Precalculus Examples

$y = - 3 \cos (π - x) - 2$

Step 1

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

$a = - 3$

$b = - 1$

$c = - π$

$d = - 2$

Step 2

Find the amplitude $| a |$ .

Amplitude: $3$

Step 3

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

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Step 3.1

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

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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 $- 1$ and $0$ is $1$ .

$\frac{2 π}{1}$

Step 3.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2

Find the period of $- 2$ .

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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 $- 1$ and $0$ 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

Dividing two negative values results in a positive value.

Phase Shift: $\frac{π}{1}$

Step 4.4

Divide $π$ by $1$ .

Phase Shift: $π$

Step 5

List the properties of the trigonometric function.

Amplitude: $3$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $- 2$

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.

$f (π) = - 3 \cos (- (π) + π) - 2$

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 $- π$ and $π$ .

$f (π) = - 3 \cos (0) - 2$

Step 6.1.2.1.2

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

$f (π) = - 3 \cdot 1 - 2$

Step 6.1.2.1.3

Multiply $- 3$ by $1$ .

$f (π) = - 3 - 2$

Step 6.1.2.2

Subtract $2$ from $- 3$ .

$f (π) = - 5$

Step 6.1.2.3

The final answer is $- 5$ .

$- 5$

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.

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

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

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

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

Step 6.2.2.1.2

Combine $π$ and $\frac{2}{2}$ .

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

Step 6.2.2.1.3

Combine the numerators over the common denominator.

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

Step 6.2.2.1.4

Simplify the numerator.

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Step 6.2.2.1.4.1

Move $2$ to the left of $π$ .

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

Step 6.2.2.1.4.2

Add $- 3 π$ and $2 π$ .

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

Step 6.2.2.1.5

Move the negative in front of the fraction.

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

Step 6.2.2.1.6

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

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

Step 6.2.2.1.7

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

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

Step 6.2.2.1.8

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

$f (\frac{3 π}{2}) = - 3 \cdot 0 - 2$

Step 6.2.2.1.9

Multiply $- 3$ by $0$ .

$f (\frac{3 π}{2}) = 0 - 2$

Step 6.2.2.2

Subtract $2$ from $0$ .

$f (\frac{3 π}{2}) = - 2$

Step 6.2.2.3

The final answer is $- 2$ .

$- 2$

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.

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

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

Multiply $2$ by $- 1$ .

$f (2 π) = - 3 \cos (- 2 π + π) - 2$

Step 6.3.2.1.2

Add $- 2 π$ and $π$ .

$f (2 π) = - 3 \cos (- π) - 2$

Step 6.3.2.1.3

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.

$f (2 π) = - 3 (- \cos (0)) - 2$

Step 6.3.2.1.4

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

$f (2 π) = - 3 (- 1 \cdot 1) - 2$

Step 6.3.2.1.5

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

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Step 6.3.2.1.5.1

Multiply $- 1$ by $1$ .

$f (2 π) = - 3 \cdot - 1 - 2$

Step 6.3.2.1.5.2

Multiply $- 3$ by $- 1$ .

$f (2 π) = 3 - 2$

Step 6.3.2.2

Subtract $2$ from $3$ .

$f (2 π) = 1$

Step 6.3.2.3

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.

$f (\frac{5 π}{2}) = - 3 \cos (- (\frac{5 π}{2}) + π) - 2$

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

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

$f (\frac{5 π}{2}) = - 3 \cos (- \frac{5 π}{2} + π \cdot \frac{2}{2}) - 2$

Step 6.4.2.1.2

Combine $π$ and $\frac{2}{2}$ .

$f (\frac{5 π}{2}) = - 3 \cos (- \frac{5 π}{2} + \frac{π \cdot 2}{2}) - 2$

Step 6.4.2.1.3

Combine the numerators over the common denominator.

$f (\frac{5 π}{2}) = - 3 \cos (\frac{- 5 π + π \cdot 2}{2}) - 2$

Step 6.4.2.1.4

Simplify the numerator.

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Step 6.4.2.1.4.1

Move $2$ to the left of $π$ .

$f (\frac{5 π}{2}) = - 3 \cos (\frac{- 5 π + 2 \cdot π}{2}) - 2$

Step 6.4.2.1.4.2

Add $- 5 π$ and $2 π$ .

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

Step 6.4.2.1.5

Move the negative in front of the fraction.

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

Step 6.4.2.1.6

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

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

Step 6.4.2.1.7

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

$f (\frac{5 π}{2}) = - 3 \cdot 0 - 2$

Step 6.4.2.1.8

Multiply $- 3$ by $0$ .

$f (\frac{5 π}{2}) = 0 - 2$

Step 6.4.2.2

Subtract $2$ from $0$ .

$f (\frac{5 π}{2}) = - 2$

Step 6.4.2.3

The final answer is $- 2$ .

$- 2$

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.

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

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

Multiply $3$ by $- 1$ .

$f (3 π) = - 3 \cos (- 3 π + π) - 2$

Step 6.5.2.1.2

Add $- 3 π$ and $π$ .

$f (3 π) = - 3 \cos (- 2 π) - 2$

Step 6.5.2.1.3

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

$f (3 π) = - 3 \cos (0) - 2$

Step 6.5.2.1.4

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

$f (3 π) = - 3 \cdot 1 - 2$

Step 6.5.2.1.5

Multiply $- 3$ by $1$ .

$f (3 π) = - 3 - 2$

Step 6.5.2.2

Subtract $2$ from $- 3$ .

$f (3 π) = - 5$

Step 6.5.2.3

The final answer is $- 5$ .

$- 5$

Step 6.6

List the points in a table.

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

Step 7

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

Amplitude: $3$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: $- 2$

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

Step 8