Precalculus Examples

$f (x) = 4 \csc (2 π x - π)$

Step 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 $y = 4 \csc (2 π x - π)$ . 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 $y = 4 \csc (2 π x - π)$ .

$2 π x - π = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $π$ to both sides of the equation.

$2 π x = π$

Step 2.2

Divide each term in $2 π x = π$ by $2 π$ and simplify.

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Step 2.2.1

Divide each term in $2 π x = π$ by $2 π$ .

$\frac{2 π x}{2 π} = \frac{π}{2 π}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{2 π x}{2 π} = \frac{π}{2 π}$

Step 2.2.2.1.2

Rewrite the expression.

$\frac{π x}{π} = \frac{π}{2 π}$

Step 2.2.2.2

Cancel the common factor of $π$ .

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Step 2.2.2.2.1

Cancel the common factor.

$\frac{π x}{π} = \frac{π}{2 π}$

Step 2.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{π}{2 π}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Cancel the common factor of $π$ .

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Step 2.2.3.1.1

Cancel the common factor.

$x = \frac{π}{2 π}$

Step 2.2.3.1.2

Rewrite the expression.

$x = \frac{1}{2}$

Step 3

Set the inside of the cosecant function $2 π x - π$ equal to $2 π$ .

$2 π x - π = 2 π$

Step 4

Solve for $x$ .

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Step 4.1

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

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Step 4.1.1

Add $π$ to both sides of the equation.

$2 π x = 2 π + π$

Step 4.1.2

Add $2 π$ and $π$ .

$2 π x = 3 π$

Step 4.2

Divide each term in $2 π x = 3 π$ by $2 π$ and simplify.

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Step 4.2.1

Divide each term in $2 π x = 3 π$ by $2 π$ .

$\frac{2 π x}{2 π} = \frac{3 π}{2 π}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{2 π x}{2 π} = \frac{3 π}{2 π}$

Step 4.2.2.1.2

Rewrite the expression.

$\frac{π x}{π} = \frac{3 π}{2 π}$

Step 4.2.2.2

Cancel the common factor of $π$ .

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Step 4.2.2.2.1

Cancel the common factor.

$\frac{π x}{π} = \frac{3 π}{2 π}$

Step 4.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{3 π}{2 π}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Cancel the common factor of $π$ .

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Step 4.2.3.1.1

Cancel the common factor.

$x = \frac{3 π}{2 π}$

Step 4.2.3.1.2

Rewrite the expression.

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

Step 5

The basic period for $y = 4 \csc (2 π x - π)$ will occur at $(\frac{1}{2}, \frac{3}{2})$ , where $\frac{1}{2}$ and $\frac{3}{2}$ are vertical asymptotes.

$(\frac{1}{2}, \frac{3}{2})$

Step 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 6.1

$2 π$ is approximately $6.2831853$ which is positive so remove the absolute value

$\frac{2 π}{2 π}$

Step 6.2

Cancel the common factor of $2$ .

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Step 6.2.1

Cancel the common factor.

$\frac{2 π}{2 π}$

Step 6.2.2

Rewrite the expression.

$\frac{π}{π}$

Step 6.3

Cancel the common factor of $π$ .

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Step 6.3.1

Cancel the common factor.

$\frac{π}{π}$

Step 6.3.2

Rewrite the expression.

$1$

Step 7

The vertical asymptotes for $y = 4 \csc (2 π x - π)$ occur at $\frac{1}{2}$ , $\frac{3}{2}$ , and every $x = \frac{1}{2} + \frac{n}{2}$ , where $n$ is an integer. This is half of the period.

$x = \frac{1}{2} + \frac{n}{2}$

Step 8

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{1}{2} + \frac{n}{2}$ where $n$ is an integer

Step 9