Pre-Algebra Examples

$y = 5 \csc (3 x - \frac{π}{2})$

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 $y = 5 \csc (3 x - \frac{π}{2})$ . 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 $y = 5 \csc (3 x - \frac{π}{2})$ .

$3 x - \frac{π}{2} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Add $\frac{π}{2}$ to both sides of the equation.

$3 x = \frac{π}{2}$

Step 1.2.2

Divide each term in $3 x = \frac{π}{2}$ by $3$ and simplify.

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Step 1.2.2.1

Divide each term in $3 x = \frac{π}{2}$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{π}{2}}{3}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{π}{2}}{3}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{3}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

Step 1.2.2.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Step 1.2.2.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

$x = \frac{π}{2 \cdot 3}$

Step 1.2.2.3.2.2

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

Step 1.3

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

$3 x - \frac{π}{2} = 2 π$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.4.1.1

Add $\frac{π}{2}$ to both sides of the equation.

$3 x = 2 π + \frac{π}{2}$

Step 1.4.1.2

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

$3 x = 2 π \cdot \frac{2}{2} + \frac{π}{2}$

Step 1.4.1.3

Combine $2 π$ and $\frac{2}{2}$ .

$3 x = \frac{2 π \cdot 2}{2} + \frac{π}{2}$

Step 1.4.1.4

Combine the numerators over the common denominator.

$3 x = \frac{2 π \cdot 2 + π}{2}$

Step 1.4.1.5

Simplify the numerator.

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Step 1.4.1.5.1

Multiply $2$ by $2$ .

$3 x = \frac{4 π + π}{2}$

Step 1.4.1.5.2

Add $4 π$ and $π$ .

$3 x = \frac{5 π}{2}$

Step 1.4.2

Divide each term in $3 x = \frac{5 π}{2}$ by $3$ and simplify.

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Step 1.4.2.1

Divide each term in $3 x = \frac{5 π}{2}$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{5 π}{2}}{3}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Cancel the common factor of $3$ .

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Step 1.4.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{5 π}{2}}{3}$

Step 1.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 π}{2}}{3}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 π}{2} \cdot \frac{1}{3}$

Step 1.4.2.3.2

Multiply $\frac{5 π}{2} \cdot \frac{1}{3}$ .

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Step 1.4.2.3.2.1

Multiply $\frac{5 π}{2}$ by $\frac{1}{3}$ .

$x = \frac{5 π}{2 \cdot 3}$

Step 1.4.2.3.2.2

Multiply $2$ by $3$ .

$x = \frac{5 π}{6}$

Step 1.5

The basic period for $y = 5 \csc (3 x - \frac{π}{2})$ will occur at $(\frac{π}{6}, \frac{5 π}{6})$ , where $\frac{π}{6}$ and $\frac{5 π}{6}$ are vertical asymptotes.

$(\frac{π}{6}, \frac{5 π}{6})$

Step 1.6

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Step 1.7

The vertical asymptotes for $y = 5 \csc (3 x - \frac{π}{2})$ occur at $\frac{π}{6}$ , $\frac{5 π}{6}$ , and every $x = \frac{π}{6} + \frac{π n}{3}$ , where $n$ is an integer. This is half of the period.

$x = \frac{π}{6} + \frac{π n}{3}$

Step 1.8

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{6} + \frac{π n}{3}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{6} + \frac{π n}{3}$ 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.

$a = 5$

$b = 3$

$c = \frac{π}{2}$

$d = 0$

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 $5 \csc (3 x - \frac{π}{2})$ .

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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 $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

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: $\frac{\frac{π}{2}}{3}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $\frac{π}{2} \cdot \frac{1}{3}$

Step 5.4

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Step 5.4.1

Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

Phase Shift: $\frac{π}{2 \cdot 3}$

Step 5.4.2

Multiply $2$ by $3$ .

Phase Shift: $\frac{π}{6}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{2 π}{3}$

Phase Shift: $\frac{π}{6}$ ( $\frac{π}{6}$ to the right)

Vertical Shift: None

Step 7

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

Vertical Asymptotes: $x = \frac{π}{6} + \frac{π n}{3}$ where $n$ is an integer

Amplitude: None

Period: $\frac{2 π}{3}$

Phase Shift: $\frac{π}{6}$ ( $\frac{π}{6}$ to the right)

Vertical Shift: None

Step 8