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.

$y=-\tan (3x+\frac{\mathrm{\pi }}{2})$

Short answer

The graph of the function $y=-\tan (3x+\frac{\mathrm{\pi }}{2})$is:

Step-by-step solution

Step 1. Given

The function$y=-\tan (3x+\frac{\mathrm{\pi }}{2})$

To graph the function with at least two cycles.

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

Compare the given function $y=-\tan \left(3\right(x-(-\frac{\mathrm{\pi }}{6})\left)\right)$with $y=A\sin (\omega x-\varphi )+B$

No amplitude

Period, $\frac{2\mathrm{\pi }}{\omega }=\frac{\mathrm{\pi }}{3}$

Phase shift, $\frac{\varphi }{\omega }=-\frac{\mathrm{\pi }}{6}$

Step 3. Determine coordinates

One cycle begins at $x=\frac{\varphi }{\omega }(-\frac{\mathrm{\pi }}{6})$and ends at

$$\begin{gathered} x=\frac{\varphi }{\omega }+\frac{2\mathrm{\pi }}{\omega } \\ =-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{3} \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$

To find the five key points, divide the interval $(-\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{6})$into four sub intervals, each of length $\frac{\mathrm{\pi }}{3}÷4=\frac{\mathrm{\pi }}{12}$

$-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{12}=-\frac{\mathrm{\pi }}{12}$

$-\frac{\mathrm{\pi }}{12}+\frac{\mathrm{\pi }}{12}=0$

$0+\frac{\mathrm{\pi }}{12}=\frac{\mathrm{\pi }}{12}$

$\frac{\mathrm{\pi }}{12}+\frac{\mathrm{\pi }}{12}=\frac{\mathrm{\pi }}{6}$

The $x-$coordinates are $-\frac{\mathrm{\pi }}{6},-\frac{\mathrm{\pi }}{12},0,\frac{\mathrm{\pi }}{12},\frac{\mathrm{\pi }}{6}$

Step 4. Determine the key points

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

The key points include $(-\frac{\mathrm{\pi }}{6},0),(-\frac{\mathrm{\pi }}{12},-1),(\frac{\mathrm{\pi }}{4},-1),(\frac{\mathrm{\pi }}{12},1),(\frac{\mathrm{\pi }}{6},0)$

Step 5. Sketch the graph

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