Answer:
Option B. y = csc(x) + 1
Explanation:
If we have to draw an asymptote graph( csc graph), we should draw the graph of sine first.
For y = a sin(bx+c) +d
Here from the given graph
a = (0+2)/2 = 1
period = (3π/2-π/2) = 2π
Therefore b = 2π/period = 2π/2π = 1
b = 1
Vertical shift d = 1
Now the equation will be y = 1×sin(1.x+0)+1
y = sinx + 1
Now we draw the sine graph y = sinx +1
Then we draw asymptotes touching the maximum and minimum values of sine graph.
Which reflects It's a graph of y = csc(x) + 1
Therefore option B. is the right answer.