Algebra Examples

$y = \csc (π - x)$

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 $y = \csc (π - x)$ . 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 $y = \csc (π - x)$ .

$π - x = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Subtract $π$ from both sides of the equation.

$- x = - π$

Step 1.2.2

Divide each term in $- x = - π$ by $- 1$ and simplify.

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Step 1.2.2.1

Divide each term in $- x = - π$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- π}{- 1}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- π}{- 1}$

Step 1.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- π}{- 1}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{π}{1}$

Step 1.2.2.3.2

Divide $π$ by $1$ .

$x = π$

Step 1.3

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

$π - x = 2 π$

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

Subtract $π$ from both sides of the equation.

$- x = 2 π - π$

Step 1.4.1.2

Subtract $π$ from $2 π$ .

$- x = π$

Step 1.4.2

Divide each term in $- x = π$ by $- 1$ and simplify.

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Step 1.4.2.1

Divide each term in $- x = π$ by $- 1$ .

$\frac{- x}{- 1} = \frac{π}{- 1}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{π}{- 1}$

Step 1.4.2.2.2

Divide $x$ by $1$ .

$x = \frac{π}{- 1}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Move the negative one from the denominator of $\frac{π}{- 1}$ .

$x = - 1 \cdot π$

Step 1.4.2.3.2

Rewrite $- 1 \cdot π$ as $- π$ .

$x = - π$

Step 1.5

The basic period for $y = \csc (π - x)$ will occur at $(π, - π)$ , where $π$ and $- π$ are vertical asymptotes.

$(π, - π)$

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

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$\frac{2 π}{1}$

Step 1.6.2

Divide $2 π$ by $1$ .

$2 π$

Step 1.7

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

$x = π + π n$

Step 1.8

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = π + π n$ 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.

$a = 1$

$b = - 1$

$c = - π$

$d = 0$

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 $\csc (π - x)$ .

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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 $- 1$ in the formula for period.

$\frac{2 π}{| - 1 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$\frac{2 π}{1}$

Step 4.4

Divide $2 π$ by $1$ .

$2 π$

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: $\frac{- π}{- 1}$

Step 5.3

Dividing two negative values results in a positive value.

Phase Shift: $\frac{π}{1}$

Step 5.4

Divide $π$ by $1$ .

Phase Shift: $π$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $2 π$

Phase Shift: $π$ ( $π$ 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: $x = π + π n$ where $n$ is an integer

Amplitude: None

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: None

Step 8