## Trigonometry (11th Edition) Clone

The function is $y=\tan (x+\frac{\pi}{4})$. When we compare $y=\tan (x+\frac{\pi}{4})$ to the general form $y=c+a\tan (bx-d)$, we find that $d=-\frac{\pi}{4}$. $d$ represents the phase shift and since $d$ is negative, the phase shift is $\frac{\pi}{4}$ units to the left. Also, since $a=1$, $b=1$ and $c=0$, we can ascertain that the graph of $y=\tan (x+\frac{\pi}{4})$ will be the same as the graph of $y=\tan x$ except that it will be shifted $\frac{\pi}{4}$ units to the left. This means that the two vertical asymptotes will now be $-\frac{\pi}{2}-\frac{\pi}{4}=-\frac{3\pi}{4}$ and $\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$. This means that the correct graph for $y=\tan (x+\frac{\pi}{4})$ is E.