Graph y=csc(x+(4pi)/5)

$y = \csc (x + \frac{4 π}{5})$

Find the asymptotes.

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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 $y = \csc (x + \frac{4 π}{5})$ . 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 $y = \csc (x + \frac{4 π}{5})$ .

$x + \frac{4 π}{5} = 0$

Subtract $\frac{4 π}{5}$ from both sides of the equation.

$x = - \frac{4 π}{5}$

Set the inside of the cosecant function $x + \frac{4 π}{5}$ equal to $2 π$ .

$x + \frac{4 π}{5} = 2 π$

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

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Subtract $\frac{4 π}{5}$ from both sides of the equation.

$x = 2 π - \frac{4 π}{5}$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$x = 2 π \cdot \frac{5}{5} - \frac{4 π}{5}$

Combine $2 π$ and $\frac{5}{5}$ .

$x = \frac{2 π \cdot 5}{5} - \frac{4 π}{5}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 5 - 4 π}{5}$

Simplify the numerator.

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Multiply $5$ by $2$ .

$x = \frac{10 π - 4 π}{5}$

Subtract $4 π$ from $10 π$ .

$x = \frac{6 π}{5}$

The basic period for $y = \csc (x + \frac{4 π}{5})$ will occur at $(- \frac{4 π}{5}, \frac{6 π}{5})$ , where $- \frac{4 π}{5}$ and $\frac{6 π}{5}$ are vertical asymptotes.

$(- \frac{4 π}{5}, \frac{6 π}{5})$

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

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The vertical asymptotes for $y = \csc (x + \frac{4 π}{5})$ occur at $- \frac{4 π}{5}$ , $\frac{6 π}{5}$ , and every $x = - \frac{4 π}{5} + π n$ , where $n$ is an integer. This is half of the period.

$x = - \frac{4 π}{5} + π n$

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{4 π}{5} + π n$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{4 π}{5} + π n$ where $n$ is an integer

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

$a = 1$

$b = 1$

$c = - \frac{4 π}{5}$

$d = 0$

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

Find the period of $\csc (x + \frac{4 π}{5})$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Find the phase shift using the formula $\frac{c}{b}$ .

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The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{- \frac{4 π}{5}}{1}$

Divide $- \frac{4 π}{5}$ by $1$ .

Phase Shift: $- \frac{4 π}{5}$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $2 π$

Phase Shift: $- \frac{4 π}{5}$ ( $\frac{4 π}{5}$ to the left)

Vertical Shift: $0$

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

Vertical Asymptotes: $x = - \frac{4 π}{5} + π n$ where $n$ is an integer

Amplitude: None

Period: $2 π$

Phase Shift: $- \frac{4 π}{5}$ ( $\frac{4 π}{5}$ to the left)

Vertical Shift: $0$

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