Problem 48 - Precalculus

Question

In Exercises \(47-56,\) graph the functions over at least one period. $$y=-3+2 \csc \left(x+\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The graph of the function is a vertically shifted and compressed cosecant curve, with period \(2\pi\), shifted left by \(\frac{\pi}{2}\), and vertical shift of \(-3\).

Step 1: Understanding the Function Form

The given function is \( y = -3 + 2 \csc\left(x + \frac{\pi}{2}\right) \). This can be identified as a transformation of the basic cosecant function \( y = a \csc(bx - c) + d \), with \( a = 2 \), \( b = 1 \), \( c = -\frac{\pi}{2} \), and \( d = -3 \).

Step 2: Determine Key Characteristics

To determine the period of the function, use the formula \( \frac{2\pi}{b} \). Since \( b = 1 \), the period is \( 2\pi \). The vertical shift is \(-3\), and the amplitude of the cosecant transformation factor (distance from the midline to a maximum/minimum point) is \( 2 \).

Step 3: Find the Critical Points

For \( \csc(x) = \frac{1}{\sin(x)} \), zeros of \( \sin(x) \) correspond to vertical asymptotes of \( \csc(x) \). The phase shift of \( \frac{-c}{b} = -\frac{\pi}{2} \) translates the graph left by \( \frac{\pi}{2} \). Determine where the basic function \( \sin(x + \frac{\pi}{2}) = 0 \): this occurs at \( x = -\frac{\pi}{2} + n\pi \), for integer values of \( n \).

Step 4: Sketching the Graph's Framework

Start by plotting vertical asymptotes at \( x = -\frac{\pi}{2} + n\pi \). Between each pair of asymptotes, plot the curve based on the transformations. The graph will cross the line \( y = -3 \) midway between each asymptote, peaking at \( y = -3 + 2 \) and bottoming at \( y = -3 - 2 \).

Step 5: Graph the Function Over One Period

Plot the function between \( x = -\frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \), completing one period of length \( 2\pi \). Ensure all transformations are applied: vertical shift, amplitude modification, phase shift, and calculate intersections accurately for a proper graph.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc(x) \), is the reciprocal of the sine function: \( \csc(x) = \frac{1}{\sin(x)} \). This function is undefined where \( \sin(x) = 0 \), meaning it has vertical asymptotes at these points. In the graph of \( \csc(x) \), you will notice this defining characteristic with curves approaching infinity or negative infinity at each asymptote. When graphing a transformed cosecant function, consider both the basic shape and these vertical asymptotes. The transformations applied will shift, stretch, or compress the basic graph. Make sure to keep track of these changes to plot the cosecant function accurately.

Amplitude

Amplitude typically describes the height from the centerline to the peak in waveforms like sine and cosine. However, for the cosecant function, this concept refers to the vertical stretch or compression of its curve. In the transformed function \( y = 2 \csc(x + \frac{\pi}{2}) \), the factor of \( 2 \) indicates that the basic cosecant curve stretches vertically by a factor of \( 2 \).

• The peaks will be at \( +2 \) units above the centerline,
• The troughs will be at \( -2 \) units below the centerline.

This vertical stretch accentuates the height of the cosecant's arms as they extend towards their respective infinities.

Vertical Shift

A vertical shift moves the graph of the function up or down along the \( y \)-axis. For the function \( y = -3 + 2 \csc\left(x + \frac{\pi}{2}\right) \), a vertical shift occurs due to the \( -3 \).This shift moves the entire graph down by 3 units. The midline of the function is now at \( y = -3 \), effectively relocating where the peaks and valleys are measured from:

• The new relative maximum is at \( -1 \) (\( -3 + 2 \)),
• And the relative minimum is at \( -5 \) (\( -3 - 2 \)).

Using this shift ensures you correctly translate the original graph vertically, positioning it within the new context of the function's range.

Phase Shift

The phase shift represents a horizontal movement left or right of the function's graph along the \( x \)-axis. This shift is essential when considering enhancements to the standard period of a wave function.In our given function \( y = 2 \csc\left(x + \frac{\pi}{2}\right) \), the \( x + \frac{\pi}{2} \) part implies a leftward shift by \( \frac{\pi}{2} \) units.

• This phase shift modifies where each cycle starts,
• It affects the positioning of the vertical asymptotes.

Knowing this shift allows for plotting the function's graph relative to the typical zero-points of sine, thus adequately reflecting the changes induced on the basic wave cycle.