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

Related questions
See all questions in Graphing Tangent, Cotangent, Secant, and Cosecant
Impact of this question
2564 views around the world
You can reuse this answer
Creative Commons License