Since #sec(x)# is, by definition, #1/cos(x)#, and #y=cos(x)# is a very familiar function with a very well known graph, let's start with a graph of #y=cos(x)#.

graph{cos(x) [-10, 10, -5, 5]}

Next step is to transform this graph into #y=1/cos(x)=sec(x)#. To accomplish this, we notice that everywhere, where #cos(x)=0#, #1/cos(x)# has vertical asymptote. The sign of #cos(x)# and #1/cos(x)# is the same, symmetry considerations and periodicity are the same as well. Also, when #cos(x)# increases in absolute value from #0# to #1#, #1/cos(x)# decreases in absolute value from #oo# to #1#.

So, the graph of #y=1/cos(x)=sec(x)# looks like this:

graph{1/cos(x) [-10, 10, -5, 5]}

Finally, if you have a graph of function #y=f(x)#, you can easily construct a graph of function #y=f(x+A)#. You just shift the graph by #A# units to the left (for positive #A#) or shift it to the right by #-A# units (for negative #A#).

So, here is a graph of #y=1/cos(x+4)=sec(x+4)#:

graph{1/cos(x+4) [-10, 10, -5, 5]}

A detailed explanation of different techniques used in creating graphs of algebraic and trigonometric functions you can find at Unizor by following the link *Algebra - Graphs*.