the table gives the wind - chill temperature when the outside temperature is 20°f. complete parts (a)…

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

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

Answer

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:

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