he table gives the​ wind-chill temperature when the outside temperature is 20​°F. Complete parts​ (a) through​ (c) below.

Question

Wind
​(mph)
Wind
Chill ​(degrees​F)
Wind
​(mph)
Wind
Chill ​(degrees​F)

5
13
35
0
10
10
40
negative 1
15
7
45
negative 2
20
5
50
negative 3
25
3
55
negative 3
30
1
60
negative 4
Question content area bottom
Part 1
a. Use x as the wind speed and create a quadratic model for the data.
yequals​(
  
enter your response here​)xsquaredplus​(
  
enter your response here​)xplus​(
  
enter your response here​)
​(Type integers or decimals rounded to three decimal places as​ needed.)

Answer 1

To create a quadratic model from the given data, we can fit a quadratic equation of the form:

\[ y = ax^2 + bx + c \]

where \(y\) is the wind chill temperature and \(x\) is the wind speed (in mph).

We can use the data points provided to create a system of equations that we will then solve for the coefficients \(a\), \(b\), and \(c\). The data points provided are:

• (5, 13)
• (10, 10)
• (15, 7)
• (20, 5)
• (25, 3)
• (30, 1)
• (35, 0)
• (40, -1)
• (45, -2)
• (50, -3)
• (55, -3)
• (60, -4)

From these points, we will select three points to create the system of equations. Let's use:

1. \( (5, 13) \)
2. \( (20, 5) \)
3. \( (35, 0) \)

Using point (5, 13):

\[ 13 = a(5^2) + b(5) + c \] \[ 13 = 25a + 5b + c \quad \text{(1)} \]

Using point (20, 5):

\[ 5 = a(20^2) + b(20) + c \] \[ 5 = 400a + 20b + c \quad \text{(2)} \]

Using point (35, 0):

\[ 0 = a(35^2) + b(35) + c \] \[ 0 = 1225a + 35b + c \quad \text{(3)} \]

Now, we have the following system of equations:

1. \( 25a + 5b + c = 13 \)
2. \( 400a + 20b + c = 5 \)
3. \( 1225a + 35b + c = 0 \)

To solve these equations, we can eliminate \(c\) by subtracting one equation from the other.

Subtracting equation (1) from equation (2):

\[ (400a + 20b + c) - (25a + 5b + c) = 5 - 13 \] \[ 375a + 15b = -8 \quad \text{(4)} \]

Subtracting equation (2) from equation (3):

\[ (1225a + 35b + c) - (400a + 20b + c) = 0 - 5 \] \[ 825a + 15b = -5 \quad \text{(5)} \]

Now we can solve equations (4) and (5).

First, let's isolate \(b\) in equation (4):

\[ 15b = -8 - 375a \] \[ b = \frac{-8 - 375a}{15} \quad \text{(6)} \]

Now substitute equation (6) into equation (5):

\[ 825a + 15\left(\frac{-8 - 375a}{15}\right) = -5 \] \[ 825a - 8 - 375a = -5 \] \[ 450a - 8 = -5 \] \[ 450a = 3 \] \[ a = \frac{1}{150} \approx 0.00667 \]

Now substitute \(a\) back into equation (6):

\[ b = \frac{-8 - 375\left(\frac{1}{150}\right)}{15} \] \[ b = \frac{-8 - 2.5}{15} = \frac{-10.5}{15} \approx -0.7 \]

Finally, substitute \(a\) and \(b\) back into one of the original equations to find \(c\), let's use equation (1):

\[ 25\left(\frac{1}{150}\right) + 5(-0.7) + c = 13 \] \[ \frac{25}{150} - 3.5 + c = 13 \] \[ c = 13 + 3.5 - \frac{25}{150} \] \[ c \approx 16.6 \]

Now rounding the coefficients to three decimal places, our quadratic model can be summarized as:

\[ y \approx 0.007x^2 - 0.700x + 16.600 \]

Thus, the quadratic model is:

\[ y = 0.007x^2 - 0.700x + 16.600 \]