Patrick B. answered 12/15/20
Math and computer tutor/teacher
One way...
Y=Ax^2+bx+c
(1,5) --> a+b+c=5
(-1,5) --> a-b+c=5
(2,7) --> 4a+2b+c=7
Subtracting the first two equations:
2b=0 --> b=0
Then a+c=5
4a+c=7
Subtracting... 3a=2--> a=2/3
C=13/3
(2/3)x^2+(13/3)=y
Completes the square to find vertex
Andra G.
I got a (0,13/3) as a vertex. Thank you so much! You're a great help, so much appreciated :)12/15/20