The graph of a secant function shows no functional values on (−7, 7) . The asymptotes of the function occur every π units.
Which function represents the secant function described?
f(x)=sec(x/2)+7
f(x)=7sec(x/2)
f(x)=7sec(x)
f(x)=sec(x)+7
f(x)=sec(x/2)+7 and f(x)=7sec(x/2) have asymptotes that occur every 2π units.
f(x)=sec(x)+7 has a functional value at on (-7, 7).
Therefore your answer is:
f(x)=7sec(x)