Question

graph two periods of the given tangent function.
y = tan(x + \frac{\pi}{4})
choose the correct graph of two periods of y = tan(x + \frac{\pi}{4}) below.

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

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 $y=\tan(x+\frac{\pi}{4})$, we set $x+\frac{\pi}{4}=(n + \frac{1}{2})\pi$. Solving for $x$ gives $x=(n+\frac{1}{2})\pi-\frac{\pi}{4}=n\pi+\frac{\pi}{4}$, $n\in\mathbb{Z}$.
When $n = - 1$, $x=-\frac{3\pi}{4}$; when $n = 0$, $x=\frac{\pi}{4}$; when $n = 1$, $x=\frac{5\pi}{4}$.

Step3: Evaluate the function at a key - point

Let $x = 0$. Then $y=\tan(0+\frac{\pi}{4})=1$.
The graph of $y = \tan(x+\frac{\pi}{4})$ is the graph of $y=\tan(x)$ shifted to the left by $\frac{\pi}{4}$ units.

Answer:

D.