Answer:
[tex]y=x^2-4x+1[/tex]
First option is correct.
Step-by-step explanation:
From the given table, we take three points (0,1), (1,-2) and (2,-3)
Now, let the quadratic equation is [tex]y=ax^2+bx+c[/tex]
Use the first point (0,1)
[tex]1=a(0)^2+b(0)+c\\\\c=1[/tex]
Thus, the equation becomes [tex]y=ax^2+bx+1[/tex]
Now, use the point (1,-2)
[tex]-2=a(1)^2+b(1)+1\\\\a+b+1=-2\\\\a+b=-3...(i)[/tex]
Now, use the point (2,-3)
[tex]-3=a(2)^2+b(2)+1\\\\4a+2b+1=-3\\\\4a+2b=-4\\\\2a+b=-2...(ii)[/tex]
Subtract (i) and (ii)
a-2a= -3+2
-a = -1
a = 1
Now, substituting this value in (i)
1 + b = -3
b = -3-1
b = -4
Therefore, the quadratic equation is [tex]y=x^2-4x+1[/tex]
First option is correct.
