Precalculus Examples

$r = - 2 + 3 \cos (x)$

Step 1

Rewrite the expression as $3 \cos (x) - 2$ .

$3 \cos (x) - 2$

Step 2

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

$a = 3$

$b = 1$

$c = 0$

$d = - 2$

Step 3

Find the amplitude $| a |$ .

Amplitude: $3$

Step 4

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

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Step 4.1

Find the period of $3 \cos (x)$ .

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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 $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.1.3

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

$\frac{2 π}{1}$

Step 4.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2

Find the period of $- 2$ .

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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 $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.3

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

$\frac{2 π}{1}$

Step 4.2.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.3

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

$2 π$

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: $\frac{0}{1}$

Step 5.3

Divide $0$ by $1$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: $3$

Period: $2 π$

Phase Shift: None

Vertical Shift: $- 2$

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.

$f (0) = - 2 + 3 \cos (0)$

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

The exact value of $\cos (0)$ is $1$ .

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

Step 7.1.2.1.2

Multiply $3$ by $1$ .

$f (0) = - 2 + 3$

Step 7.1.2.2

Add $- 2$ and $3$ .

$f (0) = 1$

Step 7.1.2.3

The final answer is $1$ .

$1$

Step 7.2

Find the point at $x = \frac{π}{2}$ .

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Step 7.2.1

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

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

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

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{π}{2}) = - 2 + 3 \cdot 0$

Step 7.2.2.1.2

Multiply $3$ by $0$ .

$f (\frac{π}{2}) = - 2 + 0$

Step 7.2.2.2

Add $- 2$ and $0$ .

$f (\frac{π}{2}) = - 2$

Step 7.2.2.3

The final answer is $- 2$ .

$- 2$

Step 7.3

Find the point at $x = π$ .

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Step 7.3.1

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

$f (π) = - 2 + 3 \cos (π)$

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

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

$f (π) = - 2 + 3 (- \cos (0))$

Step 7.3.2.1.2

The exact value of $\cos (0)$ is $1$ .

$f (π) = - 2 + 3 (- 1 \cdot 1)$

Step 7.3.2.1.3

Multiply $3 (- 1 \cdot 1)$ .

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Step 7.3.2.1.3.1

Multiply $- 1$ by $1$ .

$f (π) = - 2 + 3 \cdot - 1$

Step 7.3.2.1.3.2

Multiply $3$ by $- 1$ .

$f (π) = - 2 - 3$

Step 7.3.2.2

Subtract $3$ from $- 2$ .

$f (π) = - 5$

Step 7.3.2.3

The final answer is $- 5$ .

$- 5$

Step 7.4

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

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Step 7.4.1

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

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

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

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

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

Step 7.4.2.1.2

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{3 π}{2}) = - 2 + 3 \cdot 0$

Step 7.4.2.1.3

Multiply $3$ by $0$ .

$f (\frac{3 π}{2}) = - 2 + 0$

Step 7.4.2.2

Add $- 2$ and $0$ .

$f (\frac{3 π}{2}) = - 2$

Step 7.4.2.3

The final answer is $- 2$ .

$- 2$

Step 7.5

Find the point at $x = 2 π$ .

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Step 7.5.1

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

$f (2 π) = - 2 + 3 \cos (2 π)$

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

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

$f (2 π) = - 2 + 3 \cos (0)$

Step 7.5.2.1.2

The exact value of $\cos (0)$ is $1$ .

$f (2 π) = - 2 + 3 \cdot 1$

Step 7.5.2.1.3

Multiply $3$ by $1$ .

$f (2 π) = - 2 + 3$

Step 7.5.2.2

Add $- 2$ and $3$ .

$f (2 π) = 1$

Step 7.5.2.3

The final answer is $1$ .

$1$

Step 7.6

List the points in a table.

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

Step 8

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

Amplitude: $3$

Period: $2 π$

Phase Shift: None

Vertical Shift: $- 2$

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

Step 9