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An arcade manager finds that revenue, R, based on a per-game fee, f, for unlimited play can be modeled by the function R=−480f2+3,120f . Which of the following correctly interprets the average rate of change in revenue if the per-game fee increased from $1 to $3?(1 point) Responses Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee. Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee. Revenue would increase by an average of $1,200 per $1 increase in the per-game fee. Revenue would increase by an average of $1,200 per $1 increase in the per-game fee. The per-game fee would increase by an average of $1,200 per $1 increase in the revenue. The per-game fee would increase by an average of $1,200 per $1 increase in the revenue. The per-game fee would decrease by an average of $1,200 per $1 increase in the revenue.

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To find the average rate of change in revenue when the per-game fee increases from $1 to $3, we can use the revenue function given:

\[ R(f) = -480f^2 + 3,120f \]

First, we will calculate the revenue at \( f = 1 \) and \( f = 3 \):

  1. Calculate \( R(1) \): \[ R(1) = -480(1)^2 + 3,120(1) = -480 + 3,120 = 2,640 \]

  2. Calculate \( R(3) \): \[ R(3) = -480(3)^2 + 3,120(3) = -480(9) + 3,120(3) = -4,320 + 9,360 = 5,040 \]

Now we have:

  • Revenue at \( f = 1 \) is \( 2,640 \).
  • Revenue at \( f = 3 \) is \( 5,040 \).

Next, we find the average rate of change in revenue from \( f = 1 \) to \( f = 3 \):

Average rate of change is given by: \[ \text{Average rate of change} = \frac{R(3) - R(1)}{3 - 1} \] Substituting our values: \[ \text{Average rate of change} = \frac{5,040 - 2,640}{3 - 1} = \frac{2,400}{2} = 1,200 \]

This means that as the per-game fee increases by $1 (from $1 to $3), the revenue increases by an average of $1,200.

Thus, the correct interpretation is: Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.