Answer-
The standard form the equation is,
[tex]y=0.4x^2+5x+6[/tex]
Solution-
Hint- Putting the values of x and y in the general quadratic equation, we can get 3 equations and then solving those 3 equations we can obtain the 3 unknowns a, b, c.
For the sake of simplicity let's take those points as, (-1, 1.4), (0, 6), (1, 11.4)
Putting these values of x, y in the equation y=ax²+bx+c, we get
[tex]\Rightarrow 1.4=a(-1)^2+b(-1)+c \ \Rightarrow a-b+c=1.4 \\\\\\\Rightarrow 6=a(0)^2+b(0)+c \ \Rightarrow c=6 \\\\\\\Rightarrow 11.4=a(1)^2+b(1)+c\ \Rightarrow a+b+c=11.4[/tex]
Putting the value of c in both the equations,
[tex]\Rightarrow a-b+6=1.4\ \Rightarrow a-b=-4.6\\\\\\\Rightarrow a+b+6=11.4\ \Rightarrow a+b=5.4[/tex]
By solving these two equations (by substitution), we get
[tex]\Rightarrow a=0.4,\ b=5[/tex]
Putting all the values, i.e a, b, c in the general quadratic equation,
[tex]y=0.4x^2+5x+6[/tex]
