Answer :
To solve this problem, we will approach it step-by-step:
### Part (a): Graph the Function
1. Identify the equation:
[tex]\[ y = 0.401x^2 + 20.836x + 171.4 \][/tex]
Here, \( y \) represents the spending in millions of dollars, and \( x \) represents the number of years since 1988.
2. Determine the range for \( x \):
You are asked to graph the function for \( x \) in the interval [0, 10].
3. Calculate \( y \) values:
Substitute various values of \( x \) from 0 to 10 into the equation to find the corresponding \( y \) values. Here are some calculations for selected points:
- For \( x = 0 \): \( y = 0.401(0)^2 + 20.836(0) + 171.4 = 171.4 \)
- For \( x = 2 \): \( y = 0.401(2)^2 + 20.836(2) + 171.4 = 218.408 \)
- For \( x = 4 \): \( y = 0.401(4)^2 + 20.836(4) + 171.4 = 279.944 \)
- For \( x = 6 \): \( y = 0.401(6)^2 + 20.836(6) + 171.4 = 355.008 \)
- For \( x = 8 \): \( y = 0.401(8)^2 + 20.836(8) + 171.4 = 443.6 \)
- For \( x = 10 \): \( y = 0.401(10)^2 + 20.836(10) + 171.4 = 545.72 \)
4. Plot the function:
You can now plot these points and draw a smooth curve through them. On the x-axis, mark the years from 0 (1988) to 10 (1998), and on the y-axis, mark the spending in millions of dollars.
### Part (b): Find the Spending in 2001
1. Determine \( x \) for the year 2001:
Since \( x = 0 \) corresponds to 1988, \( x \) in 2001 would be:
[tex]\[ x = 2001 - 1988 = 13 \][/tex]
2. Substitute \( x = 13 \) into the equation to find \( y \):
[tex]\[ y = 0.401(13)^2 + 20.836(13) + 171.4 \][/tex]
3. Solve for \( y \):
[tex]\[ y = 0.401(169) + 20.836(13) + 171.4 \][/tex]
[tex]\[ y = 67.769 + 271.568 + 171.4 \][/tex]
[tex]\[ y = 510.737 \][/tex]
So, the spending by this model in 2001 is approximately \( 510.737 \) million dollars.
### Part (c): Determine if the Value is an Interpolation or an Extrapolation
1. Define interpolation and extrapolation:
- Interpolation is the process of estimating unknown values that fall within the range of known values.
- Extrapolation is the process of estimating values outside the range of known values.
2. Check the range of known values:
The model is provided for \( x \) from 0 (1988) to 10 (1998).
3. Determine the nature of the estimate:
Since the year 2001 (where \( x = 13 \)) falls outside the given range of 1988 to 1998 (where \( x \) ranges from 0 to 10), the value calculated in part (b) is an extrapolation.
In summary:
- Graph: Plot the equation over the interval [0, 10].
- Spending in 2001: Approximately \$510.737 million.
- Nature of Estimate: The value is an extrapolation.
### Part (a): Graph the Function
1. Identify the equation:
[tex]\[ y = 0.401x^2 + 20.836x + 171.4 \][/tex]
Here, \( y \) represents the spending in millions of dollars, and \( x \) represents the number of years since 1988.
2. Determine the range for \( x \):
You are asked to graph the function for \( x \) in the interval [0, 10].
3. Calculate \( y \) values:
Substitute various values of \( x \) from 0 to 10 into the equation to find the corresponding \( y \) values. Here are some calculations for selected points:
- For \( x = 0 \): \( y = 0.401(0)^2 + 20.836(0) + 171.4 = 171.4 \)
- For \( x = 2 \): \( y = 0.401(2)^2 + 20.836(2) + 171.4 = 218.408 \)
- For \( x = 4 \): \( y = 0.401(4)^2 + 20.836(4) + 171.4 = 279.944 \)
- For \( x = 6 \): \( y = 0.401(6)^2 + 20.836(6) + 171.4 = 355.008 \)
- For \( x = 8 \): \( y = 0.401(8)^2 + 20.836(8) + 171.4 = 443.6 \)
- For \( x = 10 \): \( y = 0.401(10)^2 + 20.836(10) + 171.4 = 545.72 \)
4. Plot the function:
You can now plot these points and draw a smooth curve through them. On the x-axis, mark the years from 0 (1988) to 10 (1998), and on the y-axis, mark the spending in millions of dollars.
### Part (b): Find the Spending in 2001
1. Determine \( x \) for the year 2001:
Since \( x = 0 \) corresponds to 1988, \( x \) in 2001 would be:
[tex]\[ x = 2001 - 1988 = 13 \][/tex]
2. Substitute \( x = 13 \) into the equation to find \( y \):
[tex]\[ y = 0.401(13)^2 + 20.836(13) + 171.4 \][/tex]
3. Solve for \( y \):
[tex]\[ y = 0.401(169) + 20.836(13) + 171.4 \][/tex]
[tex]\[ y = 67.769 + 271.568 + 171.4 \][/tex]
[tex]\[ y = 510.737 \][/tex]
So, the spending by this model in 2001 is approximately \( 510.737 \) million dollars.
### Part (c): Determine if the Value is an Interpolation or an Extrapolation
1. Define interpolation and extrapolation:
- Interpolation is the process of estimating unknown values that fall within the range of known values.
- Extrapolation is the process of estimating values outside the range of known values.
2. Check the range of known values:
The model is provided for \( x \) from 0 (1988) to 10 (1998).
3. Determine the nature of the estimate:
Since the year 2001 (where \( x = 13 \)) falls outside the given range of 1988 to 1998 (where \( x \) ranges from 0 to 10), the value calculated in part (b) is an extrapolation.
In summary:
- Graph: Plot the equation over the interval [0, 10].
- Spending in 2001: Approximately \$510.737 million.
- Nature of Estimate: The value is an extrapolation.
