How do you graph #y=4tan(pi/2t)# over the interval #[-2,2]#?

1 Answer
Alan P.
Feb 18, 2017

(see below)

Explanation:

If we let #theta=pi/2t#
then #tin[-2,2]#
is equivalent to #thetain[-pi,+pi]#
and we know #y=tan(theta), thetain[-pi,+pi]# looks like:
enter image source here
Replacing the horizontal axis with the #t# variable based scale:
enter image source here

Replacing #y=tan(pi/2t)# with #y=4tan(pi/2t)#
simply pushes every #y# coordinate value #4# further away from the horizontal axis;
the easiest way to show this is to modify the scale along the vertical axis to #4# times their previous values:
enter image source here

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