Answer:
x^2 - x - 12
Step-by-step explanation:
The quadratic will have the form f(x) = ax^2 + bx + c
Note that when x = 0, y = -12. Thus, we have f(0) = a(0)^2 + b(0) + c = -12, and the function (so far) is f(x) = ax^2 + bx - 12.
Arbitrarily choose two more ordered pairs from the table. I've chosen (-3, 0) and (-4, 8).
For (-3, 0), the quadratic becomes 0 = a(-3)^2 + b(-3) - 12, or:
0 = 9a - 3b, and
For (4, 8), f(4) = 8 = a(4)^2 + b(8), or 8 = 16a + 8b, or 1 = 2a + b, or b = 1 - 2a
Solve 0 = 16a + 4b and b = 1 - 2a for a and b.
Let's use substitution: Substitute b = 1 - 2a into 0 = 9a - 3b or 3a - b = 0:
3a - 1 - 2a = 0, or a = 1.
Since b = 1 - 2a, we get b = 1 - 2(1) = -1
and so a = 1, b = -1 and c = -12
Then the polynomial must be: x^2 - x - 12
