How do you graph #y = -2tan(x+(pi/4))#?

For the graphing questions, I always follow this way which is quite easy.

Firstly, Look at the trigonometric function which in this case is #tan#. You need to know the graph of #y= tanx#. graph{tanx [-8.89, 8.89, -4.444, 4.445]}

Secondly, for the period of the function, look for the coefficient of #x# which in this case is 1. So the period of this graph is same as that of #y= tanx#.

Thirdly, In # (x+(pi/4))# , if there is use of + sign, it means shifting the whole graph left. If there is - sign, then we have to shift the whole graph right. In this case, we have to shift it left.

graph{tan(x+(pi/4)) [-8.89, 8.89, -4.444, 4.445]}

Now,
Look for the coefficient of the trigonometric function which in this case is 2. So, if #tan(pi/4) # gives #y#= 1, then 2#tan(pi/4) # gives #y#= 2. Multiply every outcomes by 2. You do not have to do it for every #y# but some just to know the shape.
graph{2tan(x+(pi/4)) [-8.89, 8.89, -4.444, 4.445]}

Lastly, the minus sign will rotate the graph via x-axis. X-axis would act as the mirror.
graph {-2tan(x+(pi/4)) [-8.89, 8.89, -4.444, 4.445]}