Question

In Problems 19–26, apply the methods of this and the previous section to graph each function. Be sure to label key points and show at least two periods.

$y=-cot(2x+\frac{\mathrm{\pi }}{2})$

Short answer

The graph of the function $y=-cot(2x+\frac{\mathrm{\pi }}{2})$is:

Step-by-step solution

Step 1. Given

The function$y=-cot(2x+\frac{\mathrm{\pi }}{2})$

To graph the function with at least two cycles.

Step 2. Find the amplitude, period and phase shift.

Compare the given function $y=-cot\left(2\right(x-(-\frac{\mathrm{\pi }}{4})\left)\right)$with $y=A\sin (\omega x-\varphi )+B$

No amplitude

Period, $\frac{2\mathrm{\pi }}{\omega }=\frac{\mathrm{\pi }}{2}$

Phase shift, $\frac{\varphi }{\omega }=-\frac{\mathrm{\pi }}{4}$

Step 3. Determine x-coordinates

One cycle begins at $x=\frac{\varphi }{\omega }(-\frac{\mathrm{\pi }}{4})$and ends at

$$\begin{gathered} x=\frac{\varphi }{\omega }+\frac{2\mathrm{\pi }}{\omega } \\ =-\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{2} \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

To find the five key points, divide the interval $(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4})$into four sub intervals, each of length $\frac{\mathrm{\pi }}{2}÷4=\frac{\mathrm{\pi }}{8}$

$-\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{8}=-\frac{\mathrm{\pi }}{8}$

$-\frac{\mathrm{\pi }}{8}+\frac{\mathrm{\pi }}{8}=0$

$0+\frac{\mathrm{\pi }}{8}=\frac{\mathrm{\pi }}{8}$

$\frac{\mathrm{\pi }}{8}+\frac{\mathrm{\pi }}{8}=\frac{\mathrm{\pi }}{4}$

The $x-$coordinates are $-\frac{\mathrm{\pi }}{4},-\frac{\mathrm{\pi }}{8},0,\frac{\mathrm{\pi }}{8},\frac{\mathrm{\pi }}{4}$

Step 4. Determine the key points

Use these values of $x$to determine the key points on the graph:

The key points include $(-\frac{\mathrm{\pi }}{8},-1),(0,0),(-\frac{3\mathrm{\pi }}{8},1),(\frac{\mathrm{\pi }}{8},1),(\frac{3\mathrm{\pi }}{8},-1)$

Step 5. Sketch the graph

Plot these five points and fill in the graph of the function.