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.
