Graph f(x)=3cot(2(x-pi/2))+2

For any ${\displaystyle y=\cot \left(x\right)}$, vertical asymptotes occur at ${\displaystyle x=n\pi }$, where ${\displaystyle n}$ is an integer. Use the basic period for ${\displaystyle y=\cot \left(x\right)}$, ${\displaystyle (0,\pi )}$, to find the vertical asymptotes for ${\displaystyle y=3\cot (2x-\pi )+2}$. Set the inside of the cotangent function, ${\displaystyle bx+c}$, for ${\displaystyle y=a\cot (bx+c)+d}$ equal to ${\displaystyle 0}$ to find where the vertical asymptote occurs for ${\displaystyle y=3\cot (2x-\pi )+2}$.

Solve for ${\displaystyle x}$.

Set the inside of the cotangent function ${\displaystyle 2x-\pi }$ equal to ${\displaystyle \pi }$.

Solve for ${\displaystyle x}$.

The basic period for ${\displaystyle y=3\cot (2x-\pi )+2}$ will occur at ${\displaystyle (\frac{\pi }{2},\pi )}$, where ${\displaystyle \frac{\pi }{2}}$ and ${\displaystyle \pi }$ are vertical asymptotes.

The absolute value is the distance between a number and zero. The distance between ${\displaystyle 0}$ and ${\displaystyle 2}$ is ${\displaystyle 2}$.

The vertical asymptotes for ${\displaystyle y=3\cot (2x-\pi )+2}$ occur at ${\displaystyle \frac{\pi }{2}}$, ${\displaystyle \pi }$, and every ${\displaystyle x=\frac{\pi }{2}+\frac{\pi n}{2}}$, where ${\displaystyle n}$ is an integer.

Cotangent only has vertical asymptotes.

Find the period of ${\displaystyle 3\cot (2x-\pi )}$.

Find the period of ${\displaystyle 2}$.

The period of addition/subtraction of trig functions is the maximum of the individual periods.

The phase shift of the function can be calculated from ${\displaystyle \frac{c}{b}}$.

Replace the values of ${\displaystyle c}$ and ${\displaystyle b}$ in the equation for phase shift.