Answer:
Step-by-step explanation:
I used 3 of the coordinates to solve for the a, b, and c that we need in
[tex]y=ax^2+bx+c[/tex]
We have 3 unknowns so we need 3 equations. The coordinates I used are from your table:
(-2, 12), (-1, 6), (0, 2). Start with (0, 2). Sub in a 2 for y and a 0 for x in the standard form of the quadratic:
[tex]2=a(0)^2+b(0)+c[/tex] which simplifies very nicely to
c = 2. We will use that value now when we do this again using (-2, 12):
[tex]12=a(-2)^2+b(-2)+2[/tex] and
12 = 4a - 2b + 2 and
4a - 2b = 10 (1)
Do it again with the third coordinate (-1, 6):
[tex]6=a(-1)^2+b(-1)+2[/tex] and
6 = 1a - b + 2 and
a - b = 4 (2)
Now we have a system of equations, (1) and (2) that we will combine and solve for a:
4a - 2b = 10
a - b = 4
Multiply the second equation by -2 to get a new system:
4a - 2b = 10
-2a + 2b = -8
Add to get
2a = 2 so
a = 1
Now sub in 1 for a in (2):
1 - b = 4 and
-b = 3 so
b = -3
Now we have all that we need to write the equation:
[tex]y=x^2-3x+2[/tex]