Algebra Examples

$y = 0.25 \csc (x + π) + 1$

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 $y = 0.25 \csc (x + π) + 1$ . 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 $y = 0.25 \csc (x + π) + 1$ .

$x + π = 0$

Step 1.2

Subtract $π$ from both sides of the equation.

$x = - π$

Step 1.3

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

$x + π = 2 π$

Step 1.4

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

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Step 1.4.1

Subtract $π$ from both sides of the equation.

$x = 2 π - π$

Step 1.4.2

Subtract $π$ from $2 π$ .

$x = π$

Step 1.5

The basic period for $y = 0.25 \csc (x + π) + 1$ 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 $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.6.2

Divide $2 π$ by $1$ .

$2 π$

Step 1.7

The vertical asymptotes for $y = 0.25 \csc (x + π) + 1$ 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.

$a = 0.25$

$b = 1$

$c = - π$

$d = 1$

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 using the formula $\frac{2 π}{| b |}$ .

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Step 4.1

Find the period of $0.25 \csc (x + π)$ .

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Step 4.1.1

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

$\frac{2 π}{| b |}$

Step 4.1.2

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

$\frac{2 π}{| 1 |}$

Step 4.1.3

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

$\frac{2 π}{1}$

Step 4.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2

Find the period of $1$ .

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Step 4.2.1

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

$\frac{2 π}{| b |}$

Step 4.2.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.3

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

$\frac{2 π}{1}$

Step 4.2.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$2 π$

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

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 left)

Vertical Shift: $1$

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 left)

Vertical Shift: $1$

Step 8