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) = 7 sec(x)

Step-by-step explanation:

The range of the function is less than -7 or greater than 7.  So the midline is y=0 and the amplitude is 7.

There are two asymptotes per period, so the period of the function is 2π.

Therefore, the function is:

f(x) = A sec(2π / T x)

f(x) = 7 sec(2π / (2π) x)

f(x) = 7 sec(x)

  f(x) = 7sec(x)

Step-by-step explanation:

The parent secant function has an excluded range of (-1, 1), so the one we're seeking has been scaled vertically by a factor of 7. It will look like ...

  f(x) = 7sec( )

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The parent secant function has vertical asymptotes at odd multiples of π/2, so are separated by π units. Thus, there is no horizontal scaling.

The function described is f(x) = 7sec(x).

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.