Question

the table shows the estimated number of bees, y, in a hive x days after a pesticide is released near the hive. bee population over time
number of days | estimated number of bees
0 | 10,000
10 | 7,500
20 | 5,600
30 | 4,200
40 | 3,200
50 | 2,400
which function best models the data?
○ $y = 9,958(0.972)^x$
○ $y = 0.972(9,958)^x$
○ $y = 9,219x - 150$
○ $y = -150x + 9,219$

Explanation:

Step1: Analyze the form of the function

The data shows a decreasing trend. The general form of an exponential function is \(y = a(b)^{x}\), where \(a\) is the initial value and \(b\) is the base (\(0 < b<1\) for decay). A linear function \(y=mx + c\) has a constant rate of change.

Step2: Check the initial - value

When \(x = 0\), for an exponential function \(y=a(b)^{x}\), \(y=a\). For the linear function \(y=-150x + 9219\), when \(x = 0\), \(y = 9219\). For \(y = 9958(0.972)^{x}\), when \(x = 0\), \(y=9958(0.972)^{0}=9958\). The value of \(y\) when \(x = 0\) (initial number of bees) is close to \(10000\).

Step3: Check the nature of change

For a linear function \(y=-150x + 9219\), the rate of change \(m=-150\). Let's check the ratio of \(y\) - values for the exponential - like data.
\(\frac{7500}{10000}=0.75\), \(\frac{5600}{7500}\approx0.747\), \(\frac{4200}{5600}=0.75\), \(\frac{3200}{4200}\approx0.762\), \(\frac{2400}{3200}=0.75\). The data has a multiplicative (exponential) pattern rather than an additive (linear) pattern.

Step4: Evaluate each function

• For \(y = 9958(0.972)^{x}\):

When \(x = 10\), \(y=9958(0.972)^{10}\approx9958\times0.747\approx7440\) (close to \(7500\) considering rounding in the model).

• For \(y = 0.972(9958)^{x}\): When \(x = 0\), \(y = 0.972\), which is not correct as the initial number of bees is \(10000\).
• For \(y = 9219x-150\): When \(x = 0\), \(y=-150\), which is wrong for the initial - value.
• For \(y=-150x + 9219\): When \(x = 0\), \(y = 9219\) (not close to \(10000\)) and it is a linear function (data has an exponential - like decay).

Answer:

\(y = 9958(0.972)^{x}\)