Precalculus Examples

$y = \sin (6 x)$

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.

$a = 1$

$b = 6$

$c = 0$

$d = 0$

Step 2

Find the amplitude $| a |$ .

Amplitude: $1$

Step 3

Find the period of $\sin (6 x)$ .

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Step 3.1

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

$\frac{2 π}{| b |}$

Step 3.2

Replace $b$ with $6$ in the formula for period.

$\frac{2 π}{| 6 |}$

Step 3.3

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$\frac{2 π}{6}$

Step 3.4

Cancel the common factor of $2$ and $6$ .

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Step 3.4.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{6}$

Step 3.4.2

Cancel the common factors.

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Step 3.4.2.1

Factor $2$ out of $6$ .

$\frac{2 π}{2 \cdot 3}$

Step 3.4.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 3}$

Step 3.4.2.3

Rewrite the expression.

$\frac{π}{3}$

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: $\frac{0}{6}$

Step 4.3

Divide $0$ by $6$ .

Phase Shift: $0$

Step 5

List the properties of the trigonometric function.

Amplitude: $1$

Period: $\frac{π}{3}$

Phase Shift: None

Vertical Shift: None

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = 0$ .

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Step 6.1.1

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

$f (0) = \sin (6 (0))$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Multiply $6$ by $0$ .

$f (0) = \sin (0)$

Step 6.1.2.2

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

$f (0) = 0$

Step 6.1.2.3

The final answer is $0$ .

$0$

Step 6.2

Find the point at $x = \frac{π}{12}$ .

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Step 6.2.1

Replace the variable $x$ with $\frac{π}{12}$ in the expression.

$f (\frac{π}{12}) = \sin (6 (\frac{π}{12}))$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Cancel the common factor of $6$ .

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Step 6.2.2.1.1

Factor $6$ out of $12$ .

$f (\frac{π}{12}) = \sin (6 (\frac{π}{6 (2)}))$

Step 6.2.2.1.2

Cancel the common factor.

$f (\frac{π}{12}) = \sin (6 (\frac{π}{6 \cdot 2}))$

Step 6.2.2.1.3

Rewrite the expression.

$f (\frac{π}{12}) = \sin (\frac{π}{2})$

Step 6.2.2.2

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

$f (\frac{π}{12}) = 1$

Step 6.2.2.3

The final answer is $1$ .

$1$

Step 6.3

Find the point at $x = \frac{π}{6}$ .

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Step 6.3.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = \sin (6 (\frac{π}{6}))$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Cancel the common factor of $6$ .

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Step 6.3.2.1.1

Cancel the common factor.

$f (\frac{π}{6}) = \sin (6 (\frac{π}{6}))$

Step 6.3.2.1.2

Rewrite the expression.

$f (\frac{π}{6}) = \sin (π)$

Step 6.3.2.2

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

$f (\frac{π}{6}) = \sin (0)$

Step 6.3.2.3

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

$f (\frac{π}{6}) = 0$

Step 6.3.2.4

The final answer is $0$ .

$0$

Step 6.4

Find the point at $x = \frac{π}{4}$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = \sin (6 (\frac{π}{4}))$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Cancel the common factor of $2$ .

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Step 6.4.2.1.1

Factor $2$ out of $6$ .

$f (\frac{π}{4}) = \sin (2 (3) (\frac{π}{4}))$

Step 6.4.2.1.2

Factor $2$ out of $4$ .

$f (\frac{π}{4}) = \sin (2 \cdot (3 (\frac{π}{2 \cdot 2})))$

Step 6.4.2.1.3

Cancel the common factor.

$f (\frac{π}{4}) = \sin (2 \cdot (3 (\frac{π}{2 \cdot 2})))$

Step 6.4.2.1.4

Rewrite the expression.

$f (\frac{π}{4}) = \sin (3 (\frac{π}{2}))$

Step 6.4.2.2

Combine $3$ and $\frac{π}{2}$ .

$f (\frac{π}{4}) = \sin (\frac{3 π}{2})$

Step 6.4.2.3

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.

$f (\frac{π}{4}) = - \sin (\frac{π}{2})$

Step 6.4.2.4

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

$f (\frac{π}{4}) = - 1 \cdot 1$

Step 6.4.2.5

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = - 1$

Step 6.4.2.6

The final answer is $- 1$ .

$- 1$

Step 6.5

Find the point at $x = \frac{π}{3}$ .

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Step 6.5.1

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

$f (\frac{π}{3}) = \sin (6 (\frac{π}{3}))$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Cancel the common factor of $3$ .

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Step 6.5.2.1.1

Factor $3$ out of $6$ .

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

Step 6.5.2.1.2

Cancel the common factor.

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

Step 6.5.2.1.3

Rewrite the expression.

$f (\frac{π}{3}) = \sin (2 π)$

Step 6.5.2.2

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

$f (\frac{π}{3}) = \sin (0)$

Step 6.5.2.3

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

$f (\frac{π}{3}) = 0$

Step 6.5.2.4

The final answer is $0$ .

$0$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{12} & 1 \\ \frac{π}{6} & 0 \\ \frac{π}{4} & - 1 \\ \frac{π}{3} & 0 \end{array}$

Step 7

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

Amplitude: $1$

Period: $\frac{π}{3}$

Phase Shift: None

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{12} & 1 \\ \frac{π}{6} & 0 \\ \frac{π}{4} & - 1 \\ \frac{π}{3} & 0 \end{array}$

Step 8