Question

Tucker is a security officer responsible for monitoring the number of people in secured area. He works an 8-hour shift. At the start of his shift, there are 128 people in the secured area. Over the first 4 hours of his shift, the number of people in the secured area follows a quadratic pattern, reaching its minimum number of people, 20, in the secured area 3 hours into his shift. After this point, the number of people in the secured area declines at a decay rate of 50% each hour until his shift is over.

A graph of the function is drawn.

What features does the graph have?

(1 point)

The graph is continuous. Closed circles are drawn at (0, 128)
and at (8, 2)
. The first branch of the function ends at the point (4, 32)
and is quadratic. The second part of the graph is exponential and contains the points (4, 32)
and (8, 2)
.






The graph is continuous. Closed circles are drawn at (0, 128)
and at (8, 2)
. The first branch of the function ends at the point (3, 20)
and is quadratic. The second part of the graph is exponential and contains the points (3, 20)
and (8, 2.5)
.






The graph is not continuous. Closed circles are drawn at (0, 128)
and at (8, 2)
. The first branch of the function ends at the point (4, 32)
and is quadratic. The second part of the graph is exponential and contains the points (4, 16)
and (8, 2)
.






The graph is continuous. Closed circles are drawn at (0, 0)
and at (8, 2)
. The first branch of the function ends at the point (4, 32)
and is quadratic. The second part of the graph is exponential and contains the points (4, 32)
and (8, 2)
.





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Answer 1

The first option.

Reason: The quadratic on [0,4] with vertex (3,20) and f(0)=128 is f(t)=12(t−3)^2+20, so f(4)=12(1)+20=32. The exponential decay starts at t=4 with 32 and halves each hour, so f(8)=32(0.5)^4=2. Thus the graph is continuous, has closed dots at (0,128) and (8,2), the quadratic branch ends at (4,32), and the exponential branch contains (4,32) and (8,2).