Question

graph two periods of the given tangent function.
y = \frac{1}{2}\tan4x
choose the correct graph of two periods of y = \frac{1}{2}\tan4x below.

Explanation:

Step1: Recall period formula for tangent function

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

Step2: Analyze key - points

The tangent function $y = \tan x$ has vertical asymptotes at $x=(n+\frac{1}{2})\pi,n\in\mathbb{Z}$. For $y=\frac{1}{2}\tan(4x)$, vertical asymptotes occur when $4x=(n +\frac{1}{2})\pi$, or $x=\frac{(2n + 1)\pi}{8},n\in\mathbb{Z}$. When $x = 0$, $y=\frac{1}{2}\tan(0)=0$.

Step3: Consider the amplitude and shape

The amplitude of $y=\frac{1}{2}\tan(4x)$ is $\frac{1}{2}$, which means the values of the function are half of those of the basic tangent function $y = \tan x$. The graph of $y=\frac{1}{2}\tan(4x)$ will have the same general shape as the tangent function but with a period of $\frac{\pi}{4}$ and an amplitude of $\frac{1}{2}$.

Answer:

(Without seeing the full - set of options clearly, we can't give a specific letter - choice answer. But the graph should have vertical asymptotes at $x=\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}$ for two periods in the interval $[-\frac{\pi}{8},\frac{7\pi}{8}]$ and pass through the origin $(0,0)$ with a "flatter" shape compared to $y = \tan x$ due to the amplitude of $\frac{1}{2}$.)