Q14. - Algebra 2

Question

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.

Step-by-Step Solution

Answer

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 $\frac{\pi }{4}$.

Step 1. Write down the given information.

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

Step 2. Concept used.

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.

Step 3. Evaluating vertical shift, amplitude, period and phase shift of the given function.

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.

Step 4. Sketch the graph of the function.

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

Step 5. Conclusion.

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 $\frac{\pi }{4}$.