The first step is to substitute a value of x from the table into each equation and see if it gives us the corresponding the value of y. From the table,
For x = 0, y = 4
Substituting x = 0 into each equation, we have
A. y=2.06(0)^2 + 0.18(0)+ 2.18 = 2.18
B. y= 3.48(0)^2 + 1.22(0) +3.44 = 3.44
C. y= 1.68(0)^2 + 1.06(0) + 4.96 = 4.96
D. y=5.02(0)^2 + 3.66(0) + 4.16 = 4.16
The closest value of y to 4 is 4.16
Thus, the quadratic regression equation that best fits the data is
y = 5.02x^2 + 3.66x + 4.16
