Q. 24 - Precalculus Enhanced with Graphing Utilities

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.

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

Step-by-Step Solution

Answer

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

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.