Precalculus Examples

$y = 2 \csc (\frac{1}{3} x)$

Step 1

Find the asymptotes.

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Step 1.1

For any $y = \csc (x)$ , vertical asymptotes occur at $x = n π$ , where $n$ is an integer. Use the basic period for $y = c s c (x)$ , $(0, 2 π)$ , to find the vertical asymptotes for $y = 2 \csc (\frac{x}{3})$ . Set the inside of the cosecant function, $b x + c$ , for $y = a \csc (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = 2 \csc (\frac{x}{3})$ .

$\frac{x}{3} = 0$

Step 1.2

Set the numerator equal to zero.

$x = 0$

Step 1.3

Set the inside of the cosecant function $\frac{x}{3}$ equal to $2 π$ .

$\frac{x}{3} = 2 π$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Multiply both sides of the equation by $3$ .

$3 \frac{x}{3} = 3 (2 π)$

Step 1.4.2

Simplify both sides of the equation.

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Step 1.4.2.1

Simplify the left side.

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Step 1.4.2.1.1

Cancel the common factor of $3$ .

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Step 1.4.2.1.1.1

Cancel the common factor.

$3 \frac{x}{3} = 3 (2 π)$

Step 1.4.2.1.1.2

Rewrite the expression.

$x = 3 (2 π)$

Step 1.4.2.2

Simplify the right side.

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Step 1.4.2.2.1

Multiply $2$ by $3$ .

$x = 6 π$

Step 1.5

The basic period for $y = 2 \csc (\frac{x}{3})$ will occur at $(0, 6 π)$ , where $0$ and $6 π$ are vertical asymptotes.

$(0, 6 π)$

Step 1.6

Find the period $\frac{2 π}{| b |}$ to find where the vertical asymptotes exist. Vertical asymptotes occur every half period.

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Step 1.6.1

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{3}}$

Step 1.6.2

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 1.6.3

Multiply $3$ by $2$ .

$6 π$

Step 1.7

The vertical asymptotes for $y = 2 \csc (\frac{x}{3})$ occur at $0$ , $6 π$ , and every $3 π n$ , where $n$ is an integer. This is half of the period.

$x = 3 π n$

Step 1.8

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 3 π n$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 3 π n$ where $n$ is an integer

Step 2

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

$a = 2$

$b = \frac{1}{3}$

$c = 0$

$d = 0$

Step 3

Since the graph of the function $c s c$ does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

Step 4

Find the period of $2 \csc (\frac{x}{3})$ .

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Step 4.1

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

$\frac{2 π}{| b |}$

Step 4.2

Replace $b$ with $\frac{1}{3}$ in the formula for period.

$\frac{2 π}{| \frac{1}{3} |}$

Step 4.3

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{3}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 4.5

Multiply $3$ by $2$ .

$6 π$

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: $\frac{0}{\frac{1}{3}}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $0 \cdot 3$

Step 5.4

Multiply $0$ by $3$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $6 π$

Phase Shift: None

Vertical Shift: None

Step 7

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

Vertical Asymptotes: $x = 3 π n$ where $n$ is an integer

Amplitude: None

Period: $6 π$

Phase Shift: None

Vertical Shift: None

Step 8