Q31. - Algebra 2

Question

Graph $y=5+\tan \left(\theta +\frac{\pi }{4}\right)$. Describe the transformation to the parent graph $y=\tan \left(\theta \right)$.

Step-by-Step Solution

Answer

From the graph of the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ it can be interpreted that the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ is stretched vertically upwards 5 units and has phase shift of $-\frac{\pi }{4}$ to the left of parent function $y=\tan \left(\theta \right)$. The graph of the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ is shown below.

Step 1. Write down the given information.

The given function is $y=5+\tan \left(\theta +\frac{\pi }{4}\right)$ which can also be written as $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$. And the parent function is $y=\tan \left(\theta \right)$.

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, a period of $\frac{360°}{\left|b\right|} \text{or }\frac{2\pi }{\left|b\right|}$ for sine, cosecant, secant and cosine functions and a period of $\frac{180°}{\left|b\right|} \text{or }\frac{\pi }{\left|b\right|}$ for tangent and cotangent function. The phase shift for the functions is $\left(h\right)$.

The amplitude of secant, cosecant, tangent and cotangent functions is not defined.

The equation for the midline is written as, $y=k$. A midline is a new reference line when the parent graph is stretched vertically up or down and then the graph oscillates about new reference line called the midline.

Step 3. Explanation.

The given functions are parent function $y=\tan \left(\theta \right)$ and another function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$.

The parent function $y=\tan \left(\theta \right)$ is stretched vertically upwards 5 units up and a phase shift of $-\frac{\pi }{4}$ to the left.

It can also be seen from the graph drawn of these two functions.

Step 4. Sketch the graph of given function.

The graph of the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ is shown below.

Step 5. Interpretation from the graph of the function.

From the graph of the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ it can be interpreted that the function $y=\tan \left(\theta +\frac{\pi }{4}\right)+5$ is stretched vertically upwards 5 units and has phase shift of $-\frac{\pi }{4}$ to the left of parent function $y=\tan \left(\theta \right)$.