graphing trigonometric functions I am so confused on how to graph $$y=\tan^3x-\tan^2x+3\tan{x}-3$$
I already have the graph from wolphram however I don't know how to arrive to such graph. I believe I have to write it as a function $y=A\tan(Bx-C)+D$ but I don't know how. Please help.
 A: As others have pointed out, you can factor your function as
$$y=(\tan x-1)(\tan^2 x+3)$$
Note that the second factor is never zero, and in fact is always greater than or equal to three. Telling the difference between two large slopes, such as $3$ and $4$, is very difficult for us humans. Therefore your graph will be very similar to the graph of
$$y=3(\tan x-1)$$
You can easily graph that by hand, and there will be almost no practical difference between the graphs. Here are sample graphs from a graphing calculator (TI-Nspire CX):

It seems clear that your desired graph is not exactly equal to a linearly transformed tangent graph, so I do not see how you could do any better by hand.
A: Hint...you could factorise it as $$(\tan x-1)(\tan^2x+3)$$ which would indicate the roots etc...
A: Let $T -$ period of $y$. Then $T=\pi$. Graphing function enough to build $[-\frac{\pi}{2};\frac{\pi}{2}]$
$$y=\tan^3x-\tan^2x+3\tan{x}-3$$
$$x\not =\frac{\pi}2+\pi k$$
$$y=\tan^2x(\tan x -1)+3(\tan x-1)$$
$$y=(\tan x -1)(\tan^2x+3)$$
$$y=0; x=\frac{\pi}4+\pi n$$
If $x\rightarrow \frac{\pi}{2}$, then $\tan x\rightarrow +\infty $, then $y\rightarrow +\infty$;
If $x\rightarrow -\frac{\pi}{2}$, then $\tan x\rightarrow -\infty $, then $y\rightarrow -\infty$.
