Chapter EP, Problem 13.8.14EP

To determine

To find:The amplitude, period, graph, equation of midline and vertical shift for the given function.

Answer to Problem 13.8.14EP

The amplitude, period, graph, equation of midline and vertical shift for the given function is determined.

Explanation of Solution

Given:

$y=\frac{2}{5}\sin \left[6\left(\theta +45\right)\right]-5$

Calculation:

The general form of the phase shift is given as,

  $\begin{array}{l}y=a\sin b\left(\theta -h\right)+k\\ y=a\cos b\left(\theta -h\right)+k\\ y=a\tan b\left(\theta -h\right)+k\end{array}$

If the value of h is greater than 0, then it shift towards the right

If the value of h is less than 0, then it shift towards the left.

If the value of k is greater than 0, then the shift is up.

If the value of k is less than 0, then the shift is down.

The value of a is 2/5, b is6, h is -45 and k is -5.

The vertical shift is -5

The time period is,

$\begin{array}{c}T=\frac{2\pi }{6}\\ =\frac{\pi }{3}\end{array}$

The equation of the mid line is,

$\begin{array}{c}y=k\\ y=-5\end{array}$

The graph is plotted as,

Thus, the amplitude, period, graph, equation of midline and vertical shift for the given function is determined.