directions: determine period, vertical shift, and asymptotes for each of the following reflection and find…

directions: determine period, vertical shift, and asymptotes for each of the following reflection and find the 5 key points. then graph one period of the function. for 3. $y = 2\tan(x-\frac{pi}{4})$ period = $\frac{pi}{b}=\frac{pi}{1}=pi$ reflection = yes or no vertical shift = up or down asymptotes: and $-\frac{pi}{2}< bx + c<\frac{pi}{2}$

directions: determine period, vertical shift, and asymptotes for each of the following reflection and find the 5 key points. then graph one period of the function. for 3. $y = 2\tan(x-\frac{pi}{4})$ period = $\frac{pi}{b}=\frac{pi}{1}=pi$ reflection = yes or no vertical shift = up or down asymptotes: and $-\frac{pi}{2}< bx + c<\frac{pi}{2}$

Answer

Explanation:

Step1: Determine the period

For the tangent function (y = A\tan(Bx - C)+D), the period is given by (\frac{\pi}{|B|}). In the function (y = 2\tan(x-\frac{\pi}{4})), (B = 1). So the period (T=\frac{\pi}{|1|}=\pi).

Step2: Check for reflection

The coefficient (A = 2>0), so there is no reflection about the (x -)axis. The answer for reflection is no.

Step3: Determine the vertical - shift

The general form is (y = A\tan(Bx - C)+D). Here (D = 0), so there is no vertical shift, i.e., 0 up or down.

Step4: Find the asymptotes

For the tangent function (y=\tan x), the asymptotes are (x=-\frac{\pi}{2}+k\pi) and (x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}). For (y = 2\tan(x - \frac{\pi}{4})), we set (x-\frac{\pi}{4}=-\frac{\pi}{2}+k\pi) and (x-\frac{\pi}{4}=\frac{\pi}{2}+k\pi). Solving (x-\frac{\pi}{4}=-\frac{\pi}{2}+k\pi) gives (x=-\frac{\pi}{4}+k\pi). Solving (x-\frac{\pi}{4}=\frac{\pi}{2}+k\pi) gives (x=\frac{3\pi}{4}+k\pi). For one - period, the asymptotes are (x =-\frac{\pi}{4}) and (x=\frac{3\pi}{4}).

Step5: Find the key points

The period is (\pi). We consider the interval ((-\frac{\pi}{4},\frac{3\pi}{4})). When (x = \frac{\pi}{4}), (y=2\tan(\frac{\pi}{4}-\frac{\pi}{4})=2\tan(0)=0). When (x=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}), (y = 2\tan(\frac{\pi}{2}-\frac{\pi}{4})=2\tan(\frac{\pi}{4})=2). When (x=\frac{\pi}{4}-\frac{\pi}{4}=0), (y = 2\tan(0 - \frac{\pi}{4})=2\tan(-\frac{\pi}{4})=- 2). When (x=\frac{\pi}{4}+\frac{\pi}{6}=\frac{3\pi + 2\pi}{12}=\frac{5\pi}{12}), (y = 2\tan(\frac{5\pi}{12}-\frac{\pi}{4})=2\tan(\frac{\pi}{6})=\frac{2\sqrt{3}}{3}). When (x=\frac{\pi}{4}-\frac{\pi}{6}=\frac{3\pi - 2\pi}{12}=\frac{\pi}{12}), (y = 2\tan(\frac{\pi}{12}-\frac{\pi}{4})=2\tan(-\frac{\pi}{6})=-\frac{2\sqrt{3}}{3}).

Answer:

Period: (\pi) Reflection: no Vertical Shift: 0 up or down Asymptotes: (x =-\frac{\pi}{4}) and (x=\frac{3\pi}{4}) Key - points: ((\frac{\pi}{4},0),(\frac{\pi}{2},2),(0, - 2),(\frac{5\pi}{12},\frac{2\sqrt{3}}{3}),(\frac{\pi}{12},-\frac{2\sqrt{3}}{3}))