State the vertical shift, amplitude, period and phase shift of the function $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ and then graph the function.

The given function is $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$.

A function of the form:

$y=a\sin b\left(\theta -h\right)+k,y=a\cos b\left(\theta -h\right)+k\text{and}y=a\tan b\left(\theta -h\right)+k$ has vertical shift $\left(k\right)$. And period $\frac{360°}{\left|b\right|}\text{or}\frac{2\pi }{\left|b\right|}$ for sine and cosine functions and a period of $\frac{180°}{\left|b\right|}\text{or}\frac{\pi }{\left|b\right|}$ for tangent function. The phase shift for the functions is $\left(h\right)$.

The amplitude of tangent and cotangent functions is not defined.

With the help of concept stated above, the vertical shift, amplitude, period and phase shift of the function is evaluated as:

The vertical shift of the function $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$is 1 and a positive vertical shift of the function indicates that the vertical shift is up.

The amplitude of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$is not defined.

The period of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ is $\frac{2\pi }{\left|4\right|}=\frac{\pi }{2}$.

The phase shift of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ is $\frac{\pi }{4}$ and a positive value of phase shift indicates that the phase shift is to the right.

The graph of the function $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ is shown below.

The vertical shift of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$is 1.

The amplitude of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$is not defined.

The period of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ is $\frac{\pi }{2}$.

The phase shift of $y=\frac{1}{2}\sec \left[4\left(\theta -\frac{\pi }{4}\right)\right]+1$ is $\frac{\pi }{4}$.