Question

the parent cosecant function is shifted 4 units right and 3 units up. which of the following is the graph of the transformed function?

Explanation:

Step1: Recall Cosecant Transformations

The parent cosecant function is \( y = \csc(x) \). For horizontal shift \( h \) (right if \( h>0 \)) and vertical shift \( k \) (up if \( k>0 \)), the transformed function is \( y=\csc(x - h)+k \). Here, \( h = 4 \) (right 4) and \( k = 3 \) (up 3), so \( y=\csc(x - 4)+3 \).

Step2: Analyze Key Features

• Vertical Asymptotes: For \( y=\csc(x)=\frac{1}{\sin(x)} \), asymptotes at \( x = n\pi \). For \( y=\csc(x - 4)+3 \), asymptotes at \( x-4=n\pi\Rightarrow x = 4 + n\pi \).
• Midline: The vertical shift moves the midline (the horizontal line around which the graph oscillates) from \( y = 0 \) to \( y = 3 \).
• Behavior: The cosecant graph has "U - shaped" and "inverted U - shaped" branches. After shifting right 4 and up 3, the branches should be centered around \( y = 3 \), with asymptotes at \( x=4,4+\pi,4 + 2\pi,\dots \) and \( x=4-\pi,4 - 2\pi,\dots \).

(Note: Since the full set of graphs isn't fully visible, but based on the transformation rules, the correct graph should have vertical asymptotes shifted right by 4 units and midline at \( y = 3 \), with the cosecant - like branch shapes shifted up 3 units from the parent cosecant's midline of \( y = 0 \). If one of the graphs has asymptotes at \( x\approx4,4+\pi,\dots \) and midline \( y = 3 \), that's the transformed graph. Assuming the second graph (the lower one partially shown) has midline around \( y = 6 - 7 \)? Wait, no, correction: Wait, the vertical shift is 3 up. Wait, maybe the initial graphs had a typo, but following the transformation: parent \( \csc(x) \) has midline \( y = 0 \), after up 3, midline \( y = 3 \). So the correct graph should have its "center" (midline) at \( y = 3 \), asymptotes shifted right by 4. If the second graph (the one with y - axis labels 6,7) – no, maybe the user's image has two graphs, and the correct one is the one where the midline is 3 units above the parent's midline, and asymptotes shifted right 4. But since the problem is about identifying the graph, and based on transformation, the graph with vertical asymptotes at \( x = 4 + n\pi \) and midline \( y = 3 \) is the transformed one. If we assume the second graph (the lower one) has midline around \( y = 6 \)? No, wait, maybe the original parent in the first graph has midline \( y=-2 \) (shifted down), but our transformation is up 3. Wait, perhaps the correct graph is the one where the branches are shifted right 4 and up 3, so midline at \( y = 3 \), asymptotes at \( x = 4,4+\pi,\dots \). So the answer would be the graph (identify by position, e.g., if the second graph is the one with midline around \( y = 3 \) - like, but since the full graphs aren't clear, but following the steps, the transformed graph has \( y=\csc(x - 4)+3 \), so its key features are asymptotes at \( x = 4 + n\pi \) and midline \( y = 3 \).)

Answer:

(Assuming the second graph (the one with y - axis labels 6,7 was a mistake, and the correct graph is the one with midline \( y = 3 \) and asymptotes shifted right 4. If the options are two graphs, and the second graph (the lower one) is the transformed one, but based on standard transformation, the correct graph is the one that represents \( y=\csc(x - 4)+3 \), with vertical asymptotes at \( x = 4 + n\pi \) and midline \( y = 3 \). If we have to choose between the two, and the first graph has midline \( y=-2 \) (shifted down), the second (even partially) with midline higher, so the correct graph is the second one (the Middle or Lower graph depending on layout), but since the user's image shows two graphs, and the second is the one with y - axis 6,7 (maybe a scaling issue, but following the transformation, the answer is the graph that has been shifted right 4 and up 3, so the one with vertical asymptotes at \( x = 4 + n\pi \) and midline \( y = 3 \).)

(Note: Due to partial visibility, but based on transformation rules, the correct graph is the one that applies the horizontal shift of 4 units right and vertical shift of 3 units up to the parent cosecant graph. If the second graph (the lower one in the image) is the one with the transformed features, that's the answer. So the answer is "The second graph (the one with y - axis labels around 6 - 7, or the lower graph)".)