which is the graph of the function $y = 2\\tan(x+\frac{3\\pi}{4})$?

Explanation:

Step1: Recall the properties of the tangent - function

The general form of the tangent function is $y = A\tan(Bx - C)+D$. For the function $y = 2\tan(x+\frac{3\pi}{4})$, we have $A = 2$, $B = 1$, $C=-\frac{3\pi}{4}$, and $D = 0$. The period of the tangent function $y=\tan(Bx)$ is $T=\frac{\pi}{|B|}$. Since $B = 1$, the period of $y = 2\tan(x+\frac{3\pi}{4})$ is $\pi$.

Step2: Find the vertical asymptotes

The vertical asymptotes of the tangent function $y=\tan(x)$ occur at $x=(n+\frac{1}{2})\pi$, $n\in\mathbb{Z}$. For the function $y = 2\tan(x+\frac{3\pi}{4})$, we set $x+\frac{3\pi}{4}=(n+\frac{1}{2})\pi$. Solving for $x$ gives $x=(n+\frac{1}{2})\pi-\frac{3\pi}{4}=n\pi+\frac{\pi}{2}-\frac{3\pi}{4}=n\pi-\frac{\pi}{4}$, $n\in\mathbb{Z}$.

Step3: Analyze the phase - shift

The phase - shift of the function $y = A\tan(Bx - C)+D$ is $\frac{C}{B}$. Here, the phase - shift is $-\frac{3\pi}{4}$ (a shift to the left by $\frac{3\pi}{4}$ units). When $x =-\frac{3\pi}{4}$, $y = 2\tan(0)=0$.

We can also check some key - points: Let $x=-\frac{\pi}{4}$, then $y = 2\tan(-\frac{\pi}{4}+\frac{3\pi}{4})=2\tan(\frac{\pi}{2})$, which is undefined (vertical asymptote). Let $x = \frac{3\pi}{4}$, then $y=2\tan(\frac{3\pi}{4}+\frac{3\pi}{4})=2\tan(\frac{3\pi}{2})$, which is undefined (vertical asymptote).

Answer:

We need to check the graph that has a period of $\pi$, vertical asymptotes at $x = n\pi-\frac{\pi}{4},n\in\mathbb{Z}$, and passes through the point $(-\frac{3\pi}{4},0)$. Without seeing all the options, we can't give a specific graph as the answer, but the above - mentioned properties should be used to identify the correct graph. If you provide the other graph options, we can further determine the correct one.