Question

which of the following is the graph of $y = 0.25\csc(x + \pi) + 1)$? (image of graphs omitted)

Explanation:

Step1: Analyze the function transformation

The general form of the cosecant function is \( y = A\csc(B(x - C))+D \). For the given function \( y = 0.25\csc(x+\pi)+1 \), we can rewrite it as \( y = 0.25\csc(1\times(x - (-\pi)))+1 \). So, \( A = 0.25 \), \( B = 1 \), \( C=-\pi \), \( D = 1 \).

- The vertical shift \( D = 1 \) means the midline of the cosecant graph is \( y = 1 \).
- The amplitude (vertical stretch) \( |A|=0.25 \), so the distance from the midline to the peaks/troughs is \( 0.25 \).
- The phase shift \( C = -\pi \) means the graph is shifted \( \pi \) units to the left (since \( C\) is the shift, and here \( x - (-\pi)=x+\pi \)). The period of \( \csc(x) \) is \( 2\pi \), and since \( B = 1 \), the period remains \( 2\pi \).

Step2: Analyze the asymptotes and shape

The asymptotes of \( \csc(x) \) are at \( x = n\pi \) for integer \( n \). For \( y = 0.25\csc(x+\pi)+1 \), the asymptotes occur where \( x+\pi=n\pi \), so \( x=(n - 1)\pi \). For \( n = 0 \), \( x=-\pi \); for \( n = 1 \), \( x = 0 \); for \( n = 2 \), \( x=\pi \), etc.

The cosecant function has "U - shaped" branches above the midline and "inverted U - shaped" branches below (but since \( A = 0.25>0 \), the orientation is similar to \( \csc(x) \) but scaled and shifted). The midline is \( y = 1 \), and the vertical stretch is \( 0.25 \), so the peaks are at \( y=1 + 0.25=1.25 \) (close to \( y = 1 \)) and troughs at \( y=1-0.25 = 0.75 \) (close to \( y = 1 \)).

Looking at the first graph (the upper one with the U - shaped curves around \( y = 1 \) and inverted U - shaped below? Wait, no, the first graph shown has upper curves with minimum around \( y = 1 \) (since the midline is \( y = 1 \), and the vertical stretch is 0.25, so the minimum of the upper U - curves should be \( 1 \) (midline) plus? Wait, no, \( \csc(x) \) has values \( \geq1 \) or \( \leq - 1 \) in its domains. Wait, \( \csc(x)=\frac{1}{\sin(x)} \), so when \( \sin(x) \) is positive and near 1, \( \csc(x) \) is near 1; when \( \sin(x) \) is positive and near 0, \( \csc(x) \) goes to \(+\infty \). When \( \sin(x) \) is negative and near - 1, \( \csc(x) \) is near - 1; when \( \sin(x) \) is negative and near 0, \( \csc(x) \) goes to \( -\infty \).

For our function \( y = 0.25\csc(x+\pi)+1 \), let's substitute \( x = 0 \): \( y=0.25\csc(\pi)+1 \), but \( \csc(\pi) \) is undefined (asymptote at \( x = 0 \) since \( x+\pi=\pi \) when \( x = 0 \), so asymptote at \( x = 0 \)). At \( x=\frac{\pi}{2} \): \( x+\pi=\frac{3\pi}{2} \), \( \sin(\frac{3\pi}{2})=-1 \), so \( \csc(\frac{3\pi}{2})=-1 \), then \( y=0.25\times(-1)+1 = 0.75 \). At \( x=-\frac{\pi}{2} \): \( x+\pi=\frac{\pi}{2} \), \( \sin(\frac{\pi}{2}) = 1 \), so \( \csc(\frac{\pi}{2})=1 \), \( y=0.25\times1 + 1=1.25 \).

So the upper U - shaped curves (where \( \sin(x+\pi)>0 \)) will have minimum at \( y = 1.25 \) (wait, no, when \( \sin(x+\pi)=1 \), \( \csc(x+\pi)=1 \), so \( y = 0.25\times1+1 = 1.25 \); when \( \sin(x+\pi) \) approaches 0 from the positive side, \( \csc(x+\pi) \) approaches \(+\infty \), so the U - shaped curves go up from \( y = 1.25 \) to \(+\infty \) near the asymptotes. The inverted U - shaped curves (where \( \sin(x+\pi)<0 \)) will have maximum at \( y=0.75 \) (when \( \sin(x+\pi)=-1 \), \( \csc(x+\pi)=-1 \), \( y=0.25\times(-1)+1 = 0.75 \)) and go down to \( -\infty \) near the asymptotes.

Looking at the first graph (the one with the x - axis from \( - 2\pi \) to \( 2\pi \) and y - axis with 1 and 2 marked), the upper curves have their minimum around \( y = 1 \) (close to our calculated \( 1.25 \), maybe the graph is drawn with approximate scaling) and the lower curves have their maximum around \( y = 1 \) (close to \( 0.75 \)). The asymptotes are at \( x=-\pi,0,\pi \) etc. The first graph matches the transformation: vertical shift \( y = 1 \), vertical stretch \( 0.25 \), phase shift \( \pi \) left, and the shape of cosecant curves (U - shaped above midline, inverted U - shaped below) with asymptotes at \( x = -\pi,0,\pi \) etc.

The second graph has a different midline (around \( y = 2 - 3 \)) which does not match \( D = 1 \). So the correct graph is the first one (the upper graph with the U - shaped and inverted U - shaped curves around \( y = 1 \)).

Answer:

The graph (the upper one with x - axis from \(-2\pi\) to \(2\pi\) and y - axis with 1 and 2 marked, showing U - shaped curves above \(y = 1\) and inverted U - shaped curves below \(y = 1\) with asymptotes at \(x=-\pi,0,\pi\) etc.)