I'll do the first one to get you started. The answer is -3
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Explanation:
When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)
We'll use these three points to determine the quadratic function that goes through them all.
The general template we'll use is y = ax^2 + bx + c
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Plug in (x,y) = (0,1). Simplify
y = ax^2 + bx + c
1 = a*0^2 + b*0 + c
1 = 0a + 0b + c
1 = c
c = 1
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Plug in (x,y) = (-1,0) and c = 1
y = ax^2 + bx + c
0 = a*(-1)^2 + b(-1) + 1
0 = 1a - 1b + 1
a-b+1 = 0
a-b = -1
a = b-1
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Plug in (x,y) = (1,0), c = 1, and a = b-1
y = ax^2 + bx + c
0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1
0 = a + b + 1
0 = b-1 + b + 1 ... replace 'a' with b-1
0 = 2b
2b = 0
b = 0/2
b = 0
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If b = 0, then 'a' is...
a = b-1
a = 0-1
a = -1
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In summary so far, we found: a = -1, b = 0, c = 1
Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1
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The last thing to do is plug x = 2 into this equation and simplify
y = -x^2 + 1
y = -2^2 + 1
y = -4 + 1
y = -3
