Question

select the correct answer from each drop - down menu. the number of people who visited a state park over the last nine years is recorded in the table.

year	1	2	3	4	5	6	7	8	9
visitors	452	360	348	302	198	124	204	308	398

complete the statements describing the situation. the visitor data is best modeled by a drop - down function. based on the model, the state park management should plan for drop - down than 400 visitors next year.

Answer:

To determine the best model and the number of visitors next year, we analyze the data:

Step 1: Analyze the trend of visitor data

First, we list the number of visitors over the years:

• Year 1: 452
• Year 2: 360 (decrease from year 1)
• Year 3: 348 (decrease from year 2)
• Year 4: 302 (decrease from year 3)
• Year 5: 198 (decrease from year 4)
• Year 6: 124 (decrease from year 5)
• Year 7: 204 (increase from year 6)
• Year 8: 308 (increase from year 7)
• Year 9: 398 (increase from year 8)

The data first decreases and then increases, which is characteristic of a quadratic (parabolic) function. Linear functions have a constant rate of change (either always increasing or always decreasing), and exponential functions have a multiplicative rate of change (either increasing or decreasing by a constant factor). Since the data has a minimum point (around year 6) and then increases, a quadratic function is the best model.

Step 2: Predict the number of visitors for year 10

To predict the number of visitors for year 10, we can observe the pattern of increase after year 6:

• From year 6 (124) to year 7 (204): increase of \(204 - 124 = 80\)
• From year 7 (204) to year 8 (308): increase of \(308 - 204 = 104\)
• From year 8 (308) to year 9 (398): increase of \(398 - 308 = 90\)

The increases are relatively large and the trend after year 6 is increasing. Let's assume the quadratic trend continues. The value at year 9 is 398. To find the next value, we can look at the pattern of the increases. The increase from year 8 to 9 is 90. If we assume a similar or slightly larger increase (since the quadratic function would be increasing at an increasing rate after the minimum), the number of visitors next year (year 10) is likely to be more than 400? Wait, no. Wait, let's check the data again. Wait, year 9 is 398. Let's see the quadratic model. Alternatively, maybe I made a mistake. Wait, let's list the differences:

First, let's compute the first differences (change from year to year):

• Year 1 to 2: \(360 - 452 = -92\)
• Year 2 to 3: \(348 - 360 = -12\)
• Year 3 to 4: \(302 - 348 = -46\)
• Year 4 to 5: \(198 - 302 = -104\)
• Year 5 to 6: \(124 - 198 = -74\)
• Year 6 to 7: \(204 - 124 = 80\)
• Year 7 to 8: \(308 - 204 = 104\)
• Year 8 to 9: \(398 - 308 = 90\)

Now, the second differences (change of the first differences):

• \(-12 - (-92) = 80\)
• \(-46 - (-12) = -34\)
• \(-104 - (-46) = -58\)
• \(-74 - (-104) = 30\)
• \(80 - (-74) = 154\)
• \(104 - 80 = 24\)
• \(90 - 104 = -14\)

The second differences are not constant, but the data has a clear minimum at year 6 (124) and then increases. This is a parabolic (quadratic) shape. Now, to predict year 10, let's see the trend from year 6 to 9:

Year 6: 124

Year 7: 204 (increase by 80)

Year 8: 308 (increase by 104)

Year 9: 398 (increase by 90)

The total increase from year 6 to 9 is \(398 - 124 = 274\) over 3 years. The average increase per year from year 6 to 9 is \(274 / 3 \approx 91.33\). If we continue this trend, year 10 would be \(398 + 91.33 \approx 489.33\), which is more than 400. Wait, but maybe my initial analysis is wrong. Wait, let's check the data again. Wait, year 1: 452, year 2: 360 (down), year 3: 348 (down), year 4: 302 (down), year 5: 198 (down), year 6: 124 (down), year 7: 204 (up), year 8: 308 (up), year 9: 398 (up). So the data decreases to a minimum at year 6 and then increases, which is a quadratic function (since it has a vertex at year 6). Now, to find the next value, let's assume the quadratic function. Let's model the data with a quadratic function. Let's let \(x\) be the year (1 to 9) and \(y\) be the number of visitors. The vertex is at \(x = 6\) (year 6), so the quadratic function can be written as \(y = a(x - 6)^2 + k\), where \(k = 124\) (the minimum value at \(x = 6\)). Now, we can use another point to find \(a\). Let's use year 7 (\(x = 7\), \(y = 204\)):

\(204 = a(7 - 6)^2 + 124\)

\(204 = a(1) + 124\)

\(a = 204 - 124 = 80\)

So the quadratic function is \(y = 80(x - 6)^2 + 124\). Let's check year 8 (\(x = 8\)):

\(y = 80(8 - 6)^2 + 124 = 80(4) + 124 = 320 + 124 = 444\). But the actual value is 308. So my assumption of \(a\) is wrong. Let's use year 9 (\(x = 9\), \(y = 398\)):

\(398 = a(9 - 6)^2 + 124\)

\(398 = 9a + 124\)

\(9a = 398 - 124 = 274\)

\(a = 274 / 9 \approx 30.44\)

Now, check year 8 (\(x = 8\)):

\(y = 30.44(8 - 6)^2 + 124 = 30.44(4) + 124 = 121.76 + 124 = 245.76\). But the actual value is 308. So the quadratic model with vertex at year 6 is not a perfect fit, but the trend is still quadratic (since it has a minimum and then increases). Alternatively, maybe the data is better modeled by a quadratic function, and the prediction for year 10:

Using the quadratic function with \(x = 10\):

\(y = 30.44(10 - 6)^2 + 124 = 30.44(16) + 124 = 487.04 + 124 = 611.04\), which is more than 400. Wait, but maybe I made a mistake. Alternatively, let's look at the increase from year 8 to 9: 398 - 308 = 90. If we assume the increase from year 9 to 10 is similar, say 90, then year 10 would be 398 + 90 = 488, which is more than 400. So the state park management should plan for more than 400 visitors next year.

Final Answers

The visitor data is best modeled by a \(\boldsymbol{\text{quadratic}}\) function.
Based on the model, the state park management should plan for \(\boldsymbol{\text{more}}\) than 400 visitors next year.