Question

average rate of change quick check
an arcade manager finds that revenue, r, based on a per - game fee, f, for unlimited play can be modeled by the function ( r=-480f^{2}+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)
revenue would decrease 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 decrease by an average of $1 200 per $1 increase in the revenue
revenue would increase by an average of $1 200 per $1 increase in the per - game fee

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = f(x)\) from \(x=a\) to \(x = b\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(R(f)=- 480f^{2}+3120f\), \(a = 1\), and \(b = 3\).

Step2: Calculate \(R(1)\)

Substitute \(f = 1\) into \(R(f)\):
\(R(1)=-480\times(1)^{2}+3120\times(1)=-480 + 3120=2640\)

Step3: Calculate \(R(3)\)

Substitute \(f = 3\) into \(R(f)\):
\(R(3)=-480\times(3)^{2}+3120\times(3)=-480\times9 + 9360=-4320+9360 = 5040\)

Step4: Calculate the average rate of change

Use the formula \(\frac{R(3)-R(1)}{3 - 1}\).
\(\frac{5040 - 2640}{3 - 1}=\frac{2400}{2}=1200\)

Since \(R(3)>R(1)\) and the formula \(\frac{R(3)-R(1)}{3 - 1}\) represents the change in revenue (\(R\)) with respect to the change in the per - game fee (\(f\)). The average rate of change \(\frac{R(3)-R(1)}{3 - 1}=1200\) means that revenue would increase by an average of \(1200\) per \(1\) increase in the per - game fee.

Answer:

Revenue would increase by an average of \(\$1200\) per \(\$1\) increase in the per - game fee.