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A probable passes through (3,0) (-2,3) and (-1,4) what function does this represent

User Erict
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Thought you'd want to know: If you're talking about parabolas, it's parabolas, not probables. ;)

The standard equation of a a quadratic is y = ax^2 + bx + c. We need to find the values of the coefficients a, b and c.

Taking the first point: When x=3, y=0, so write 0 = a(3)^2 + b(3) + c, or
0 = 9a + 3b + 1c

Do the same for points (-2,3) and (-1,4).

You will have obtained three linear equations in a, b and c:

3= a(-2)^2 + b(-2) + c, or 3 = 4a - 2b + 1c, also

4 = a(-1)^2 + b(-1) + 1c, or 1a - 1b + 1c.

I used matrix operations to solve this system. The results are:

a= -2/5, b= 1/5, c= 21/5

and so the function f(x) is f(x) = (-2/5)x^2 + (1/5)x + 21/5.
User Rob N
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