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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

 Feb 28, 2020
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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)

 Jun 1, 2020

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