Add $1$ and $2$ .

$f (\frac{3 π}{2}) = \frac{3}{2}$

Step 6.2.2.5

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

$\frac{3}{2}$

Step 6.3

Find the point at $x = 2 π$ .

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Step 6.3.1

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

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

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Subtract $π$ from $2 π$ .

$f (2 π) = \frac{\sin (π)}{2} + 1$

Step 6.3.2.2

Simplify each term.

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Step 6.3.2.2.1

Simplify the numerator.

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Step 6.3.2.2.1.1

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

$f (2 π) = \frac{\sin (0)}{2} + 1$

Step 6.3.2.2.1.2

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

$f (2 π) = \frac{0}{2} + 1$

Step 6.3.2.2.2

Divide $0$ by $2$ .

$f (2 π) = 0 + 1$

Step 6.3.2.3

Add $0$ and $1$ .

$f (2 π) = 1$

Step 6.3.2.4

The final answer is $1$ .

$1$

Step 6.4

Find the point at $x = \frac{5 π}{2}$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{5 π}{2}$ in the expression.

$f (\frac{5 π}{2}) = \frac{\sin ((\frac{5 π}{2}) - π)}{2} + 1$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{5 π}{2}) = \frac{\sin (\frac{5 π}{2} - π \cdot \frac{2}{2})}{2} + 1$

Step 6.4.2.2

Combine fractions.

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Step 6.4.2.2.1

Combine $- π$ and $\frac{2}{2}$ .

$f (\frac{5 π}{2}) = \frac{\sin (\frac{5 π}{2} + \frac{- π \cdot 2}{2})}{2} + 1$

Step 6.4.2.2.2

Combine the numerators over the common denominator.

$f (\frac{5 π}{2}) = \frac{\sin (\frac{5 π - π \cdot 2}{2})}{2} + 1$

Step 6.4.2.3

Simplify each term.

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Step 6.4.2.3.1

Simplify the numerator.

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Step 6.4.2.3.1.1

Multiply $2$ by $- 1$ .

$f (\frac{5 π}{2}) = \frac{\sin (\frac{5 π - 2 π}{2})}{2} + 1$

Step 6.4.2.3.1.2

Subtract $2 π$ from $5 π$ .

$f (\frac{5 π}{2}) = \frac{\sin (\frac{3 π}{2})}{2} + 1$

Step 6.4.2.3.2

Simplify the numerator.

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Step 6.4.2.3.2.1

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.

$f (\frac{5 π}{2}) = \frac{- \sin (\frac{π}{2})}{2} + 1$

Step 6.4.2.3.2.2

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

$f (\frac{5 π}{2}) = \frac{- 1 \cdot 1}{2} + 1$

Step 6.4.2.3.2.3

Multiply $- 1$ by $1$ .

$f (\frac{5 π}{2}) = \frac{- 1}{2} + 1$

Step 6.4.2.3.3

Move the negative in front of the fraction.

$f (\frac{5 π}{2}) = - \frac{1}{2} + 1$

Step 6.4.2.4

Simplify the expression.

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Step 6.4.2.4.1

Write $1$ as a fraction with a common denominator.

$f (\frac{5 π}{2}) = - \frac{1}{2} + \frac{2}{2}$

Step 6.4.2.4.2

Combine the numerators over the common denominator.

$f (\frac{5 π}{2}) = \frac{- 1 + 2}{2}$

Step 6.4.2.4.3

Add $- 1$ and $2$ .

$f (\frac{5 π}{2}) = \frac{1}{2}$

Step 6.4.2.5

The final answer is $\frac{1}{2}$ .

$\frac{1}{2}$

Step 6.5

Find the point at $x = 3 π$ .

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Step 6.5.1

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

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

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Subtract $π$ from $3 π$ .

$f (3 π) = \frac{\sin (2 π)}{2} + 1$

Step 6.5.2.2

Simplify each term.

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Step 6.5.2.2.1

Simplify the numerator.

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Step 6.5.2.2.1.1

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

$f (3 π) = \frac{\sin (0)}{2} + 1$

Step 6.5.2.2.1.2

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

$f (3 π) = \frac{0}{2} + 1$

Step 6.5.2.2.2

Divide $0$ by $2$ .

$f (3 π) = 0 + 1$

Step 6.5.2.3

Add $0$ and $1$ .

$f (3 π) = 1$

Step 6.5.2.4

The final answer is $1$ .

$1$

Step 6.6

List the points in a table.

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

Step 7

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

Amplitude: $\frac{1}{2}$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $1$

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

Step 8