## Precalculus (6th Edition) Blitzer

The two consecutive asymptotes occur at $\left( x+\frac{\pi }{2} \right)=0\text{ and }\left( x+\frac{\pi }{2} \right)=\pi$. We solve $\left( x+\frac{\pi }{2} \right)=0$ to get \begin{align} & \left( x+\frac{\pi }{2} \right)=0 \\ & x=\left( 0-\frac{\pi }{2} \right) \\ & x=-\frac{\pi }{2} \end{align} Again, solve $\left( x+\frac{\pi }{2} \right)=\pi$ to get \begin{align} & \left( x+\frac{\pi }{2} \right)=\pi \\ & x=\pi -\frac{\pi }{2} \\ & x=\frac{\pi }{2} \end{align} Now, the x-intercept is in between the two consecutive asymptotes. Therefore, the x-intercept is given as follows: \begin{align} & x\text{-intercept = }\frac{\left( -\frac{\pi }{2}+\frac{\pi }{2} \right)}{2} \\ & =\frac{0}{2} \\ & =0 \end{align} Thus, the graph passes through $\left( 0,0 \right)$ and the x-intercept is $0$. As the coefficient of the provided cotangent function is $2$, the points on the graph midway between the x-intercept and the asymptotes have y-coordinates of $2$ and $-2$. We use the two consecutive asymptotes, $x=-\frac{\pi }{2}$ and $x=\frac{\pi }{2}$, to graph one full period of $y=2\cot \left( x+\frac{\pi }{2} \right)$ from $-\frac{\pi }{2}\text{ to }\frac{\pi }{2}$.