To find a quadratic function that models the data, we can use the following general form: y = ax^2 + bx + c
We can use the values from the table to set up a system of equations to solve for the coefficients a, b, and c.
Using the values (15, 21.4), (25, 26.6), and (35, 27.9), we get the following equations:
21.4 = a(15)^2 + b(15) + c -> Equation 1
26.6 = a(25)^2 + b(25) + c -> Equation 2
27.9 = a(35)^2 + b(35) + c -> Equation 3
Simplifying these equations, we get:
1) 225a + 15b + c = 21.4
2) 625a + 25b + c = 26.6
3) 1225a + 35b + c = 27.9
We can solve this system using any method (substitution, elimination, or matrices). Using the elimination method, we can eliminate c by subtracting Equation 1 from Equations 2 and 3, resulting in:
2) 625a + 25b + c = 26.6
- 1) 225a + 15b + c = 21.4
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400a + 10b = 5.2 -> Equation 4
3) 1225a + 35b + c = 27.9
- 1) 225a + 15b + c = 21.4
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1000a + 20b = 6.5 -> Equation 5
We now have a system of equations:
4) 400a + 10b = 5.2
5) 1000a + 20b = 6.5
Multiplying Equation 4 by 5 and Equation 5 by 2 to eliminate b, we get:
20(400a + 10b) = 20(5.2)
2(1000a + 20b) = 2(6.5)
Simplifying, we have:
8000a + 200b = 104
2000a + 40b = 13
Subtracting Equation 6 from Equation 7, we get:
8000a + 200b - 2000a - 40b = 104 - 13
Simplifying, we have:
6000a + 160b = 91
Dividing the equation by 4 to simplify, we have:
1500a + 40b = 22.75
Now, we have a system of equations:
6) 1500a + 40b = 22.75
7) 1000a + 20b = 6.5
Multiplying Equation 6 by 2 to eliminate b, we get:
2(1500a + 40b) = 2(22.75)
Simplifying, we have:
3000a + 80b = 45.5
Subtracting Equation 8 from Equation 7, we get:
3000a + 80b - 1000a - 20b = 45.5 - 6.5
Simplifying, we have:
2000a + 60b = 39
Dividing the equation by 20 to simplify, we have:
100a + 3b = 1.95
We can now solve this system of equations:
6) 1500a + 40b = 22.75
8) 100a + 3b = 1.95
Multiplying Equation 6 by 3 and Equation 8 by 40, we get:
3(1500a + 40b) = 3(22.75)
40(100a + 3b) = 40(1.95)
Simplifying, we have:
4500a + 120b = 68.25
4000a + 120b = 78
Subtracting Equations 9 and 10, we get:
(4500a + 120b) - (4000a + 120b) = 68.25 - 78
Simplifying, we have:
500a = -9.75
Dividing both sides by 500, we have:
a = -9.75/500 = -0.0195
Plugging this value of a into Equation 9, we can solve for b:
100(-0.0195) + 3b = 1.95
Simplifying, we have:
-1.95 + 3b = 1.95
Adding 1.95 to both sides, we get:
3b = 3.9
Dividing both sides by 3, we have:
b = 1.3
Now that we have the values of a and b, we can use Equation 1 or any other equation to solve for c. Let's use Equation 1:
21.4 = a(15)^2 + b(15) + c
Substituting the values of a, b, and x, we get:
21.4 = -0.0195(15)^2 + 1.3(15) + c
Simplifying, we have:
21.4 = -4.3875 + 19.5 + c
Adding 4.3875 to both sides, we get:
25.7875 = 19.5 + c
Subtracting 19.5 from both sides, we get:
c = 6.2875
Therefore, the quadratic function that models the data is:
y = -0.0195x^2 + 1.3x + 6.2875
To predict the fuel economy for 60 mph, we can substitute x = 60 into the equation:
y = -0.0195(60)^2 + 1.3(60) + 6.2875
y = -0.0195(3600) + 78 + 6.2875
y = -70.2 + 78 + 6.2875
y = 14.0875
Therefore, the predicted fuel economy for 60 mph is approximately 14.0875 mpg.
The table shows the fuel economy y (mpg) at different speeds x (mph) for a new car prototype.
Speed (mph)
15
25
35
45
55
Fuel Economy (mpg)
21.4
26.6
27.9
29.0
30.5
Use a quadratic function that models the data to predict the fuel economy for 60 mph.
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