Function: $y = 5 \csc(2x)$

We are tasked with graphing two full periods of this function and finding various characteristics like the amplitude, period, phase shift, vertical shift, asymptotes, and extremum points.

1. Amplitude or Stretching Factor

For a cosecant function $y = A \csc(Bx + C) + D$, the coefficient $A$ affects the vertical stretch of the graph.

• Here, $A = 5$, so the stretching factor is 5.

Cosecant functions do not have an amplitude like sine or cosine functions because they have vertical asymptotes and no maximum or minimum values. However, the absolute value of $A$ represents the vertical stretching.

2. Period

The period of a cosecant function is given by: $\text{Period} = \frac{2\pi}{B}$ In this case, $B = 2$, so the period is: $\text{Period} = \frac{2\pi}{2} = \pi$ Thus, one period of this function occurs over an interval of length $\pi$.

3. Phase Shift

The phase shift is determined by $C$, but since there is no horizontal shift (i.e., $C = 0$), the phase shift is: $\text{Phase Shift} = \frac{-C}{B} = 0$

4. Vertical Shift

The vertical shift is given by $D$, but in this case, $D = 0$, so there is no vertical shift.

5. Asymptotes

The cosecant function has vertical asymptotes wherever the sine function is zero, because $\csc(x) = \frac{1}{\sin(x)}$.

The function $\sin(2x) = 0$ when: $2x = n\pi, \text{ where } n \text{ is any integer}.$ Thus, the vertical asymptotes occur at: $x = \frac{n\pi}{2}, \text{ where } n \text{ is any integer}.$

For two full periods, the vertical asymptotes will appear at:

• $x = 0$
• $x = \frac{\pi}{2}$
• $x = \pi$
• $x = \frac{3\pi}{2}$
• $x = 2\pi$

6. Maximum and Minimum Points

The cosecant function attains its maximum and minimum values where the sine function reaches its maximum and minimum values. The sine function attains its maximum at $\sin(2x) = 1$ and its minimum at $\sin(2x) = -1$.

For $y = 5 \csc(2x)$, the maximum value of $\csc(2x)$ occurs at the points where $\sin(2x) = 1$, which gives: $y = 5 \csc(2x) = 5 \times 1 = 5$ The minimum value of $\csc(2x)$ occurs at the points where $\sin(2x) = -1$, which gives: $y = 5 \csc(2x) = 5 \times (-1) = -5$

This will happen at the points:

• Maximum at $x = \frac{\pi}{4}$ with $y = 5$
• Minimum at $x = \frac{3\pi}{4}$ with $y = -5$

Graphing Two Full Periods

To graph two full periods of $y = 5 \csc(2x)$:

• The period is $\pi$, so two periods span from $x = 0$ to $x = 2\pi$.
• Vertical asymptotes occur at $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$.
• The maximum occurs at $x = \frac{\pi}{4}$ with $y = 5$.
• The minimum occurs at $x = \frac{3\pi}{4}$ with $y = -5$.

Would you like a visual graph of this, or further details on any specific part?

Additional Questions:

1. How is the vertical asymptote derived in a cosecant function?
2. Why does the cosecant function have no amplitude?
3. What is the relationship between sine and cosecant in terms of their graph shapes?
4. How does the period change if the function is $y = 5 \csc(3x)$?
5. How would a phase shift affect the location of the asymptotes and extrema?

Tip: Vertical asymptotes for $\csc(x)$ occur where the sine function equals zero, as the function is undefined at these points.