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.

$y=\frac{1}{2}sec(3x-\mathrm{\pi })$

Short answer

The graph of the function $y=\frac{1}{2}sec(3x-\mathrm{\pi })$is:

Step-by-step solution

Step 1. Given

The function$y=\frac{1}{2}sec(3x-\mathrm{\pi })$

To graph the function with at least two cycles.

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

Compare the given function $y=\frac{1}{2}sec\left(3\right(x-\frac{\mathrm{\pi }}{3}\left)\right)$with $y=A\sin (\omega x-\varphi )+B$

No amplitude

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

Phase shift, $\frac{\varphi }{\omega }=\frac{\mathrm{\pi }}{3}$

Step 3. Determine x-coordinates

One cycle begins at $x=\frac{\varphi }{\omega }\left(\frac{\mathrm{\pi }}{3}\right)$and ends at

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

To find the five key points, divide the interval $(\frac{\mathrm{\pi }}{3},\mathrm{\pi })$into four sub intervals, each of length $\frac{2\mathrm{\pi }}{3}÷4=\frac{\mathrm{\pi }}{6}$

$\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{2}$

$\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{6}=\frac{2\mathrm{\pi }}{3}$

$\frac{2\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}=\frac{5\mathrm{\pi }}{6}$

$\frac{5\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{6}=\mathrm{\pi }$

The $x-$coordinates are $\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2},\frac{2\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{6},\mathrm{\pi }$

Step 4. Determine the key points

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

The key points include $(\frac{\mathrm{\pi }}{3},\frac{1}{2}),(-\frac{\mathrm{\pi }}{3},\frac{1}{2}),(\frac{2\mathrm{\pi }}{3},-\frac{1}{2}),(-\frac{2\mathrm{\pi }}{3},-\frac{1}{2}),(\mathrm{\pi },\frac{1}{2})$

Step 5. Sketch the graph

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