Understanding function transformations I am trying to wrap my head around some function transformation concepts. Take the following equation for instance:
$y = \arctan(Bx - C) + D$


*

*In this equation, $B$ is the period where $P = \frac{\pi}{B}$. Is $x$ a part of the period? It is kind of sitting there and I'm not entirely sure what to make of it. My assumption is that it is but no one has really discussed it and it's all a bit vague how to correlate the two.

*The $C$ variable dictates the phase shift. What confuses me is $(x - \frac{\pi}{2})$ would result in a shift to the right - why? we are clearly doing a subtraction here so shouldn't it move to the left of the $x$ axis?

*If we plot $x=\frac{\pi}{2}$ for the above equation, how is that to be understood ? If my assumption in question 1 is correct and $x$ is a part of the period, and $B=1$ then $P=\frac{\pi}{2}$ and therefore the asymptotes are $-\frac{\pi}{8}$ and $\frac{\pi}{8}$.
 A: For your first question, x is the independent variable. Think about graphing, the horizontal x is what you’re changing, and what you’re “actually graphing” is the y. So x is not part of the period
Second question, think of it as moving the x axis and the grid paper to the left instead. When you move the grid to the left, relatively the graph is moving to the right. Hope that makes sense haha.
Lastly, if B=1, that’s equal to tan(x) shifted around I.e.not horizontally stretched. Then the asymptote of tan(x) is $-\pi$/2 and $\pi$/2. However since we shifted the graph to the right by $\pi$/2, it’s now at $-\pi$,0, $\pi$ instead (and every $\pi$ there’s one)
Hope this clarify your questions :)
Also I’m on phone so typing the latex is really troublesome. Would some nice person help format the pi? Thank you so much!
A: I will answer to all your questions below:
1) We can't consider $x$ to be part of the period if the function $f(x)$ because the period must havr a costant value, not dependent on $x$. When you plot $f(x)$, the variable $x$ is indipendent, so it isn'tpart of the period.
2) When we are cosidering shifted functon we have always thonk that given a vector $\vec{v}(x_0,y_0)$, the new coords are obtained from:
$$\left\{\begin{matrix}
x'=x-x_0
\\y'=y-y_0
\end{matrix}\right.$$
And so:
$$\left\{\begin{matrix}
x=x'-x_0
\\y=y'-y_0
\end{matrix}\right.$$
From here it's very simple to understand what you have noted.
3)Because you assumption in question $(1)$ are not correct, you conclusion here is wrong. In particular because $Bx$ is not the period, you have:
$$f\left(\frac{\pi}{2}\right)=\arctan\left(B\cdot\frac{\pi}{2}-C\right)+D$$
