Graph y=csc((pix)/2)

Problem

y=csc((π*x)/2)

Solution

1. Identify the parent function and its properties. The function is y=csc(θ) which is the reciprocal of sin(θ) It has vertical asymptotes where sin(θ)=0

2. Determine the period of the function. The period P is calculated using the coefficient of x which is b=π/2

P=(2*π)/|b|

P=(2*π)/π/2

P=4

3. Find the vertical asymptotes by setting the argument of the cosecant function equal to n*π where n is an integer.

(π*x)/2=n*π

x=2*n

The asymptotes occur at x=0,±2,±4,…

4. Locate key points by finding the relative extrema. These occur halfway between the asymptotes, where the sine function reaches 1 or −1
   For n=0 and n=1 the midpoint is x=1

y=csc((π*(1))/2)=csc(π/2)=1

For n=1 and n=2 the midpoint is x=3

y=csc((π*(3))/2)=csc((3*π)/2)=−1

5. Sketch the graph by drawing the vertical asymptotes at even integers and plotting the local minimums at (1,1),(5,1),… and local maximums at (3,−1),(7,−1),… The curves approach the asymptotes in a U-shape or inverted U-shape.

Final Answer

y=csc((π*x)/2)* has period *4, asymptotes at *x=2*n, and range *(−∞,−1]∪[1,∞)

Want more problems? Check here!