Answer:
[tex]y=-4x^2+8x-1[/tex]
Step-by-step explanation:
I am assuming the parabola has his axis parallel to the y axis. You should remember from theory that the axis (and thus the first coordinate of the vertex) is [tex]-\frac b{2a}[/tex] and then you have two points of passage.
[tex]-b/2a = 1\\a(1)^2 +b(1)+c =3\\a(2)^2+b(2)+c=-1\\[/tex]
Let's solve subtracting the 3rd from the second and keeping the first as is (ok, slightly rewriting it).
[tex]2a=-b\rightarrow 2a+b =0 \\a+b+c=3\\4a+2b+c=-1\\III-II: 4a+2b+c-a-b-c=-1-3 \rightarrow 3a+b=-4 \rightarrow a +(2a+b)=-4\\a= -4; b=8; c =3 +4 -8 =-1[/tex]
My answer was incorrect and I don't know how to delete an answer
