Graph y=-1/2csc(x+pi/2)

$y = - \frac{1}{2} \csc (x + \frac{π}{2})$

Step 1

Find the asymptotes.

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

$x + \frac{π}{2} = 0$

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = - \frac{π}{2}$

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

$x + \frac{π}{2} = 2 π$

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

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = 2 π - \frac{π}{2}$

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

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

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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

$(- \frac{π}{2}, \frac{3 π}{2})$

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

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

$x = - \frac{π}{2} + π n$

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{2} + π n$ 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.

$a = - \frac{1}{2}$

$b = 1$

$c = - \frac{π}{2}$

$d = 0$

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 $- \frac{\csc (x + \frac{π}{2})}{2}$ .

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 $1$ .

$2 π$

Step 5

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

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: $\frac{- \frac{π}{2}}{1}$

Divide $- \frac{π}{2}$ by $1$ .

Phase Shift: $- \frac{π}{2}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $2 π$

Phase Shift: $- \frac{π}{2}$ ( $\frac{π}{2}$ to the left)

Vertical Shift: None

Step 7

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

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

Amplitude: None

Period: $2 π$

Phase Shift: $- \frac{π}{2}$ ( $\frac{π}{2}$ to the left)

Vertical Shift: None

Step 8

Graph y=-1/2csc(x+pi/2)

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