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( )

__

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).

table 1

step-by-step explanation:

1-16

2-32

4-64

8-128

all multiples of 16