Question

the table gives the wind - chill temperature when the outside temperature is 20°f. complete parts (a) through (c) below.
a. use x as the wind speed and create a quadratic model for the data.
y = ( )x²+( )x+( )
(type integers or decimals rounded to three decimal places as needed.)

Explanation:

Step1: Set up the system of equations

Let the quadratic model be $y = ax^{2}+bx + c$. We can choose three data - points from the table. Let's take $(x_1,y_1)=(5,13)$, $(x_2,y_2)=(10,10)$ and $(x_3,y_3)=(15,6)$.
Substitute these points into the quadratic equation:
For $(5,13)$: $13=a(5)^{2}+b(5)+c = 25a + 5b + c$.
For $(10,10)$: $10=a(10)^{2}+b(10)+c=100a + 10b + c$.
For $(15,6)$: $6=a(15)^{2}+b(15)+c = 225a+15b + c$.

Step2: Subtract equations to eliminate $c$

Subtract the first equation from the second equation:
$(100a + 10b + c)-(25a + 5b + c)=10 - 13$.
$100a+10b + c - 25a - 5b - c=-3$.
$75a + 5b=-3$.
Subtract the second equation from the third equation:
$(225a+15b + c)-(100a + 10b + c)=6 - 10$.
$225a+15b + c - 100a - 10b - c=-4$.
$125a+5b=-4$.

Step3: Solve the new - formed system of equations

Subtract the equation $75a + 5b=-3$ from $125a+5b=-4$:
$(125a+5b)-(75a + 5b)=-4+3$.
$125a+5b - 75a - 5b=-1$.
$50a=-1$, so $a=-\frac{1}{50}=-0.020$.
Substitute $a = - 0.020$ into $75a + 5b=-3$:
$75\times(-0.020)+5b=-3$.
$-1.5 + 5b=-3$.
$5b=-3 + 1.5=-1.5$.
$b=-0.300$.
Substitute $a=-0.020$ and $b = - 0.300$ into the first equation $25a + 5b + c=13$:
$25\times(-0.020)+5\times(-0.300)+c=13$.
$-0.5-1.5 + c=13$.
$c=13 + 2=15.000$.

Answer:

$y=-0.020x^{2}-0.300x + 15.000$