How do you graph #y=secpix#?

1 Answer
Oct 24, 2016

#sec (pix) = 1/cos(pi x) #


To graph any inverse function like this it is often easier to graph the normal function lightly and then draw the inverse over the top.

so for this graph #cos(pi*x)#

which goes through the y-axis at (0,1) has min at (1,-1) and max at (2,1)

with a period of 2. giving you a graph like this,

graph{cos(pix) [-2, 2, -4, 4]}

now let's look at some of the features of the graph and decide what will happen when we do #1/f(x)#.

  1. any points where y=1 or y=-1 will remain the same
  2. any points whey y=0 will become an asymptote
  3. points at #y= 1/2# will go to# y=2 #, points at #y=2# will go to #y = 1/2# ect.

now we can do a rough sketch of what the graph will look like.

graph{1/(cos(pix)) [-2, 2, -4, 4]}

It is easier to see the relationship if you superimpose them on one another