Problem 48 - Precalculus Essentials

Question

Graph two periods of each function. $$y=\csc \left(2 x-\frac{\pi}{2}\right)+1$$

Step-by-Step Solution

Answer

With a period of \(\pi\), phase shift of \(\frac{\pi}{4}\) to the right, and vertical shift upward by 1 unit, the two periods of the function \(y = \csc(2x - \frac{\pi}{2})+1\) can be accurately graphed.

Step 1: Finding the period

The standard formula for the period of a transformed trigonometric function is given by \( P = \frac{2\pi}{|b|} \), where \(b\) is a coefficient bound with the argument of the considered function. In this case, \(b = 2\), so the period of the function \( y = \csc(2x - \frac{\pi}{2})+1 \) will be \( P = \frac{2\pi}{2} = \pi \) - the function repeats every \( \pi \) units.

Step 2: Identifying the phase shift

The phase shift of the function can be found by setting the argument of the function equal to zero. In this case, setting \(2x - \frac{\pi}{2}\) equal to zero gives \(x = \frac{\pi}{4}\). Therefore the function is shifted \(\frac{\pi}{4}\) units to the right.

Step 3: Identifying the vertical shift

The function is shifted upwards by 1 unit, as indicated by the '+1' at the end of the function.

Step 4: Sketching the function

Utilizing the information, it is known that every \(\pi\) units the function will repeat itself. Moreover, we know that it has been horizontally shifted by \(\frac{\pi}{4}\) to the right and vertically upward by 1 unit. Now, draw a clear and comprehensible graph accordingly. Remember that cosecant function will have asymptotes where corresponding sine function crosses x-axis. Use correct labeling.